User manual | Chapter P Prerequisites

Section P.1
Real Numbers
1
Chapter P
Prerequisites
■ Section P.1 Real Numbers
15. –1<x<2; all numbers between –1 and 2, excluding
both –1 and 2
Quick Review P.1
16. 5 x 6 q , or x 5; all numbers greater than or equal
to 5
1. {1, 2, 3, 4, 5, 6}
17. 1 -3, q 2 ; all numbers greater than –3
2. {–2, –1, 0, 1, 2, 3, 4, 5, 6}
18. (–7, –2); all numbers between –7 and –2, excluding both
–7 and –2
3. {–3, –2, –1}
4. {1, 2, 3, 4}
5. (a) 1187.75
(b) –4.72
6. (a) 20.65
(b) 0.10
7. (–2)3-2(–2)+1=–3; (1.5)3-2(1.5)+1=1.375
8. (–3)2+(–3)(2)+22=7
10. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
23. The real numbers greater than 4 and less than or equal to 9.
1. –4.625 (terminating)
25. The real numbers greater than or equal to –3, or the real
numbers which are at least –3.
2. 0.15 (repeating)
3. -2.16 (repeating)
26. The real numbers between –5 and 7, or the real numbers
greater than –5 and less than 7.
4. 0.135 (repeating)
5.
0
1
2
3
4
5
all real numbers less than or equal to 2 (to the left of and
including 2)
6.
0
1
2
3
4
5
6
all real numbers between –2 and 5, including –2 and
excluding 5
7.
2 1
0
1
2
3
4
5
6
7
8
all real numbers less than 7 (to the left of 7)
8.
5 4 3 2 1
0
1
2
3
4
5
all real numbers between –3 and 3, including both –3 and 3
9.
5 4 3 2 1
0
1
2
3
4
5
all real numbers less than 0 (to the left of 0)
10.
1
0
1
2
3
21. (–3, 4]; all numbers between –3 and 4, excluding –3 and
including 4
24. The real numbers greater than or equal to –1, or the real
numbers which are at least –1.
Section P.1 Exercises
4 3 2 1
20. 3 -1, q 2 ; all numbers greater than or equal to –1
22. 1 0, q 2 ; all numbers greater than 0
9. 0, 1, 2, 3, 4, 5, 6
5 4 3 2 1
19. (–2, 1); all numbers between –2 and 1, excluding both –2
and 1
4
5
6
7
8
9
all real numbers between 2 and 6, including both 2 and 6
27. The real numbers greater than –1.
28. The real numbers between –3 and 0 (inclusive), or
greater than or equal to –3 and less than or equal to 0.
29. - 3 6 x 4; endpoints –3 and 4; bounded; half-open
30. - 3 6 x 6 -1; endpoints –3 and –1; bounded; open
31. x 6 5; endpoint 5; unbounded; open
32. x -6; endpoint –6; unbounded; closed
33. His age must be greater than or equal to 29: x 29 or
3 29, q 2 ; x=Bill’s age
34. The costs are between 0 and 2 (inclusive): 0 x 2 or
[0, 2]; x=cost of an item
35. The prices are between $1.099 and $1.399 (inclusive):
1.099 x 1.399 or [1.099, 1.399]; x=$ per gallon of
gasoline
36. The raises are between 0.02 and 0.065: 0.02 6 x 6 0.065
or (0.02, 0.065); x=average percent of all salary raises
37. a(x2+b)=a # x2+a # b=ax2+ab
38. (y-z3)c=y # c-z3 # c=yc-z3c
11. –1 x<1; all numbers between –1 and 1 including –1
and excluding 1
39. ax2+dx2=a # x2+d # x2=(a+d)x2
12. - q 6 x 4, or x 4; all numbers less than or equal
to 4
41. The opposite of 6-∏, or –(6-∏)=–6+∏
=∏-6
13. - q <x<5, or x<5; all numbers less then 5
42. The opposite of –7, or –(–7)=7
14. -2 x 6 2; all numbers between –2 and 2, including
–2 and excluding 2
43. In –52, the base is 5.
40. a3z+a3w=a3 # z+a3 # w=a3(z+w)
44. In (–2)7, the base is –2.
2
Chapter P
Prerequisites
66. (a)
45. (a) Associative property of multiplication
(b) Commutative property of multiplication
(c) Addition inverse property
(d) Addition identity property
(e) Distributive property of multiplication over addition
46. (a) Multiplication inverse property
(b) Multiplication identity property, or distributive
property of multiplication over addition, followed by
the multiplication identity property. Note that we also
use the multiplicative commutative property to say
that 1 # u = u # 1 = u.
(c) Distributive property of multiplication over
subtraction
47.
Remainder
1
0
1
2
0
10
3
5
15
4
8
14
5
8
4
6
2
6
7
3
9
8
5
5
9
2
16
10
9
7
11
4
2
(e) Associative property of multiplication; multiplicative
inverse; multiplicative identity
12
1
3
13
1
13
14
7
11
15
6
8
16
4
12
17
7
1
x2
y2
1 3x 2 y
2
3 1x 2 y
2
=
2 2 4
9x y
=
2
3y
4 4
2
3y
3y
= 3x4y2
42
16
4 2
= 4
49. a 2 b =
2 2
x
1x 2
x
xy 3
x3y3
x3y3
2 -3
50. a
b = a
b = 3 =
xy
2
8
2
51.
Quotient
(d) Definition of subtraction; associative property of
addition; definition of subtraction
2 2 4
48.
Step
1 x-3y2 2 -4
1 y6x-4 2 -2
52. a
x12y-8
=
y-12x8
(b) When the remainder is repeated, the quotients
generated in the long division process will also repeat.
(c) When any remainder is first repeated, the next
quotient will be the same number as the quotient
resulting after the first occurrence of the remainder,
since the decimal representation does not terminate.
x4
= x4y4
y-4
=
3
4a b
3b2
4a
3
12a
6
b a 2 4b = a 2 b a 2 2b =
=
2 3
ab
2a b
b
2a b
2a2b4
ab4
67. False. If the real number is negative, the additive inverse
is positive. For example, the additive inverse of –5 is 5.
57. 4.839μ10
58. –1.6μ10–19
68. False. If the positive real number is less than 1, the
reciprocal is greater than 1. For example, the reciprocal
1
of is 2.
2
59. 0.000 000 033 3
60. 673,000,000,000
69. [–2, 1) corresponds to -2 x 6 1. The answer is E.
53. 3.6930338μ1010
54. 2.21802107μ1011
11
11
55. 1.93175805μ10
56. 4.51908251μ10
8
61. 5,870,000,000,000
70. (–2)4=(–2)(–2)(–2)(–2)=16. The answer is A.
62. 0.000 000 000 000 000 000 000 001 674 7 (23 zeros between
the decimal point and the 1)
71. In –7¤=–(72), the base is 7. The answer is B.
63.
1 1.352 1 2.41 2 * 10
1.25 * 109
=
64.
-7 + 8
=
3.2535 * 10
1.25 * 109
3.2535
* 101 - 9 = 2.6028 * 10-8
1.25
1 3.72 1 4.3 2 * 10
2.5 * 107
=
1
-7 + 6
-1
=
15.91 * 10
2.5 * 107
15.91
* 10-1 - 7 = 6.364 * 10-8
2.5
65. (a) When n=0, the equation aman=am+n becomes
ama0=am+0. That is, ama0=am. Since a Z 0, we can
divide both sides of the equation by am. Hence a0=1.
(b) When n=–m, the equation aman=am+n becomes
ama–m=am+(–m). That is am-m=a0. We know from
part (a) that a0=1. Since a Z 0, we can divide both
sides of the equation ama–m=1 by am. Hence
1
a-m = m .
a
x2 # x4
x6
=
= x4. The answer is D.
x2
x2
73. The whole numbers are 0, 1, 2, 3, . . ., so the whole
numbers with magnitude less than 7 are 0, 1, 2, 3, 4, 5, 6.
72.
74. The natural numbers are 1, 2, 3, 4, . . ., so the natural
numbers with magnitude less than 7 are 1, 2, 3, 4, 5, 6.
75. The integers are . . ., –2, –1, 0, 1, 2, . . ., so the integers
with magnitude less than 7 are –6, –5, –4, –3, –2, –1, 0,
1, 2, 3, 4, 5, 6.
Section P.2
■ Section P.2 Cartesian Coordinate System
0.5
1
1.5
2
2.5
3
Distance: ƒ 17 - 12 ƒ = 17 - 12 L 1.232
2
1.75
1.5
5
9
25
27 2
2
2
Distance: 2 - - a - b 2 = 2 +
= 2 2 =
3
5
15
15
15
15
or 0.13
4.
5
5
5
= 2.5, 2 15 - 2 = - 15 or
2
2
2
2.5 - 15.
1.
3.
3
9. Since ∏≠3.14<4, ƒ ∏-4 ƒ =4-∏.
10. Since 15 L 2.236 and
Quick Review P.2
2.
Cartesian Coordinate System
5 4 3 2 1
0
1
2
3
4
5
5 4 3 2 1
0
1
2
3
4
5
y
5.
A
5
11. ƒ 10.6 - 1 -9.32 ƒ = ƒ 10.6 + 9.3 ƒ = 19.9
12. |–17-(–5)|=|–17+5|=|–12|=12
13. 2 1 -3 - 52 2 + 3 -1 - 1 -12 4 2 = 21 - 82 2 + 02 = 164
=8
14. 2 1 -4 - 12 2 + 1 -3 - 1 2 2 = 21 -52 2 + 1 - 42 2
= 125 + 16 = 141≠6.403
15. 2 10 - 3 2 2 + 1 0 - 4 2 2 = 232 + 42 = 19 + 16
= 125=5
16. 2 1 -1 - 22 2 + 32 - 1 -3 2 4 2 = 21 -32 2 + 52
= 19 + 25 = 134 L 5.830
17. 2 1 -2 - 52 2 + 10 - 02 2 = 2 1 -7 2 2 + 02 = 149=7
B
C
5
18. 2 10 - 0 2 2 + 3 -8 - 1 -1 2 4 2 = 202 + 1 - 72 2= 149
=7
x
19. An isosceles triangle
y
D
5
y
6.
5 C
5
D
5
x
x
2 10 - 1 -52 2 2 + 1 -1 - 32 2 = 252 + 1 -42 2 = 141
B
A
7. 5.5
8. 21.40
9. 10
10. 18.44
2 10 - 1 -42 2 2 + 1 -1 - 42 2 = 2 1 -42 2 + 1 -52 2
= 141
2 1 -5 - 42 2 + 13 - 42 2 = 2 1 -9 2 2 + 1 -12 2 = 182
1. A(1, 0); B(2, 4); C(–3, –2); D(0, –2)
Perimeter = 2141 + 182 L 21.86
Since 1 1412 2 + 1 1412 2 = 1 1822 2, this is a right triangle.
2. A(0, 3); B(–3, 1); C(–2, 0); D(4, –1)
Area =
Section P.2 Exercises
3. (a) First quadrant
1
1 1412 1 1412 = 20.5
2
20. A square
y
(b) On the y-axis, between quadrants I and II
(c) Second quadrant
3
(d) Third quadrant
4. (a) First quadrant
(b) On the x-axis, between quadrants II and III
3
x
(c) Third quadrant
(d) Third quadrant
5. 3 + ƒ -3 ƒ = 3 + 3 = 6
6. 2 - ƒ -2 ƒ = 2 - 2 = 0
7. ƒ 1 -2 23 ƒ = ƒ -6 ƒ = 6
8.
-2
-2
= -1
=
2
ƒ -2 ƒ
This is a square with sides of length 4. Perimeter=16;
Area=16
4
Chapter P
Prerequisites
21. A parallelogram
31.
y
5
7
[1995, 2005] by [0, 150]
x
32.
23 3 - 1 -12 4 2 + 3 -1 - 1 -3 2 4 2 = 242 + 22 = 220
This is a parallelogram with base 8 units and height
4 units. Perimeter = 2 120 + 16 L 24.94;
Area = 8 # 4 = 32
[1995, 2005] by [0, 100]
33.
22. A rectangle
y
7
[1995, 2005] by [0, 50]
7
x
This is a rectangle with length 6 units and height 5 units.
Perimeter=22; Area=30
23.
10.6 + 1 -9.3 2
24.
- 17 + 1 -5 2
2
2
1.3
=
= 0.65
2
=
-22
= -11
2
25. a
-1 + 5 3 + 9
4 12
,
b = a ,
b = 1 2, 6 2
2
2
2 2
26. a
3 + 6 12 + 2
9 2 + 12
,
b = a ,
b
2
2
2
2
27.
°
28. a
-
9
7
5 3
2
6
+ a- b
+
- 4 ¢
° 3
3
3 4
4¢
1 3
,
=
,
= a- , - b
2
2
2
2
3 4
5 + 1 -1 2 -2 + 1 -4 2
4 -6
,
b = a ,
b =(2, –3)
2
2
2 2
29.
34.
[1995, 2005] by [0, 150]
35. (a) about $183,000
(b) about $277,000
36. (a) 1996: about $144,000; 1997: about $183,000
183 - 144
L 0.27
144
An increase of about 27%
(b) 2000: about $277,000; 2001: about $251,000
251 - 277
L -0.09
277
A decrease of about 9%
(c) 1995: about $120,000; 2004: about $311,000
311 - 120
L 1.59
120
An increase of about 159%
37. The three side lengths (distances between pairs of points)
are
2 14 - 1 2 2 + 1 7 - 3 2 2 = 232 + 42 = 19 + 16
= 125=5
[1995, 2005] by [0, 10]
30.
2 18 - 4 2 2 + 1 4 - 7 2 2 = 242 + 1 -32 2 = 116 + 9
= 125=5
2 18 - 1 2 2 + 1 4 - 3 2 2 = 272 + 12 = 149 + 1
= 150 = 5 12.
Since two sides of the triangle formed have the same
length, the triangle is isosceles.
[1995, 2005] by [0, 5]
Section P.2
38. (a) Midpoint of diagonal from (–7, –1) to (3, –1) is
-7 + 3 -1 + 1 -1 2
a
,
b =(–2, –1)
2
2
Midpoint of diagonal from (–2, 4) to (–2, –6) is
-2 + 1 -2 2 4 + 1 -6 2
,
b =(–2, –1)
2
2
Both diagonals have midpoint (–2, –1), so they bisect
each other.
a
(b) Midpoint of diagonal from (–2, –3) to (6, 7) is
-2 + 6 -3 + 7
,
b =(2, 2)
a
2
2
Midpoint of diagonal from (0, 1) to (4, 3) is
0 + 4 1 + 3
a
,
b =(2, 2)
2
2
Both diagonals have midpoint (2, 2), so they bisect
each other.
39. (a) Vertical side: length=6-(–2)=8; horizontal side:
length=3-(–2)=5; diagonal side:
length = 2 3 6 - 1 - 22 4 2 + 3 3 - 1 - 22 4 2
= 282 + 52 = 189
(b) 82+52=64+25=89= 1 1892 2, so the
Pythagorean Theorem implies the triangle is a right
triangle.
40. (a) 214 - 02 2 + 1 -4 - 0 2 2 = 242 + 1 -4 2 2 = 132
214 - 32 2 + 1 -4 - 3 2 2 = 212 + 1 -7 2 2 = 150
213 - 02 2 + 13 - 0 2 2 = 232 + 32 = 118
(b) Since 1 1322 2 + 1 1182 2 = 1 1502 2, the triangle is a
right triangle.
41. (x-1)2+(y-2)2=52, or (x-1)2+(y-2)2=25
42. [x-(–3)]2+(y-2)2=12, or
(x+3)2+(y-2)2=1
2
2
2
43. [x-(–1)] +[y-(–4)] =3 , or
(x+1)2+(y+4)2=9
44. (x-0)2+(y-0)2= 1 132 2, or x2+y2=3
45. (x-3)2+(y-1)2=62, so the center is (3, 1) and the
radius is 6.
46. [x-(–4)]2+(y-2)2=112, so the center is (–4, 2)
and the radius is 11.
47. (x-0)2+(y-0)2= 1 152 2, so the center is (0, 0) and
the radius is 15.
48. (x-2)2+[(y-(–6)]2=52, so the center is (2, –6)
and the radius is 5.
49. ƒ x - 4 ƒ = 3
50. ƒ y - 1 -22 ƒ 4, or ƒ y + 2 ƒ 4
51. ƒ x - c ƒ 6 d
52. The distance between y and c is greater than d, so
|y-c|>d.
53.
1 + a
= 4
2
1+a=8
a=7
and
2 + b
= 4
2
2+b=8
b=6
Cartesian Coordinate System
5
54. Show that two sides have the same length, but not all
three sides have the same length:
2 33 - 1 -1 2 4 2 + 12 - 0 2 2 = 242 + 22 = 116 + 4
= 120 = 2 15
2 35 - 1 -1 2 4 2 + 14 - 2 2 2 = 262 + 22 = 136 + 4
= 140 = 2 110
2 15 - 3 2 2 + 1 4 - 0 2 2 = 222 + 42 = 14 + 16
= 120 = 2 15.
5 + 0 0 + 7
55. The midpoint of the hypotenuse is a
,
b
2
2
5 7
= a , b = 12.5, 3.5 2 . The distances from this point to
2 2
the vertices are:
2 12.5 - 02 2 + 1 3.5 - 0 2 2 = 22.52 + 3.52
= 16.25 + 12.25 = 118.5
2 12.5 - 52 2 + 1 3.5 - 0 2 2 = 2 1 -2.52 2 + 3.52
= 16.25 + 12.25 = 118.5
2 12.5 - 02 2 + 1 3.5 - 7 2 2 = 22.52 + 1 -3.52 2
= 16.25 + 12.25 = 118.5.
56. |x-2|<3 means the distance from x to 2 must be less
than 3. So x must be between –1 and 5. That is,
–1<x<5.
57. |x+3| 5 means the distance from x to –3 must be
5 or more. So x can be 2 or more, or x can be –8 or less.
That is, x -8 or x 2.
58. True. An absolute value is always greater than or equal to
zero. If a 7 0, then ƒ a ƒ = a 7 0. If a 6 0, then
ƒ a ƒ = -a 7 0. If a = 0, then ƒ a ƒ = 0.
length of AM
1
= because M is the midpoint of AB.
59. True.
length of AB
2
By similar triangles,
length of AM
length of AM¿
length of AC
=
length of AB
=
1
,
2
so M is the midpoint of AC.
60. 1 6 13, so 1 - 13 6 0 and
ƒ 1 - 13 ƒ = - 11 - 13 2 = 13 - 1. The answer is B.
61. For a segment with endpoints at a=–3 and b=2, the
a + b
-3 + 2
-1
1
=
=
= - .
midpoint lies at
2
2
2
2
The answer is C.
62. (x-3)2+(y+4)2=2 corresponds to
(x-h)2+(y-k)2= 1 122 2, with h=3 and k=–4.
So the center, (h, k), is (3, –4). The answer is A.
63. In the third quadrant, both coordinates are negative. The
answer is E.
1
1
64. (a) 2 + 18 - 22 = 2 + 162 = 2 + 2 = 4;
3
3
2
2 + 16 2 = 2 + 4 = 6
3
1
1
1 7 - 1 -32 2 = -3 + 1102
3
3
1
2
= ; - 3 + 17 - 1 -32 2 = -3 +
3
3
20
11
= -3 +
=
3
3
(b) - 3 +
= -3 +
2
1102
3
10
3
6
Chapter P
Prerequisites
1
1
1b - a2 = a + b 3
3
1
2a + b
;
= 12a + b 2 =
3
3
2
2
a + 1b - a2 = a + b 3
3
1
a + 2b
= 1a + 2b 2 =
3
3
(c) a +
1
2
1
a = a + b
3
3
3
70. Let the points on the number line be (a, 0) and (b, 0).
The distance between them is
2 1a - b 2 2 + 10 - 02 2 = 21a - b 2 2 = ƒ a - b ƒ .
2
1
2
a = a + b
3
3
3
■ Section P.3 Linear Equations and
Inequalities
2 11 2 + 7 2 1 22 + 11
9 15
(d) a
,
b = a ,
b = 1 3, 5 2 ;
3
3
3 3
1 + 21 72 2 + 2 111 2
15 24
a
,
b = a ,
b = 1 5, 82
3
3
3 3
2a + c 2b + d
a + 2c b + 2d
(e) a
,
b; a
,
b
3
3
3
3
65. If the legs have lengths a and b, and the hypotenuse is
c units long, then without loss of generality, we can
assume the vertices are (0, 0), (a, 0), and (0, b). Then the
midpoint
a + 0 b + 0
a b
of the hypotenuse is a
,
b = a , b . The
2
2
2 2
distance to the other vertices is
a 2
b 2
a2
b2
c
1
a b + a b =
+
= = c.
B 2
2
B4
4
2
2
66.
Quick Review P.3
1. 2x + 5x + 7 + y - 3x + 4y + 2
= 12x + 5x - 3x2 + 1 y + 4y2 + 17 + 22
= 4x + 5y + 9
2. 4 + 2x - 3z + 5y - x + 2y - z - 2
= 12x - x2 + 15y + 2y2 + 1 -3z - z2 + 14 - 22
= x + 7y - 4z + 2
3. 312x - y 2 + 41y - x2 + x + y
= 6x - 3y + 4y - 4x + x + y = 3x + 2y
4. 512x + y - 1 2 + 41 y - 3x + 22 + 1
= 10x + 5y - 5 + 4y - 12x + 8 + 1
= -2x + 9y + 4
5.
6.
y
a
B
C
(
a
Q 4,0
(
)
D
a
31y - 1 2
y - 2
1
3
+
=
+
y - 1
y - 2
1y - 1 2 1y - 22
1y - 12 1 y - 22
4y - 5
y - 2 + 3y - 3
=
=
1y - 12 1y - 22
1 y - 1 2 1 y - 22
1
2x
1
2x + 1
=
+
=
x
x
x
x
y
x2y
y + x - x2y
1
1
x
+
- x =
+
=
8.
x
y
xy
xy
xy
xy
5
1x
+
42
213x
12
x + 4
3x - 1
+
=
+
9.
2
5
10
10
11x + 18
5x + 20 + 6x - 2
=
=
10
10
7. 2 +
a
P a, 2
A
2
3
5
+
=
y
y
y
)
x
(a) Area of ^BPQ = area of ABCD - area of
^BCP - area of ^BAQ - area of ^DPQ
1
a
1
a
1 a
3
= a2 - 1 a2 a b - 1 a2 a b - a b a a b
2
2
2
4
2 2
4
a2
a2
3a2
= a2 4
8
16
7a2
=
16
7
# 1 area of ABCD2 , which is
(b) Area of ^BPQ =
16
just under half the area of the square ABCD.
Note that the result is the same if a 6 0, but the location
of the points in the plane is different.
For #67–69, note that since P(a, b) is in the first quadrant,
then a and b are positive. Hence, –a and –b are negative.
67. Q(a, –b) is in the fourth quadrant and, since P and Q both
have first coordinate a, PQ is perpendicular to the x-axis.
68. Q(–a, b) is in the second quadrant and, since P and Q both
have second coordinate b, PQ is perpendicular to the y-axis.
69. Q(–a, –b) is in the third quadrant, and the midpoint of
a + 1 -a 2 b + 1 -b2
,
b = 1 0, 0 2 .
PQ is a
2
2
10.
x
4x
3x
7x
x
+
=
+
=
3
4
12
12
12
Section P.3 Exercises
1. (a) and (c): 2(–3)2+5(–3)=2(9)-15
1
1
5
1 2
=18-15=3, and 2 a b + 5 a b = 2 a b +
2
2
4
2
1
5
6
1
= + = = 3. Meanwhile, substituting x = 2
2
2
2
gives –2 rather than 3.
-1
-1
1
3
1
2
1
1
+
= - +
= - = - and
= - .
2
6
6
6
6
3
3
3
x
1
x
Or: Multiply both sides by 6: 6 a b + 6 a b = 6 a b ,
2
6
3
so 3x+1=2x. Subtract 2x from both sides: x+1=0.
Subtract 1 from both sides: x=–1.
2. (a):
3. (b): 21 - 02 + 2 = 11 + 2 = 1 + 2 = 3.
Meanwhile, substituting x=–2 or x=2 gives
11 - 4 + 2 = 1- 3 + 2, which is undefined.
Section P.3
4. (c): (10-2)1/3=81/3=2. Meanwhile, substituting
x=–6 gives –2 rather than 2; substituting x=8 gives
61/3≠1.82 rather than 2.
5. Yes: –3x+5=0.
Linear Equations and Inequalities
z - 17
z - 17
z - 8
-8
8
z =
19
23. 6 - 8z - 10z - 15
- 18z - 9
-18z
-19z
6. No: There is no variable x in the equation.
7. No: Subtracting x from both sides gives 3=–5, which is
false and does not contain the variable x.
8. No: The highest power of x is 2, so the equation is
quadratic and not linear.
24. 15z - 9 - 8z - 4
7z - 13
7z
2z
5z
5z
5z
11
11
z =
2
9. No: The equation has a root in it, so it is not linear.
1
10. No. The equation has = x-1 in it, so it is not linear.
x
11. 3x=24
x=8
12. 4x=–16
x=–4
13. 3t=12
t=4
14. 2t=12
t=6
15. 2x - 3
2x
-2x
x
=
=
=
=
4x - 5
4x - 2
-2
1
16. 4 - 2x
- 2x
- 5x
x
=
=
=
=
3x - 6
3x - 10
-10
2
17. 4 - 3y = 2y + 8
- 3y = 2y + 4
- 5y = 4
4
y = - = -0.8
5
18. 4y = 5y + 8
-y = 8
y = -8
1
7
19. 2 a x b = 2 a b
2
8
7
x = = 1.75
4
2
4
20. 3 a x b = 3 a b
3
5
12
2x =
5
12
x =
10
6
x = = 1.2
5
1
1
21. 2 a x + b = 21 12
2
3
2
x + = 2
3
4
x =
3
1
1
22. 3 a x + b = 31 12
3
4
3
x + = 3
4
9
x = = 2.25
4
=
=
=
=
25. 4 a
2x - 3
+ 5b
4
2x - 3 + 20
2x + 17
17
=
=
=
=
- 2
- 2
+ 11
= 5.5
= 413x2
= 12x
= 12x
= 10x
17
x =
= 1.7
10
4x - 5
b
3
4x - 5
4x + 7
7
7
= 3.5
2
26. 312x - 4 2 = 3 a
6x - 12 =
6x =
2x =
x =
t - 2
t + 5
b
8
2
3 1t + 5 2 - 121 t - 2 2
3t + 15 - 12t + 24
- 9t + 39
-9t
1
= 24 a b
3
= 8
= 8
= 8
= -31
31
t =
9
27. 24 a
t + 5
t - 1
+
b
3
4
41t - 1 2 + 31t + 52
4t - 4 + 3t + 15
7t + 11
7t
28. 12 a
1
= 12 a b
2
= 6
= 6
= 6
= -5
5
t = 7
29. (a) The figure shows that x = -2 is a solution of the
equation 2x2 + x - 6 = 0.
3
is a solution of the
2
equation 2x2 + x - 6 = 0.
(b) The figure shows that x =
30. (a) The figure shows that x = 2 is not a solution of the
equation 7x + 5 = 4x - 7.
(b) The figure shows that x = -4 is a solution of the
equation 7x + 5 = 4x - 7.
31. (a): 2(0)-3=0-3=–3<7. Meanwhile,
substituting x=5 gives 7 (which is not less than 7);
substituting x=6 gives 9.
32. (b) and (c): 3(3)-4=9-4=5 5, and
3(4)-4=12-4=8 5.
7
8
Chapter P
Prerequisites
33. (b) and (c): 4(2)-1=8-1=7 and –1<7 11,
and also 4(3)-1=12-1=11 and –1<11 11.
Meanwhile, substituting x=0 gives –1 (which is not
greater than –1).
34. (a), (b), and (c): 1-2(–1)=1+2=3 and –3 3 3; 1-2(0)=1-0=1 and –3 1 3;
1 - 2 12 2 = 1 - 4 = - 3 and –3 –3 3.
44. 5 a
45. 314 2 3 a
35.
1
0
1
2
3
4
5
6
7
8
9
1
0
1
2
3
4
5
6
7
8
9
5 4 3 2 1
0
1
2
3
4
5
12
17
17
2
1
2
36.
37.
2x-1
2x
–2x
x
4x+3
4x+4
4
–2
5 4 3 2 1
0
1
2
3
4
47.
39.
0
1
2
3
4
5
2 x + 6 6 9
-4 x <3
40.
5 4 3 2 1
0
1
2
3
4
5
- 1 3x - 2 6 7
1 3x <9
1
x <3
3
0
1
2
3
4
10-6x+6x-3
7
6
3
x
5
6
7
8
2x+1
2x+1
2x
x
3
0
1
2
3
4
5
6
7
4-4x+5+5x>3x-1
9+x>3x-1
10+x>3x
10>2x
5>x
x 6 5
43. 4 a
0 2z+5<8
–5 2z <3
5
3
z <
- 2
2
48. –6<5t-1<0
–5< 5t <1
1
–1< t <
5
5x + 7
b 4 1 -3 2
4
5x + 7 -12
5x - 19
19
x 5
8
x - 5
3 - 2x
+
b 6 12 1 -22
4
3
3(x-5)+4(3-2x)<–24
3x-15+12-8x<–24
–5x-3<–24
–5x<–21
21
x 7
5
50. 6 a
42.
1
49. 12 a
41.
2 1
3y - 1
b 7 4 1 -1 2
4
4> 3y-1 >–4
5>
3y
>–3
5
>
y
>–1
3
5
–1<
y
<
3
5
6x+8
6x+9
9
–3
5 4 3 2 1
2y - 5
b 31 - 22
3
2y-5
–6
2y
–1
1
y
2
17
y
2
46. 4 11 2 7 4 a
38.
3x-1
3x
–3x
x
3x - 2
b 7 5 1 -12
5
3x-2>–5
3x>–3
x>–1
9
3 - x
5x - 2
+
b 6 6 1 -12
2
3
3(3-x)+2(5x-2)<–6
9-3x+10x-4<–6
7x+5<–6
7x<–11
11
x 6 7
2y - 3
3y - 1
51. 10 a
+
b 6 10 1y - 1 2
2
5
5(2y-3)+2(3y-1)<10y-10
10y-15+6y-2<10y-10
16y-17<10y-10
16y<10y+7
6y<7
7
y 6
6
Section P.3
2y - 3
3 - 4y
b
6
8
4(3-4y)-3(2y-3)
12-16y-6y+9
–22y+21
–22y
2y
52. 24 a
24 12 - y2
64. True. 2 65. 3x+5=2x+1
Subtracting 5 from each side gives 3x=2x-4.
The answer is E.
48-24y
48-24y
48-24y
27-24y
27
27
y 2
1
53. 2 c 1x - 4 2 - 2x d
2
x-4-4x
–3x-4
–3x
7x
Linear Equations and Inequalities
2 35 1 3 - x2 4
10(3-x)
30-10x
34-10x
34
34
x 7
1
1
54. 6 c 1x + 3 2 + 2 1x - 42 d <6 c 1x - 3 2 d
2
3
3(x+3)+12(x-4)<2(x-3)
3x+9+12x-48<2x-6
15x-39<2x-6
15x<2x+33
13x<33
33
x<
13
55. x2 - 2x 6 0 for x = 1
66. –3x<6
Dividing each side by –3 and reversing the<gives
x>–2.
The answer is C.
67.
68.
800
799
7
801
800
103
102
7 102
101
(e) If your calculator returns 0 when you enter
2x + 1 6 4, you can conclude that the value stored in
x is not a solution of the inequality 2x + 1 6 4.
70.
P=2(L+W)
1
P=L+W
2
1
P-L=W
2
P - 2L
1
W= P-L=
2
2
58. x2 - 2x 0 for x = 0, 1, 2
59. Multiply both sides of the first equation by 2.
60. Divide both sides of the first equation by 2.
63. False. 6>2, but –6<–2 because –6 lies to the left of
–2 on the number line.
1
x
1
2x
+ =
3
2
4
3
Multiplying each side by 12 gives 8x+6=3x-4.
The answer is B.
(d) -
57. x2 - 2x 7 0 for x = 3, 4, 5, 6
(b) No: they have different solutions.
2x+5=x-7
2x=x-7
2x=x-12
x=–7
x=–12
x(x+1)=0
x=0 or x+1=0
x=–1
The answer is A.
69. (c)
56. x - 2x = 0 for x = 0, 2
(b) Yes: the solution to both equations is x=4.
6x+2=4x+10 3x+1=2x+5
6x=4x+8
3x=2x+4
2x=8
x=4
x=4
9
62. (a) Yes: the solution to both equations is x = .
2
3x+2=5x-7 –2x+2=–7
3x=5x-9
–2x=–9
9
–2x=–9
x=
2
9
x=
2
6
6
includes the possibility that 2 = , and this is
3
3
the case.
2
61. (a) No: they have different solutions.
3x=6x+9
x=2x+9
–3x=9
–x=9
x=–3
x=–9
9
71.
72.
1
A= h(b1+b2)
2
h(b1+b2)=2A
2A
b1+b2=
h
2A
b1=
- b2
h
V =
4 3
∏r
3
3
V = r3
4∏
3 3V
= r
B 4∏
r =
73.
3 3V
B 4∏
C =
5
1 F - 322
9
9
C = F - 32
5
9
C + 32 = F
5
9
F = C + 32
5
10
Chapter P
Prerequisites
■ Section P.4 Lines in the Plane
6.
Exploration 1
1. The graphs of y=mx+b and y=mx+c have the
same slope but different y-intercepts.
2.
[–4.7, 4.7] by [–3.1, 3.1]
m=2
The angle between the two lines appears to be 90°.
3.
7. 2x+y=17+2(x-2y)
2x+y=17+2x-4y
y=17-4y
5y=17
17
y=
5
8. x2+y=3x-2y
y=3x-2y-x2
3y=3x-x2
1
y=x - x2
3
9.
[–4.7, 4.7] by [–3.1, 3.1]
m=1
[–4.7, 4.7] by [–3.1, 3.1]
m=3
1
1
x + y=2
3
4
1
1
12 a x + y b =12(2)
3
4
4x+3y=24
3y=–4x+24
4
y= - x + 8
3
10.
9 - 5
4
2
= =
-2 - 1 -8 2
6
3
-4 - 6
-10
5
=
=
-14 - 1 -2 2
-12
6
Section P.4 Exercises
1. m=–2
[–4.7, 4.7] by [–3.1, 3.1]
m=4
[–4.7, 4.7] by [–3.1, 3.1]
m=5
In each case, the two lines appear to be at right angles to
one another.
3. m =
9 - 5
4
=
4 + 3
7
4. m =
-3 - 1
4
= 5 + 2
7
5. m =
3 + 5
= 8
-1 + 2
6. m =
12 + 3
5
= -4 - 5
3
Quick Review P.4
1. –75x+25=200
–75x=175
7
x= 3
2. 400-50x=150
–50x=–250
x=5
3. 3(1-2x)+4(2x-5)=7
3-6x+8x-20=7
2x-17=7
2x=24
x=12
4. 2(7x+1)=5(1-3x)
14x+2=5-15x
29x+2=5
29x=3
3
x=
29
5. 2x-5y=21
–5y=–2x+21
21
2
y= x 5
5
7. 2 =
2. m =
2
3
6
9 - 3
=
, so x=2
5 - x
5 - x
y - 3
y - 3
=
, so y=–15
4 + 2
6
y + 5
y + 5
=
9. 3 =
, so y=16
4 + 3
7
8. - 3 =
10.
2 + 2
4
1
=
=
, so x=0
2
x + 8
x + 8
11. y-4=2(x-1)
2
12. y - 3 = - 1x + 4 2
3
13. y+4=–2(x-5)
14. y-4=3(x+3)
15. Since m=1, we can choose A=1 and B=–1. Since
x=–7, y=–2 solves x-y+C=0, C must equal 5:
x-y+5=0. Note that the coefficients can be
multiplied by any non-zero number, e.g., another answer
would be 2x-2y+10=0. This comment also applies
to the following problems.
Section P.4
16. Since m=1, we can choose A=1 and B=–1. Since
x=–3, y=–8 solves x-y+C=0, C must equal –5:
x-y-5=0. See comment in #15.
Lines in the Plane
11
29. Graph y=(429-123x)/7; window should include
(3.488, 0) and (0, 61.29), for example, [–1, 5]*[–10, 80].
17. Since m=0, we can choose A=0 and B=1. Since
x=1, y=–3 solves 0x+y+C=0. C must equal 3:
0x+y+3=0, or y+3=0. See comment in #15.
18. Since m=–1, we can choose A=1 and B=1. Since
x=–1, y=–5 solves x+y+C=0, C must equal 6:
x+y+6=0. See comment in #15.
19. The slope is m=1=–A/B, so we can choose A=1
and B=–1. Since x=–1, y=2 solves
x-y+C=0, C must equal 3: x-y+3=0.
See comment in #15.
[–1, 5] by [–10, 80]
30. Graph y=(3540-2100x)/12=295-175x; window
should include (1.686, 0) and (0, 295), for example,
[–1, 3]*(–50, 350].
20. Since m is undefined we must have B=0, and we can
choose A=1. Since x=4, y=5 solves
x+0y+C=0, C must equal –4: x-4=0. See
comment in #15.
21. Begin with point-slope form: y-5=–3(x-0), so
y=–3x+5.
1
22. Begin with point-slope form: y - 2 = 1x - 12 , so
2
3
1
y = x + .
2
2
1
1
23. m = - , so in point-slope form, y - 5 = - 1x + 42 ,
4
4
1
and therefore y = - x + 4.
4
1
1
, so in point-slope form, y - 2 = 1 x - 42 , and
7
7
10
1
therefore y = x +
.
7
7
24. m =
25. Solve for y: y = -
2
12
.
x +
5
5
7
26. Solve for y: y =
x - 8.
12
27. Graph y=49-8x; window should include (6.125, 0)
and (0, 49), for example, [–5, 10]*[–10, 60].
[–1, 3] by [–50, 350]
31. (a): The slope is 1.5, compared to 1 in (b).
32. (b): The slopes are
7
and 4, respectively.
4
33. Substitute and solve: replacing y with 14 gives x = 4, and
replacing x with 18 gives y = 21.
34. Substitute and solve: replacing y with 14 gives x = 2, and
replacing x with 18 gives y = -18.
35. Substitute and solve: replacing y with 14 gives x = -10,
and replacing x with 18 gives y = -7.
36. Substitute and solve: replacing y with 14 gives x = 14,
and replacing x with 18 gives y = 20.
37. Ymin = -30, Ymax = 30, Yscl = 3
38. Ymin = -50, Ymax = 50, Yscl = 5
39. Ymin = -20>3, Ymax = 20>3, Yscl = 2>3
40. Ymin = -12.5, Ymax = 12.5, Yscl = 1.25
In #41–44, use the fact that parallel lines have the same slope;
while the slopes of perpendicular lines multiply to give –1.
41. (a) Parallel: y - 2 = 3 1x - 12 , or y = 3x - 1.
1
(b) Perpendicular: y - 2 = - 1x - 12 , or
3
1
7
y = - x + .
3
3
[–5, 10] by [–10, 60]
28. Graph y=35-2x; window should include (17.5, 0) and
(0, 35), for example , [–5, 20]*[–10, 40].
42. (a) Parallel: y - 3 = -21x + 2 2 , or y = -2x - 1.
(b) Perpendicular: y - 3 =
1
1
1x + 2 2 , or y = x + 4.
2
2
2
43. (a) Parallel: 2x + 3y = 9, or y = - x + 3.
3
(b) Perpendicular: 3x - 2y = 7, or y =
44. (a) Parallel: 3x - 5y = 13, or y =
[–5, 20] by [–10, 40]
3
7
x - .
2
2
3
13
x .
5
5
5
(b) Perpendicular: 5x + 3y = 33, or y = - x + 11.
3
12
Chapter P
Prerequisites
45. (a) m=(67,500-42,000)/8=3187.5, the y-intercept is
b=42,000 so V=3187.5t+42,000.
(d)
(b) The house is worth about $72,500 after 9.57 years.
[1995, 2005] by [5, 10]
[0, 15] by [40,000, 100,000]
(c) 3187.5t+42,000=74,000: t=10.04.
(d) t=12 years.
46. (a) 0 x 18000
52. (a) Slope of the line between the points (1997, 85.9) and
131.3 - 85.9
45.4
=
= 11.35.
(2001, 131.3) is m =
2001 - 1997
4
Using the point-slope form equation for the line,
we have y-85.9=11.35(x-1997), so
y=11.35x-22580.05.
(b)
(b) I=0.05x+0.08(18,000-x)
(c) x=14,000 dollars.
[1995, 2005] by [50, 180]
(c) Using y=11.35x-22580.05 and x=2006, the
model predicts U.S. imports from Mexico in 2006 will
be approximately $188.1 billion.
[0, 18,000] by [0, 1500]
53. (a)
(d) x=8500 dollars.
3
x, where y is altitude and x is horizontal distance.
8
The plane must travel x=32,000 ft horizontally–just
over 6 miles.
47. y =
48. (a) m =
6 ft
= 0.06.
100 ft
(b) 4166.6 ft, or about 0.79 mile.
(c) 2217.6 ft.
[0, 15] by [5000, 7000]
(b) Slope of the line between the points (7, 5852) and
6377 - 5852
525
(14, 6377) is m =
=
= 75.
14 - 7
7
Using the point-slope form equation for the line, we
have y-5852=75(x-7), so y=75x+5327.
3
4
= 0.375 7
= 0.33, so asphalt shingles are
8
12
acceptable.
49. m =
50. We need to find the value of y when x=2000, 2002, and
2003 using the equation y=0.4x-791.8.
y=0.4(2000)-791.8=800-791.8=8.2
y=0.4(2002)-791.8=800.8-791.8=9
y=0.4(2003)-791.8=801.2-791.8=9.4
Americans’ income in the years 2000, 2002, and 2003 was,
respectively, 8.2, 9, and 9.4 trillion dollars.
51. (a) Slope of the line between the points (1998, 5.9) and
6.3 - 5.9
0.4
=
= 0.4.
(1999, 6.3) is m =
1999 - 1998
1
Using the point-slope form equation for the line, we
have y-5.9=0.4(x-1998), so y=0.4x-793.3.
(b) Using y=0.4x-793.3 and x=2002, the model
estimates Americans’ expenditures in 2002 were $7.5
trillion.
(c) Using y=0.4x-793.3 and x=2006, the model
predicts Americans’ expenditures in 2006 will be $9.1
trillion.
[0, 15] by [5000, 7000]
(c) The year 2006 is represented by x=16. Using
y=75x+5327 and x=16, the model predicts the
midyear world population in 2006 will be
approximately 6527 million, which is a little larger
than the Census Bureau estimate of 6525 million.
54. (a)
[0, 15] by [0, 100]
Section P.4
(b) Slope of the line between the points (6, 67.6) and
52.1 - 67.6
-15.5
(13, 52.1) is m =
=
= -2.2143.
13 - 6
7
Using the point-slope form equation for the line,
we have y-67.6=–2.2143(x-6), so
y=–2.2143x+80.8857.
Lines in the Plane
62. True. If b=0, then a Z 0 and the graph of x =
13
c
is a
a
vertical line. If b Z 0, then the graph of
a
a
c
y = - x + is a line with slope - and y-intercept
b
b
b
c
c
. If b Z 0 and a=0, y= , which is a horizontal line.
b
b
An equation of the form ax+by=c is called linear
for this reason.
63. With (x1, y1)=(–2, 3) and m=4, the point-slope form
equation y-y1=m(x-x1) becomes y-3
[0, 15] by [0, 100]
(c) The year 2006 is represented by x=16. Using
y=2.2143x+80.8857 and x=16, the model
predicts U.S. exports to Japan in 2006 will be
approximately $45.5 billion.
55.
56.
8 - 0
4 - 0
=
a - 3
3 - 0
8
4
=
a - 3
3
24=4(a-3)
6=a-3
9=a
a - 0
2 - 0
=
5 - 3
1 - 0
a
= 2
2
a=4
57. AD ß BC 1 b = 5;
5
5
AB ß DC 1
= 1a = 6
a - 4
2
=4[x-(–2)] or y-3=4(x+2). The answer is A.
64. With m=3 and b=–2, the slope-intercept form
equation y=mx+b becomes y=3x+(–2) or
y=3x-2. The answer is B.
65. When a line has a slope of m1=–2, a perpendicular line
1
1
= . The answer is E.
must have a slope of m2 = m1
2
66. The line through (x1, y1)=(–2, 1) and (x2, y2)=(1, –4)
y2 - y1
-4 - 1
-5
5
has a slope of m =
=
=
= - .
x2 - x1
1 - 1 -2 2
3
3
The answer is C.
67. (a)
[–5, 5] by [–5, 5]
(b)
58. BC ßAD 1 b = 4;
4
4 - 0
AB ß CD 1 =
1a = 3
a
8 - 5
59. (a) No, it is not possible for two lines with positive slopes
to be perpendicular, because if both slopes are
positive, they cannot multiply to –1.
(b) No, it is not possible for two lines with negative slopes
to be perpendicular, because if both slopes are
negative, they cannot multiply to –1.
[–5, 5] by [–5, 5]
(c)
60. (a) If b=0, both lines are vertical; otherwise, both have
slope m=–a/b, and are, therefore, parallel. If c=d,
the lines are coincident.
(b) If either a or b equals 0, then one line is horizontal and
the other is vertical. Otherwise, their slopes are –a/b and
b/a, respectively. In either case, they are perpendicular.
61. False. The slope of a vertical line is undefined. For
example, the vertical line through (3, 1) and (3, 6) would
5
6 - 1
have a slope of
= , which is undefined.
3 - 3
0
[–5, 5] by [–5, 5]
(d) From the graphs, it appears that a is the x-intercept
and b is the y-intercept when c=1.
Proof: The x-intercept is found by setting y=0.
x
x
0
When c=1, we have +
= 1. Hence = 1 so
a
b
a
x=a. The y-intercept is found by setting x=0.
y
y
0
When c=1, we have +
= 1. Hence = 1, so
a
b
b
y=b.
14
Chapter P
Prerequisites
(e)
[–10, 10] by [–10, 10]
[–10, 10] by [–10, 10]
[–10, 10] by [–10, 10]
From the graphs, it appears that a is half the
x-intercept and b is half the y-intercept when c=2.
Proof: When c=2, we can divide both sides by 2
y
x
and we have
+
= 1. By part (d) the x-intercept
2a
2b
is 2a and the y-intercept is 2b.
4
70. The line from the origin to (3, 4) has slope , so the
3
3
tangent line has slope - , and in point-slope form,
4
3
the equation is y - 4 = - 1 x - 32 .
4
b c
a + b c
, b , so the
71. A has coordinates a , b , while B is a
2 2
2
2
line containing A and B is the horizontal line y=c/2,
a + b
b
a
and the distance from A to B is 2
- 2 = .
2
2
2
■ Section P.5 Solving Equations Graphically,
Numerically, and Algebraically
Exploration 1
1.
(f) By a similar argument, when c=–1, a is the opposite
of the x-intercept and b is the opposite of the y-intercept.
[–1, 4] by [–5, 10]
68. (a)
2. Using the numerical zoom, we find the zeros to be 0.79
and 2.21.
3.
[–8, 8] by [–5, 5]
These graphs all pass through the origin. They have
different slopes.
(b) If m>0, then the graphs of y=mx and y=–mx
have the same steepness, but one increases from left
to right, and the other decreases from left to right.
(c)
[–1, 4] by [–5, 10]
[–1, 4] by [–5, 10]
By this method we have zeros at 0.79 and 2.21.
4.
[2.05, 2.36] by [–0.5, 0.43]
[0.63, 0.94] by [–0.39, 0.55]
[–8, 8] by [–5, 5]
Zooming in and tracing reveals the same zeros, correct to
two decimal places.
These graphs have the same slope, but different
y-intercepts.
69. As in the diagram, we can choose one point to be the
origin, and another to be on the x-axis. The midpoints of
the sides, starting from the origin and working around
counterclockwise in the diagram, are then
a
a + b c
b + d c + e
Aa , 0b, Ba
, b, Ca
,
b , and
2
2
2
2
2
d e
D a , b .The opposite sides are therefore parallel, since
2 2
the slopes of the four lines connecting those points are:
c
e
c
e
mAB = ; mBC =
; m = ; mDA =
.
b
d - a CD
b
d - a
5. The answers in parts 2, 3, and 4 are the same.
6. On a calculator, evaluating 4x2-12x+7 when x=0.79
gives y=0.0164 and when x=2.21 gives y=0.0164, so
the numbers 0.79 and 2.21 are approximate zeros.
7.
[2.17, 2.24] by [–0.12, 0.11]
[0.75, 0.83] by [–0.11, 0.12]
Zooming in and tracing reveals zeros of 0.792893 and
2.207107 accurate to six decimal places. If rounded to two
decimal places, these would be the same as the answers
found in part 3.
Section P.5
Solving Equations Graphically, Numerically, and Algebraically
Quick Review P.5
2
2.
2
2
1. (3x-4) =9x -12x-12x+16=9x -24x+16
2. (2x+3)2=4x2+6x+6x+9=4x2+12x+9
3. (2x+1)(3x-5)=6x2-10x+3x-5
=6x2-7x-5
4. (3y-1)(5y+4)=15y2+12y-5y-4
=15y2+7y-4
[–5, 5] by [–10, 10]
x=–3 or x=0.5
The left side factors to (x+3)(2x-1)=0:
x+3=0
or 2x-1=0
x=–3
2x=1
x=0.5
5. 25x2-20x+4=(5x-2)(5x-2)=(5x-2)2
6. 15x3-22x2+8x=x(15x2-22x+8)
=x(5x-4)(3x-2)
7. 3x3+x2-15x-5=x2(3x+1)-5(3x+1)
=(3x+1)(x2-5)
3.
8. y4-13y2+36=(y2-4)(y2-9)
=(y-2)(y+2)(y-3)(y+3)
9.
10.
x
2
2x + 1
x + 3
2 12x + 1 2
x1 x + 3 2
=
12x + 12 1x + 3 2
12x + 12 1x + 3 2
x2 + 3x - 4x - 2
x2 - x - 2
=
=
12x + 12 1x + 32
12x + 12 1 x + 3 2
1x - 2 2 1x + 1 2
=
12x + 12 1x + 3 2
3x + 11
x + 1
- 2
x - 5x + 6
x - x - 6
x + 1
3x + 11
=
1x - 32 1x - 2 2
1x - 32 1 x + 2 2
13x + 11 2 1x - 22
1x + 12 1x + 2 2
=
1x - 32 1x - 2 2 1x + 2 2
1 x - 3 2 1 x - 22 1 x + 22
1x2 + 3x + 2 2 - 13x2 + 5x - 22 2
=
1 x - 32 1 x - 2 2 1 x + 2 2
[–3, 3] by [–2, 2]
x=0.5 or x=1.5
The left side factors to (2x-1)(2x-3)=0:
2x-1=0
or 2x-3=0
2x=1
2x=3
x=0.5
x=1.5
4.
2
=
=
-2x2 - 2x + 24
1x - 32 1x - 2 2 1x + 2 2
-2 1x2 + x - 122
1x - 32 1x - 2 2 1x + 2 2
[–6, 6] by [–4, 4]
x=3 or x=5
Rewrite as x2-8x+15=0; the left side factors to
(x-3)(x-5)=0:
x-3=0 or x-5=0
x=3
x=5
5.
-2 1x + 4 2 1x - 3 2
1x - 32 1x - 2 2 1x + 2 2
-2 1x + 4 2
if x 3
=
1x - 22 1x + 2 2
=
Section P.5 Exercises
1.
[–10, 10] by [–30, 30]
x=–4 or x=5
The left side factors to (x+4)(x-5)=0:
x+4=0
or x-5=0
x=–4
x=5
[–6, 6] by [–20, 20]
2
x= - or x=3
3
Rewrite as 3x2-7x-6=0; the left side factors to
(3x+2)(x-3)=0:
3x+2=0
or x-3=0
2
x = x=3
3
15
16
Chapter P
Prerequisites
6.
16.
x2+6x=4
6 2
6 2
x2 + 6x + a b = 4 + a b
2
2
(x+3)2=4+9
x + 3 = ; 113
x = -3 ; 113
x = -3 - 113 L -6.61 or x = - 3 + 113 L 0.61
[–10, 10] by [–30, 30]
4
3
Rewrite as 3x2+11x-20=0; the left side factors to
(3x-4)(x+5)=0:
3x-4=0 or x+5=0
4
x=–5
x =
3
x=–5 or x=
5
7. Rewrite as (2x)2=52; then 2x=—5, or x = ; .
2
2
8. Divide both sides by 2 to get (x-5) =8.5. Then
x - 5 = ; 18.5 and x = 5 ; 18.5.
9. Divide both sides by 3 to get 1x + 4 2 2 =
x + 4 = ;
8
. Then
3
8
8
and x = -4 ;
.
B3
B3
10. Divide both sides by 4 to get (u+1)2=4.5. Then
u + 1 = ; 14.5 and u = -1 ; 14.5.
17.
2x2-7x+9=x2-2x-3+3x
2x2-7x+9=x2+x-3
x2-8x=–12
x2-8x+(–4)2=–12+(–4)2
(x-4)2=4
x-4=—2
x=4_2
x=2 or x=6
18.
3x2-6x-7=x2+3x-x2-x+3
3x2-8x=10
8
10
x2 - x =
3
3
8
4 2
10
4 2
2
+ a- b
x - x + a- b =
3
3
3
3
4 2
10
16
ax - b =
+
3
3
9
x -
11. Adding 2y2+8 to both sides gives 4y2=14. Divide both
7
7
sides by 4 to get y2 = , so y = ;
.
B2
2
12. 2x+3=_13 so x =
1
1 -3 ; 13 2 , which gives
2
x =
x =
x=–8 or x=5.
13. x2+6x+32=7+32
(x+3)2=16
x+3= ; 116
x=–3_4
x=–7 or x=1
14.
x2+5x=9
5 2
5 2
x2 + 5x + a b = 9 + a b
2
2
(x+2.5)2=9+6.25
x+2.5= ; 115.25
x = -2.5 - 115.25 L - 6.41 or
x = -2.5 + 115.25 L 1.41
5
15.
x - 7x = 4
7 2
5
7 2
x2 - 7x + a - b = - + a - b
2
4
2
7 2
a x - b = 11
2
7
x - = ; 111
2
7
x = ; 111
2
2
x =
7
7
- 111 L 0.18 or x = + 111 L 6.82
2
2
4
46
= ;
3
B9
1
4
; 146
3
3
4
4
1
1
- 146 L -0.93 or x = + 146 L 3.59
3
3
3
3
19. a=1, b=8, and c=–2:
-8 ; 282 - 41 12 1 - 22
=
-8 ; 172
2
=
3 ; 11
3
1
= ;
4
4
4
2 11 2
-8 ; 612
= -4 ; 312
=
2
x≠–8.24 or x≠0.24
x =
20. a=2, b=–3, and c=1:
x =
x =
3 ; 2 1 -3 2 2 - 4122 11 2
21 22
1
or x=1
2
21. x2-3x-4=0, so
a=1, b=–3, and c=–4:
3 ; 2 1 -3 2 2 - 4112 1 -42
2 11 2
x=–1 or x=4
x =
=
3
5
3 ; 125
= ;
2
2
2
22. x2 - 13x - 5 = 0, so
a=1, b = - 13, and c=–5:
x =
13 ; 21 - 132 2 - 41 12 1 -5 2
21 12
13 ; 123
1
1
=
= 13 ; 123
2
2
2
x≠–1.53 or x≠3.26
Section P.5
- 5 ; 215 2 2 - 4 11 2 1 -12 2
2 11 2
5
173
-5 ; 173
= - ;
=
2
2
2
x≠–6.77 or x≠1.77
[–5, 5] by [–5, 5]
34.
24. x2-4x-32=0, so
a=1, b=–4, c=–32:
- 1 - 4 2 ; 21 - 42 2 - 4 11 2 1 -32 2
2 11 2
4 ; 1144
= 2 ; 6
=
2
x=–4 or x=8
x =
17
33.
23. x2+5x-12=0, so
a=1, b=5, c=–12
x =
Solving Graphically, Numerically, and Algebraically
[–5, 5] by [–5, 5]
2
35. x +2x-1=0; x≠0.4
25. x-intercept: 3; y-intercept: –2
36. x3-3x=0; x≠–1.73
26. x-intercepts: 1, 3; y-intercept: 3
37. Using TblStart =1.61 and Tbl=0.001 gives a zero
at 1.62.
Using TblStart =–0.62 and Tbl=0.001 gives a zero
at –0.62.
27. x-intercepts: –2, 0, 2; y-intercept: 0
28. no x-intercepts; no y-intercepts
29.
38. Using TblStart=1.32 and Tbl=0.001 gives a zero
at 1.32.
39. Graph y = ƒ x - 8 ƒ and y=2: t=6 or t=10
40. Graph y = ƒ x + 1 ƒ and y=4: x=–5 or x=3
41. Graph y = ƒ 2x + 5 ƒ and y=7: x=1 or x=–6
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
30.
1
7
42. Graph y = ƒ 3 - 5x ƒ and y=4: x = - or x =
5
5
43. Graph y = ƒ 2x - 3 ƒ and y=x2: x=–3 or x=1
44. Graph y = ƒ x + 1 ƒ and y=2x-3: x=4
45. (a) The two functions are y1 = 3 1x + 4 (the one that
begins on the x-axis) and y2=x2-1.
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
(b) This is the graph of y = 31x + 4 - x2 + 1.
(c) The x-coordinates of the intersections in the first
picture are the same as the x-coordinates where the
second graph crosses the x-axis.
31.
46. Any number between 1.324 and 1.325 must have the digit
4 in its thousandths position. Such a number would round
to 1.32.
47. The left side factors to (x+2)(x-1)=0:
x+2=0
or x-1=0
x=–2
x=1
[–5, 5] by [–5, 5]
32.
48. Graphing y=x2-18 in (e.g.) [–10, 10]*[–20, 10] and
looking for x-intercepts gives x≠–4.24 or x≠4.24.
x2-3x=12-3x+6
x2-18=0
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
49. 2x-1=5 or 2x-1=–5
2x=6
2x=–4
x=3
x=–2
50. x + 2 = 21x + 3
x2+4x+4=4(x+3)
x2=8
x = - 18 or x = 18
- 18 is an extraneous solution, x = 18 L 2.83
[–5, 5] by [–5, 5]
18
Chapter P
Prerequisites
51. From the graph of y=x3+4x2-3x-2 on
[–10, 10]*[–10, 10], the solutions of the equation
(x-intercepts of the graph) are x≠–4.56, x≠–0.44,
x=1.
3
52. From the graph of y=x -4x+2 on
[–10, 10]*[–10, 10], the solutions of the equation
(x-intercepts of the graph) are x≠–2.21, x≠0.54,
and x≠1.68.
53. x2+4x-1=7
x2+4x-8=0
x =
or x2+4x-1=–7
x2+4x+6=0
- 4 ; 116 + 32
2
x = -2 ; 213
x =
-4 ; 116 - 24
2
— no real solutions to
this equation.
54. Graph y = ƒ x + 5 ƒ - ƒ x - 3 ƒ : x=–1
55. Graph y = ƒ 0.5x + 3 ƒ and y=x2-4:
x≠–2.41 or x≠2.91
56. Graph y = 1x + 7 and y=–x2+5:
x≠–1.64 or x≠1.45
57. (a) There must be 2 distinct real zeros, because
b2-4ac>0 implies that ; 2b2 - 4ac are 2
distinct real numbers.
(b) There must be 1 real zero, because b2-4ac=0
implies that ; 2b2 - 4ac = 0, so the root must
b
be x = - .
a
(c) There must be no real zeros, because b2-4ac<0
implies that ; 2b2 - 4ac are not real numbers.
62. True. If 2 is an x-intercept of the graph of
y=ax2+bx+c, then y=0 when x=2. That is,
ax2+bx+c=0 when x=2.
63. False. Notice that for x=–3, 2x2=2(–3)2=18. So x
could also be –3.
64. x(x-3)=0 when x=0 and when x-3=0 or
x=3. The answer is D.
65. For x2-5x+? to be a perfect square, ? must be
2
5
replaced by the square of half of –5, which is a - b .
2
The answer is B.
66. By the quadratic formula, the solutions are
- 1 - 32 ; 2 1 -3 2 2 - 4 12 2 1 -12
3 ; 117
x =
=
2 122
4
The answer is B.
67. Since an absolute value cannot be negative, there are no
solutions. The answer is E.
68. (a) ax2+bx+c=0
ax2+bx=–c
b
c
x2 + x = a
a
b
1 b 2
c
1 b 2
(b) x2 + x + a # b = - + a # b
a
2 a
a
2 a
x2 +
ax +
b
b ¤
c
b2
x + ¢ ≤ =- +
a
2a
a
4a2
b
4ac
b2
b
b ax +
b = - 2 +
2a
2a
4a
4a2
¢x +
58. For (a)–(c), answers may vary.
(a) x2+2x-3 has discriminant (2)2-4(1)(–3)=16,
so it has 2 distinct real zeros. The graph (or factoring)
shows the zeros are at x=–3 and x=1.
(b) x2+2x+1 has discriminant (2)2-4(1)(1)=0, so
it has 1 real zero. The graph (or factoring) shows the
zero is at x=–1.
(c) x2+2x+2 has discriminant (2)2-4(1)(2)=–4,
so it has no real zeros. The graph lies entirely above
the x-axis.
59. Let x be the width of the field (in yd); the length is
x+30. Then the field is 80 yd wide and
80+30=110 yd long.
8800=x(x+30)
0=x2+30x-8800
0=(x+110)(x-80)
0=x+110 or 0=x-80
x=–110 or x=80
2
2
2
2
60. Solving x +(x+5) =18 , or 2x +10x-299=0,
gives x≠9.98 or x≠–14.98. The ladder is about
x+5≠14.98 ft up the wall.
61. The area of the square is x2. The area of the semicircle is
2
1 2
1
1
1
pr = p a x b since the radius of the semicircle is x.
2
2
2
2
2
1
1
Then 200 = x2 + p a x b . Solving this (graphically is
2
2
easiest) gives x≠11.98 ft (since x must be positive).
b ¤
b2 - 4ac
≤ =
2a
4a2
b
b2 - 4ac
= ;
2a
B 4a2
b
; 2b2 - 4ac
x +
=
2a
2a
(c) x +
x = x =
2b2 - 4ac
b
;
2a
2a
-b ; 2b2 - 4ac
2a
69. Graph y = ƒ x2 - 4 ƒ and y=c for several values of c.
(a) Let c=2. The graph suggests y=2 intersects
y = ƒ x2 - 4 ƒ four times.
ƒ x2 - 4 ƒ = 2 1 x2 - 4 = 2 or x2 - 4 = -2
x2=6
x2=2
x = ; 12
x = ; 16
ƒ x2 - 4 ƒ = 2 has four solutions: 5 ; 12, ; 166 .
(b) Let c=4. The graph suggests y=4 intersects
y = ƒ x2 - 4 ƒ three times.
ƒ x2 - 4 ƒ = 4 1 x2 - 4 = 4 or x2-4=–4
x2=8
x2=0
x=0
x = ; 18
(c) Let c=5. The graph suggest y=5 intersects
y = ƒ x2 - 4 ƒ twice.
ƒ x2 - 4 ƒ = 5 1 x2 - 4 = 5 or x2-4=–5
x2=9
x2=–1
x=—3
no solution
ƒ x2 - 4 ƒ = 5 has two solutions: {—3}.
Section P.6
(d) Let c=–1. The graph suggests y=–1 does not
intersect y = ƒ x2 - 4 ƒ . Since absolute value is never
negative, ƒ x2 - 4 ƒ =–1 has no solutions.
(e) There is no other possible number of solutions of this
equation. For any c, the solution involves solving two
quadratic equations, each of which can have 0, 1, or 2
solutions.
70. (a) Let D=b2-4ac. The two solutions are
- b ; 1D
;
2a
adding them gives
- b + 1D
- b - 1D
-2b + 1D - 1D
+
=
2a
2a
2a
-2b
b
= =
2a
a
(b) Let D=b2-4ac. The two solutions are
- b ; 1D
;
2a
multiplying them gives
1 -b 2 2 - 1 1D2 2
- b + 1D # -b - 1D
=
2a
2a
4a2
2
b - 1b2 - 4ac2
c
=
=
2
a
4a
b
71. From #70(a), x1 + x2 = - = 5. Since a=2, this means
a
c
= 3; since a=2, this
b=–10. From #70(b), x1 # x2 =
a
10 ; 1100 - 48
means c=6. The solutions are
; this
4
1
reduces to 2.5 ; 113, or approximately 0.697 and 4.303.
2
Complex Numbers
7. (i2+3)-(7+i3)=(–1+3)-(7-i)
=(2-7)+i=–5+i
8. 1 17 + i2 2 - 16 - 1- 812 = 1 17 - 12
-(6-9i)= 1 17 - 1 - 6 2 +9i= 1 17 - 72 +9i
In #9–16, multiply out and simplify, recalling that i2=–1.
9. (2+3i)(2-i)=4-2i+6i-3i2
=4+4i+3=7+4i
10. (2-i)(1+3i)=2+6i-i-3i2
=2+5i+3=5+5i
11. (1-4i)(3-2i)=3-2i-12i+8i2
=3-14i-8=–5-14i
12. (5i-3)(2i+1)=10i2+5i-6i-3
=–10-i-3=–13-i
13. (7i-3)(2+6i)=14i+42i2-6-18i
=–42-6-4i=–48-4i
14. 1 1-4 + i2 (6-5i)=(3i)(6-5i)=18i-15i2
= 15+18i
15. (–3-4i)(1+2i)=–3-6i-4i-8i2
=–3-10i+8=5-10i
16. 1 1-2 + 2i2 (6+5i)= 1 12 + 22 i(6+5i)
=612 + 122 i+512 + 122 i2
=– 110 + 512 2 + 112 + 6 122 i
17. 1-16 = 4i
18. 1-25 = 5i
19. 1-3 = 13i
20. 1-5 = 15i
In #21–24, equate the real and imaginary parts.
21. x=2, y=3
■ Section P.6 Complex Numbers
22. x=3, y=–7
Quick Review P.6
24. x=7, y=–7/2
23. x=1, y=2
1. x+9
In #25–28, multiply out and simplify, recalling that i2=–1.
2. x+2y
25. (3+2i)2=9+12i+4i2=5+12i
3. a+2d
26. (1-i)3=(1-2i+i2)(1-i)=(–2i)(1-i)
=–2i+2i2=–2-2i
4. 5z-4
5. x2-x-6
6. 2x2+5x-3
7. x2-2
8. x2-12
27. a
12
12 4
12 4
+
ib =a
b (1+i)4
2
2
2
1
1
1
= (1+2i+i2)2= 12i2 2= (–4)=–1
4
4
4
2. (2-3i)+(3-4i)=(2+3)+(–3-4)i=5-7i
13
1 3
1 3
+ i b = a b 1 13 + i2 3
2
2
2
1
= 13 + 213 i + i2 2 1 13 + i2
8
1
= 4 11 + 13 i2 1 13 + i2
1
1
= 1 13 + i + 3i + 13 i2 2 = 14i2 =i
4
4
3. (7-3i)+(6-i)=(7+6)+(–3-1)i=13-4i
In #29–32, recall that (a+bi)(a-bi)=a2+b2.
4. (2+i)-(9i-3)=(2+3)+(1-9)i=5-8i
29. 22+32=13
9. x2-2x-1
10. x2-4x+1
Section P.6 Exercises
In #1–8, add or subtract the real and imaginary parts separately.
1. (2-3i)+(6+5i)=(2+6)+(–3+5)i=8+2i
5. (2-i)+ 13 - 1-32 =(2+3)+ 1 - 1 - 132i
=5- 11 + 132 i
6. 1 15 - 3i2 + 1 -2 + 1-92 = 1 15 - 2 2 + (–3+3)i
= 15 - 2
19
28. a
30. 52+62=61
31. 32+42=25
32. 12+ 1 122 2=3
20
Chapter P
Prerequisites
In #33–40, multiply both the numerator and denominator by
the complex conjugate of the denominator, recalling that
(a+bi)(a-bi)=a2+b2.
33.
1
# 2 - i = 2 - i = 2 - 1i
2 + i 2 - i
5
5
5
34.
2
i
# 2 + i = 2i + i = - 1 + 2 i
2 - i 2 + i
5
5
5
35.
2 + i 2 + i
4 + 4i + i2
3
4
#
=
= + i
2 - i 2 + i
5
5
5
2 + i -3i
- 6i - 3i2
1
2
#
=
= - i
36.
3i
-3i
9
3
3
1 2 + i2 2 1 - i2 1 - i
14 + 4i + i2 2 1 - i + i2 2
#
=
37.
1 + i
1 - i
2
13 + 4i2 1 - 1 - i2
-3 - 3i - 4i - 4i2
1
7
=
= - i
=
2
2
2
2
1 2 - i2 11 + 2i2 5 - 2i
12 + 4i - i - 2i2 2 1 5 - 2i2
#
=
38.
5 + 2i
5 - 2i
29
14 + 3i2 15 - 2i2
20 - 8i + 15i - 6i2
=
=
29
29
26
7
+
i
=
29
29
39.
1 1 - i2 12 - i2 1 + 2i
12 - i - 2i + i2 2 1 1 + 2i2
#
=
1 - 2i
1 + 2i
5
11 - 3i2 11 + 2i2
1 + 2i - 3i - 6i2
7
1
=
= - i
=
5
5
5
5
1 1 - 12i2 11 + i2 1 - 12i
#
40.
1 + 12i
1 - 12i
11 + i - 12i - 12i2 2 11 - 12i2
=
3
31 + 12 + 1 1 - 122i4 1 1 - 12i2
=
3
1 + 12 - 1 12 + 22 i + 1 1 - 122 i - 1 12 - 2 2i2
=
3
1 + 12 + 12 - 2 + 1 -212 - 1 2i
=
3
=
2 12 - 1
2 12 + 1
i
3
3
In #41–44, use the quadratic formula.
41. x=–1 ; 2i
42. x = 43. x =
1
123
;
i
6
6
7
115
;
i
8
8
44. x = 2 ; 115i
45. False. When a=0, z=a+bi becomes z=bi, and then
- z=–(–bi)=bi=z.
46. True. Because i2=–1, i3=i(i2)=–i, and i4=(i2)2=1,
we obtain i+i2+i3+i4=i+(–1)+(–i)+1=0.
47. (2+3i)(2-3i) is a product of conjugates and equals
22+32=13. The answer is E.
48.
1
1 -i
-i
= #
=
= -i. The answer is D.
i
i -i
1
49. Complex, nonreal solutions of polynomials with real
coefficients always come in conjugate pairs. So another
solution is 2+3i, and the answer is A.
50. 11 - i2 3 = 1 -2i2 11 - i2 = - 2i + 2i2 = -2 - 2i. The
answer is C.
51. (a) i=i
i2=–1
i3=(–1)i=–i
i4=(–1)2=1
i5=i # i4=i
i6=i2 # i4=–1
i7=i3 # i4=–i
i8=i4 # i4=1 # 1=1
1
1 i
1 1
1
(b) i–1= = # =–i i–5= # 4 = =–i
i
i i
i i
i
1
1#1
–2
–6
i = 2 4 =–1
i = 2 =–1
i
i i
1 1
1
1 1
1
i–7= 3 # 4 = - =i
i–3= # 2 = - =i
i i
i
i
i i
1 1
i–4= 2 # 2=(–1)(–1)=1
i i
(c) i0=1
i–8=
1 1
# =1 # 1=1
i4 i4
(d) Answers will vary.
52. Answers will vary. One possibility: the graph has the
shape of a parabola, but does not cross the x-axis when
plotted in the real plane, beacuse it does not have any real
zeros. As a result, the function will always be positive or
always be negative.
53. Let a and b be any two real numbers. Then (a+bi)
-(a-bi)=(a-a)+(b+b)i=0+2bi=2bi.
54. 1a + bi2 # 1a + bi2 = 1a + bi2 # 1a - bi2 = a2 + b2,
imaginary part is zero.
55. 1a + bi2 # 1c + di2 = 1ac - bd2 + 1ad + bc2i =
1ac - bd2 - 1ad + bc2i and 1a + bi2 # 1c + di2 =
1a - bi2 # 1c - di2 = 1ac - bd2 - 1 ad + bc2i are equal.
56. 1a + bi2 + 1c + di2 = 1 a + c2 + 1b + d2 i =
1a + c2 - 1b + d2i and 1a + bi2 + 1c + di2 =
1a - bi2 + 1c - di2 = 1a + c2 - 1b + d2i are equal.
57. 1 -i2 2 - i1 -i2 + 2 = 0 but 1 i2 2 - i1i2 + 2 Z 0.
Because the coefficient of x in x2-ix+2=0 is not a
real number, the complex conjugate, i, of –i, need not be
a solution.
■ Section P.7 Solving Inequalities
Algebraically and Graphically
Quick Review P.7
1. –7<2x-3<7
–4< 2x <10
–2< x <5
2. 5x-2 7x+4
–2x 6
x –3
3. |x+2|=3
x+2=3 or x+2=–3
x=1 or
x=–5
4. 4x2-9=(2x-3)(2x+3)
5. x3-4x=x(x2-4)=x(x-2)(x+2)
6. 9x2-16y2=(3x-4y)(3x+4y)
Section P.7
1z - 5 2 1 z +
z2 - 25
=
2
z1 z - 5 2
z - 5z
1x +
x2 + 2x - 35
=
8. 2
1x x - 10x + 25
x
x + 1
+
9.
x - 1
3x - 4
x1 3x - 4 2
=
+
1x - 12 13x - 4 2
2
4x - 4x - 1
=
1x - 12 13x - 4 2
7.
10.
52
z + 5
z
7 2 1 x - 52
x + 7
=
5 2 1 x - 52
x - 5
=
1x + 1 2 1x - 1 2
1x - 12 1 3x - 4 2
2x - 1
x - 3
+
1 x - 2 2 1x + 1 2
1x - 2 2 1x - 12
12x - 12 1x - 1 2 + 1x - 3 2 1x + 12
=
1 x - 2 2 1x + 1 2 1x - 1 2
12x2 - 3x + 1 2 + 1x2 - 2x - 3 2
=
1 x - 2 2 1 x + 1 2 1x - 1 2
13x + 1 2 1x - 22
3x2 - 5x - 2
=
=
1x - 22 1x + 1 2 1x - 1 2
1 x - 2 2 1 x + 12 1 x - 12
13x + 12
if x Z 2 =
1x + 1 2 1x - 1 2
Solving Inequalities Algebraically and Graphically
7. 1 - q, -114 ´ 37, q 2 :
1210 8 6 4 2
0
2
4
6
8
5 4 3 2 1
1
2
3
4
5
6
7
8
6
8
4
5
3. (1, 5):
2 1
4. [–8, 2]:
–2<x-3<2
1< x <5
0
1
2
3
4
5
-5 x + 3 5
-8 x
2
1210 8 6 4 2
0
2
4
2 10
5. a - ,
b : |4-3x|<6
3 3
–6<4-3x<6
–10< –3x <2
10
2
> x
>3
3
5 4 3 2 1
0
1
2
3
6. 1 - q, 02 ´ 13, q 2 : |3-2x|>3
3-2x>3 or 3-2x<–3
–2x>0
–2x<–6
x<0
x>3
5 4 3 2 1
0
1
2
3
4
5
x + 2
3
3
x + 2 9
x 7
1210 8 6 4 2
0
2
2x - 52
4
x - 5
-6 4
-24 x-5
-19 x
4
6
8
6
6
24
29
5040302010 0 10 20 30 40 50
9.
2x2+17x+21=0
(2x+3)(x+7)=0
2x+3=0
or x+7=0
3
x= or
x=–7
2
2
The graph of y=2x +17x+21 lies below the x-axis
3
3
for –7 6 x 6 - . Hence c - 7, - d is the solution since
2
2
the endpoints are included.
10.
6x2-13x+6=0
(2x-3)(3x-2)=0
2x-3=0 or 3x-2=0
3
2
x=
or
x=
2
3
The graph of y=6x2-13x+6 lies above the x-axis for
2
3
2
3
x 6 and for x 7 . Hence a - q, d ´ c , q b is the
3
2
3
2
solution since the endpoints are included.
2. 1 - q, - 1.3 2 ´ 12.3, q 2 :
2x-1>3.6 or 2x-1<–3.6
2x>4.6
2x<–2.6
x>2.3
x<–1.3
0
x + 2
–3 or
3
x + 2 –9
x –11
8. [–19, 29]:
Section P.7 Exercises
1. 1 - q, - 9 4 ´ 3 1, q 2 :
x+4 5 or x+4 –5
x 1
x –9
21
2x2+7x-15=0
(2x-3)(x+5)=0
2x-3=0 or x+5=0
3
x=
or
x=–5
2
2
The graph of y=2x +7x-15 lies above the x-axis for
3
3
x<–5 and for x 7 . Hence 1 - q, -52 ´ a , q b is
2
2
the solution.
12.
4x2-9x+2=0
(4x-1)(x-2)=0
4x-1=0 or x-2=0
1
x=
or
x=2
4
The graph of y=4x2-9x+2 lies below the x-axis for
1
1
6 x 6 2. Hence a , 2 b is the solution.
4
4
11.
13.
2-5x-3x2=0
(2+x)(1-3x)=0
2+x=0
or 1-3x=0
1
x=–2 or
x=
3
The graph of y=2-5x-3x2 lies below the x-axis for
1
1
x<–2 and for x 7 . Hence 1 - q, -22 ´ a , q b is
3
3
the solution.
22
Chapter P
Prerequisites
14.
21+4x-x2=0
(7-x)(3+x)=0
7-x=0 or 3+x=0
x=7 or
x=–3
The graph of y=21+4x-x2 lies above the x-axis for
–3<x<7. Hence (–3, 7) is the solution.
15.
x3-x=0
x(x2-1)=0
x(x+1)(x-1)=0
x=0 or x+1=0
or x-1=0
x=0 or
x=–1 or
x=1
The graph of y=x3-x lies above the x-axis for
x>1 and for –1<x<0. Hence [–1, 0] ´ [1, q ) is the
solution.
16.
x3-x2-30x=0
x(x2-x-30)=0
x(x-6)(x+5)=0
x=0 or x-6=0 or x+5=0
x=0 or
x=6 or
x=–5
The graph of y=x3-x2-30x lies below the
x-axis for x<–5 and for 0<x<6. Hence
(– q , –5] ´ [0, 6] is the solution.
4x2-4x+1=0
(2x-1)(2x-1)=0
(2x-1)2=0
2x-1=0
1
x=
2
The graph of y=4x2-4x+1 lies entirely above the
1
1
1
x-axis, except at x = . Hence a - q, b ´ a , q b is
2
2
2
the solution set.
24.
x2-6x+9=0
(x-3)(x-3)=0
(x-3)2=0
x-3=0
x=3
The graph of y=x2-6x+9 lies entirely above the
x-axis, except at x=3. Hence x=3 is the only
solution.
23.
25.
x2-8x+16=0
(x-4)(x-4)=0
(x-4)2=0
x-4=0
x=4
The graph of y=x2-8x+16 lies entirely above the
x-axis, except at x=4. Hence there is no solution.
26.
9x2+12x+4=0
(3x+2)(3x+2)=0
(3x+2)2=0
3x+2=0
17. The graph of y=x2-4x-1 is zero for x≠–0.24 and
x≠4.24, and lies below the x-axis for –0.24<x<4.24.
Hence (–0.24, 4.24) is the approximate solution.
4
and
3
3
3
4
x = and lies above the x-axis for x 6 and for x 7 .
4
4
3
4
3
Hence a - q, d ´ c , q b is the solution.
4
3
18. The graph of y=12x2-25x+12 is zero for x =
19.
20.
6x2-5x-4=0
(3x-4)(2x+1)=0
3x-4=0 or 2x+1=0
4
1
x=
or
x= 3
2
The graph of y=6x2-5x-4 lies above the
1
4
x-axis for x< - and for x> . Hence
2
3
4
1
a - q, - b ´ a , q b is the solution.
2
3
4x2-1=0
(2x+1)(2x-1)=0
2x+1=0
or 2x-1=0
1
1
x= or
x=
2
2
The graph of y=4x2-1 lies below the x-axis for
1 1
1
1
- <x< Hence c - , d is the solution.
2
2
2 2
21. The graph of y=9x2+12x-1 appears to be zero for
x≠–1.41 and x≠0.08. and lies above the x-axis for
x<–1.41 and x>0.08. Hence (– q , –1.41] ´ [0.08, q )
is the approximate solution.
22. The graph of y=4x2-12x+7 appears to be zero for
x≠0.79 and x≠2.21. and lies below the x-axis for
0.79<x<2.21. Hence (0.79, 2.21) is the approximate
solution.
2
3
The graph of y=9x2+12x+4 lies entirely above the
2
x-axis, except at x = - . Hence every real number
3
satisfies the inequality. The solution is 1 - q, q 2 .
x= -
27. The graph of y=3x3-12x+2 is zero for x≠–2.08,
x≠0.17, and x≠1.91 and lies above the x-axis for
–2.08<x<0.17 and x>1.91. Hence, [–2.08, 0.17] ´
[1.91, q ) is the approximate solution.
28. The graph of y=8x-2x3-1 is zero for x≠–2.06,
x≠0.13, and x≠1.93 and lies below the x-axis for
–2.06<x<0.13 and x>1.93. Hence, (–2.06, 0.13) ´
(1.93, q ) is the approximate solution.
29. 2x3+2x>5 is equivalent to 2x3+2x-5>0. The
graph of y=2x3+2x-5 is zero for x≠1.11 and lies
above the x-axis for x>1.11. So, (1.11, q ) is the
approximate solution.
30. 4 2x3 + 8x is equivalent to 2x3 + 8x - 4 0. The
graph of y=2x3+8x-4 is zero for x≠0.47 and lies
above the x-axis for x>0.47. So, [0.47, q ) is the
approximate solution.
31. Answers may vary. Here are some possibilities.
(a) x2+1>0
(b) x2+1<0
(c) x2 0
(d) (x+2)(x-5) 0
(e) (x+1)(x-4)>0
(f) x(x-4) 0
Section P.7
32.
–16t2+288t-1152=0
t2-18t+72=0
(t-6)(t-12)=0
t-6=0 or t-12=0
t=6 or
t=12
The graph of –16t2+288t-1,152 lies above the t-axis
for 6<t<12. Hence [6, 12] is the solution. This agrees
with the result obtained in Example 10.
33. s=–16t2+256t
(a)
–16t2+256t=768
–16t +256t-768=0
t2-16t+48=0
(t-12)(t-4)=0
t-12=0 or t-4=0
t=12 or
t=4
The projectile is 768 ft above ground twice: at t=4
sec, on the way up, and t=12 sec, on the way down.
2
(b) The graph of s=–16t2+256t lies above the graph of
s=768 for 4<t<12. Hence the projectile’s height
will be at least 768 ft when t is in the interval [4, 12].
(c) The graph of s=–16t2+256t lies below the graph
of s=768 for 0<t<4 and 12<t<16. Hence the
projectile’s height will be less than or equal to 768 ft
when t is in the interval (0, 4] or [12, 16).
34. s=–16t2+272t
(a)
2
–16t +272t=960
–16t2+272t-960=0
t2-17t+60=0
(t-12)(t-5)=0
t-12=0 or t-5=0
t=12 or
t=5
The projectile is 960 ft above ground twice: at t=5
sec, on the way up, and t=12 sec, on the way down.
(b) The graph of s=–16t2+272t lies above the graph
of s=960 for 5<t<12. Hence the projectile’s
height will be more than 960 ft when t is in the
interval (5, 12).
(c) The graph of s=–16t2+272t lies below the graph
of s=960 for 0<t<5 and 12<t<17. Hence the
projectile’s height will be less than or equal to 960 ft
when t is in the interval (0, 5] or [12, 17).
35. Solving the corresponding equation in the process of
solving an inequality reveals the boundaries of the
solution set. For example, to solve the inequality
x2-4 0, we first solve the corresponding equation
x2-4=0 and find that x=—2. The solution, [–2, 2],
of inequality has —2 as its boundaries.
36. Let x be her average speed; then 105<2x. Solving this
gives x>52.5, so her least average speed is 52.5 mph.
37. (a) Let x>0 be the width of a rectangle; then the length
is 2x-2 and the perimeter is P=2[x+(2x-2)].
Solving P<200 and 2x-2>0 gives
1 in.<x<34 in.
2[x+(2x-2)]<200 and 2x-2>0
2(3x-2)<200
2x>2
6x-4<200
x>1
6x<204
x<34
Solving Inequalities Algebraically and Graphically
23
(b) The area is A=x(2x-2). We already know x>1
from part (a). Solve A 1200.
x(2x-2)=1200
2x2-2x-1200=0
x2-x-600=0
(x-25)(x+24)=0
x-25=0 or x+24=0
x=25 or
x=–24
The graph of y=2x2-2x-1200 lies below the
x-axis for 1<x<25, so A 1200 when x is in the
interval (1, 25].
400
to find
V
400
400
= 20 and P =
= 10. The
the range for P: P =
20
40
pressure can range from 10 to 20, or 10 P 20.
400
Alternatively, solve graphically: graph y =
on
x
[20, 40]μ[0, 30] and observe that all y-values are
between 10 and 20.
38. Substitute 20 and 40 into the equation P =
200,000 + x
2.
50,000 + x
Solving for x reveals that the company can borrow no
more than $100,000.
39. Let x be the amount borrowed; then
40. False. If b is negative, there are no solutions, because the
absolute value of a number is always nonnegative and
every nonnegative real number is greater than any
negative real number.
41. True. The absolute value of any real number is always
nonnegative, i.e., greater than or equal to zero.
42. ƒ x - 2 ƒ 6 3
-3 6 x - 2 6 3
-1 6 x 6 5
1 -1, 52
The answer is E.
43. The graph of y=x2-2x+2 lies entirely above the
x-axis so x2 - 2x + 2 0 for all real numbers x. The
answer is D.
44. x2>x is true for all negative x, and for positive x when
x>1. So the solution is 1 - q, 02 ´ 11, q 2 . The answer
is A.
45. x2 1 implies -1 x 1, so the solution is [–1, 1]. The
answer is D.
46. (a) The lengths of the sides of the box are x, 12-2x, and
15-2x, so the volume is x(12-2x)(15-2x). To
solve x(12-2x)(15-2x)=125, graph
y=x(12-2x)(15-2x) and y=125 and find
where the graphs intersect: Either x≠0.94 in. or
x≠3.78 in.
(b) The graph of y=x(12-2x)(15-2x) lies above
the graph of y=125 for 0.94<y<3.78
(approximately). So choosing x in the interval (0.94,
3.78) will yield a box with volume greater than 125 in3.
(c) The graph of y=x(12-2x)(15-2x) lies below
the graph of y=125 for 0<y<0.94 and for
3.78<x<6 (approximately). So choosing x in either
interval (0, 0.94) or interval (3.78, 6) will yield a box
with volume at most 125 in3.
24
Chapter P
Prerequisites
47. 2x2+7x-15=10 or 2x2+7x-15=–10
2x2+7x-25=0
2x2+7x-5=0
The graph of
The graph of
y=2x2+7x-25
y=2x2+7x-5
appears to be zero for
appears to be zero for
x≠–5.69 and x≠2.19
x≠–4.11 and x≠0.61
Now look at the graphs of y=|2x2+7x-15| and
y=10. The graph of y=|2x2+7x-15| lies below the
graph of y=10 when –5.69<x<–4.11 and when
0.61<x<2.19. Hence (–5.69, –4.11) ´ (0.61, 2.19) is
the approximate solution.
48. 2x2+3x-20=10 or 2x2+3x-20=–10
2x2+3x-30=0
2x2+3x-10=0
The graph of
The graph of
y=2x2+3x-30
y=2x2+3x-10
appears to be zero for
appears to be zero for
x≠–4.69 and x≠3.19
x≠–3.11 and x≠1.61
Now look at the graphs of y=|2x2+3x-20| and
y=10. The graph of y=|2x2+7x-20| lies above the
graph of y=10 when x<–4.69, –3.11<x<1.61, and
x>3.19. Hence (– q , –4.69] ´ [–3.11, 1.61] ´ [3.19, q )
is the (approximate) solution.
15. The three side lengths (distances between pairs of points)
are
2 33 - 1 -2 2 4 2 + 111 - 1 2 2 = 252 + 102
= 125 + 100 = 1125 = 5 15
2 17 - 3 2 2 + 1 9 - 11 2 2 = 242 + 1 -22 2
= 116 + 4 = 120 = 2 15
2 37 - 1 -2 2 4 2 + 19 - 1 2 2 = 292 + 82
= 181 + 64 = 1145.
Since 12 152 2 + 15152 2 = 20 + 125 = 145 = 1 11452 2
—the sum of the squares of the two shorter side lengths
equals the square of the long side length—the points
determine a right triangle.
16. The three side lengths (distances between pairs of points)
are
2 14 - 0 2 2 + 1 1 - 1 2 2 = 242 + 02 = 116 = 4
212 - 0 2 2 + 3 11 - 2 132 - 14 2 = 222 + 1 - 213 2 2
= 14 + 12 = 116 = 4
214 - 2 2 2 + 3 1 - 11 - 2 132 4 2 = 222 + 12132 2
= 14 + 12 = 116 = 4.
Since all three sides have the same length, the figure is an
equilateral triangle.
■ Chapter P Review
1. Endpoints 0 and 5; bounded
2. Endpoint 2; unbounded
3. 2(x2-x)=2x2-2x
3
2
4. 2x +4x =2x
5.
1 uv2 2 3
2 3
2
# x+2x2 # 2=2x2(x+2)
u3v6
= v4
u3v2
=
vu
6. (3x2y3)–2 =
9
7. 3.68*10
1
1
1
= 2 2 2 3 2 =
13x2y3 2 2
3 1 x 2 1y 2
9x4y6
17. (x-0)2+(y-0)2=22, or x2+y2=4
18. (x-5)2+[y-(–3)]2=42, or
(x-5)2+(y+3)2=16
19. [x-(–5)]2+[y-(–4)]2=32, so the center is
(–5, –4) and the radius is 3.
20. (x-0)2+(y-0)2=12, so the center is (0, 0) and the
radius is 1.
21. (a) Distance between (–3, 2) and (–1, –2):
2 1 -2 - 22 2 + 3 -1 - 1 -32 4 2 = 21 - 42 2 + 122 2
= 116 + 4 = 120 L 4.47
Distance between (–3, 2) and (5, 6):
8. 7*10–6
2 16 - 2 2 2 + 3 5 - 1 -3 2 4 2 = 242 + 82
= 116 + 64 = 180 L 8.94
Distance between (5, 6) and (–1, –2):
9. 5,000,000,000
10. 0.000 000 000 000 000 000 000 000 000 910 94
(27 zeros between the decimal point and the first 9)
2 1 -2 - 62 2 + 1 -1 - 5 2 2 = 21 -82 2 + 1 - 62 2
11. (a) 5.0711*1010
(b) 4.63*109
= 164 + 36 = 1100 = 10
(b) 1 1202 2 + 1 1802 2 = 20 + 80 = 100 = 102, so the
Pythagorean Theorem guarantees the triangle is a
right triangle.
(c) 5.0*108
(d) 3.995*109
(e) 1.4497*1010
12. -0.45 (repeating)
13. (a) Distance: ƒ 14 - 1 -5 2 ƒ = ƒ 19 ƒ = 19
-5 + 14
9
=
= 4.5
(b) Midpoint:
2
2
14. (a) Distance:
23 5 - 1 -42 4 2 + 1 -1 - 32 2 = 292 + 1 - 42 2
= 181 + 16 = 197 L 9.85
(b) Midpoint:
-4 + 5 3 + 1 -1 2
1 2
1
a
,
b = a , b = a , 1b
2
2
2 2
2
22. ƒ z - 1 -32 ƒ 1, or ƒ z + 3 ƒ 1
23.
-1 + a
1 + b
= 3 and
= 5
2
2
–1+a=6
1+b=10
a=7
b=9
24. m =
-5 + 2
3
= 4 + 1
5
2
25. y + 1 = - 1x - 2 2
3
Chapter P Review
9
A
26. The slope is m=– =– , so we can choose A=9
7
B
and B=7. Since x=–5, y=4 solves 9x+7y+C=0,
C must equal 17: 9x+7y+17=0. Note that the
coefficients can be multiplied by any non-zero number, e.g.,
another answer would be 18x+14y+34=0.
4
27. Beginning with point-slope form: y + 2 = 1x - 32 , so
5
4
y = x - 4.4.
5
2 + 4
3
28. m =
= , so in point-slope form,
3 + 1
2
3
3
5
y + 4 = 1 x + 1 2 , and therefore y = x - .
2
2
2
29. y=4
30. Solve for y: y =
7
3
x - .
4
4
31. The slope of the given line is the same as the line we
2
2
want: m = - , so y + 3 = - 1 x - 2 2 , and therefore
5
5
2
11
y = - x .
5
5
2
32. The slope of the given line is - , so the slope of the line
5
5
5
we seek is m = . Then y + 3 = 1x - 2 2 , and
2
2
5
therefore y = x - 8.
2
33. (a)
35. m =
25
5
=
= 2.5
10
2
36. Both graphs look the same, but the graph on the left has
2
slope —less than the slope of the one on the right,
3
12
4
which is
= . The different horizontal and vertical
15
5
scales for the two windows make it difficult to judge by
looking at the graphs.
37. 3x-4=6x+5
–3x=9
x=–3
x + 5
1
x - 2
+
=
3
2
3
2(x-2)+3(x+5)=2
2x-4+3x+15=2
5x+11=2
5x=–9
9
x = 5
39. 2(5-2y)-3(1-y)=y+1
10-4y-3+3y=y+1
7-y=y+1
–2y=–6
y=3
38.
40. 3(3x-1)2=21
(3x-1)2=7
3x-1= ; 17
3x-1= - 17 or 3x-1= 17
1
17
1
17
x = L -0.55 x = +
L 1.22
3
3
3
3
41.
[0, 15] by [500, 525]
(b) Slope of the line between the points (5, 506) and
514 - 506
8
(10, 514) is m =
= = 1.6.
10 - 5
5
Using the point-slope form equation for the line, we
have y-506=1.6(x-5), so y=1.6x+498.
x2-4x-3=0
x2-4x=3
x2-4x+(2)2=3+(2)2
(x-2)2=7
x-2= ; 17
x-2= - 17 or x - 2 = 17
x = 2 - 17 L -0.65 x = 2 + 17 L 4.65
42. 16x2-24x+7=0
Using the quadratic formula:
x =
24 ; 2242 - 4 116 2 17 2
21 162
3
12
24 ; 1128
= ;
=
32
4
4
[0, 15] by [500, 525]
(c) The year 1996 is represented by x=6. Using
y=1.6x+498 and x=6, we estimate the average
SAT math score in 1996 to be 507.6, which is very
close to the actual value 508.
(d) The year 2006 is represented by x=16. Using
y=1.6x+498 and x=16, we predict the average
SAT math score in 2006 will be 524.
4
34. (a) 4x-3y=–33, or y = x + 11
3
3
3
(b) 3x+4y=–6, or y = - x 4
2
25
x =
43.
3
12
12
3
L 0.40 or x =
+
L 1.10
4
4
4
4
6x2+7x=3
6x +7x-3=0
(3x-1)(2x+3)=0
3x-1=0 or 2x+3=0
3
1
or x = x =
3
2
2
26
Chapter P
Prerequisites
44. 2x2+8x=0
2x(x+4)=0
2x=0 or x+4=0
x=0 or
x=–4
45.
x(2x+5)=4(x+7)
2x2+5x=4x+28
2
2x +x-28=0
(2x-7)(x+4)=0
2x-7=0 or x+4=0
7
or
x=–4
x =
2
46. 4x+1=3
4x=2
1
x=
2
47.
48.
49.
53. 2x2-3x-1=0
3
1
x2 - x - = 0
2
2
3
3
1
3
x2 - x + a - b 2 =
+ a - b2
2
4
2
4
3
17
ax - b2 =
4
16
x -
x =
or 4x+1=–3
or
4x=–4
x =
x=–1
or
57. The graph of y=x3-2x2-2 is zero for x≠2.36.
58. The graph of y = ƒ 2x - 1 ƒ - 4 + x2 is zero for x=–1
and for x≠1.45.
50. Solving 4x -4x+2=0 by using the quadratic
formula with a=4, b=–4, and c=2 gives
=
4 ; 1- 16
8
4 ; 4i
1
1
=
=
; i
8
2
2
51. Solving x2-6x+13=0 by using the quadratic
formula with a=1, b=–6, and c=13 gives
2112
=
6 ; 1- 16
2
6 ; 4i
=
= 3 ; 2i
2
=
2 ; 2 1 - 2 2 2 - 4 11 2 1 4 2
2112
2 ; 2i13
= 1 ; 13i
2
59. –2<x+4 7
–6<x 3
Hence (–6, 3] is the solution.
10 8 6 4 2
=
2 ; 1- 12
2
0
2
4
6
8 10
60. 5x+1 2x-4
3x –5
5
x 3
5
Hence c - , q b is the solution.
3
5 4 3 2 1
61.
52. Solving x2-2x+4=0 by using the quadratic formula
with a=1, b=–2, and c=4 gives
x =
2
2
17
17
L -1.55 or x = - +
L 0.22
3
3
3
3
56. The graph of y=x3+2x2-4x-8 is zero for x=–2,
and x=2.
2
6 ; 2 1 -6 2 2 - 4 11 2 113 2
3
117
+
L 1.78
4
4
55. The graph of y=3x3-19x2-14x is zero for x=0,
2
x = - , and x=7.
3
x2=3x
x -3x=0
x(x-3)=0
x=0 or x-3=0
x=0 or
x=3
x =
x =
or
-4 ; 2 14 2 2 - 4132 1 -12
x = -
2
2142
3
117
L -0.28
4
4
2 13 2
2
17
-4 ; 128
= - ;
=
6
3
3
x =
–9x2+12x-4=0
9x2-12x+4=0
(3x-2)(3x-2)=0
(3x-2)2=0
3x-2=0
2
x =
3
4 ; 2 1 - 4 2 2 - 4 14 2 1 2 2
3
117
;
4
4
54. 3x2+4x-1=0
4x2-20x+25=0
(2x-5)(2x-5)=0
(2x-5)2=0
2x-5=0
5
x =
2
x =
3
117
= ;
4
4
0
1
2
3
4
3x - 5
-1
4
3x-5 –4
3x 1
1
x 3
1
Hence a - q, d is the solution.
3
62. –7<2x-5<7
–2<2x<12
–1<x<6
Hence (–1, 6) is the solution.
5
Chapter P Review
63. 3x+4 2 or 3x+4 –2
72.
3x –2 or 3x –6
2
x - or x –2
3
2
Hence 1 - q, -2 4 ´ c - , q b is the solution.
3
64.
4x2+3x-10=0
(4x-5)(x+2)=0
4x-5=0 or x+2=0
5
or
x=–2
x =
4
2
The graph of y=4x +3x-10 lies above the x-axis for
5
5
x<–2 and for x 7 . Hence 1 - q, - 22 ´ a , q b is
4
4
the solution.
65. The graph of y=2x2-2x-1 is zero for x≠–0.37
and x≠1.37, and lies above the x-axis for x<–0.37 and
for x>1.37. Hence 1 - q, - 0.37 2 ´ 11.37, q 2 is the
approximate solution.
66. The graph of y=9x2-12x-1 is zero for x≠–0.08,
and x≠1.41, and lies below the x-axis for
–0.08<x<1.41. Hence [–0.08, 1.41] is the
approximate solution.
67. x3 - 9x 3 is equivalent to x3 - 9x - 3 0. The
graph of y=x3-9x-3 is zero for x≠–2.82,
x≠–0.34, and x≠3.15, and lies below the x-axis for
x<–2.82 and for –0.34<x<3.15. Hence the
approximate solution is 1 - q, -2.824 ´ 3 -0.34, 3.154 .
68. The graph of y=4x3-9x+2 is zero for x≠–1.60,
x≠0.23, and x≠1.37, and lies above the x-axis for
–1.60<x<0.23 and for x>1.37. Hence the
approximate solution is 1 -1.60, 0.23 2 ´ 1 1.37, q 2 .
69.
70.
71.
x + 7
x + 7
>2 or
<–2
5
5
x+7>10
or x+7<–10
x>3
or
x<–17
Hence 1 - q, -17 2 ´ 13, q 2 is the solution.
2x2+3x-35=0
(2x-7)(x+5)=0
2x-7=0 or x+5=0
7
x=
or
x=–5
2
2
The graph of y=2x +3x-35 lies below the x-axis for
7
7
-5 6 x 6 . Hence a -5, b is the solution.
2
2
4x2+12x+9=0
(2x+3)(2x+3)=0
(2x+3)2=0
2x+3=0
3
2
The graph of y=4x2+12x+9 lies entirely above the
3
x-axis except for x = - . Hence all real numbers satisfy
2
the inequality. So 1 - q, q 2 is the solution.
x= -
27
x2-6x+9=0
(x-3)(x-3)=0
(x-3)2=0
x-3=0
x=3
The graph of y=x2-6x+9 lies entirely above the
x-axis except for x=3. Hence no real number satisfies
the inequality. There is no solution.
73. 1 3 - 2i2 + 1 -2 + 5i2 = 13 - 2 2 + 1 -2 + 52i
= 1 + 3i
74. 15 - 7i2 - 1 3 - 2i2 = 15 - 32 + 1 -7 + 2 2i
= 2 - 5i
75. 1 1 + 2i2 13 - 2i2 = 3 - 2i + 6i - 4i2
= 3 + 4i + 4
= 7 + 4i
76. 11 + i2 3 = 1 11 + i2 11 + i2 2 11 + i2
= 11 + 2i + i2 2 11 + i2 = 2i1 1 + i2
= 2i + 2i2 = -2 + 2i
77. 11 + 2i2 2 11 - 2i2 2 = 11 + 4i + 4i2 2 11 - 4i + 4i2 2
= 1 -3 + 4i2 1 -3 - 4i2
29
78. i
= i i = 1i 2
28
2 14
= 9 - 12i + 12i - 16i2 = 25
i = 1 -12 14 i = i
79. 1-16 = 11162 1 -12 = 4 1-1 = 4i
80.
2 + 3i # 1 + 5i
2 + 10i + 3i + 15i2
2 + 3i
=
=
1 - 5i
1 - 5i 1 + 5i
1 + 5i - 5i - 25i2
1
1
-13 + 13i
= - + i
26
2
2
81. s=–16t2+320t
=
(a) –16t2+320t=1538
–16t2+320t-1538=0
The graph of s=–16t2+320t-1538 is zero at
t =
-320 ; 23202 - 4 1 -162 1 -15382
21 -16 2
40 ; 162
-320 ; 13968
.
=
=
-32
4
40 - 162
So t =
L 8.03 sec or
4
40 + 162
t =
L 11.97.
4
The projectile is 1538 ft above ground twice: at
t≠8 sec, on the way up, and at t≠12 sec, on the
way down.
(b) The graph of s=–16t2+320t lies below the graph
of s=1538 for 0<t<8 and for 12<t<20
(approximately). Hence the projectile’s height will be
at most 1538 ft when t is in the interval (0, 8] or
[12, 20) (approximately).
(c) The graph of s=–16t2+320t lies above the graph
of s=1538 for 8<t<12 (approximately). Hence
the projectile’s height will be greater than or equal to
1538 when t is in the interval [8, 12] (approximately).
28
Chapter P
Prerequisites
82. Let the take-off point be located at (0, 0). We want the
4
slope between (0, 0) and (d, 20,000) to be .
9
20,000 - 0
4
=
d - 0
9
180,000=4d
45,000=d
The airplane must fly 45,000 ft horizontally to reach an
altitude of 20,000 ft.
83. (a) Let w>0 be the width of a rectangle; the length is
3w+1 and the perimeter is P=2[w+(3w+1)].
Solve P 150.
2[w+(3w+1)] 150
2(4w+1) 150
8w+2 150
8w 148
w 18.5
Thus P 150 cm when w is in the interval (0, 18.5].
(b) The area is A=w(3w+1). Solve A>1500.
w(3w+1)>1500
3w2+w-1550>0
The graph of A=3w2+w-1500 appears to be
zero for w≠22.19 when w is positive, and lies above
the w-axis for w>22.19. Hence, A>1500 when w is
in the interval (22.19, q ) (approximately).
Section 1.1
Modeling and Equation Solving
29
Chapter 1
Functions and Graphs
■ Section 1.1 Modeling and Equation Solving
8. x2-3x+4
9. (2x-1)(x-5)
Exploration 1
1. k =
10. (x2+5)(x2-4)=(x2+5)(x+2)(x-2)
100 - 25
75
d
=
=
= 0.75
m
100
100
Section 1.1 Exercises
1. (d) (q)
2. t=6.5%+0.5%=7% or 0.07
2. (f) (r)
d
, s = d + td
k
s = pm
3. m =
3. (a) (p)
4. (h) (o)
d1 1 + t2
s
d + td
d + td # k
p =
=
=
=
m
d
1
d
1
k
k11 + t 2
= 1 0.75 2 11.072 = 0.8025
=
1
#
k
d
4. Yes, because $36.99 *0.8025=$29.68.
5. $100÷0.8025=$124.61
1. Because the linear model maintains a constant positive
slope, it will eventually reach the point where 100% of the
prisoners are female. It will then continue to rise, giving
percentages above 100%, which are impossible.
2. Yes, because 2009 is still close to the data we are modeling. We would have much less confidence in the linear
model for predicting the percentage 25 years from 2000.
3. One possible answer: Males are heavily dominant in
violent crime statistics, while female crimes tend to be
property crimes like burglary or shoplifting. Since property crimes rates are senstive to economic conditions, a
statistician might look for adverse economic factors in
1990, especially those that would affect people near or
below the poverty level.
4. Yes. Table 1.1 shows that the minimum wage worker had
less purchasing power in 1990 than in any other year since
1955, which gives some evidence of adverse economic
conditions among lower-income Americans that year.
Nonetheless, a careful sociologist would certainly want to
look at other data before claiming a connection between
this statistic and the female crime rate.
Quick Review 1.1
2. (x+5)(x+5)
3. (9y+2)(9y-2)
4. 3x1 x2 - 5x + 6 2 = 3x(x-2)(x-3)
5. (4h +9) 14h - 92 =(4h +9)(2h+3)(2h-3)
6. (x+h)(x+h)
7. (x+4)(x-1)
7. (g) (t)
8. (j) (k)
9. (i) (m)
10. (c) (n)
(b) The greatest increase occurred between 1974 and
1979.
12. (a) Except for some minor fluctuations, the percentage
has been decreasing overall.
(b) The greatest decrease occurred between 1979 and
1984.
13. Women (), Men 1 + 2
[–5, 55] by [23, 92]
14. Vice versa: The female percentages are increasing faster
than the male percentages are decreasing.
15. To find the equation, first find the slope.
change in y
58.5 - 32.3
26.2
=
=
Women: Slope=
change in x
1999 - 1954
45
= 0.582. The y-intercept is 32.3, so the equation of the
line is y = 0.582x + 32.3.
-9.5
74.0 - 83.5
=
= -0.211. The
1999 - 1954
45
y-intercept is 83.5, so the equation of the line is
y = -0.211x + 83.5.
Men: Slope =
1. (x+4)(x-4)
2
6. (b) (s)
11. (a) The percentage increased from 1954 to 1999 and then
decreased slightly from 1999 to 2004.
Exploration 2
2
5. (e) (l)
2
In both cases, x represents the number of years after 1954.
16. 2009 is 55 years since 1954, so x = 55.
Women: y = 10.5822 1 552 + 32.3 L 64.3%
Men: y = 1 -0.211 2 155 2 + 83.5 L 71.9%
30
Chapter 1
Functions and Graphs
17. For the percentages to be the same, we need to set the
two equations equal to each other.
0.582x + 32.3 = -0.211x + 83.5
0.793x = 51.2
x L 64.6
(b) To find the equation, first find the slope:
213.9
666.1 - 452.2
=
L 23.77.
Slope =
2000 - 1991
9
The y-intercept is 452.3, so the equation of the line is
y = 23.77x + 452.3.
So, approximately 65 years after 1954 (2018), the models
predict that the percentages will be about the same. To
check:
Males: y = 1 - 0.211 2 165 2 + 83.5 L 69.9%
Females: y = 10.582 2 165 2 + 32.3 L 69.9%
18. The linear equations will eventually give percentages
above 100% for women and below 0% for men, neither
of which is possible.
19.
20. Let h be the height of the rectangular cake in inches.
The volume of the rectangular cake is
V1 = 9 # 13 # h = 117h in.3
The volume of the round cake is
V2 = ∏1 42 2 12h2 L 3.14 # 16 # 2h = 100.48h in.3
The rectangular cake gives a greater amount of cake for
the same price.
21. Because all stepping stones have the same thickness, what
matters is area.
The area of a square stepping stone is
A1 = 12 # 12 = 144 in.2
The area of a round stepping stone is
13 2
A2=∏ a b ≠3.14(6.5)2=132.665 in.2
2
The square stones give a greater amount of rock for the
same price.
22. (a) t=14 1180≠3.35 sec
(b) d=16(12.5)2=2500 ft
23. A scatter plot of the data suggests a parabola with its
vertex at the origin.
[–1, 6] by [–5, 35]
The model y=1.2t2 fits the data.
[–1, 15] by [400, 750]
(c) To find the year the number of passengers should
reach 900, let y=900, and solve the equation for x.
900=23.77x+452.3; x L 19, so by the model, the
number of passengers should reach 900 million by
2010 (1991+19).
(d) The terrorist attacks on September 11, 2001, caused a
major disruption in American air traffic from which
the airline industry was slow to recover.
25. The lower line shows the minimum salaries, since they are
lower than the average salaries.
26. The points that show the 1990 salaries are the Year 10
points. Both graphs show unprecedented increases in that
year. Note: At year 10 the minimum salary jumps, but at
year 11 the average salary jumps.
27. The 1995 points are third from the right, Year 15, on both
graphs. There is a clear drop in the average salary right
after the 1994 strike.
28. One possible answer: (a) The players will be happy to see
the average salary continue to rise at this rate. The discrepancy between the minimum salary and the average salary
will not bother baseball players like it would factory workers, because they are happy to be in the major leagues with
the chance to become a star. (b) The team owners are not
happy with this graph because it shows that their top players are being paid more and more money, forcing them to
pay higher salaries to be competitive. This benefits the
wealthiest owners. (c) Fans are unhappy with the higher
ticket prices and with the emphasis on money in baseball
rather than team loyalty. Fans of less wealthy teams are
unhappy that rich owners are able to pay high salaries to
build super-teams filled with talented free agents.
29. Adding 2v2+5 to both sides gives 3v2=13. Divide both
13
13
sides by 3 to get v2 =
.
, so v=—
3
A3
2
2
3v =13 is equivalent to 3v -13=0. The graph of
y=3v2-13 is zero for v≠–2.08 and for v≠2.08.
24. (a)
[–1, 15] by [400, 750]
[–5, 5] by [–15, 15]
Section 1.1
Modeling and Equation Solving
31
30. x+11=—11 so x=–11 — 11, which gives x=–22 or
x=0.
(x+11)2=121 is equivalent to (x+11)2-121=0.
The graph of y=(x+11)2-121 is zero for x=–22
and for x=0.
[–10, 10] by [–15, 15]
[–30, 30] by [–150, 150]
31.
2x2-5x+2=x2-5x+6+3x
x2-3x-4=0
(x-4)(x+1)=0
x-4=0 or x+1=0
x=4 or
x=–1
2x2-5x+2=(x-3)(x-2)+3x is equivalent
to 2x2-8x+2-(x-3)(x-2)=0. The graph
of y=2x2-8x+2-(x-3)(x-2) is zero for
x=–1 and for x=4.
34. Rewrite as 2x2-x-10=0; the left side factors to
(x+2)(2x-5)=0:
x+2 =0
or 2x-5=0
x =–2
2x=5
x=2.5
The graph of y=2x2-x-10 is zero for x=–2 and
for x=2.5.
[–10, 10] by [–10, 10]
2
35. x +7x-14=0, so
a=1, b=7, and c=–14:
[–10, 10] by [–10, 10]
32.
x2-7x=
3
4
-7 ; 272 - 4112 1 -142
-7 ; 2105
2
7
1
=– _ 2105
2
2
The graph of y=x2+7x-14 is zero for x≠–8.62
and for x≠1.62.
x=
2 11 2
=
7 2
7 2
x2-7x+ a - b =0.75+ a - b
2
2
(x-3.5)2=0.75+12.25
x-3.5=— 213
x=3.5_ 213
3
The graph of y=x2-7x- is zero for x≠–0.11 and
4
for x≠7.11.
[–20, 20] by [–30, 30]
2
36. x -4x-12=0, so
a=1, b=–4, and c=–12:
[–10, 10] by [–15, 15]
33. Rewrite as 2x2-5x-12=0; the left side factors to
(2x+3)(x-4)=0:
2x+3=0 or x-4=0
2x=–3
x=4
x=–1.5
The graph of y=2x2-5x-12 is zero for x=–1.5
and for x=4.
4 ; 21 -4 2 2 - 411 2 1 -122
4 ; 264
2
2 11 2
8
=2_ =2_4
2
x=–2 or x=6
The graph of y=x2-4x-12 is zero for x=–2 and
for x=6.
x=
[–20, 20] by [–30, 30]
=
32
Chapter 1
Functions and Graphs
37. Change to x2-2x-15=0 (see below); this factors to
(x+3)(x-5)=0, so x=–3 or x=5. Substituting
the first of these shows that it is extraneous.
x+1=2 1x + 4
(x+1)2=22 1 1x + 42 2
x2+2x+1=4x+16
x2-2x-15=0
The graph of y=x+1-2 1x + 4 is zero for x=5.
41. x≠1.33 or x=4
[–10, 10] by [–10, 10]
42. x≠2.66
[–10, 10] by [–10, 10]
38. Change to x2-3x+1=0 (see below);
3
1
3 ; 19 - 4
then x =
= ; 15, so
2
2
2
3
15
x= . Substituting the second of these shows that
2
2
it is extraneous.
1x=1-x
1 1x2 2=(1-x)2
x=1-2x+x2
0=x2-3x+1
1x+x=1 is equivalent to x+ 1x-1=0.
[–10, 10] by [–2, 2]
43. x≠1.77
[–5, 5] by [–10, 10]
44. x≠2.36
The graph of y=x+ 1x-1 is zero for x≠0.38.
[–5, 5] by [–10, 10]
[–3, 3] by [–2, 2]
45. x≠–1.47
39. x≠3.91
[–4, 4] by [–10, 10]
[–10, 10] by [–10, 10]
46. {0, 1, –1}
40. x≠–1.09 or x≠2.86
[–3, 3] by [–1, 4]
[–10, 10] by [–10, 10]
47. Model the situation using C=0.18x+32, where x is the
number of miles driven and C is the cost of a day’s rental.
(a) Elaine’s cost is 0.18(83)+32=$46.94.
(b) If for Ramon C=$69.80, then
69.80 - 32
x=
=210 miles.
0.18
Section 1.1
48. (a) 4x+5-(x3+2x2-x+3)=0 or
–x3-2x2+5x+2 =0
(b) –x3-2x2+5x+2=0
(c) A vertical line through the x-intercept of y3 passes
through the point of intersection of y1 and y2.
(d) At x=1.6813306, y1=y2=11.725322.
At x=–0.3579264, y1=y2=3.5682944.
At x=–3.323404, y1=y2=–8.293616.
49. (a) y = 1x200 2 1>200 = x200>200 = x1 = x for all x 0.
(b) The graph looks like this:
Modeling and Equation Solving
33
53. Let n be any integer.
n2+2n=n(n+2), which is either the product of two
odd integers or the product of two even integers.
The product of two odd integers is odd.
The product of two even integers is a multiple of 4, since
each even integer in the product contributes a factor of 2
to the product.
Therefore, n2+2n is either odd or a multiple of 4.
54. One possible story: The jogger travels at an approximately constant speed throughout her workout. She jogs
to the far end of the course, turns around and returns to
her starting point, then goes out again for a second trip.
55. False. A product is zero if any factor is zero. That is, it
takes only one zero factor to make the product zero.
[0, 1] by [0, 1]
(c) Yes, this is different from the graph of y=x.
(d) For values of x close to 0, x200 is so small that the
calculator is unable to distinguish it from zero. It
returns a value of 01/200 =0 rather than x.
50. The length of each side of the square is x+b, so the area
of the whole square is (x+b)2. The square is made up
of one square with area x # x=x2, one square with area
b # b=b2, and two rectangles, each with area
b # x=bx. Using these four figures, the area of the square
is x2+2bx+b2.
51. (a) x=–3 or x=1.1 or x=1.15.
56. False. Predictions are always fallible, and in particular
an algebraic model that fits the data well for a certain
range of input values may not work for other input
values.
57. This is a line with a negative slope and a y-intercept of 12.
The answer is C. (The graph checks.)
58. This is the graph of a square root function, but flipped
left-over-right. The answer is E. (The graph checks.)
59. As x increases by ones, the y-values get farther and farther apart, which implies an increasing slope and suggests
a quadratic equation. The answer is B. (The equation
checks.)
60. As x increases by 2’s, y increases by 4’s, which implies a
constant slope of 2. The answer is A. (The equation
checks.)
61. (a) March
(b) $120
(c) June, after three months of poor performance
[–5, 5] by [–200, 500]
(b) x=–3 only.
(d) Ahmad paid (100)($120)=$12,000 for the stock and
sold it for (100)($100)=$10,000. He lost $2,000 on
the stock.
(e) After reaching a low in June, the stock climbed back
to a price near $140 by December. LaToya’s shares
had gained $2000 by that point.
(f) One possible graph:
Stock Index
120
100
80
60
[–10, 10] by [–5, 5]
40
b
b
52. (a) Area: x + x a b + x a b = x2 + bx
2
2
2
#
b
b 2
= a b
2
2
b 2
b 2
(c) x + bx + a b = a x + b is the algebraic
2
2
formula for completing the square, just as the area
b 2
a b completes the area x2 + bx to form the area
2
b 2
ax + b .
2
0
Jan
Fe .
b
Ma .
Apr.
Mar.
Juny
Jul e
Au y
Se g.
p
Oct.
No t.
Dev.
c.
b
(b)
2
20
62. (a)
2
[–4, 4] by [–10, 10]
34
Chapter 1
Functions and Graphs
(b) Factoring, we find y=(x+2)(x-2)(x-2).
There is a double zero at x=2, a zero at x=–2,
and no other zeros (since it is a cubic).
(c) Same visually as the graph in (a).
(d) b2-4ac is the discriminant. In this case,
b2-4ac=(–4)2-4(1)(4.01)=–0.04, which is
negative. So the only real zero of the product
y=(x+2)(x2-4x+4.01) is at x=–2.
(e) Same visually as the graph in (a).
(f) b2-4ac=(–4)2-4(1)(3.99)=0.04, which is positive. The discriminant will provide two real zeros of
the quadratic, and (x+2) provides the third. A cubic
equation can have no more than three real roots.
63. (a)
Subscribers
Monthly Bills
(f) In 1995, cellular phone technology was still emerging,
so the growth rate was not as fast as it was in more
recent years. Thus, the slope from 1995 (t=5) to 1998
(t=8) is lower than the slope from 1998 to 2004.
Cellular technology was more expensive before competition brought prices down. This explains the anomaly on the monthly bill scatter plot.
64. One possible answer: The number of cell phone users is
increasing steadily (as the linear model shows), and the
average monthly bill is climbing more slowly as more
people share the industry cost. The model shows that the
number of users will continue to rise, although the linear
model cannot hold up indefinitely.
■ Section 1.2 Functions and Their Properties
Exploration 1
1. From left to right, the tables are (c) constant,
(b) decreasing, and (a) increasing.
2.
[7, 15] by [50, 200]
[7, 15] by [35, 55]
(b) The graph for subscribers appears to be linear. Since
time t=the number of years after 1990, t=8 for
1998 and t=14 for 2004. The slope of the line is
111.2
180.4 - 69.2
=
L 18.53.
14 - 8
6
Use the point-slope form to write the equation:
y - 69.2 = 18.531 x - 8 2.
Solve for y: y - 69.2 = 18.53x - 148.24
y = 18.53x - 79.04
The linear model for subscribers as a function of years
is y = 18.53x - 79.04.
(c) The fit is very good. The line goes through or is close
to all the points.
X
X
X
moves ≤X ≤Y1 moves ≤X ≤Y2 moves ≤X≤Y3
from
from
from
–2 to –1 1
0 –2 to –1 1 –2 –2 to –1 1
2
–1 to 0
1
0 –1 to 0
1 –1 –1 to 0
1
2
0 to 1
1
0
0 to 1
1 –2
0 to 1
1
2
1 to 3
2
0
1 to 3
2 –4
1 to 3
2
3
3 to 7
4
0
3 to 7
4 –6
3 to 7
4
6
3. For an increasing function, ≤Y/≤X is positive. For a
decreasing function, ≤Y/≤X is negative. For a constant
function, ≤Y/≤X is 0.
4. For lines, ≤Y/≤X is the slope. Lines with positive slope
are increasing, lines with negative slope are decreasing,
and lines with 0 slope are constant, so this supports our
answers to part 3.
Quick Review 1.2
1. x2 - 16 = 0
x2 = 16
x = ;4
[7, 15] by [50, 200]
(d) The monthly bill scatter plot has a curved shape that
could be modeled more effectively by a function with
a curved graph. Some possibilities include a quadratic
function (parabola), a logarithmic function, a power
function (e.g., square root), a logistic function, or a
sine function.
(e)
Subscribers
Monthly Bills
2. 9 - x2 = 0
9 = x2
;3 = x
3. x - 10 6 0
x 6 10
4. 5 - x 0
-x -5
x 5
5. As we have seen, the denominator of a function cannot
be zero.
We need x - 16 = 0
x = 16
[4, 15] by [10, 200]
[4, 15] by [30, 60]
6. We need
x2 - 16 = 0
x2 = 16
x = ;4
7. We need
x - 16 6 0
x 6 16
Section 1.2
8. We need
9. We need
x2 - 1 = 0
x2 = 1
x = ;1
3-x 0
and
10. We need
35
12. We need x Z 0 and x-3 Z 0. Domain:
(–q, 0) ª (0, 3) ª (3, q).
x+2<0
3 x
x<–2
Functions and Their Properties
x<–2
and
x 3
2
x - 4 = 0
x2 = 4
x = ;2
Section 1.2 Exercises
1. Yes, y = 2x - 4 is a function of x, because when a
number is substituted for x, there is at most one value
produced for 2x - 4.
[–10, 10] by [–10, 10]
13. We notice that g1 x2 =
x
x
=
.
x1x - 5 2
x2 - 5x
As a result, x - 5 Z 0 and x Z 0.
Domain: (–q, 0) ª (0, 5) ª (5, q).
2. No, y=x2 — 3 is not a function of x, because when a
number is substituted for x, y can be either 3 more or 3
less than x2.
3. No, x=2y2 does not determine y as a function of x,
because when a positive number is substituted for x, y can
be either
x
x
or –
.
B2
B2
4. Yes, x=12-y determines y as a function of x,
because when a number is substituted for x, there is
exactly one number y which, when subtracted from 12,
produces x.
[–10, 10] by [–5, 5]
14. We need x-3 Z 0 and 4-x2 0. This means x Z 3
and x2 4; the latter implies that –2 x 2, so the
domain is [–2, 2].
5. Yes
6. No
7. No
8. Yes
9. We need x2+4 0; this is true for all real x.
Domain: (–q, q).
[–3, 3] by [–2, 2]
15. We need x+1 Z 0, x2+1 Z 0, and 4-x 0.
The first requirement means x Z –1, the second is true
for all x, and the last means x 4. The domain is therefore (–q, –1) ª (–1, 4].
[–5, 5] by [–5, 15]
10. We need x-3 Z 0. Domain: (–q, 3) ª (3, q).
[–5, 5] by [–5, 5]
16. We need
x4 - 16x2
x 1 x2 - 162
or x2 - 16
x2
2
x2=0
[–5, 15] by [–10, 10]
11. We need x+3 Z 0 and x-1 Z 0. Domain:
(–q, –3) ª (–3, 1) ª (1, q).
[–10, 10] by [–10, 10]
x=0
0
0
0
16
or x 4, x -4
Domain: (–q, –4] ª {0} ª [4, q)
[–5, 5] by [0, 16]
36
Chapter 1
Functions and Graphs
17. f(x)=10-x2 can take on any negative value. Because
x2 is nonnegative, f(x) cannot be greater than 10. The
range is (–q, 10].
24. Yes, non-removable
18. g1 x2 = 5 + 24 - x can take on any value 5, but
because 24 - x is nonnegative, g(x) cannot be less than 5.
The range is [5, q).
19. The range of a function is most simply found by
graphing it. As our graph shows, the range of f(x) is
(–q, –1) ª [0, q).
[–5, 5] by [–5, 5]
25. Local maxima at (–1, 4) and (5, 5), local minimum at
(2, 2). The function increases on (–q, –1], decreases on
[–1, 2], increases on [2, 5], and decreases on [5, q).
26. Local minimum at (1, 2), (3, 3) is neither, and (5, 7) is a
local maximum. The function decreases on (–q, 1],
increases on [1, 5], and decreases on [5, q).
[–10, 10] by [–10, 10]
20. As our graph illustrates, the range of g(x) is
(–q, –1) ª [0.75, q).
27. (–1, 3) and (3, 3) are neither. (1, 5) is a local maximum,
and (5, 1) is a local minimum. The function increases on
(–q, 1], decreases on [1, 5], and increases on [5, q).
28. (–1, 1) and (3, 1) are local minima, while (1, 6) and (5, 4)
are local maxima. The function decreases on (–q, –1],
increases on [–1, 1], decreases on (1, 3], increases on
[3, 5], and decreases on [5, q).
29. Decreasing on (–q, –2]; increasing on [–2, q)
[–10, 10] by [–10, 10]
21. Yes, non-removable
[–10, 10] by [–2, 18]
30. Decreasing on (–q, –1]; constant on [–1, 1];
increasing on [1, q)
[–10, 10] by [–10, 10]
22. Yes, removable
[–10, 10] by [–2, 18]
31. Decreasing on (–q, –2]; constant on [–2, 1];
increasing on [1, q)
[–5, 5] by [–10, 10]
23. Yes, non-removable
[–10, 10] by [0, 20]
32. Decreasing on (–q, –2]; increasing on [–2, q)
[–10, 10] by [–2, 2]
[–7, 3] by [–2, 13]
Section 1.2
33. Increasing on (–q, 1]; decreasing on [1, q)
Functions and Their Properties
37
43. Local minimum: y≠–4.09 at x≠–0.82.
Local maximum: y≠–1.91 at x≠0.82.
[–4, 6] by [–25, 25]
34. Increasing on (–q, –0.5]; decreasing on [–0.5, 1.2],
increasing on [1.2, q). The middle values are approximate
—they are actually at about –0.549 and 1.215. The values
given are what might be observed on the decimal window.
[–2, 3] by [–3, 1]
35. Constant functions are always bounded.
36.
2
x 7 0
[–5, 5] by [–50, 50]
44. Local maximum: y≠9.48 at x≠–1.67.
Local minimum: y=0 when x=1.
[–5, 5] by [–50, 50]
45. Local maximum: y≠9.16 at x≠–3.20.
Local minima:
y=0 at x=0 and y=0 at x=–4.
-x2 6 0
2 - x2 6 2
y is bounded above by y=2.
37. 2x>0 for all x, so y is bounded below by y=0.
38. 2–x=
1
7 0 for all x, so y is bounded below by y=0.
2x
39. Since y = 21 - x2 is always positive, we know that
y 0 for all x. We must also check for an upper bound:
x2 7 0
-x2 6 0
1 - x2 6 1
21 - x2 6 21
21 - x2 6 1
Thus, y is bounded.
40. There are no restrictions on either x or x3, so y is not
bounded above or below.
41. f has a local minimum when x=0.5, where y=3.75.
It has no maximum.
[–5, 5] by [0, 80]
46. Local maximum: y=0 at x=–2.5.
Local minimum: y≠–3.13 at x=–1.25.
[–5, 5] by [–10, 10]
47. Even: f1 -x2 = 2 1 -x2 4 = 2x4 = f1x2
48. Odd: g1 -x2 = 1 -x2 3 = -x3 = -g1 x2
49. Even: f1 -x2 = 21 - x2 2 + 2 = 2x2 + 2 = f1 x2
50. Even: g1 -x2 =
3
3
=
= g1x2
1 + 1 - x2 2
1 + x2
51. Neither: f(–x)=–(–x)2+0.03(–x)+5=
–x2-0.03x+5, which is neither f(x) nor –f(x).
[–5, 5] by [0, 36]
42. Local maximum: y≠4.08 at x≠–1.15.
Local minimum: y≠–2.08 at x≠1.15.
52. Neither: f(–x)=(–x)3+0.04(–x)2+3=
–x3+0.04x2+3, which is neither f(x) nor –f(x).
53. Odd: g1 -x2 = 2 1 -x2 3 - 31 -x2
=–2x3+3x=–g(x)
54. Odd: h1 -x2 =
[–5, 5] by [–50, 50]
1
1
= - = - h1x2
-x
x
38
Chapter 1
Functions and Graphs
x
is undefined at x=1, indicating that
x - 1
x
x=1 is a vertical asymptote. Similarly, lim
= 1,
xS∞ x - 1
indicating a horizontal asymptote at y=1. The graph
confirms these.
55. The quotient
x2 + 2
is undefined at x=1 and x=–1.
x2 - 1
So we expect two vertical asymptotes. Similarly, the
x2 + 2
= 1, so we expect a horizontal asymptote
lim 2
xS∞ x - 1
at y=1. The graph confirms these asymptotes.
59. The quotient
[–10, 10] by [–10, 10]
x - 1
is undefined at x=0, indicating
x
a possible vertical asymptote at x=0. Similarly,
x - 1
lim
= 1, indicating a possible horizontal asympxS∞
x
tote at y=1. The graph confirms those asymptotes.
56. The quotient
[–10, 10] by [–10, 10]
x + 2
is undefined at x=3, indicating
3 - x
a possible vertical asymptote at x=3. Similarly,
x + 2
lim
= -1, indicating a possible horizontal asympxS∞ 3 - x
tote at y=–1. The graph confirms these asymptotes.
57. The quotient
[–8, 12] by [–10, 10]
58. Since g(x) is continuous over –q<x<q,
we do not expect a vertical asymptote. However,
1
lim 1.5x = lim 1.5–x = lim
= 0, so we expect a
xS-∞
xS∞
xS∞ 1.5x
horizontal asymptote y=0. The graph confirms this
asymptote.
[–10, 10] by [–10, 10]
60. We note that x2 + 1 7 0 for –q<x<q, so we do
not expect a vertical asymptote. However,
4
= 0, so we expect a horizontal asymptote at
lim
xS∞ x2 + 1
y=0. The graph confirms this.
[–5, 5] by [0, 5]
4x - 4
does not exist at x=2,
x3 - 8
so we expect a vertical asymptote there. Similarly,
4x - 4
= 0, so we expect a horizontal asymptote
lim
xS∞ x3 + 8
at y=0. The graph confirms these asymptotes.
61. The quotient
[–4, 6] by [–5, 5]
21x - 22
2
2x - 4
. Since
=
=
1 x - 22 1x + 22
x + 2
x2 - 4
x=2 is a removable discontinuity, we expect a vertical
2
asymptote at only x=–2. Similarly, lim
= 0, so
xS∞ x - 2
we expect a horizontal asymptote at y=0. The graph
confirms these asymptotes.
62. The quotient
[–10, 10] by [–10, 10]
[–6, 4] by [–10, 10]
Section 1.2
1
63. The denominator is zero when x = - , so there is a
2
1
vertical asymptote at x = - . When x is very large,
2
x + 2
1
x
behaves much like
= , so there is a horizontal
2x + 1
2x
2
1
asymptote at y = . The graph matching this description
2
is (b).
1
64. The denominator is zero when x = - , so there is a
2
1
vertical asymptote at x = - . When x is very large,
2
x
x2 + 2
x2
x
behaves much like
= , so y = is a slant
2x + 1
2x
2
2
asymptote. The graph matching this description is (c).
65. The denominator cannot equal zero, so there is no vertical
x + 2
asymptote. When x is very large, 2
behaves much
2x + 1
1
x
like 2 =
, which for large x is close to zero. So there
2x
2x
is a horizontal asymptote at y=0. The graph matching
this description is (a).
66. The denominator cannot equal zero, so there is no vertical
x3 + 2
asymptote. When x is very large, 2
behaves much
2x + 1
3
x
x
x
like 2 = , so y = is a slant asymptote. The graph
2
2
2x
matching this description is (d).
x
= 0, we expect a horizontal
x2 - 1
asymptote at y=0. To find where our function
crosses y=0, we solve the equation
x
= 0
2
x - 1
x = 0 # 1x2 - 1 2
x = 0
The graph confirms that f(x) crosses the horizontal
asymptote at (0, 0).
67. (a) Since, lim
xS∞
Functions and Their Properties
39
[–10, 10] by [–5, 5]
x2
= 0, we expect a horizontal
x + 1
asymptote at y=0. To find where h(x) crosses
y=0, we solve the equation
x2
= 0
x3 + 1
2
x = 0 # 1x3 + 12
x2 = 0
x = 0
The graph confirms that h(x) intersects the horizontal
asymptote at (0, 0).
(c) Since lim
xS∞
3
[–5, 5] by [–5, 5]
68. We find (a) and (c) have graphs with more than one horizontal asymptote as follows:
(a) To find horizontal asymptotes, we check limits, at
x S q and x S -q. We also know that our
numerator |x3+1|, is positive for all x, and that our
denominator, 8-x3, is positive for x<2 and negative for x>2. Considering these two statements, we
find
0 x3 + 1 0
0 x3 + 1 0
and
lim
=
-1
lim
= 1.
xS∞ 8 - x3
xS–∞ 8 - x3
The graph confirms that we have horizontal asymptotes at y=1 and y=–1.
[–10, 10] by [–5, 5]
[–10, 10] by [–10, 10]
x
= 0, we expect a horizontal
x2 + 1
asymptote at y=0. To find where our function
crosses y=0, we solve the equation:
x
= 0
x2 + 1
x = 0 # 1x2 + 1 2
x = 0
The graph confirms that g(x) crosses the horizontal
asymptote at (0, 0).
(b) Since lim
xS∞
(b) Again, we see that our numerator, @ x - 1 @ , is positive
for all x. As a result, g(x) can be negative only when
x2-4<0, and g(x) can be positive only when
x2-4>0. This means that g(x) can be negative
only when –2<x<2; if x<–2 or x>2, g(x) will
be positive. As a result, we know that
0x - 10
0x - 10
lim
= lim 2
= 0, giving just one
xS∞ x2 - 4
xS–∞ x - 4
horizontal asymptote at y=0. Our graph confirms
this asymptote.
40
Chapter 1
Functions and Graphs
76. The height of a swinging pendulum goes up and down
over time as the pendulum swings back and forth. The
answer is E.
77. (a)
[–5, 5] by [–5, 15]
(c) As we demonstrated earlier, we need x2-4>0
otherwise our function is not defined within the
real numbers. As a result, we know that our
denominator, 2x2 - 4, is always positive [and that
h(x) is defined only in the domain (–q, –2) ª (2, q)].
x
Checking limits, we find lim
= 1 and
xS∞
2
2x - 4
x
lim
= -1 . The graph confirms that we
xS–∞
2x2 - 4
have horizontal asymptotes at y=1 and y=–1.
[–3, 3] by [–2, 2]
k=1
x
(b)
6 1 3 x 6 1 + x2 3 x2 - x + 1 7 0
1 + x2
But the discriminant of x2-x+1 is negative (–3),
so the graph never crosses the x-axis on the interval
(0, q).
(c) k=–1
(d)
x
7 - 1 3 x 7 -1 - x2 3 x2 + x + 1 7 0
1 + x2
But the discriminant of x2+x+1 is negative (–3),
so the graph never crosses the x-axis on the interval
(–q, 0).
78. (a) Increasing
[–10, 10] by [–10, 10]
(b)
≤y
1
1.05
0.52
0.43
0.36
0.33
0.31
0.28
(c)
≤≤y
0.05
–0.53
–0.09
–0.07
–0.03
–0.02
–0.03
69. (a) The vertical asymptote is x=0, and this function is
undefined at x=0 (because a denominator can’t be
zero).
(b)
[–10, 10] by [–10, 10]
Add the point (0, 0).
(c) Yes. It passes the vertical line test.
70. The horizontal asymptotes are determined by the two
limits, lim f1 x2 and lim f1x2 . These are at most two
xS–∞
xS + ∞
different numbers.
71. True. This is what it means for a set of points to be the
graph of a function.
72. False. There are many function graphs that are symmetric
with respect to the x-axis. One example is f(x)=0.
73. Temperature is a continuous variable, whereas the other
quantities all vary in steps. The answer is B.
74. “Number of balls” represents a whole number, so that the
quantity changes in jumps as the ball radius is altered. The
answer is C.
75. Air pressure drops with increasing height. All the other
functions either steadily increase or else go both up and
down. The answer is C.
≤y is none of these since it first increases from 1 to
1.05 and then decreases.
(d) The graph rises, but bends downward as it rises.
(e) An example:
y
5
5
x
Section 1.2
79. One possible graph:
y
4x
4x
= lim
=
xS∞
x + 2x + 1
1x + 12 2
x
1
4
ba
b = lim
lim 4 a
xS∞
xS∞ x + 1
x + 1
x + 1
(since x+1≠x for very large x)=0.
4x
= 0] As a result, we
[Similarly, lim 2
xS–∞ x + x + 1
know that g(x) is bounded by y=0 as x goes to q
and –q.
However, g(x)>0 for all x>0 (since (x+1)2>0
always and 4x>0 when x>0), so we must check
points near x=0 to determine where the function is
at its maximum. [Since g(x)<0 for all x<0 (since
(x+1)2>0 always and 4x<0 when x<0) we can
ignore those values of x since we are concerned only
with the upper bound of g(x).] Examining our graph,
we see that g(x) has an upper bound at y=1, which
occurs when x=1. The least upper bound of
g(x)=1, and it is in the range of g(x).
(e) lim
xS∞
x
5
–5
80. One possible graph:
y
5
–5
41
(d) For all values of x, we know that sin(x) is bounded
above by y=1. Similarly, 2 sin(x) is bounded above
by y=2 # 1=2. It is in the range.
5
–5
Functions and Their Properties
5
y=1
x
2
y
x=0
81. One possible graph:
y
5
5
x
82. Answers vary.
83. (a)
x2 7 0
- 0.8x2 6 0
2 - 0.8x2 6 2
f(x) is bounded above by y=2. To determine if
y=2 is in the range, we must solve the equation for
x: 2 = 2 - 0.8x2
0 = -0.8x2
x
84. As the graph moves continuously from the point (–1, 5)
down to the point (1, –5), it must cross the x-axis somewhere along the way. That x-intercept will be a zero of the
function in the interval [–1, 1].
85. Since f is odd, f(–x)=–f(x) for all x. In particular,
f(–0)=–f(0). This is equivalent to saying that
f(0)=–f(0), and the only number which equals its
opposite is 0. Therefore, f(0)=0, which means the
graph must pass through the origin.
86.
2
0 = x
0 = x
Since f(x) exists at x=0, y=2 is in the range.
3x2
3x2
=
lim
= lim 3 = 3. Thus, g(x) is
xS∞ 3 + x2
xS∞ x2
xS∞
bounded by y=3. However, when we solve for x,
3x2
we get
3 =
3 + x2
2
313 + x 2 = 3x2
9 + 3x2 = 3x2
9 = 0
Since 9 Z 0, y=3 is not in the range of g(x).
(b) lim
(c) h(x) is not bounded above.
[–6, 6] by [–2, 2]
(a) y=1.5
(b) [–1, 1.5]
3x2 - 1
1.5
2x2 + 1
3x2 - 1
0 1+ 2
2.5
2x + 1
0 2x2+1+3x2-1 5x2+2.5
0 5x2 5x2+2.5
True for all x.
(c) –1 Chapter 1
42
Functions and Graphs
■ Section 1.3 Twelve Basic Functions
Exploration 1
1
1. The graphs of f(x)= and f(x)=ln x have vertical
x
asymptotes at x=0.
1
2. The graph of g(x)= +ln x (shown below) does have
x
a vertical asymptote at x=0.
Section 1.3 Exercises
1. y=x3+1; (e)
2. y = 0 x 0 - 2; (g)
3. y= - 1x; (j)
4. y=–sin x or y=sin(–x); (a)
5. y=–x; (i)
6. y=(x-1)2; (f)
7. y=int(x+1); (k)
1
8. y = - ; (h)
x
9. y=(x+2)3; (d)
10. y=ex-2 ; (c)
11. 2-
[–2.7, 6.7] by [–1.1, 5.1]
1
1
3. The graphs of f(x)= , f(x)=ex, and f(x)=
x
1 + e-x
have horizontal asymptotes at y=0.
1
4. The graph of g(x)= +ex (shown below) does have a
x
horizontal asymptote at y=0.
4
; (l)
1 + e-x
12. y=cos x+1; (b)
13. Exercise 8
14. Exercise 3
15. Exercises 7, 8
16. Exercise 7 (Remember that a continuous function is one
that is continuous at every point in its domain.)
17. Exercises 2, 4, 6, 10, 11, 12
18. Exercises 3, 4, 11, 12
1
19. y=x, y=x3, y= , y=sin x
x
20. y=x, y=x3, y= 1x, y=ex, y=ln x, y=
[–3, 3] by [–5, 5]
5.
1
1 + e-x
1
21. y=x2, y= , y= 0 x 0
x
22. y=sin x, y=cos x, y=int(x)
1
1
23. y= , y=ex, y=
x
1 + e-x
24. y=x, y=x3, y=ln x
[–4.7, 4.7] by [–3.1, 3.1]
1
1
1
Both f(x)= and g(x)= 2
=
have
x
x1 2x - 12
2x - x
vertical asymptotes at x=0, but h(x)=f(x)+g(x)
does not; it has a removable discontinuity.
Quick Review 1.3
1. 59.34
2. 5 - p
1
1
25. y= , y=sin x, y=cos x, y=
x
1 + e-x
26. y=x, y=x3, y=int(x)
1
27. y=x, y=x3, y= , y=sin x
x
28. y=sin x, y=cos x
29. Domain: All reals
Range: [–5, q)
30. Domain: All reals
Range: [0, q)
3. 7 - p
4. 3
5. 0
6. 1
7. 3
8. –15
9. –4
10. @ 1-∏ @ -∏=(∏-1)-∏=∏-1-∏=–1
[–10, 10] by [–10, 10]
[–10, 10] by [–10, 10]
Section 1.3
31. Domain: (–6, q)
Range: All reals
32. Domain: (–q, 0) ª (0, q)
Range: (–q, 3) ª (3, q)
Twelve Basic Functions
43
37.
[–5, 5] by [–1, 4]
(a) f(x) is increasing on (–q, q).
[–10, 10] by [–10, 10]
33. Domain: All reals
Range: All integers
[–5, 5] by [–2, 8]
(b) f(x) is neither odd nor even.
34. Domain: All reals
Range: [0, q)
(c) There are no extrema.
3
1
is the logistic function,
,
1 + e-x
1 + e-x
stretched vertically by a factor of 3.
(d) f1x2 =
38.
[–10, 10] by [–10, 10]
[–10, 10] by [–10, 10]
35.
[–11.4, 7.4] by [–2.2, 10.2]
(a) q(x) is increasing on (–q, q).
(b) q(x) is neither odd nor even.
(c) There are no extrema.
(d) q(x)=ex+2 is the exponental function, ex, shifted 2
units up.
[0, 20] by [–5, 5]
(a) r(x) is increasing on [10, q).
(b) r(x) is neither odd nor even.
39.
(c) The one extreme is a minimum value of 0 at x=10.
(d) r(x)= 2x - 10 is the square root function, shifted
10 units right.
36.
[–15, 15] by [–20, 10]
(a) h(x) is increasing on [0, q) and decreasing on (–q, 0].
(b) h(x) is even, because it is symmetric about the y-axis.
(c) The one extremum is a minimum value of –10 at
x=0.
[0, 7] by [2, 7]
(a) f(x) is increasing on c 12k - 1 2
∏
∏
, 1 2k + 12 d and
2
2
∏
∏
decreasing on c 12k + 12 , 1 2k + 3 2 d , where k is
2
2
an even integer.
(d) h(x)= 0 x 0 - 10 is the absolute value function, 0 x 0 ,
shifted 10 units down.
40.
(b) f(x) is neither odd nor even.
∏
(c) There are minimum values of 4 at x = 1 2k - 12
2
∏
and maximum values of 6 at x = 1 2k + 1 2 , where k
2
is an even integer.
(d) f(x)=sin(x)+5 is the sine function, sin x, shifted 5
units up.
[0, 7] by [–5, 5]
(a) g(x) is increasing on [(2k-1)∏, 2k∏] and decreasing on [2k∏, (2k+1)∏], where k is an integer.
(b) g(x) is even, because it is symmetric about the y-axis.
(c) There are minimum values of –4 at x=(2k-1)∏
and maximum values of 4 at x=2k∏, where k is an
integer.
(d) g(x)=4 cos (x) is the cosine function, cos x,
stretched vertically by a factor of 4.
44
Chapter 1
Functions and Graphs
41.
47.
y
5
5
[–2.7, 6.7] by [–1.1, 5.1]
x
(a) s(x) is increasing on [2, q) and decreasing on (–q, 2].
(b) s(x) is neither odd nor even.
(c) The one extremum is a minimum value of 0 at x=2.
(d) s(x)= 0 x - 2 0 is the absolute value function, 0 x 0 ,
shifted 2 units to the right.
There are no points of discontinuity.
48.
42.
y
5
5
x
[–10, 10] by [–10, 10]
(a) f(x) is increasing on (–q, 0] and decreasing on [0, q).
(b) f(x) is even, because it is symmetric about the y-axis.
(c) The one extremum is a maximum value of 5 at x=0.
There is a point of discontinuity at x=0.
49.
(d) f(x)=5-abs(x) is the absolute value function,
abs(x), reflected across the x-axis and then shifted 5
units up.
y
5
43. The end behavior approaches the horizontal asymptotes
y=2 and y=–2.
5
44. The end behavior approaches the horizontal asymptotes
y=0 and y=3.
45.
x
y
5
There are no points of discontinuity.
50.
5
y
5
x
5
x
There are no points of discontinuity.
46.
y
5
There are no points of discontinuity.
51.
5
x
y
5
5
x
There is a point of discontinuity at x=0.
There is a point of discontinuity at x=0.
Section 1.3
52.
Twelve Basic Functions
45
(b) One possible answer: It is similar because it has discontinuities spaced at regular intervals. It is different
because its domain is the set of positive real numbers,
and because it is constant on intervals of the form
(k, k+1] instead of [k, k+1), where k is an integer.
y
5
5
57. The Greatest Integer Function f(x)=int (x)
x
There are points of discontinuity at x=2, 3, 4, 5, . . . .
53. (a)
[–4.7, 4.7] by [–3.1, 3.1]
[–5, 5] by [–5, 5]
This is g(x)= 0 x 0 .
(b) Squaring x and taking the (positive) square root has
the same effect as the absolute value function.
54. (a)
f1 x2 = 2x2 = 2 0 x 0 2 = 0 x 0 = g1 x2
Domain: all real numbers
Range: all integers
Continuity: There is a discontinuity at each integer
value of x.
Increasing/decreasing behavior: constant on intervals of
the form [k, k+1), where k is an integer
Symmetry: none
Boundedness: not bounded
Local extrema: every non-integer is both a local minimum
and local maximum
Horizontal asymptotes: none
Vertical asymptotes: none
End behavior: int(x) S –q as x S –q and
int(x) S q as x S q.
58. False. Because the greatest integer function is not one-toone, its inverse relation is not a function.
59. True. The asymptotes are x=0 and x=1.
[–5, 5] by [–5, 5]
This appears to be f(x)= 0 x 0 .
(b) For example, g1 1 2 L 0.99 Z f11 2 = 1.
1
5
60. Because 3- Z 3, 0<
<5, –4 4 cos x 4,
x
1 + e-x
and int(x-2) takes only integer values. The answer is A.
55. (a)
61. 3<3+
1
<4. The answer is D.
1 + e-x
62. By comparison of the graphs, the answer is C.
63. The answer is E. The others all have either a restricted
domain or intervals where the function is decreasing or
constant.
[–5, 5] by [–5, 5]
64. (a) Answers will vary.
This is the function f(x)=x.
(b) The fact that lnA ex B = x shows that the natural
logarithm function takes on arbitrarily large values.
In particular, it takes on the value L when x=eL.
56. (a)
(b) In this window, it appears that 1x 6 x 6 x2:
y
Cost ($)
1.29
[0, 30] by [0, 20]
1.06
(c)
0.83
0.60
0.37
1
2
3
4
Weight (oz)
5
x
[0, 2] by [0, 1.5]
46
Chapter 1
Functions and Graphs
(d) On the interval (0, 1), x2 6 x 6 1x.
On the interval (1, q), 1x 6 x 6 x2.
All three functions equal 1 when x=1.
65. (a) A product of two odd functions is even.
(b) A product of two even functions is even.
(c) A product of an odd function and an even function is
odd.
If f = x - 3 and g=ln(e3 x), then f g=ln(e3 x)-3=
ln(e3)+ln x-3=3 ln e+ln x-3=3+ln x-3=
ln x.
x
x
x
If f=2 sin x cos x and g = , then f g=2 sin cos =
2
2
2
x
sin a 2 a b b = sin x. This is the double angle formula
2
(see Section 5.4). You can see this graphically.
66. Answers vary.
67. (a) Pepperoni count ought to be proportional to the area
of the pizza, which is proportional to the square of the
radius.
(b) 12 = k142 2
12
3
k =
= = 0.75
16
4
(c) Yes, very well.
[0, 2␲] by [–2, 2]
x
If f = 1 - 2x and g = sin a b ,
2
x
2 x
then f g=1-2 a sin a b b = cos a 2 a b b = cos x.
2
2
(The double angle formula for cos 2x is cos 2x=cos2 xsin2 x=(1-sin2 x)-sin2 x=1-2 sin2 x. See Section 5.3.)
This can be seen graphically:
2
(d) The fact that the pepperoni count fits the expected
quadratic model so perfectly suggests that the pizzeria
uses such a chart. If repeated observations produced
the same results, there would be little doubt.
68. (a) y=ex and y=ln x
1
(b) y=x and y=
x
(c) With domain [0, q), y=x2 becomes the inverse of
y= 1x.
1
69. (a) At x=0, does not exist, ex=1, ln x is not defined,
x
1
= 1.
cos x=1, and
1 + e-x
[0, 2␲] by [–2, 2]
(b) for f(x)=x, f(x+y)=x+y=f(x)+f(y)
f
(c) for f(x)=ex, f(xy)=exy=exey=f(x) # f(y)
2x 3
(d) for f(x)=ln x, f(x+y)=ln(xy)=ln(x)+ln(y)
=f(x)+f(y)
1
(e) The odd functions: x, x3, , sin x
x
2x 4
x
■ Section 1.4 Building Functions from
Functions
5
x
x3
2 sin x cos x
Exploration 1
If f=2x-3 and g =
x + 3
, then
2
f g=2 a
x + 3
b - 3 = x + 3 - 3 = x.
2
1x - 2 2 1 x + 22
If f = 0 2x + 4 0 and g =
,
2
1x - 2 2 1x + 22
then f g=2 a
b + 4
2
= 1x - 22 1x + 2 2 + 4 = x2 - 4 + 4 = x2.
If f = 1x and g = x2, then f g= 2x2 = 0 x 0 . Note, we
use the absolute value of x because g is defined for
- q 6 x 6 +q, while f is defined only for positive values of
x. The absolute value function is always positive.
If f = x5 and g = x0.6, then f g= 1x0.6 2 5 = x3.
1 2x2
Quick Review 1.4
1. (–q, –3) ª (–3, q)
2. (1, q)
3. (–q, 5]
4. (1/2, q)
5. [1, q)
6. [–1, 1]
7. (–q, q)
8. (–q, 0) ª (0, q)
9. (–1, 1)
10. (–q, q)
g
x3
2
(x 2)(x 2)
2
x2
0.6
x
ln(e3x)
x
2
x
sin 2
f
g
x
x2
x
x3
ln x
sin x
cos x
Section 1.4
Section 1.4 Exercises
Building Functions from Functions
9.
47
10.
1. (f+g)(x)=2x-1+x2; (f-g)(x)=2x-1-x2;
(fg)(x)=(2x-1)(x2)=2x3-x2.
There are no restrictions on any of the domains, so all
three domains are (–q, q).
2. (f+g)(x)=(x-1)2+3-x=
x2-2x+1+3-x=x2-3x+4;
(f-g)(x)=(x-1)2-3+x=
x2-2x+1-3+x=x2-x-2;
(fg)(x)=(x-1)2(3-x)=(x2-2x+1)(3-x)
=3x2-x3-6x+2x2+3-x
=–x3+5x2-7x+3.
There are no restrictions on any of the domains, so all
three domains are (–q, q).
3. (f+g)(x)= 1x+sin x; (f-g)(x)= 1x-sin x;
(fg)(x)= 1x sin x.
Domain in each case is [0, q). For 1x, x 0. For sin x,
–q<x<q.
4. (f+g)(x)= 1x + 5 + @ x + 3 @ ;
(f-g)(x)= 1x + 5 - @ x + 3 @ ;
(fg)(x)= 1x + 5 @ x + 3 @ .
All three expressions contain 1x + 5, so x+5 0
and x –5; all three domains are [–5, q). For |x+3|,
–q<x<q.
1x + 3
; x + 3 0 and x Z 0,
x2
so the domain is [–3, 0) ª (0, q).
x2
(g/f)(x)=
; x + 3 7 0,
1x + 3
so the domain is (–3, q).
5. (f/g)(x)=
6. (f/g)(x)=
2x - 2
=
x - 2
; x - 2 0 and
Bx + 4
2x + 4
x + 4 7 0, so x 2 and x 7 -4; the domain is [2, q).
2x + 4
x + 4
; x + 4 0 and
Bx - 2
2x - 2
x - 2 7 0, so x - 4 and x 7 2; the domain is (2, q).
(g/f)(x)=
7. 1f>g2 1x2 =
=
x2
. The denominator cannot be zero
21 - x2
and the term under the square root must be positive, so
1 - x2 7 0. Therefore, x2 6 1, which means that
-1 6 x 6 1. The domain is 1 - 1, 1 2 .
21 - x2
. The term under the square root
x2
must be nonnegative, so 1 - x2 0 (or x2 1). The
denominator cannot be zero, so x Z 0. Therefore,
–1 x<0 or 0 6 x 1. The domain is 3 -1, 02 ´ 1 0, 14 .
1g>f2 1x2 =
8. 1f>g2 1x2 =
x3
. The denominator cannot be 0, so
3
21
- x3
1 - x3 Z 0 and x3 Z 1. This means that x Z 1. There are
no restrictions on x in the numerator. The domain is
1 - q, 1 2 ´ 11, q 2 .
3
21
- x3
. The denominator cannot be 0, so
x3
x3 Z 0 and x Z 0. There are no restrictions on x in the
numerator. The domain is 1 - q, 0 2 ´ 10, q 2 .
1g>f2 1x2 =
[0, 5] by [0, 5]
[–5, 5] by [–10, 25]
11. (f g)(3)=f(g(3))=f(4)=5;
(g f)(–2)=g(f(–2))=g(–7)=–6
12. (f g)(3)=f(g(3))=f(3)=8;
(g f)(–2)=g(f(–2))=g(3)=3
13. (f g)(3)=f(g(3))=f1 13 + 12 =f(2)=
22+4=8;
(g f)(–2)=g(f(–2))=g((–2)2+4)
=g(8)= 18 + 1=3
0
= 0;
14. (f g)(3)=f(g(3))=f(9-32)=f(0)=
0 + 1
-2
(g f)(–2)=g(f(–2))=g a
b
-2 + 1
2
=g(2)=9-2 =5
15. f(g(x))=3(x-1)+2=3x-3+2=3x-1.
Because both f and g have domain (–q, q), the domain
of f(g(x)) is (–q, q).
g(f(x))=(3x+2)-1=3x+1; again, the domain is
(–q, q).
2
1
1
b - 1=
- 1. The domain
x - 1
1x - 1 2 2
of g is x Z 1, while the domain of f is (–q, q), so the
domain of f(g(x)) is x Z 1, or (–q, 1) ª (1, q).
1
1
g(f(x))= 2
.
= 2
1x - 12 - 1
x - 2
The domain of f is (–q, q), while the domain of g is
(–q, 1) ª (1, q), so g(f(x)) requires that f1x2 Z 1. This
means x2 - 1 Z 1, or x2 Z 2, so the domain of g(f(x)) is
x Z ; 12, or 1 -q, - 122 ´ 1 - 12, 122 ´ 1 12, q2 .
16. f(g(x))= a
17. f(g(x))= 1 1x + 12 2 -2=x+1-2=x-1. The
domain of g is x -1, while the domain of f is (–q, q),
so the domain of f(g(x)) is x -1, or [–1, q).
g(f(x))= 2 1x2 - 2 2 + 1 = 2x2 - 1 . The domain of
f is (–q, q), while the domain of g is [–1, q), so
g(f(x)) requires that f1 x2 -1 .
This means x2 - 2 -1 , or x2 1 , which means
x -1 or x 1 . Therefore the domain of g(f(x)) is
(–q, –1] ª [1, q).
1
. The domain of g is x 0 , while the
1x - 1
domain of f is (–q, 1) ª (1, q), so f(g(x)) requires that
x 0 and g1x2 Z 1, or x 0, and x Z 1. The domain of
f(g(x)) is [0, 1) ª (1, q).
18. f(g(x))=
1
1
g(f(x))=
. The domain of f is
=
Bx - 1
1x - 1
x Z 1 , while the domain of g is [0, q), so g(f(x))
requires that x Z 1 and f1x2 0, or x Z 1 and
1
0 . The latter occurs if x - 1 7 0 , so the
x - 1
domain of g(f(x)) is (1, q).
48
Chapter 1
Functions and Graphs
19. f1 g1x2 2 = f1 21 - x2 2 = 1 21 - x2 2 2 = 1 - x2;
the domain is [–1, 1].
g1 f1x2 2 = g1 x2 2 = 21 - 1x2 2 2 = 21 - x4;
the domain is [–1, 1].
3
3
20. f1 g1x2 2 = f1 21 - x3 2 = 1 21 - x3 2 3 = 1 - x3;
the domain is 1 - q, q 2 .
3
3
g1 f1x2 2 = g1 x3 2 = 21
- 1x3 2 3 = 21
- x9;
the domain is 1 - q, q 2 .
21. f1 g1 x2 2 = f a
1
1
1
3x
b =
=
=
;
3x
2 11>3x2
2>3x
2
the domain is 1 - q, 0 2 ´ 10, q 2.
g1 f1x2 2 = g a
1
1
1
2x
b =
=
=
;
2x
3 11>2x2
3>2x
3
the domain is 1 - q, 0 2 ´ 10, q 2 .
1
1
b =
=
x - 1
11> 1x - 1 2 2 + 1
1
1
x - 1
=
=
;
11 + 1 x - 1 2 2 > 1x - 1 2
x> 1 x - 1 2
x
the domain is all reals except 0 and 1.
22. f1 g1x2 2 = f a
1
1
b =
=
x + 1
11> 1x + 1 2 2 - 1
1
x + 1
1
=
=
;
11 + 1 x - 1 2 2 > 1x + 1 2
x> 1 x + 1 2
x
g1 f1x2 2 = g a
the domain is all reals except 0 and 1.
23. One possibility: f1x2 = 1x and g1x2 = x2 - 5x
24. One possibility: f1x2 = 1x + 1 2 2 and g1x2 = x3
25. One possibility: f1x2 = 0 x 0 and g1x2 = 3x - 2
26. One possibility: f1x2 = 1>x and g1x2 = x3 - 5x + 3
27. One possibility: f1x2 = x5 - 2 and g1 x2 = x - 3
28. One possibility: f1 x2 = ex and g1x2 = sin x
29. One possibility: f1x2 = cos x and g1 x2 = 1x.
30. One possibility: f1x2 = x2 + 1 and g1x2 = tan x.
31. r = 48 + 0.03t in., so V =
4
4 3
pr = p148 + 0.03t2 3;
3
3
when t=300,
4
V = p1 48 + 9 2 3 = 246,924p L 775,734.6 in3.
3
32. The original diameter of each snowball is 4 in., so
the original radius is 2 in. and the original volume
4
V = pr3 L 33.5 in3. The new volume is V = 33.5 - t,
3
where t is the number of 40-day periods. At the end
of 360 days, the new volume is V = 33.5 - 9 = 24.5.
4
3 3V
Since V = pr3, we know that r=
≠1.8 in.
3
A 4∏
The diameter, then, is 2 times r, or≠3.6 in.
2
33. The initial area is (5)(7)=35 km . The new length and
width are l=5+2t and w=7+2t, so A=lw=
(5+2t)(7+2t). Solve (7+2t)(5+2t)=175
(5 times its original size), either graphically or
algebraically: the positive solution is t≠3.63 seconds.
34. The initial volume is (5)(7)(3)=105 cm3. The new
length, width, and height are l=5+2t, w=7+2t,
and h=3+2t, so the new volume is V=
(5+2t)(7+2t)(3+2t). Solve graphically
(5+2t)(7+2t)(3+2t) 525 (5 times the
original volume): t≠1.62 sec.
35. 3(1)+4(1)=3+4=7 5
3(4)+4(–2)=12-8=4 5
3(3)+4(–1)=9-4=5
The answer is (3, –1).
36. (5)2+(1)2=25+1=26 25
(3)2+ (4)2=9+16=25
(0)2+(–5)2=0+25=25
The answer is (3, 4) and (0, –5).
37. y2=25-x2, y = 225 - x2 and y = - 225 - x2
38. y2=25-x, y = 125 - x and y = - 125 - x
39. y2=x2-25, y = 2x2 - 25 and y = - 2x2 - 25
40. y2=3x2-25, y = 23x2 - 25 and y = - 23x2 - 25
41. x + 0 y 0 = 1 1 0 y 0 = -x + 1 1 y = -x + 1 or
y = - 1 -x + 12 . y = 1 - x and y = x - 1
42. x - 0 y 0 = 1 1 0 y 0 = x - 1 1 y = x - 1 or
y = - 1x - 1 2 = -x + 1. y = x - 1 and y = 1 - x
43. y2 = x2 1 y = x and y = - x or y = 0 x 0 and y = - 0 x 0
44. y2 = x 1 y = 1x and y = - 1x
f
45. False. If g(x)=0, then a b (x) is not defined and 0 is
g
f
not in the domain of a b 1 x2 , even though 0 may be in
g
the domains of both f(x) and g(x).
46. False. For a number to be in the domain of (fg)(x), it
must be in the domains of both f(x) and g(x), so that
f(x) and g(x) are both defined.
47. Composition of functions isn’t necessarily commutative.
The answer is C.
48. g1x2 = 14 - x cannot equal zero and the term under
the square root must be positive, so x can be any real
number less than 4. The answer is A.
49. (f f)(x)=f(x2+1)=(x2+1)2+1=
(x4+2x2+1)+1=x4+2x2+2. The answer is E.
50. y = 0 x 0 1 y = x, y = -x; y = - x 1 x = -y; x = - y
or x = y 1 x2 = y2. The answer is B.
51. If f1 x2 = ex and g1x2 = 2 ln x, then f1g1x2 2 = f12 ln x2
= e2 ln x = 1eln x 2 2 = x2. The domain is 10, q 2 .
If f1 x2 = 1x2 + 2 2 2 and g1 x2 = 1x - 2, then
f1g1 x2 2 = f1 1x - 22 = 1 1 1x - 22 2 + 2 2 2 =
1 x - 2 + 2 2 2 = x2. The domain is 32, q 2 .
If f1 x2 = 1x2 - 2 2 2 and g1 x2 = 12 - x, then
f1g1 x2 2 = f1 12 - x2 = 1 1 12 - x2 2 - 2 2 2 =
12 - x - 2 2 2 = x2. The domain is 1 - q, 2 4 .
Section 1.5
1
x + 1
and g1 x2 =
, then
x
1x - 1 2 2
x + 1
1
f1 g1 x2 2 = f a
b =
=
2
x
x + 1
a
- 1b
x
If f1 x2 =
1
1
=
= x2. The domain is x Z 0.
x + 1 - x 2
1
a
b
x
x2
2
If f1 x2 = x - 2x + 1 and g1x2 = x + 1, then
f1 g1 x2 2 = f1 x + 1 2 = 1x + 1 2 2 - 2 1 x + 1 2 + 1 =
1 1x + 1 2 - 12 2 = x2. The domain is 1 - q, q 2 .
x + 1 2
1
b and g1x2 =
, then
x
x - 1
2
1
+ 1
1
x - 1
f1 g1 x2 2 = f a
b = ±
≤ =
x - 1
1
x - 1
Parametric Relations and Inverses
49
55. y2+x2y-5=0. Using the quadratic formula,
y =
-x2 ; 2 1x2 2 2 - 4 1 1 2 1 -52
2
-x2 ; 2x4 + 20
=
2
-x2 + 2x4 + 20
2
-x2 - 2x4 + 20
and y2 =
.
2
so,
y1 =
If f1 x2 = a
1 + x - 1 2
x - 1
≤ = x2. The domain is x Z 1.
±
1
x - 1
[–9.4, 9.4] by [–6.2, 6.2]
■ Section 1.5 Parametric Relations
and Inverses
Exploration 1
f
g
D
ex
2 ln x
(0, )
(x 2)2
x
2
[2, )
(x2 2)2
2
x
(, 2]
1
2
(x 1)
x1
x
x0
x2 2x 1
x1
(, )
x1
1
x1
x1
2
x
2
52. (a) 1fg2 1x2 = x4 - 1 = 1x2 + 1 2 1x2 - 12 =
f1 x2 # 1 x2 - 1 2 , so g1x2 = x2 - 1.
(b) 1 f + g2 1 x2 = 3x2 1 3x2 - 1x2 + 1 2 = 2x2 - 1 =
g1x2 .
(c) 1f>g2 1x2 = 1 1 f1x2 = g1x2 . So g1 x2 = x2 + 1.
(d) f1 g1 x2 2 = 9x + 1 and f1x2 = x + 1. If g(x)=
3x2, then f1 g1 x2 2 = f13x2 2 = 13x2 2 2 + 1 = 9x4 + 1.
4
2
(e) g1 f1 x2 2 = 9x4 + 1 and f1 x2 = x2 + 1. Then
g1 x2 + 12 = 9x4 + 1 = 91 1x2 + 1 2 - 12 2 + 1,
so g1x2 = 9 1x - 1 2 2 + 1.
53. (a) (f+g)(x)=(g+f)(x)=f(x) if g(x)=0.
(b) (fg)(x)=(gf)(x)=f(x) if g(x)=1.
(c) (f g)(x)=(g f)(x)=f(x) if g(x)=x.
54. Yes, by definition, function composition is associative.
That is, (f (g h))(x)=f(g(h))(x) and
((f g) h)(x)=f(g(h))(x).
1. T starts at –4, at the point (8, –3). It stops at T=2, at
the point (8, 3). 61 points are computed.
2. The graph is smoother because the plotted points are
closer together.
3. The graph is less smooth because the plotted points are
further apart. In CONNECT mode, they are connected by
straight lines.
4. The smaller the Tstep, the slower the graphing proceeds.
This is because the calculator has to compute more X
and Y values.
5. The grapher skips directly from the point (0, –1) to the
point (0, 1), corresponding to the T-values T=–2 and
T=0. The two points are connected by a straight line,
hidden by the Y-axis.
6. With the Tmin set at –1, the grapher begins at the point
(–1, 0), missing the bottom of the curve entirely.
7. Leave everything else the same, but change Tmin back to
–4 and Tmax to –1.
Quick Review 1.5
1. 3y=x+6, so y=
x + 6
1
= x + 2
3
3
2. 0.5y=x-1, so y=
x - 1
= 2x - 2
0.5
3. y2=x-4, so y= ; 1x - 4
4. y2=x+6, so y= ; 1x + 6
5. x(y+3) =y-2
xy+3x =y-2
xy-y =–3x-2
y(x-1) =–(3x+2)
3x + 2
3x + 2
y =–
=
x - 1
1 - x
50
Chapter 1
Functions and Graphs
6. x(y+2) =3y-1
xy+2x =3y-1
xy-3y =–2x-1
y(x-3) =–(2x+1)
2x + 1
2x + 1
y =–
=
x - 3
3 - x
7. x(y-4) =2y+1
xy-4x =2y+1
xy-2y =4x+1
y(x-2) =4x+1
4x + 1
y=
x - 2
8. x(3y-1)=4y+3
3xy-x =4y+3
3xy-4y =x+3
y(3x-4) =x+3
x + 3
y=
3x - 4
6. (a)
t
(x, y)=(t+1, t2-2t)
–3
(–2, 15)
–2
(–1, 8)
–1
(0, 3)
0
(1, 0)
1
(2, –1)
2
(3, 0)
3
(4, 3)
(b) t=x-1, y=(x-1)2-2(x-1)
=x2-2x+1-2x+2
=x2-4x+3
This is a function.
(c)
9. x = 1y + 3, y -3 3and x 0 4
x2 = y + 3, y -3, and x 0
y = x2 - 3, y -3, and x 0
10. x = 1y - 2, y 2 3and x 0 4
x2 = y - 2, y 2, and x 0
y = x2 + 2, y 2, and x 0
[–1, 5] by [–2, 6]
7. (a)
Section 1.5 Exercises
1. x = 3 12 2 = 6, y = 2 2 + 5 = 9. The answer is (6, 9).
2. x = 5 1 -2 2 - 7 = -17, y = 17 - 3 1 -2 2 = 23. The
answer is (–17, 23).
3. x = 3 - 4 1 3 2 = 15, y = 13 + 1 = 2. The answer is
(15, 2).
1
1
= 4. x = @ - 8 + 3 @ = 5, y =
-8
8
3
5. (a)
t
(x, y)=(2t, 3t-1)
–3
(–6, –10)
–2
(–4, –7)
–1
(–2, –4)
0
(0, –1)
1
(2, 2)
2
(4, 5)
3
(6, 8)
t (x, y)=(t2, t-2)
–3
(9, –5)
–2
(4, –4)
–1
(1, –3)
0
(0, –2)
1
(1, –1)
2
(4, 0)
3
(9, 1)
(b) t=y 2, x=(y+2)2. This is not a function.
(c)
[–1, 5] by [–5, 1]
8. (a)
x
x
(b) t= , y=3 a b -1=1.5x-1. This is a
2
2
function.
(c)
t (x, y)=( 1 t, 2t-5)
–3
1-3 not defined
–2
1-2 not defined
–1
1-1 not defined
0
(0, –5)
1
(1, –3)
2
( 1 2, –1)
3
( 1 3, 1)
(b) t=x2, y=2x2-5. This is a function.
(c)
[–5, 5] by [–3, 3]
[–2, 4] by [–6, 4]
Section 1.5
9. (a) By the vertical line test, the relation is not a function.
Parametric Relations and Inverses
3
x= 1y - 2
3
x = y - 2
f - 1 1x2 = y = x3 + 2; (–q, q)
3
22. y= 1x - 2 1
(b) By the horizontal line test, the relation’s inverse is a
function.
10. (a) By the vertical line test, the relation is a function.
23. One-to-one
y
(b) By the horizontal line test, the relation’s inverse is not
a function.
11. (a) By the vertical line test, the relation is a function.
3
(b) By the horizontal line test, the relation’s inverse is a
function.
x
5
12. (a) By the vertical line test, the relation is not a function.
(b) By the horizontal line test, the relation’s inverse is a
function.
13. y = 3x - 6 1
x = 3y - 6
3y = x + 6
x + 6
1
f - 1 1x2 = y =
= x + 2; (–q, q)
3
3
x = 2y + 5
2y = x - 5
x - 5
1
5
f - 1 1x2 = y =
= x - ;
2
2
2
(–q, q)
2y - 3
2x - 3
x =
1
15. y =
x + 1
y + 1
x1 y + 1 2 = 2y - 3
xy + x = 2y - 3
xy - 2y = -x - 3
y1x - 2 2 = - 1 x + 3 2
x + 3
x + 3
-1
f 1x2 = y = =
x - 2
2 - x;
(–q, 2) ª (2, q)
24. Not one-to-one
25. One-to-one
14. y = 2x + 5 1
16. y =
y + 3
y - 2
x1 y - 2 2 = y + 3
xy - 2x = y + 3
xy - y = 2x + 3
y1x - 1 2 = 2x + 3
2x + 3
f - 1 1x2 = y =
;
x - 1
x Z 1 or (–q, 1) ª (1, q)
x + 3
1
x - 2
x =
17. y = 1x - 3, x 3, y 0 1
x = 1y - 3 , x 0, y 3
x2 = y - 3 , x 0, y 3
f - 1 1x2 = y = x2 + 3 , x 0
18. y = 1x + 2, x -2, y 0 1
x = 1y + 2 , x 0, y -2
x2 = y + 2 ,
x 0, y -2
f - 1 1x2 = y = x2 - 2 , x 0
19. y = x3 1
20. y = x3 + 5 1
x = y3
3
f - 1 1x2 = y = 1x
; (–q, q)
x = y3 + 5
x - 5 = y3
3
f - 1 1x2 = y = 1x
- 5; (–q, q)
3
x= 1y + 5
3
21. y=1x + 5 1
f
-1
x3 = y + 5
1x2 = y = x3 - 5; (–q, q)
y
3
5
x
26. Not one-to-one
1
27. f1g1 x2 2 = 3 c 1x + 22 d - 2 = x + 2 - 2 = x ;
3
1
1
g1f1x2 2 = 3 13x - 2 2 + 24 = 1 3x2 = x
3
3
1
1
3 14x - 32 + 3 4 = 1 4x2 = x ;
4
4
1
g1f1x2 2 = 4 c 1 x + 32 d - 3 = x + 3 - 3 = x
4
28. f1g1 x2 2 =
29. f1g1 x2 2 = 3 1x - 1 2 1>3 4 3 + 1 = 1x - 12 1 + 1
= x - 1 + 1 = x;
g1f1x2 2 = 3 1x3 + 1 2 - 14 1>3 = 1x3 2 1>3 = x1 = x
30. f1g1 x2 2 =
31. f1g1 x2 2 =
=
g1f1x2 2 =
=
7
7
=
7
1
x
#x
7
1
+
x - 1
1
x - 1
1 + x 1
x + 1
x
x
x + 1 -
1
= x ; g1 f1x2 2 =
= 1x - 12 a
7
7
=
7
1
x
7
1
+ 1b
x - 1
1 = x;
1
¢
°x + 1
1
- 1
x
x
=
= x
1
x
=
#x
#
x
x
= x
51
52
Chapter 1
Functions and Graphs
2x + 3
2x + 3
+ 3
+ 3
x - 1
x - 1
= ±
≤
32. f1 g1x2 2 =
2x + 3
2x + 3
- 2
- 2
x - 1
x - 1
2x + 3 + 3 1x - 1 2
5x
=
=
= x;
2x + 3 - 2 1x - 1 2
5
g1 f1x2 2 =
a
x - 1
b
x - 1
x + 3
b + 3
x - 2
x + 3
- 1
x - 2
2a
= ≥
=
#
x + 3
b + 3
x - 2
¥
x + 3
- 1
x - 2
2a
# ax - 2b
21x + 32 + 3 1 x - 2 2
x + 3 - 1x - 2 2
x - 2
=
5x
= x
5
33. (a) y=(1.08)(100)=108 euros
(b) x=
y
25
= y. This converts euros (x) to dollars (y).
1.08
27
(c) x=(0.9259)(48)=$44.44
34. (a) 9c(x)=5(x-32)
9
c1 x2 = x - 32
5
9
c1 x2 + 32 = x
5
In this case, c(x) becomes x, and x becomes c–1(x) for
9
the inverse. So, c–1(x)= x+32. This converts
5
Celsius temperature to Fahrenheit temperature.
5
(b) (k c)(x)=k(c(x))=k a 1x - 32 2 b
9
5
5
(x-32)+273.16= x+255.38. This is used to
9
9
convert Fahrenheit temperature to Kelvin temperature.
35. y=ex and y=ln x are inverses. If we restrict the domain
of the function y=x2 to the interval 30, q 2 , then the
restricted function and y = 1x are inverses.
42. The inverse of the relation given by xy2-3x=12 is the
relation given by yx2-3y=12.
1 -4 2 10 2 2 - 3 1 -4 2 = 0 +
112 14 2 2 - 3 11 2 = 16 - 3
122 13 2 2 - 3 12 2 = 18 - 6
1122 12 2 2 - 3 112 2 = 48 1 -6 2 11 2 2 - 3 1 -6 2 = -6
The answer is B.
12 = 12
= 13 Z 12
= 12
36 = 12
+ 18 = 12
43. f(x)=3x-2
y=3x-2
The inverse relation is
x=3y-2
x+2=3y
x + 2
=y
3
x + 2
f–1(x)=
3
The answer is C.
44. f1x2 = x3 + 1
y = x3 + 1
The inverse relation is
x = y3 + 1
x-1=y3
2x - 1 = y
3
f–1(x)= 2x - 1
The answer is A.
3
45. (Answers may vary.)
(a) If the graph of f is unbroken, its reflection in the line
y=x will be also.
(b) Both f and its inverse must be one-to-one in order to
be inverse functions.
(c) Since f is odd, (–x, –y) is on the graph whenever
(x, y) is. This implies that (–y, –x) is on the graph of
f–1 whenever (x, y) is. That implies that f–1 is odd.
(d) Let y=f(x). Since the ratio of y to x is positive,
the ratio of x to y is positive. Any ratio of y
to x on the graph of f–1 is the same as some ratio
of x to y on the graph of f, hence positive. This
implies that f–1 is increasing.
46. (Answers may vary.)
36. y=x and y=1/x are their own inverses.
(a) f(x)=ex has a horizontal asymptote;
f–1(x)=ln x does not.
38. y=x
(b) f(x)=ex has domain all real numbers;
f–1(x)=ln x does not.
37. y = 0 x 0
39. True. All the ordered pairs swap domain and range values.
40. True. This is a parametrization of the line y=2x+1.
41. The inverse of the relation given by x2y+5y=9 is the
relation given by y2x+5x=9.
11 2 2 122 + 51 22 = 2 + 10 = 12 Z 9
11 2 2 1 -2 2 + 51 - 2 2 = - 2 - 10 = -12 Z 9
12 2 2 1 -1 2 + 51 - 1 2 = - 4 - 5 = -9 Z 9
1 - 1 2 2 122 + 51 2 2 = 2 + 10 = 12 Z 9
1 - 2 2 2 112 + 51 1 2 = 4 + 5 = 9
The answer is E.
(c) f(x)=ex has a graph that is bounded below;
f–1(x)=ln x does not.
x2 - 25
has a removable discontinuity at
x - 5
x=5 because its graph is the line y=x+5 with the
point (5, 10) removed. The inverse function is the line
y=x-5 with the point (10, 5) removed. This function has a removable discontinuity, but not at x=5.
(d) f(x)=
Section 1.6
47. (a)
(c)
(b) To find the inverse, we substitute y for x and x for y,
and then solve for y:
x = 0.75y + 31
x - 31 = 0.75y
4
1 x - 31 2
3
The inverse function converts scaled scores to raw
scores.
48. The function must be increasing so that the order of the
students’ grades, top to bottom, will remain the same after
scaling as it is before scaling. A student with a raw score
of 136 gets dropped to 133, but that will still be higher
than the scaled score for a student with 134.
[65, 100] by [65, 100]
The composition function of (y ø y–1)(x)is y=x, so
they are inverses.
51. When k=1, the scaling function is linear. Opinions will
vary as to which is the best value of k.
■ Section 1.6 Graphical Transformations
Exploration 1
1.
49. (a) It does not clear the fence.
[–5, 5] by [–5, 15]
They raise or lower the parabola along the y-axis.
2.
[0, 350] by [0, 300]
(b) It still does not clear the fence.
[–5, 5] by [–5, 15]
They move the parabola left or right along the x-axis.
3.
[0, 350] by [0, 300]
(c) Optimal angle is 45°. It clears the fence.
[–5, 5] by [–5, 15]
[–5, 5] by [–5, 15]
[–5, 5] by [–5, 15]
[–5, 5] by [–5, 15]
[0, 350] by [0, 300]
50. (a)
x = a
x - 1 = a
1x - 1 2 1.7 = a
1
31.7
1y - 65 2 b 1.7 + 1
30
1
31.7
1y - 65 2 b 1.7
30
31.7
1y - 65 2 b
30
30
1x - 1 2 1.7 = y - 65
31.7
30
y = 1.7 1 x - 1 2 1.7 + 65
3
This can be use to convert GPA’s to percentage grades.
53
(b) Yes; x is restricted to the domain [1, 4.28].
¢y
97 - 70
27
=
=
= 0.75, which gives us the
¢x
88 - 52
36
slope of the equation. To find the rest of the
equation, we use one of the initial points
y - 70 = 0.75 1 x - 52 2
y = 0.75x - 39 + 70
y = 0.75x + 31
y =
Graphical Transformations
[–5, 5] by [–5, 15]
Yes
[–3.7, 5.7] by [–1.1, 5.1]
54
Chapter 1
Functions and Graphs
Exploration 2
Quick Review 1.6
1. (x+1)2
1.
2. (x-3)2
3. (x+6)2
4. (2x+1)2
5. (x-5/2)2
Graph C. Points with positive y-coordinates remain
unchanged, while points with negative y-coordinates are
reflected across the x-axis.
2.
6. (2x-5)2
7. x2-4x+4+3x-6+4=x2-x+2
8. 2(x2+6x+9)-5x-15-2=
2x2+12x+18 -5x-17=2x2+7x+1
9. (x3-3x2+3x-1)+3(x2-2x+1)-3x+3
=x3-3x2+2+3x2-6x+3=x3-6x+5
10. 2(x3+3x2+3x+1)-6(x2+2x+1)+
6x+6-2=2x3+6x2+6x+2-6x2-12x
-6+6x+6-2=2x3
Graph A. Points with positive x-coordinates remain
unchanged. Since the new function is even, the graph for
negative x-values will be a reflection of the graph for
positive x-values.
3.
Section 1.6 Exercises
1. Vertical translation down 3 units
2. Vertical translation up 5.2 units
3. Horizontal translation left 4 units
4. Horizontal translation right 3 units
5. Horizontal translation to the right 100 units
6. Vertical translation down 100 units
Graph F. The graph will be a reflection across the x-axis
of graph C.
4.
7. Horizontal translation to the right 1 unit, and vertical
translation up 3 units
8. Horizontal translation to the left 50 units and vertical
translation down 279 units
9. Reflection across x-axis
10. Horizontal translation right 5 units
11. Reflection across y-axis
Graph D. The points with negative y-coordinates in graph
A are reflected across the x-axis.
Exploration 3
1.
[–4.7, 4.7] by [–1.1, 5.1]
The 1.5 and the 2 stretch the graph vertically; the 0.5 and
the 0.25 shrink the graph vertically.
2.
12. This can be written as y = 1- 1 x - 3 2 or
y = 1-x + 3 . The first of these can be interpreted as
reflection across the y-axis followed by a horizontal translation to the right 3 units. The second may be viewed as a
horizontal translation left 3 units followed by a reflection
across the y-axis.
Note that when combining horizontal changes (horizontal
translations and reflections across the y-axis), the order is
“backwards” from what one may first expect: With
y = 1- 1 x - 3 2 , although we first subtract 3 from x
then negate, the order of transformations is reflect then
translate. With y = 1-x + 3 , although we negate x
then add 3, the order of transformations is translate then
reflect.
For #13–20, recognize y=c # x3 (c>0) as a vertical stretch
(if c>1) or shrink (if 0<c<1) of factor c, and y=(c # x)3
as a horizontal shrink (if c>1) or stretch (if 0<c<1)
of factor 1/c. Note also that y=(c # x)3=c3x3, so that for this
function, any horizontal stretch/shrink can be interpreted as an
equivalent vertical shrink/stretch (and vice versa).
13. Vertically stretch by 2
[–4.7, 4.7] by [–1.1, 5.1]
The 1.5 and the 2 shrink the graph horizontally; the 0.5
and the 0.25 stretch the graph horizontally.
14. Horizontally shrink by 1>2, or vertically stretch by 23=8
15. Horizontally stretch by 1>0.2 = 5, or vertically shrink by
0.23=0.008
Section 1.6
16. Vertically shrink by 0.3
17. g(x)= 1x - 6 + 2 = f1x - 6 2 ; starting with f,
translate right 6 units to get g.
18. g(x)=–(x+4-1)2=–f(x+4); starting with f,
translate left 4 units, and reflect across the x-axis to get g.
Graphical Transformations
55
27. The graph is reflected across the x-axis, translated left
2 units, and translated up 3 units. y = - 1x would be
reflected across the x-axis, y = - 1x + 2 adds the
horizontal translation, and finally, the vertical translation
gives f1x2 = - 1x + 2 + 3 = 3 - 1x + 2 .
19. g(x)=–(x+4-2)3=–f(x+4); starting with f,
translate left 4 units, and reflect across the x-axis to get g.
28. The graph is vertically stretched by 2, translated left
5 units, and translated down 3 units. y = 2 1x would be
vertically stretched, y = 21x + 5 adds the horizontal
translation, and finally, the vertical translation gives
f1x2 = 21x + 5 - 3 .
21.
29. (a) y=–f(x)=–(x3-5x2-3x+2)
=–x3+5x2+3x-2
20. g(x)=2 @ 2x @ =2f(x); starting with f, vertically stretch
by 2 to get g.
y
10
f
(b) y=f(–x)=(–x)3-5(–x)2-3(–x)+2
=–x3-5x2+3x+2
g
–2
6
30. (a) y=–f(x)= - 12 1x + 3 - 4 2 = -2 1x + 3 + 4
x
(b) y=f(–x)=21- x + 3 - 4 = 213 - x - 4
3
3
31. (a) y=–f(x)=–( 18x)=–2 1x
h
(b) y=f(–x)= 18 1 -x2 = 1-8x=–2 1x
3
3
3
32. (a) y=–f(x)=–3 @ x+5 @
22.
(b) y=f(–x)=3 @ –x+5 @ =3 @ 5-x @
y
33. Let f be an odd function; that is, f(–x)=–f(x) for all
x in the domain of f. To reflect the graph of y=f(x)
across the y-axis, we make the transformation y=f(–x).
But f(–x) =–f(x) for all x in the domain of f, so this
transformation results in y =–f(x). That is exactly the
translation that reflects the graph of f across the x-axis, so
the two reflections yield the same graph.
10
g
h
–7
3
x
f
23.
34. Let f be an odd function; that is, f(–x)=–f(x) for all
x in the domain of f. To reflect the graph of y=f(x)
across the y-axis, we make the transformation y=f(–x).
Then, reflecting across the x-axis yields y=–f(–x). But
f(–x)=–f(x) for all x in the domain of f, so we have
y=–f(–x)=–[–f(x)]=f(x); that is, the original
function.
y
h
3
–6
6
x
35.
f
y
g
–6
x
24.
y
10
g
h
36.
–5
5
y
x
f
–10
25. Since the graph is translated left 5 units, f1x2 = 1x + 5.
26. The graph is reflected across the y-axis and translated
right 3 units. y = 1-x would be reflected across the
y-axis; the horizontal translation gives
f(x)= 1- 1x - 32 = 13 - x .
See also Exercise 12 in this section, and note accompanying that solution.
x
56
Chapter 1
37.
y
Functions and Graphs
51. Translate left 1 unit, then vertically stretch by 3, and
finally translate up 2 units. The four vertices are
transformed to (–3, –10), (–1, 2), (1, 8), and (3, 2).
y
x
x
38.
y
x
52. Translate left 1 unit, then reflect across the x-axis, and
finally translate up 1 unit. The four vertices are transformed to (–3, 5), (–1, 1), (1, –1), and (3, 1).
y
39. (a) y1=2y=2(x3-4x)=2x3-8x
x
(b) y2=f a 1 b =f(3x)=(3x)3-4(3x)=27x3-12x
3
40. (a) y1=2y=2 @ x+2 @
x
(b) y2=f(3x)= @ 3x+2 @
41. (a) y1=2y=2(x2+x-2)=2x2+2x-4
(b) y2=f(3x)=(3x)2+3x-2=9x2+3x-2
42. (a) y1=2y=2 a
1
2
b=
x + 2
x + 2
1
(b) y2=f(3x)=
3x + 2
1
53. Horizontally shrink by . The four vertices are
2
transformed to (–1, –4), (0, 0), (1, 2), (2, 0).
y
2
43. Starting with y=x , translate right 3 units, vertically
stretch by 2, and translate down 4 units.
44. Starting with y= 1x, translate left 1 unit, vertically
stretch by 3, and reflect across x-axis.
x
1
45. Starting with y=x , horizontally shrink by and
3
translate down 4 units.
2
46. Starting with y= 0 x 0 , translate left 4 units, vertically
stretch by 2, reflect across x-axis, and translate up 1 unit.
47. First stretch (multiply right side by 3): y=3x2, then
translate (replace x with x-4): y=3(x-4)2.
48. First translate (replace x with x-4): y=(x-4)2, then
stretch (multiply right side by 3): y=3(x-4)2.
49. First translate left (replace x with x+2): y=|x+2|,
then stretch (multiply right side by 2): y=2|x+2|,
then translate down (subtract 4 from the right side):
y=2|x+2|-4.
50. First translate left (replace x with x+2): y=|x+2|,
then shrink (replace x with 2x): y=|2x+2|,
then translate down (subtract 4 from the right side):
y=|2x+2|-4. This can be simplified to
y=|2(x+1)|-4=2|x+1|-4.
To make the sketches for #51–54, it is useful to apply the
described transformations to several selected points on the
graph. The original graph here has vertices (–2, –4), (0, 0),
(2, 2), and (4, 0); in the solutions below, the images of these
four points are listed.
54. Translate right 1 unit, then vertically stretch by 2, and
finally translate up 2 units. The four vertices are transformed to (–1, –6), (1, 2), (3, 6), and (5, 2).
y
x
55. Reflections have more effect on points that are farther
away from the line of reflection. Translations affect the
distance of points from the axes, and hence change the
effect of the reflections.
Section 1.6
56. The x-intercepts are the values at which the function equals
zero. The stretching (or shrinking) factors have no effect on
the number zero, so those y-coordinates do not change.
Graphical Transformations
(b) The original graph is on the left; the graph of
y = fA @ [email protected] B is on the right.
9
57. First vertically stretch by , then translate up 32 units.
5
9
5
58. Solve for C: F= C+32, so C= (F-32)=
5
9
5
160
5
F. First vertically shrink by , then translate
9
9
9
160
down
= 17.7 units.
9
[–5, 5] by [–10, 10]
(c)
[–5, 5] by [–10, 10]
y
59. False. y=f(x+3) is y=f(x) translated 3 units to the
left.
60. True. y=f(x) –c represents a translation down by
c units. (The translation is up when c<0.)
x
61. To vertically stretch y=f(x) by a factor of 3, multiply
the f(x) by 3. The answer is C.
62. To translate y=f(x) 4 units to the right, subtract 4 from
x inside the f(x). The answer is D.
63. To translate y=f(x) 2 units up, add 2 to f(x):
y=f(x) ± 2. To reflect the result across the y-axis,
replace x with –x. The answer is A.
(d)
y
64. To reflect y=f(x) across the x-axis, multiply f(x) by
–1: y=–f(x). To shrink the result horizontally by a
1
factor of , replace x with 2x. The answer is E.
2
65. (a)
x
y
Price (dollars)
36
68. (a)
35
34
33
32
31
x
1
2
3 4 5
Month
6
7
8
[–4.7, 4.7] by [–3.1, 3.1]
(b) Change the y-value by multiplying by the conversion
rate from dollars to yen, a number that changes
according to international market conditions. This
results in a vertical stretch by the conversion rate.
x=2 cos t
(b)
y=sin t
66. Apply the same transformation to the Ymin, Ymax, and
Yscl as you apply to transform the function.
67. (a) The original graph is on the left; the graph of
y= @ f(x) @ is on the right.
[–4.7, 4.7] by [–3.1, 3.1]
x=3 cos t
(c)
y=3 sin t
[–4.7, 4.7] by [–3.1, 3.1]
[–5, 5] by [–10, 10]
[–5, 5] by [–10, 10]
57
58
Chapter 1
Functions and Graphs
x=4 cos t
(d)
y=2 sin t
3. Linear: r2=0.9758
Power: r2=0.9903
Quadratic: R2=1
Cubic: R2=1
Quartic: R2=1
4. The best-fit curve is quadratic: y=0.5x2-1.5x. The
cubic and quartic regressions give this same curve.
[–4.7, 4.7] by [–3.1, 3.1]
■ Section 1.7 Modeling with Functions
Exploration 1
5. Since the quadratic curve fits the points perfectly, there is
nothing to be gained by adding a cubic term or a quartic
term. The coefficients of these terms in the regressions
are zero.
6. y=0.5x2-1.5x. At x=128,
y=0.5(128)2-1.5(128)=8000
1.
Quick Review 1.7
1. h=2(A/b)
2. h=2A/(b1+b2)
n = 3; d = 0
n = 5; d = 5
3. h=V/(pr2)
n = 4; d = 2
n = 6; d = 9
4. h=3V/(pr2)
5. r=
3 3V
A 4p
6. r=
A
B 4p
7. h =
A - 2pr2
A
=
- r
2pr
2pr
8. t=I/(Pr)
A
r -nt
b
nt = A a 1 +
11 + r>n2
n
21 H - s2
9. P =
10. t=
B
g
Section 1.7 Exercises
n = 7; d = 14
n = 8; d = 20
1. 3x+5
2. 3(x+5)
3. 0.17x
4. 0.05x+4
5. A=/w=(x+12)(x)
1
1
6. A= bh= (x)(x+2)
2
2
n = 9; d = 27
n = 10; d = 35
7. x+0.045x=(1+0.045)x=1.045x
8. x-0.03x=(1-0.03)x=0.97x
9. x-0.40x=0.60x
2.
10. x+0.0875x=1.0875x
11. Let C be the total cost and n be the number of items
produced; C=34,500+5.75n.
12. Let C be the total cost and n be the number of items
produced; C=(1.09)28,000+19.85n.
[3, 11] by [0, 40]
13. Let R be the revenue and n be the number of items sold;
R=3.75n.
14. Let P be the profit, and s be the amount of sales; then
P=200,000+0.12s.
Section 1.7
15. The basic formula for the volume of a right circular
cylinder is V = pr2h, where r is the radius and h is
height. Since height equals diameter (h=d) and the
diameter is two times r (d=2r), we know h=2r.
Then, V = pr2 1 2r2 = 2pr3.
r
16. Let c=hypotenuse, a=“short” side, and b=“long”
side. Then c2=a2+b2=a2+(2a)2=a2+4a2=5a2,
so c= 25a2 = a15 .
59
P
r
Q
2r
Modeling with Functions
l
r
l R
s
S
19. Let r be the radius of the sphere. Since the sphere is
tangent to all six faces of the cube, we know that the
height (and width, and depth) of the cube is equal to the
sphere’s diameter, which is two times r (2r). The surface
area of the cube is the sum of the area of all six faces,
which equals 2r # 2r=4r2. Thus, A=6 # 4r2=24r2.
r
r
a 5
2a
2r
r
2r
a
2r
17. Let a be the length of the base. Then the other two sides
of the triangle have length two times the base, or 2a. Since
the triangle is isoceles, a perpendicular dropped from the
“top” vertex to the base is perpendicular. As a result,
a 2
a2
16a2 - a2
h2 + a b = 12a2 2, or h2 = 4a2 =
2
4
4
15a2
a 115
15a2
, so h =
. The triangle’s area is
=
=
4
2
B 4
1
1
a215
a2 215
.
A = bh = 1a2 a
b =
2
2
2
4
20. From our graph, we see that y provides the height of our
b
triangle, i.e., h=y when x= . Since y=6-x2
2
b 2
b2
24 - b2
24 - b2
=
=6- a b =6 , h=
.
2
4
4
4
1 24 - b2
1
The area of the triangle is A = bh = b a
b
2
2
4
3
24b - b
=
.
8
y
6
(0, 6)
y = 6 – x2
2a
2a
b 24 – b2
a ,
b
2
4
h
a
2
a
2
a
h=
a 15
2
18. Since P lies at the center of the square and the circle,
we know that segment PR = QR = RS . Let / be the
length of these segments. Then, /2 + /2 = r2, 2/2 = r2,
r2
r
r22
r2
.
=
=
,/ =
2
B2
2
12
Since each side of the square is two times /,
r12
we know that s = 2/ = a
b 2 = r12 .
2
2
2
As a result, A = s = 1r122 = r2 # 2 = 2r2 .
/2 =
(0, 0)
h
a 6 , 0b
b b
b (b, 0)
2 a 2 , 0b 2
6
x
b
2
h = 24 – b
4
21. Solving x+4x=620 gives x=124, so 4x=496. The
two numbers are 124 and 496.
22. x+2x+3x=714, so x=119; the second and third
numbers are 238 and 357.
23. 1.035x=36,432, so x=35,200
24. 1.023x=184.0, so x=179.9.
25. 182=52t, so t=3.5 hr.
26. 560=45t+55(t+2), so t=4.5 hours on local highways.
27. 0.60(33)=19.8; 0.75(27)=20.25. The $33 shirt sells for
$19.80. The $27 shirt sells for $20.25. The $33 shirt is a better bargain, because the sale price is cheaper.
60
Chapter 1
Functions and Graphs
28. Let x be gross sales. For the second job to be more
attractive than the first, we need
20,000+0.07x>25,000+0.05x, 0.02x>5000,
5000
x>
=$250,000.
0.02
Gross sales would have to exceed $250,000.
29. 71 065 000 11 + x2 = 82 400 000
71 065 000x = 82 400 000 - 71 065 000
82 400 000 - 71 065 000
L 0.1595
x =
71 065 000
There was a 15.95% increase in sales.
30. 26 650 000 11 + x2 = 30 989 000
26 650 000x = 30 989 000 - 26 650 000
30 989 000 - 26 650 000
x =
L 0.1628
26 650 000
Shipments of personal computers grew 16.28%.
31. (a) 0.10x+0.45(100-x)=0.25(100).
(b) Graph y1=0.1x+0.45(100-x) and y2=25.
Use x≠57.14 gallons of the 10% solution and about
42.86 gal of the 45% solution.
[0, 100] by [0, 50]
36. Solve 2x+2(x+3)=54. This gives x=12; the room
is 12 ft*15 ft.
37. Original volume of water:
1
1
V0= ∏r2h= ∏(9)2(24)≠2035.75 in.3
3
3
Volume lost through faucet:
Vl=time*rate=(120 sec)(5 in.3/sec)=600 in.3
Find volume:
Vf=V0-Vl=2035.75-600=1435.75
Since the final cone-shaped volume of water has radius
3
and height in a 9-to-24 ratio, or r= h:
8
1
3 2
3
Vf= ∏ a h≤ h =
∏h3=1435.75
3
8
64
Solving, we obtain h≠21.36 in.
38. Solve 900=0.07x+0.085(12,000-x).
x=8000 dollars was invested at 7%; the other $4000 was
invested at 8.5%.
39. Bicycle’s speed in feet per second:
(2*∏*16 in./rot)(2 rot/sec)=64∏ in./sec
Unit conversion:
1
1
(64∏ in./sec) a
ft/in. b a
mi/ft b (3600 sec/hr)
12
5280
≠11.42 mi/hr
40. Solve 1571=0.055x+0.083(25,000-x).
x=18,000 dollars was invested at 5.5%; the other $7000
was invested at 8.3%.
32. Solve 0.20x+0.35(25-x)=0.26(25). Use
x=15 liters of the 20% solution and 10 liters of the 35%
solution.
41. True. The correlation coefficient is close to 1 (or –1) if
there is a good fit. A correlation coefficient near 0 indicates a very poor fit.
33. (a) The height of the box is x, and the base measures
10-2x by 18-2x.
V(x)=x(10-2x)(18-2x)
42. False. The graph over time of the height of a freely falling
object is a parabola. A quadratic regression is called for.
(b) Because one side of the original piece of cardboard
measures 10 in., 2x must be greater than 0 but less
than 10, so that 0<x<5. The domain of V(x) is
(0, 5).
(c) Graphing V(x) produces a cubic-function curve that
between x=0 and x=5 has a maximum at approximately (2.06, 168.1). The cut-out squares should measure approximately 2.06 in. by 2.06 in.
34. Solve 2x+2(x+16)=136. Two pieces that are
x=26 ft long are needed, along with two 42 ft pieces.
35. Equation of the parabola, to pass through (–16, 8) and
(16, 8):
y=kx2
8=k (—16)2
8
1
k=
=
256
32
1
y= x2
32
y-coordinate of parabola 8 in. from center:
1
y= (8)2=2
32
From that point to the top of the dish is 8-2=6 in.
43. The pattern of points is S-shaped, which suggests a cubic
model. The answer is C.
44. The points appear to lie along a straight line. The answer
is A.
45. The points appear to lie along an upward-opening
parabola. The answer is B.
46. The pattern of points looks sinusoidal. The answer is E.
47. (a) C=100,000+30x
(b) R=50x
(c) 100,000+30x=50x
100,000=20x
x=5000 pairs of shoes
(d) Graph y1=100,000+30x and y2=50x; these
graphs cross when x=5000 pairs of shoes. The point
of intersection corresponds to the break-even point,
where C=R.
[–10, 10] by [–2, 18]
Chapter 1
48. Solve 48,814.20=x+0.12x+0.03x+0.004x. Then
48,814.20=1.154x, so x=42,300 dollars.
Review
61
(c) The regression equation is y=118.07 * 0.951x. It fits
the data extremely well.
49. (a) y1=u(x)=125,000+23x.
(b) y2=s(x)=125,000+23x+8x=125,000+31x.
(c) y3=ru(x)=56x.
(d) y4=Rs(x)=79x.
(e)
[0, 22] by [100, 200]
52. Answers will vary in (a)–(e), depending on the conditions
of the experiment.
[–10, 10] by [–2, 18]
(f) You should recommend stringing the rackets; fewer
strung rackets need to be sold to begin making a profit
(since the intersection of y2 and y4 occurs for smaller x
than the intersection of y1 and y3).
50. (a)
(f) Some possible answers: the thickness of the liquid,
the darkness of the liquid, the type of cup it is in, the
amount of surface exposed to the air, the specific heat
of the substance (a technical term that may have been
learned in physics), etc.
■ Chapter 1 Review
1. (d)
2. (f)
3. (i)
4. (h)
[–1, 15] by [9, 16]
5. (b)
6. (j)
(b) y=0.409x+9.861
7. (g)
(c) r=0.993, so the linear model is appropriate.
8. (c)
(d) y=0.012x2+0.247x+10.184
9. (a)
(e) r2=0.998, so a quadratic model is appropriate.
10. (e)
(f) The linear prediction is 18.04 and the quadratic prediction is 19.92. Despite the fact that both models
look good for the data, the predictions differ by 1.88.
One or both of them must be ineffective, as they both
cannot be right.
11. (a) All reals
(b) All reals
12. (a) All reals
(b) All reals
(g) The linear regression and the quadratic regression are
very close from x=0 to x=13. The quadratic
regression begins to veer away from the linear regression at x=13. Since there are no data points beyond
x=13, it is difficult to know which is accurate.
51. (a)
13. (a) All reals
(b) g1 x2 = x2 + 2x + 1 = 1x + 12 2.
At x = -1, g1x2 = 0, the function’s minimum.
The range is 30, q 2.
14. (a) All reals
(b) 1x - 22 2 0 for all x, so 1x - 22 2 + 5 ≥ 5 for all x.
The range is 35, q 2 .
15. (a) All reals
(b) 0 x 0 0 for all x, so 3 0 x 0 0 and 3 0 x 0 + 8 8 for
all x. The range is 38, q 2.
16. (a) We need 24 - x2 0 for all x, so 4 - x2 0,
4 x2 , -2 x 2. The domain is [–2, 2].
[0, 22] by [100, 200]
(b) List L3={112.3, 106.5, 101.5, 96.6, 92.0, 87.2, 83.1,
79.8, 75.0, 71.7, 68, 64.1, 61.5, 58.5, 55.9, 53.0, 50.8,
47.9, 45.2, 43.2}
(b) 0 24 - x2 2 for all x, so
–2 24 - x2 - 2 0 for all x. The range is
3 -2, 04.
x
x
=
. x Z 0 and
x2 - 2x
xA x - 2 B
x - 2 Z 0, x Z 2. The domain is all reals except
0 and 2.
17. (a) f1 x2 =
(b) For x 7 2, f1x2 7 0 and for x 6 2, f1x2 6 0. f1x2
does not cross y = 0, so the range is all reals except
f1 x2 = 0.
62
Chapter 1
Functions and Graphs
18. (a) We need
29 - x2 7 0, 9 - x2 7 0, 9 7 x2, - 3 6 x 6 3.
The domain is 1 - 3, 3 2.
1
7 0. On the domain
(b) Since 29 - x2 7 0,
29 - x2
1
1 -3, 3 2 , k1 02 = , a minimum, while k(x)
3
approaches q when x approaches both -3 and 3,
1
maximums for k(x). The range is c , q b .
3
19. Continuous
23. (a) None
7x
= 7 and
2x2 + 10
7x
lim
= - 7, we expect horizontal
xS–∞
2
2x + 10
asymptotes at y = 7 and y = - 7.
(b) Since lim
xS∞
[–15, 15] by [–10, 10]
24. (a) x + 1 Z 0, x Z -1, so we expect a vertical
asymptote at x = -1.
[–7, 3] by [–12, 8]
(b) lim
20. Continuous
[–5, 5] by [–8, 12]
21. (a) x2 - 5x Z 0, x1x - 5 2 Z 0, so x Z 0 and x Z 5.
We expect vertical asymptotes at x = 0 and x = 5.
0x0
= 1 and lim
0x0
= -1, so we can
x + 1
x + 1
expect horizontal asymptotes at y = 1 and y = - 1.
xS∞
25. 1 - q, q 2
xS–∞
[–6, 4] by [–5, 5]
(b) y=0
[–4.7, 4.7] by [–3.1, 3.1]
[–7, 13] by [–10, 10]
22. (a) x - 4 Z 0, x Z 4, so we expect a vertical asymptote
at x = 4.
26. 0 x - 1 0 = 0 when x = 1, which is where the function’s
minimum occurs. y increases over the interval 31, q 2.
(Over the interval 1 - q, 1 4, it is decreasing.)
3x
3x
= 3 and lim
= 3, we also
xS–∞ x - 4
x - 4
expect a horizontal asymptote at y = 3.
(b) Since lim
xS∞
[–3.7, 5.7] by [0, 6.2]
27. As the graph illustrates, y is increasing over the intervals
1 - q, - 12, 1 -1, 1 2, and 11, q 2 .
[–15, 15] by [–15, 15]
[–4.7, 4.7] by [–3.1, 3.1]
Chapter 1
28. As the graph illustrates, y is increasing over the intervals
1 - q, -2 2 and 1 -2, 0 4 .
34. (a) 2, at x = -1
Review
63
(b) -2, at x = 1
[–5, 5] by [–10, 10]
[–4.7, 4.7] by [–3.1, 3.1]
35. (a) -1, at x = 0
(b) None
29. - 1 sin x 1, but - q 6 x 6 q, so f1x2 is not
bounded.
[–5, 5] by [–10, 10]
36. (a) 1, at x = 2 (b) –1, at x=–2
[–5, 5] by [–5, 5]
30. g1 x2 = 3 at x = 1, a maximum and g1 x2 = -3,
a minimum, at x = -1. It is bounded.
[–10, 10] by [–5, 5]
37. The function is even since it is symmetrical about the y-axis.
[–10, 10] by [–5, 5]
31. ex 7 0 for all x, so - ex 6 0 and 5 - ex 6 5 for all x.
h1 x2 is bounded above.
[–4.7, 4.7] by [–3.1, 3.1]
38. Since the function is symmetrical about the origin, it is odd.
[–5, 5] by [–10, 10]
32. The function is linear with slope
1
and y-intercept
1000
1000. Thus k(x) is not bounded.
[–4.7, 4.7] by [–3.1, 3.1]
39. Since no symmetry is exhibited, the function is neither.
[–5, 5] by [–999.99, 1000.01]
33. (a) None
(b) -7, at x = - 1
[–1.35, 3.35] by [–1.55, 1.55]
[–6, 4] by [–10, 20]
64
Chapter 1
Functions and Graphs
40. Since the function is symmetrical about the origin, it is odd.
50. No
51.
[–9.4, 9.4] by [–6.2, 6.2]
x - 3
41. x = 2y + 3, 2y = x - 3, y =
, so
2
x - 3
.
f - 1 1x2 =
2
3
42. x= 1y - 8, x3 = y - 8, y = x3 + 8, so
f - 1 1x2 = x3 + 8.
43. x =
2
2
2
, xy = 2, y = , so f - 1 1x2 = .
y
x
x
6
, 1y + 4 2 x = 6, xy + 4x = 6, xy = 6 - 4x,
y + 4
6 - 4x
6
y =
- 4.
, so f-1 1 x2 =
x
x
44. x =
45.
[–5, 5] by [–5, 5]
52. f1x2 = e
x + 3 if x -1
x2 + 1 if x 7 -1
53. (f g)(x)=fA g1x2 B = f1x2 - 4 2 = 2x2 - 4.
Since x2 - 4 0, x2 4, x -2 or x 2.
The domain is 1 -q, -24 ´ 32, q 2 .
54. (g f)(x)=g1 f1 x2 2 = g1 2x2 = 1 2x2 2 - 4 =
x - 4. Since 2x 0, x 0. The domain is 3 0, q 2 .
55. 1f # g2 1x2 = f1x2 # g1 x2 = 2x
# 1x2 - 4 2.
Since 2x 0, the domain is 30, q 2.
f1x2
f
1x
= 2
56. a b 1 x2 =
Since x2 - 4 Z 0,
g
g1x2
x - 4
1x + 22 1 x - 2 2 Z 0, x Z -2, x Z 2. Also since
2x 0, x 0. The domain is 30, 2 2 ´ 12, q 2 .
57. lim 2x = q . (Large negative values are not in the
xSq
domain.)
58. lim 2x2 - 4 = q. (The graph resembles the line
xS ; q
[–5, 5] by [–5, 5]
y=x.)
2s2
s ¤
s ¤
2s2
s22
=
.
59. r¤= ¢ ≤ + ¢ ≤ = , r =
B 4
2
2
4
2
The area of the circle is
s22 2
2ps2
ps2
A = pr2 = p a
b =
=
2
4
2
46.
[–5, 5] by [–5, 5]
47.
s
r
s
2
s
2
s
s 2
2
r=
[–5, 5] by [–5, 5]
48.
s 2
ps2
60. A = pr2 = p a b =
2
4
s
s
2
s
2
[–5, 5] by [–5, 5]
49.
[–5, 5] by [–5, 5]
s
Chapter 1
d
, so the radius of the tank is 10 feet.
2
Volume is V = pr2 # h = p 110 2 2 # h = 100ph
61. d=2r, r =
Review
65
(b) The regression line is y=61.133x+725.333.
20 ft
[4, 15] by [940, 1700]
h
(c) 61.133(20)+725.333≠1948 (thousands of barrels)
66. (a)
62. The volume of oil in the tank is the amount of original oil
1pr2h2 minus the amount of oil drained.
V = pr2h - 2t = p 110 2 2 140 2 - 2t = 4000p - 2t
20 ft
[–10, 10] by [–2, 18]
(b) The linear model would eventually intersect the
x-axis, which would represent a swimmer covering
100 meters in a time of 0.00. This is clearly impossible.
h = 40 ft
(c) Based on the data, 52 seconds represents the limit of
women’s capability in this race. The addition of future
data could determine a different model.
(d) The regression curve is y=(97.100)(0.9614x).
63. Since V = 4000p - 2t, we know that
pr2h = 4000p - 2t. In this case, r = 10¿ , so
4000p - 2t
t
100ph = 4000p - 2t, h =
= 40 100p
50p
20 ft
[–10, 10] by [–2, 18]
h1
h = 40 ft
h2
64. Since the depth of the tank is decreasing by 2 feet per
hour, we know that the tank is losing a total volume of
V = pr2h = p110 2 2 1 2 2 = 200p cubic feet per hour.
The volume of remaining oil in the tank is the amount
of original oil subtracting the amount which has been
drained, or V = 4000p - 200pt. This is a significantly
higher loss than our solution in #68!
(e) (97.100)(0.9614108)≠1.38. Add 52 to find the projected
winning time in 2008: 1.38+52=53.38 seconds.
h 2
67. (a) r 2 + a b = 1 232 2,
2
h2
= 3 - r2, h2 = 12 - 4r2, h = 212 - 4r2,
4
h = 223 - r2
3
h
65. (a)
h
2
3
r
3
r
h = 2 3 – r2
[4, 15] by [940, 1700]
(b) V = pr2h = 1pr2 2 1223 - r2 2 = 2pr2 23 - r2
(c) Since 23 - r2 0, 3 - r2 0
3 r2, - 23 r 23.
However, r 6 0 are invalid values, so the domain is
30, 234.
66
Chapter 1
Functions and Graphs
Chapter 1 Project
(d)
1.
[0, 13] by [0, 20]
[–1, 13] by [–100, 2600]
(e) 12.57 in3
68.
2. The exponential regression produces
y L 21.956(1.511)x.
y
3. 2000: For x=13, y L 4690
2001: For x=14, y L 7085
y = 36 – x2
y
x
x
x
(a) A = 2xy = 2x136 - x2 2 = 72x - 2x3
(b) 36 - x2 0, 16 - x2 16 + x2 0, -6 x 6.
However, x<0 are invalid values, so the domain is
3 0, 64 .
(c)
4. The model, which is based on data from the early, highgrowth period of Starbucks Coffee’s company history,
does not account for the effects of gradual market saturation by Starbucks and its competitors. The actual growth
in the number of locations is slowing while the model
increases more rapidly.
5. The logistic regression produces
y L
4914.198
.
1 + 269.459 e-0.486x
6. 2000: For x=13, y L 3048
2001: For x=14, y L 3553
These predictions are less than the actual numbers, but
are not off by as much as the numbers derived from the
exponential model were. For the year 2020 (x=33), the
logistic model predicts about 4914 locations. (This prediction is probably too conservative.)
[0, 6] by [0, 180]
(d) The maximum area occurs when x L 3.46, or an area
of approximately 166.28 square units.
Section 2.1
Linear and Quadratic Functions and Modeling
67
Chapter 2
Polynomial, Power, and Rational Functions
■ Section 2.1 Linear and Quadratic Functions
and Modeling
Exploration 1
1. –$2000 per year
2. The equation will have the form v(t)=mt+b. The
value of the building after 0 year is
v(0)=m(0)+b=b=50,000.
The slope m is the rate of change, which is –2000 (dollars
per year). So an equation for the value of the building (in
dollars) as a function of the time (in years) is
v(t)=–2000t+50,000.
3. v(0)=50,000 and
v(16)=–2000(16)+50,000=18,000 dollars
4. The equation v(t)=39,000 becomes
–2000t+50,000=39,000
–2000t=–11,000
t=5.5 years
6. (x-4)2=(x-4)(x-4)=x2-4x-4x+16
=x2-8x+16
7. 3(x-6)2=3(x-6)(x-6)=(3x-18)(x-6)
=3x2-18x-18x+108=3x2-36x+108
8. –3(x+7)2=–3(x+7)(x+7)
=(–3x-21)(x+7)=–3x2-21x-21x-147
=–3x2-42x-147
9. 2x2-4x+2=2(x2-2x+1)=2(x-1)(x-1)
=2(x-1)2
10. 3x2+12x+12=3(x2+4x+4)=3(x+2)(x+2)
=3(x+2)2
Section 2.1 Exercises
1. Not a polynomial function because of the exponent –5
2. Polynomial of degree 1 with leading coefficient 2
3. Polynomial of degree 5 with leading coefficient 2
4. Polynomial of degree 0 with leading coefficient 13
5. Not a polynomial function because of cube root
Quick Review 2.1
6. Polynomial of degree 2 with leading coefficient –5
1. y=8x+3.6
5
5
5
18
7. m= so y-4= (x-2) ⇒ f(x)= x+
7
7
7
7
2. y=–1.8x-2
y
3
3. y-4=– (x+2), or y=–0.6x+2.8
5
5
y
(2, 4)
7
3
x
(–5, –1)
(2, 4)
(3, 1)
5
x
7
7
7
8
8. m= - so y-5= - (x+3) ⇒ f(x)=– x+
9
9
9
3
7
8
8
4. y-5= (x-1), or y= x+
3
3
3
y
10
y
(–3, 5)
6
(1, 5)
10
(6, –2)
5
x
(2, 3)
5. (x+3)2=(x+3)(x+3)=x2+3x+3x+9
=x2+6x+9
x
68
Chapter 2
Polynomial, Power, and Rational Functions
4
4
4
2
9. m= - so y-6= - (x+4) ⇒ f(x)= - x+
3
3
3
3
y
10
16. (f)—the vertex is in Quadrant I, at (1, 12), meaning it
must be either (b) or (f). Since f(0)=10, it cannot be (b):
if the vertex in (b) is (1, 12), then the intersection with the
y-axis occurs considerably lower than (0, 10). It must be (f).
17. (e)—the vertex is at (1, –3) in Quadrant IV, so it must
be (e).
(–4, 6)
(–1, 2)
10
x
18. (c)—the vertex is at (–1, 12) in Quadrant II and the
parabola opens down, so it must be (c).
19. Translate the graph of f(x)=x2 3 units right to obtain
the graph of h(x)=(x-3)2, and translate this graph
2 units down to obtain the graph of g(x)=(x-3)2-2.
y
3
5
5
5
10. m= so y-2= (x-1) ⇒ f(x)= x+
4
4
4
4
10
y
10
10
(5, 7)
x
(1, 2)
10
x
11. m=–1 so y-3=–1(x-0) ⇒ f(x)=–x+3
y
20. Vertically shrink the graph of f(x)=x2 by a factor of
1
1
to obtain the graph of g(x)= x2, and translate this
4
4
1
graph 1 unit down to obtain the graph of h(x)= x2-1.
4
y
5
10
(0, 3)
(3, 0)
x
5
1
1
1
12. m= so y-2= (x-0) ⇒ f(x)= x+2
2
2
2
y
10
(0, 2)
(–4, 0)
10
x
10
x
21. Translate the graph of f(x)=x2 2 units left to obtain the
graph of h(x)=(x+2)2, vertically shrink this graph by
1
1
a factor of to obtain the graph of k(x)= (x+2)2,
2
2
and translate this graph 3 units down to obtain the graph
1
of g(x)= (x+2)2-3.
2
y
10
13. (a)—the vertex is at (–1, –3), in Quadrant III, eliminating all but (a) and (d). Since f(0)=–1, it must be (a).
14. (d)—the vertex is at (–2, –7), in Quadrant III, eliminating
all but (a) and (d). Since f(0)=5, it must be (d).
15. (b)—the vertex is in Quadrant I, at (1, 4), meaning it
must be either (b) or (f). Since f(0)=1, it cannot be (f):
if the vertex in (f) is (1, 4), then the intersection with the
y-axis would be about (0, 3). It must be (b).
10
x
Section 2.1
22. Vertically stretch the graph of f(x)=x2 by a factor of 3
to obtain the graph of g(x)=3x2, reflect this graph
across the x-axis to obtain the graph of k(x)=–3x2, and
translate this graph up 2 units to obtain the graph of
h(x)=–3x2+2.1
y
10
10
x
For #23–32, with an equation of the form
f(x)=a(x-h)2+k, the vertex is (h, k) and the axis is
x=h.
23. Vertex: (1, 5); axis: x=1
24. Vertex: (–2, –1); axis: x=–2
25. Vertex: (1, –7); axis: x=1
Linear and Quadratic Functions and Modeling
32. h(x)=–2 a x2 +
7
x b -4
2
7
49
49
=–2 a x2 + 2 # x +
b -4+
4
16
8
7 2
17
=–2 a x + b +
4
8
7
7 17
Vertex: a - , b ; axis: x = 4 8
4
33. f(x)=(x2-4x+4)+6-4=(x-2)2+2.
Vertex: (2, 2); axis: x=2; opens upward; does not intersect x–axis.
[–4, 6] by [0, 20]
34. g(x)=(x2-6x+9)+12-9=(x-3)2+3.
Vertex: (3, 3); axis: x=3; opens upward; does not intersect x–axis.
26. Vertex: ( 13, 4); axis: x= 13
5
x b -4
3
5
25
25
5 2
73
b -4- =3 a x + b =3 a x2 + 2 # x +
6
36
12
6
12
5
5 73
Vertex: a - , - b ; axis: x=–
6 12
6
27. f(x)=3 a x2 +
7
x b -3
2
7
49
49
b -3+
=–2 a x2 - 2 # x +
4
16
8
7 2
25
=–2 a x - b +
4
8
7
7 25
Vertex: a , b ; axis: x=
4 8
4
28. f(x)=–2 a x2 -
29. f(x)=–(x2-8x)+3
=–(x2-2 # 4x+16)+ 3+16=–(x-4)2+19
Vertex: (4, 19); axis: x=4
1
30. f(x)=4 a x2 - x b +6
2
2
23
1
1
1
1
=4 a x2 - 2 # x +
b +6- =4 a x - b +
4
16
4
4
4
1 23
1
Vertex: a , b ; axis: x =
4 4
4
6
31. g(x)=5 a x2 - x b +4
5
3
9
9
3 2
11
=5 a x2 - 2 # x +
b +4- =5 a x - b +
5
25
5
5
5
3
3 11
Vertex: a , b ; axis: x =
5 5
5
69
[–2, 8] by [0, 20]
35. f(x)=–(x2+16x)+10
=–(x2+16x+64)+10+64=–(x+8)2+74.
Vertex: (–8, 74); axis: x=–8; opens downward; intersects x–axis at about –16.602 and 0.602 1 -8 ; 1742 .
[–20, 5] by [–100, 100]
36. h(x)=–(x2-2x)+8=–(x2-2x+1)+8+1
=–(x-1)2+9
Vertex: (1, 9); axis: x=1; opens downward; intersects
x–axis at –2 and 4.
[–9, 11] by [–100, 10]
70
Chapter 2
Polynomial, Power, and Rational Functions
37. f(x)=2(x2+3x)+7
9
3 2
9
5
=2 a x2 + 3x + b +7- =2 a x + b +
4
2
2
2
3
3 5
Vertex: a - , b ; axis: x = - ; opens upward; does not
2 2
2
intersect the x-axis; vertically stretched by 2.
49. (a)
[15, 45] by [20, 50]
(b) Strong positive
50. (a)
[–3.7, 1] by [2, 5.1]
38. g(x)=5(x2-5x)+12
25
125
=5 a x2 - 5x +
b +124
4
77
5 2
=5 a x - b 2
4
5 77
5
Vertex: a , - b ; axis: x = ; opens upward; intersects
2
4
2
x–axis at about 0.538 and
5
1
4.462 ¢ or ;
1385 b ; vertically stretched by 5.
2
10
[0, 90] by [0, 70]
(b) Strong negative
2350
=–470 and b=2350,
5
so v(t)=–470t+2350.
At t=3, v(3)=(–470)(3)+2350=$940.
51. m=–
52. Let x be the number of dolls produced each week and y
be the average weekly costs. Then m=4.70, and b=350,
so y=4.70x+350, or 500=4.70x+350:
x=32; 32 dolls are produced each week.
53. (a) y L 0.541x + 4.072. The slope, m L 0.541, represents
the average annual increase in hourly compensation
for production workers, about $0.54 per year.
(b) Setting x=40 in the regression equation leads to
y L $25.71.
[–5, 10] by [–20, 100]
For #39–44, use the form y=a(x-h)2+k, taking the
vertex (h, k) from the graph or other given information.
54. If the length is x, then the width is 50-x, so
A(x)=x(50-x); maximum of 625 ft2 when x=25
(the dimensions are 25 ft*25 ft).
39. h=–1 and k=–3, so y=a(x+1)2-3. Now
substitute x=1, y=5 to obtain 5=4a-3,
so a=2: y=2(x+1)2-3.
55. (a) [0, 100] by [0, 1000] is one possibility.
40. h=2 and k=–7, so y=a(x-2)2-7. Now
substitute x=0, y=5 to obtain 5=4a-7,
so a=3: y=3(x-2)2-7.
56. The area of the picture and the frame, if the width of the
picture is x ft, is A(x)=(x+2)(x+5) ft2. This equals
208 when x=11, so the painting is 11 ft*14 ft.
41. h=1 and k=11, so y=a(x-1)2+11. Now
substitute x=4, y=–7 to obtain –7=9a+11,
so a=–2: y=–2(x-1)2+11.
57. If the strip is x feet wide, the area of the strip is
A(x)=(25+2x)(40+2x)-1000 ft2.
This equals 504 ft2 when x=3.5 ft.
42. h=–1 and k=5, so y=a(x+1)2+5. Now
substitute x=2, y=–13 to obtain –13=9a+5,
so a=–2: y=–2(x+1)2+5.
58. (a) R(x)=(800+20x)(300-5x).
43. h=1 and k=3, so y=a(x-1)2+3. Now
substitute x=0, y=5 to obtain 5=a+3,
so a=2: y=2(x-1)2+3.
(b) When x≠107.335 or x≠372.665 — either 107, 335
units or 372, 665 units.
(b) [0, 25] by [200,000, 260,000] is one possibility
(shown).
44. h=–2 and k=–5, so y=a(x+2)2-5. Now
substitute x=–4, y=–27 to obtain –27=4a-5,
11
11
so a=– : y=– (x+2)2-5.
2
2
45. Strong positive
46. Strong negative
47. Weak positive
48. No correlation
[0, 25] by [200,000, 260,000]
(c) The maximum income — $250,000 — is achieved
when x=10, corresponding to rent of $250 per
month.
Section 2.1
59. (a) R(x)=(26,000-1000x)(0.50+0.05x).
(b) Many choices of Xmax and Ymin are reasonable.
Shown is [0, 15] by [10,000, 17,000].
[0, 15] by [10,000, 17,000]
(c) The maximum revenue — $16,200 — is achieved
when x=8; that is, charging 90 cents per can.
60. Total sales would be S(x)=(30+x)(50-x)
thousand dollars, when x additional salespeople are hired.
The maximum occurs when x=10 (halfway between the
two zeros, at –30 and 50).
61. (a) g L 32 ft>sec2. s0 = 83 ft and v0 = 92 ft>sec. So the
models are height=s1 t2 = - 16t2 + 92t + 83 and
vertical velocity=v1t2 = -32t + 92. The maximum
height occurs at the vertex of s1t2 .
92
b
= = 2.875, and
h = 2a
2 1 -16 2
k = s12.8752 = 215.25. The maximum height of the
baseball is about 215 ft above the field.
(b) The amount of time the ball is in the air is a zero of
s1t2 . Using the quadratic formula, we obtain
-92 ; 292 - 4 1 -16 2 183 2
2
2 1 -16 2
-92 ; 113,776
L -0.79 or 6.54. Time is not
=
-32
negative, so the ball is in the air about 6.54 seconds.
t =
(c) To determine the ball’s vertical velocity when it hits
the ground, use v 1t2 = - 32t + 92, and solve for
t = 6.54. v 16.542 = -32 16.54 2 + 92 L - 117 ft>sec
when it hits the ground.
2
62. (a) h=–16t +48t+3.5.
(b) The graph is shown in the window [0, 3.5] by
[0, 45]. The maximum height is 39.5 ft, 1.5 sec after
it is thrown.
Linear and Quadratic Functions and Modeling
71
64. The exact answer is 32 13, or about 55.426 ft/sec. In
addition to the guess-and-check strategy suggested, this
can be found algebraically by noting that the vertex of
the parabola y=ax2+bx+c has y coordinate
b2
b2
c=
(note a=–16 and c=0), and setting
4a
64
this equal to 48.
65. The quadratic regression is y≠0.449x2+0.934x+
114.658. Plot this curve together with the curve y=450,
and then find the intersection to find when the number of
patent applications will reach 450,000. Note that we use
y=450 because the data were given as a number of
thousands. The intersection occurs at x L 26.3, so the
number of applications will reach 450,000 approximately
26 years after 1980—in 2006.
[1, 8.25] by [0, 5]
66. (a) m=
6 ft
=0.06
100 ft
(b) r≠4167 ft, or about 0.79 mile.
(c) 2217.6 ft
67. (a)
[15, 45] by [20, 40]
(b) y≠0.68x+9.01
(c) On average, the children gained 0.68 pound
per month.
(d)
[15, 45] by [20, 40]
(e) ≠29.41 lbs
[0, 3.5] by [0, 45]
63. (a) h=–16t2+80t-10. The graph is shown in the
window [0, 5] by [–10, 100].
[0, 5] by [–10, 100]
(b) The maximum height is 90 ft, 2.5 sec after it is shot.
68. (a) The linear regression is y L 548.30x + 21,027.56,
where x represents the number of years since 1940.
(b) 2010 is 70 years after 1940, so substitute 70 into the
equation to predict the median U.S. family income in
2010.
y = 548.301702 + 21,027.56 L $59,400.
72
Chapter 2
Polynomial, Power, and Rational Functions
For #75–76, f(x)=2(x+3)2-5 corresponds to
f(x)=a(x-h)2+k with a=2 and (h, k)=(–3, –5).
69. The Identity Function f(x)=x
75. The axis of symmetry runs vertically through the vertex:
x=–3. The answer is B.
76. The vertex is (h, k)=(–3, –5). The answer is E.
[–4.7, 4.7] by [–3.1, 3.1]
Domain: 1 - q, q 2
Range: 1 - q, q 2
Continuity: The function is continuous on its domain.
Increasing–decreasing behavior: Increasing for all x
Symmetry: Symmetric about the origin
Boundedness: Not bounded
Local extrema: None
Horizontal asymptotes: None
Vertical asymptotes: None
End behavior: lim f1x2 = - q and lim f1x2 = q
xS-q
77. (a) Graphs (i), (iii), and (v) are linear functions. They can
all be represented by an equation y=ax+b, where
a Z 0.
(b) In addition to graphs (i), (iii), and (v), graphs (iv) and
(vi) are also functions, the difference is that (iv) and (vi)
are constant functions, represented by y=b, b Z 0.
(c) (ii) is not a function because a single value x
(i.e., x=–2) results in a multiple number of y-values.
In fact, there are infinitely many y-values that are
valid for the equation x=–2.
78. (a)
xSq
70. The Squaring Function f(x)=x2
(b)
(c)
(d)
[–4.7, 4.7] by [–1, 5]
Domain: 1 - q, q 2
Range: 3 0, q 2
Continuity: The function is continuous on its domain.
Increasing–decreasing behavior: Increasing on 30, q 2 ,
decreasing on 1 - q, 0 4 .
Symmetry: Symmetric about the y-axis
Boundedness: Bounded below, but not above
Local extrema: Local minimum of 0 at x=0
Horizontal asymptotes: None
Vertical asymptotes: None
End behavior: lim f1x2 = lim f1x2 = q
xSq
xS-q
2
71. False. For f(x)=3x +2x-3, the initial value is
f(0)=–3.
72. True. By completing the square, we can rewrite f(x)
1
1
so that f(x)= a x2 - x + b +14
4
3
1 2
3
= a x - b + . Since f1 x2 , f(x)>0 for all x.
2
4
4
73. m =
1 - 3
-2
1
=
= - . The answer is E.
4 - 1 -22
6
3
74. f(x)=mx+b
1
3= - (–2)+b
3
2
3= +b
3
2
7
b=3- =
3
3
The answer is C.
(e)
(f)
(g)
(h)
f1 32 - f1 12
3 - 1
f1 52 - f1 22
5 - 2
f1 c2 - f1a2
c - a
g1 32 - g112
3 - 1
g1 42 - g112
4 - 1
=
9 - 1
=4
2
=
25 - 4
=7
3
=
1c - a2 1c + a2
c2 - a2
=
=c+a
c - a
c - a
=
11 - 5
=3
2
=
14 - 5
=3
3
13c + 2 2 - 13a + 2 2
=
c - a
c - a
3c - 3a
=
=3
c - a
g1 c2 - g1a2
h1c2 - h1 a2
=
17c - 32 - 17a - 32
=
1mc + b2 - 1 ma + b 2
c - a
7c - 7a
=
=7
c - a
k 1c2 - k 1a 2
c - a
c - a
c - a
mc - ma
=
=m
c - a
l1c2 - l1a2
c3 - a 3
-2b
-b
b
(i)
=
=
=
= c - a
c - a
2a
a
a
1c - a2 1c2 + ac + a2 2
=
=c2+ac+a2
1c - a2
79. Answers will vary. One possibility: When using the leastsquares method, mathematicians try to minimize the
residual yi-(axi+b), i.e., place the “predicted”
y-values as close as possible to the actual y-values. If
mathematicians reversed the ordered pairs, the objective
would change to minimizing the residual xi-(cyi+d),
i.e., placing the “predicted” x-values as close as possible to
the actual x-values. In order to obtain an exact inverse,
the x- and y-values for each xy pair would have to be
almost exactly the same distance from the regression
line—which is statistically impossible in practice.
Section 2.2
80. (a)
Power Functions with Modeling
73
83. Multiply out f(x) to get x2-(a+b)x+ab. Complete
1a + b22
a + b 2
b +abthe square to get a x . The
2
4
a + b
and
2
1a + b 2 2
1a - b 2 2
k=ab=.
4
4
vertex is then (h, k) where h=
[0, 17] by [2, 16]
(b) y≠0.115x+8.245
84. x1 and x2 are given by the quadratic formula
b
-b ; 2b2 - 4ac
; then x1+x2= - , and the line of
2a
a
x1 + x2
b
symmetry is x= - , which is exactly equal to
.
2a
2
[0, 17] by [2, 16]
(c) y≠0.556x+6.093
[0, 15] by [0, 15]
(d) The median–median line appears to be the better fit,
because it approximates more of the data values more
closely.
81. (a) If ax2+bx2+c=0, then
- b ; 2b2 - 4ac
by the quadratic
2a
-b + 2b2 - 4ac
and
formula. Thus, x1=
2a
x=
x2=
- b - 2b2 - 4ac
and
2a
x1+x2=
-b + 2b2 - 4ac - b - 2b2 - 4ac
2a
85. The Constant Rate of Change Theorem states that a
function defined on all real numbers is a linear function
if and only if it has a constant nonzero average rate of
change between any two points on its graph. To prove
this, suppose f1x2 = mx + b with m and b constants
and m Z 0. Let x1 and x2 be real numbers with x1 Z x2.
Then the average rate of change is
f1 x2 2 - f1x1 2
1mx2 + b2 - 1mx1 + b2
=
=
x2 - x1
x2 - x1
m 1x2 - x1 2
mx2 - mx1
=
= m, a nonzero constant.
x2 - x1
x2 - x1
Now suppose that m and x1 are constants, with m Z 0.
Let x be a real number such that x Z x1, and let f be a
function defined on all real numbers such that
f1 x2 - f1x1 2
= m. Then f1x2 - f1x1 2 = m1x - x1 2 =
x - x1
mx - mx1, and f1 x2 = mx + 1f1 x1 2 - mx1 2 .
f1 x1 2 - mx1 is a constant; call it b. Then
f1 x1 2 - mx1 = b; so, f1x1 2 = b + mx1 and
f1 x2 = b + mx for all x Z x1. Thus, f is a linear function.
■ Section 2.2 Power Functions with Modeling
Exploration 1
1.
-2b
-b
b
=
=
= - .
2a
a
a
(b) Similarly,
-b + 2b2 - 4ac
-b - 2b2 - 4ac
ba
b
2a
2a
b2 - 1b2 - 4ac2
4ac
c
=
= 2= .
2
a
4a
4a
82. f(x)=(x-a)(x-b)=x2-bx-ax+ab
x1 # x2= a
[–2.35, 2.35] by [–1.5, 1.5]
=x2+(–a-b)x+ab. If we use the vertex form of
-a - b
a quadratic function, we have h= - a
b
2
=
a + b
a + b
. The axis is x=h=
.
2
2
[–5, 5] by [–15, 15]
74
Chapter 2
Polynomial, Power, and Rational Functions
6.
1
2m3
7. 3x3>2
8. 2x5>3
9. L 1.71x-4>3
[–20, 20] by [–200, 200]
The pairs (0, 0), (1, 1), and (–1, –1) are common
to all three graphs. The graphs are similar in that if
x 6 0, f1 x2 , g1 x2 , and h1 x2 6 0 and if x>0, f(x), g(x),
and h(x)>0. They are different in that if 0 x 0 6 1,
f1 x2, g1 x2 , and h1 x2 S 0 at dramatically different rates,
and if 0 x 0 7 1, f1x2, g1x2, and h1x2 S q at dramatically
different rates.
2.
10. L 0.71x-1>2
Section 2.2 Exercises
1. power=5, constant= -
1
2
5
2. power= , constant=9
3
3. not a power function
4. power=0, constant=13
5. power=1, constant=c2
k
2
g
7. power=2, constant=
2
6. power=5, constant=
[–1.5, 1.5] by [–0.5, 1.5]
8. power=3, constant=
4p
3
9. power=–2, constant=k
10. power=1, constant=m
11. degree=0, coefficient=–4
12. not a monomial function; negative exponent
[–5, 5] by [–5, 25]
13. degree=7, coefficient=–6
14. not a monomial function; variable in exponent
15. degree=2, coefficient=4∏
16. degree=1, coefficient=l
17. A=ks¤
18. V=kr¤
[–15, 15] by [–50, 400]
The pairs (0, 0), (1, 1), and (–1, 1) are common to all
three graphs. The graphs are similar in that for x Z 0,
f1x2, g1 x2 , and h1 x2 7 0. They are diffferent in that if
0 x 0 6 1, f1x2, g1x2 , and h1x2 S 0 at dramatically
different rates, and if 0 x 0 7 1, f1 x2, g1x2 , and h1x2 S q
at dramatically different rates.
Quick Review 2.2
3 2
1. 2x
2. 2p5
1
3. 2
d
1
4. 7
x
1
5. 5
2q4
19. I=V/R
20. V=kT
21. E=mc¤
22. p = 12gd
23. The weight w of an object varies directly with its mass m,
with the constant of variation g.
24. The circumference C of a circle is proportional to its
diameter D, with the constant of variation ∏.
25. The refractive index n of a medium is inversely proportional to v, the velocity of light in the medium, with constant of variation c, the constant velocity of light in free
space.
26. The distance d traveled by a free-falling object dropped
from rest varies directly with the square of its speed p,
1
with the constant of variation .
2g
Section 2.2
27. power=4, constant=2
Domain: 1 - q, q 2
Range: 3 0, q 2
Continuous
Decreasing on 1 - q, 0 2 . Increasing on 1 0, q 2 .
Even. Symmetric with respect to y-axis.
Bounded below, but not above
Local minimum at x=0.
Asymptotes: None
End Behavior: lim 2x4 = q , lim 2x4 = q
xS-q
Power Functions with Modeling
75
Discontinuous at x=0
Increasing on 1 - q, 0 2 . Increasing on 10, q 2 .
Odd. Symmetric with respect to origin
Not bounded above or below
No local extrema
Asymptotes at x=0 and y=0
End Behavior: lim -2x-3 = 0, lim -2x-3 = 0.
xS-q
xSq
xSq
[–5, 5] by [–5, 5]
[–5, 5] by [–1, 49]
28. power=3, constant=–3
Domain: 1 - q, q 2
Range: 1 - q, q 2
Continuous
Decreasing for all x
Odd. Symmetric with respect to origin
Not bounded above or below
No local extrema
Asymptotes: None
End Behavior: lim -3x3 = q , lim - 3x3 = - q
xS-q
xSq
2
31. Start with y = x4 and shrink vertically by . Since
3
2
2
f(–x)= 1 - x2 4 = x4, f is even.
3
3
[–5, 5] by [–1, 19]
32. Start with y = x3 and stretch vertically by 5. Since
f1 -x2 = 51 -x2 3 = -5x3 = -f1x2 , f is odd.
[–5, 5] by [–20, 20]
1
1
29. power= , constant=
4
2
Domain: 3 0, q 2
Range: 3 0, q 2
Continuous
Increasing on 30, q 2
Bounded below
Neither even nor odd
Local minimum at (0, 0)
Asymptotes: None
1 4
End Behavior: lim 1x
= q
xSq 2
[–5, 5] by [–20, 20]
33. Start with y = x5, then stretch vertically by 1.5 and reflect
over the x-axis. Since f1 -x2 = - 1.51 -x2 5 = 1.5x5
=–f(x), f is odd.
[–5, 5] by [–20, 20]
34. Start with y = x6, then stretch vertically by 2 and reflect
over the x-axis. Since f1 -x2 = - 21 -x2 6 = -2x6
=f(x), f is even.
[–1, 99] by [–1, 4]
30. power=–3, constant=–2
Domain: 1 - q, 0 2 ´ 10, q 2
Range: 1 - q, 0 2 ´ 10, q 2
[–5, 5] by [–19, 1]
76
Chapter 2
Polynomial, Power, and Rational Functions
1
35. Start with y = x8, then shrink vertically by . Since
4
1
1
f(–x)= 1 -x2 8 = x8 = f1 x2, f is even.
4
4
4
45. k=–2, a= . In the fourth quadrant, f is decreasing
3
3
and concave down. f1 -x2 = -2 1 2 1 -x2 4 2
3
=–2( 2x4)= -2x4>3=f(x), so f is even.
[–5, 5] by [–1, 49]
1
36. Start with y = x7, then shrink vertically by . Since
8
1
1
f(–x)= 1 -x2 7 = - x7 = -f1 x2 , f is odd.
8
8
[–10, 10] by [–29, 1]
2
5
46. k= , a= . In the first quadrant, f is increasing and
5
2
concave up. f is undefined for x 6 0.
[–2, 8] by [–1, 19]
[–5, 5] by [–50, 50]
37. (g)
38. (a)
39. (d)
40. (g)
1
47. k= , a=–3. In the first quadrant, f is decreasing and
2
1
1
1
= - x-3
concave up. f(–x)= 1 -x2 -3 =
2
2
2 1 -x2 3
=–f(x), so f is odd.
41. (h)
42. (d)
1
43. k=3, a= . In the first quadrant, the function is
4
increasing and concave down. f is undefined for x 6 0.
[–5, 5] by [–20, 20]
[–1, 99] by [–1, 10]
48. k=–1, a=–4. In the fourth quadrant, f is increasing
1
and concave down. f(–x)= - 1 -x2 -4 = 1 -x2 4
1
= - 4 = -x-4 = f1x2 , so f is even.
x
2
44. k=–4, a= . In the fourth quadrant, the function is
3
3
decreasing and concave up. f1 -x2 = -4 1 2 1 -x2 2 2
3 2
= -4 2x
= -4x2>3=f(x), so f is even.
[–5, 5] by [–19, 1]
49. y =
[–10, 10] by [–29, 1]
8
, power=–2, constant=8
x2
1
50. y = -2 1x, power= , constant=–2
2
Section 2.2
1 0.926 atm2 13.46 L 2
kT
PV
, so k =
=
P
T
302°K
atm–L
=0.0106
K
0.0106 atm–L
b 1 302 °K2
a
K
At P=1.452 atm, V=
1.452 atm
=2.21 L
51. V=
52. V=kPT, so k =
=0.0124
L
atm–K
1 3.46 L2
V
=
PT
10.926 atm2 1302 °K2
At T=338°K, V= a 0.0124
Power Functions with Modeling
77
57. (a)
[0.8, 3.2] by [–0.3, 9.2]
(b) y L 7.932 # x-1.987; yes
(c)
L
b (0.926 atm)
atm–K
(338°K)=3.87 L
[0.8, 3.2] by [0.3, 9.2]
3.00 * 108 m
a
b
sec
c
c
m
53. n = , so v= =
=1.24 * 108
v
n
2.42
sec
P
15 w
54. P = kv , so k = 3 =
= 1.5 * 10-2
v
1 10 mph 2 3
3
Wind Speed (mph)
Power (W)
10
15
20
120
40
960
80
7680
Since P = kv3 is a cubic, power will increase significantly
with only a small increase in wind speed.
55. (a)
(d) Approximately 2.76
W
W
and 0.697 2 , respectively.
2
m
m
58. True, because f(–x)=(–x)-2/3=[(–x)2]-1/3
=(x2)-1/3=x-2/3 =f(x).
59. False. f(–x)=(–x)1/3=–(x1/3)=–f(x) and so the
function is odd. It is symmetric about the origin, not the
y-axis.
60. f1 42 = 2 14 2 -1>2 =
2
2
2
=
=
= 1.
2
14
41>2
The answer is A.
61. f1 02 = -3102 -1>3 = -3 #
1
1
= -3 # is undefined.
0
01>3
Also, f(1)=–3(–1)–1/3=–3(–1)=3,
f(1)=–3(1)–1/3=–3(1)=–3, and
f(3)=–3(3)–1/3≠–2.08. The answer is E.
62. f(–x)=(–x)2/3=[(–x)2]1/3=(x2)1/3=x2/3=f(x)
The function is even. The answer is B.
[–2, 71] by [50, 450]
(b) r L 231.204 # w-0.297
(c)
63. f(x)=x3/2=(x1/2)3= 1 1x2 3 is defined for x 0.
The answer is B.
64. Answers will vary. In general, however, students will find
n
n even: f1x2 = k # xm>n = k # 2xm, so f is
undefined for x<0.
n m
m even, n odd: f1x2 = k # xm>n = k # 2x
; f1 -x2
n
n m
= k# 2
1 -x2 m = k # 2x
= f1x2,
so f is even.
n m
m odd, n odd: f1 x2 = k # xm>n = k # 2x
; f1 -x2
[–2, 71] by [50, 450]
(d) Approximately 37.67 beats/min, which is very close to
Clark’s observed value.
56. Given that n is an integer, n 1:
If n is odd, then f1 - x2 = 1 - x2 n = - 1 xn 2 = -f1 x2 and
so f(x) is odd.
If n is even, then f(–x)=(–x)n=xn=f(x) and so
f(x) is even.
n
n
= k # 21 - x2 m = -k # 2xm
= - k # xm>n = -f1x2 , so f is odd.
78
Chapter 2
Polynomial, Power, and Rational Functions
65. (a)
[0, 1] by [0, 5]
[0, 3] by [0, 3]
f
[–2, 2] by [–2, 2]
g
h
k
Domain
x0
x0
x0
x0
Range
y0
y0
y0
y0
Continuous
yes
yes
yes
yes
Increasing
(, 0)
(, 0)
Decreasing
(, 0), (0, ) (0, )
(, 0), (0, ) (0, )
Symmetry
w.r.t. origin
w.r.t. y-axis
w.r.t. origin
w.r.t. y-axis
Bounded
not
below
not
below
Extrema
none
none
none
none
Asymptotes
x-axis, y-axis
x-axis, y-axis
x-axis, y-axis
x-axis, y-axis
End Behavior
lim f(x) 0
x → lim g(x) 0
x → lim h(x) 0
x → lim k(x) 0
x → (b)
[0, 1] by [0, 1]
[0, 3] by [0, 2]
f
[–3, 3] by [–2, 2]
g
h
k
Domain
[0, )
(, )
[0, )
(, )
Range
y0
(, )
y0
(, )
Continuous
yes
yes
yes
yes
Increasing
[0, )
(, )
[0, )
(, )
Symmetry
none
w.r.t. origin
none
w.r.t. origin
Bounded
below
not
below
not
Extrema
min at (0, 0)
none
min at (0, 0)
none
Asymptotes
none
none
none
none
End
behavior
lim f(x) lim g(x) lim h(x) Decreasing
x→
x→
lim g(x) x → The graphs of f(x)=x–1 and
h(x)=x–3 are similar and appear in
the 1st and 3rd quadrants only. The
graphs of g(x)=x–2 and k(x)=x–4
are similar and appear in the 1st and
2nd quadrants only. The pair (1, 1) is
common to all four functions.
x→
lim k(x) x→
lim k(x) x → The graphs of f(x)=x1/2 and
h(x)=x1/4 are similar and appear in
the 1st quadrant only. The graphs of
g(x)=x1/3 and k(x)=x1/5 are similar and appear in the 1st and 3rd
quadrants only. The pairs (0, 0), (1, 1)
are common to all four functions.
Section 2.3
66.
y
5
Polynomial Functions of Higher Degree with Modeling
79
69. If f is even,
x– π
1
1
, 1 f1 x2 Z 02.
=
f1x2
f1 -x2
1
1
Since g(x)=
=g(–x), g is also even.
=
f1 x2
f1 -x2
f1 x2 = f1 -x2 , so
xπ
x
1
a b
π
x
9
–xπ
If g is even,
1
a b
π
g(x)=g(–x), so g(–x)=
–x
–x–π
Since
The graphs look like those shown in Figure 2.14 on page
192.
f1 x2 = xp looks like the red graph in Figure 2.14(a)
because k=1>0 and a = p 7 1.
f1 x2 = x1>p looks like the blue graph in Figure 2.14(a)
because k=1>0 and 0<a=1/p<1.
1
1
=g(x)=
.
f1 -x2
f1x2
1
1
=
, f(–x)=f(x), and f is even.
f1 -x2
f1 x2
If f is odd,
1
1
, f(x) Z 0.
= f1 x2
f1x2
1
1
Since g(x)=
=–g(x), g is also odd.
= f1 x2
f1x2
f1 x2 = -f1 x2 , so
If g is odd,
f1 x2 = x -p looks like the green graph in Figure 2.14(a)
because k=1<0 and a = -p 6 0.
g(x)=g(–x), so g(–x)=
f1 x2 = - xp looks like the red graph in Figure 2.14(b)
because k=–1<0 and a = p 7 1.
Since
f1 x2 = - x1>p looks like the blue graph in Figure 2.14(b)
because k=–1<0 and a = -p 6 0.
f1 x2 = - x -p looks like the green graph in Figure 2.14(b)
because k=–1<0 and a = -p 6 0.
67. Our new table looks like:
Table 2.10 (revised) Average Distances and Orbit
Periods for the Six Innermost Planets
Average Distance
from Sun (Au)
Period of
Orbit (yrs)
Mercury
0.39
0.24
Venus
0.72
0.62
Earth
1
1
Mars
1.52
1.88
Planet
Jupiter
5.20
11.86
Saturn
9.54
29.46
then, y = x
1T = a
1 1T2 = 1a
2
(c) The force of gravity F acting on two objects varies
jointly as their masses m1 and m2 and inversely as the
square of the distance r between their centers, with
the constant of variation G, the universal gravitational
constant.
2 1T = a .
3>2 2
70. Let g1x2 = x -a and f1x2 = xa. Then g1x2 = 1>xa =
1>f1 x2 . Exercise 69 shows that g1x2 = 1>f1x2 is even if
and only if f1x2 is even, and g1 x2 = 1>f1 x2 is odd if and
only if f1x2 is odd. Therefore, g1x2 = x -a is even if and
only if f1x2 = xa is even, and that g1x2 = x -a is odd if
and only if f1x2 = xa is odd.
(b) The kinetic energy KE of an object varies jointly as
the mass m of the object and the square of the
velocity v of the object.
Using these new data, we find a power function model of:
y L 0.99995 # x1.50115 L x1.5. Since y represents years, we
set y = T and since x represents distance, we set x = a
3>2
1
1
=, f(–x)=–f(x), and f is odd.
f1 -x2
f1x2
71. (a) The force F acting on an object varies jointly as the
mass m of the object and the acceleration a of the
object.
Source: Shupe, Dorr, Payne, Hunsiker, et al., National Geographic Atlas of the World
(rev. 6th ed.). Washington, DC: National Geographic Society, 1992, plate 116.
1.5
1
1
=–g(x)= .
f1 -x2
f1x2
2
3
68. Using the free-fall equations in Section 2.1, we know that
1
s1t2 = - gt2 + v0 t + s0 and that v 1 t 2 = -gt + v0. If
2
t=0 is the time at which the object is dropped, then
1
1
v0 = 0. So d = s0 - s = s0 - a - gt2 + s0 b = gt2
2
2
1 2
and p = 0 v 0 = 0 - gt 0 . Solving d = gt for t, we have
2
2dg2
2d
2d
. Then p = ` -g
t =
` =
= 12dg.
A g
B g
A g
The results of Example 6 approximate this formula.
■ Section 2.3 Polynomial Functions
of Higher Degree with Modeling
Exploration 1
1. (a) lim 2x3 = q , lim 2x3 = - q
xSq
xS-q
[–5, 5] by [–15, 15]
80
Chapter 2
Polynomial, Power, and Rational Functions
(b) lim 1 - x3 2 = - q , lim 1 -x3 2 = q
xSq
xS-q
(d) lim 1 -0.5x2 2 = - q , lim 1 -0.5x2 2 = - q
xSq
[–5, 5] by [–15, 15]
[–5, 5] by [–15, 15]
5
3. (a) lim 1 -0.3x5 2 = - q , lim 1 -0.3x5 2 = q
5
(c) lim x = q , lim x = - q
xSq
xSq
xS-q
(d) lim 1 - 0.5x 2 = - q , lim 1 - 0.5x 2 = q
7
7
xS-q
(b) lim 1 -2x2 2 = - q , lim 1 - 2x2 2 = - q
xSq
[–5, 5] by [–15, 15]
2. (a) lim 1 - 3x 2 = - q , lim 1 -3x 2 = - q
4
xS-q
(c) lim 3x4 = q , lim 3x4 = q
xSq
[–5, 5] by [–15, 15]
4
xS-q
(d) lim 2.5x3 = q , lim 2.5x3 = - q
xSq
[–5, 5] by [–15, 15]
(c) lim 2x6 = q , lim 2x6 = q
xSq
xS-q
xS-q
[–5, 5] by [–15, 15]
4
(b) lim 0.6x = q , lim 0.6x = q
xSq
xS-q
[–5, 5] by [–15, 15]
4
xSq
xS-q
[–5, 5] by [–15, 15]
[–5, 5] by [–15, 15]
xSq
xS-q
xS-q
[–5, 5] by [–15, 15]
If an 7 0 and n>0, lim f1x2 = q and
xSq
lim f1x2 = - q . If an 6 0 and n>0,
xS-q
lim f1x2 = - q and lim f1x2 = q .
xSq
[–5, 5] by [–15, 15]
xS-q
Section 2.3
Polynomial Functions of Higher Degree with Modeling
81
y
Exploration 2
3
2
1. y=0.0061x +0.0177x -0.5007x+0.9769 It is an
exact fit, which we expect with only 4 data points!
200
15
x
[–5, 10] by [–5, 5]
2. y=–0.375x4+6.917x3-44.125x2+116.583x-111
It is an exact fit, exactly what we expect with only 5 data
points!
3. Start with y=x3, shift to the left by 1 unit, vertically
1
shrink by , reflect over the x-axis, and then vertically
2
3
shift up 2 units. y-intercept: a 0, b
2
y
5
[2.5, 8.5] by [–18, 15]
5
x
Quick Review 2.3
1. (x-4)(x+3)
2. (x-7)(x-4)
3. (3x-2)(x-3)
4. Start with y=x3, shift to the right by 3 units, vertically
2
shrink by , and vertically shift up 1 unit. y-intercept:
3
(0, –17)
4. (2x-1)(3x-1)
5. x(3x-2)(x-1)
6. 2x(3x-2)(x-3)
7. x=0, x=1
y
8. x=0, x=–2, x=5
20
9. x=–6, x=–3, x=1.5
10. x=–6, x=–4, x=5
Section 2.3 Exercises
10
x
1. Start with y=x3, shift to the right by 3 units, and then
stretch vertically by 2. y-intercept: (0, –54)
y
10
5. Start with y=x4, shift to the left 2 units, vertically stretch
by 2, reflect over the x-axis, and vertically shift down
3 units. y-intercept: (0, –35)
10
x
y
40
2. Start with y=x3, shift to the left by 5 units, and then
reflect over the x-axis. y-intercept: (0, –125)
5
x
82
Chapter 2
Polynomial, Power, and Rational Functions
6. Start with y=x4, shift to the right 1 unit, vertically
stretch by 3, and vertically shift down 2 units. y-intercept:
(0, 1)
14. One possibility:
y
5
[–50, 50] by [–1000, 1000]
5
x
15. One possibility:
7. local maximum:≠(0.79, 1.19), zeros: x=0 and
x≠1.26. The general shape of f is like y=–x4, but
near the origin, f behaves a lot like its other term, 2x. f is
neither even nor odd.
[–50, 50] by [–1000, 1000]
16. One possibility:
[–5, 5] by [–5, 2]
8. local maximum at (0, 0) and local minima at (1.12, –3.13)
and (–1.12, –3.13), zeros: x=0, x≠1.58, x≠–1.58.
f behaves a lot like y=2x4 except in the interval
[–1.58, 1.58], where it behaves more like its second
building block term, –5x2.
[–100, 100] by [–2000, 2000]
For #17–24, when one end of a polynomial function’s graph
curves up into Quadrant I or II, this indicates a limit at q . And
when an end curves down into Quadrant III or IV, this indicates
a limit at - q .
17.
[–5, 5] by [–5, 15]
[–5, 3] by [–8, 3]
9. Cubic function, positive leading coefficient. The answer
is (c).
10. Cubic function, negative leading coefficient. The answer
is (b).
lim f1x2 = q
xSq
lim f1x2 = - q
xS-q
18.
11. Higher than cubic, positive leading coefficient. The answer
is (a).
12. Higher than cubic, negative leading coefficient. The
answer is (d).
13. One possibility:
[–5, 5] by [–15, 15]
lim f1x2 = - q
xSq
lim f1x2 = q
xS-q
[–100, 100] by [–1000, 1000]
Section 2.3
Polynomial Functions of Higher Degree with Modeling
83
24.
19.
[–8, 10] by [–120, 100]
lim f1x2 = - q
xSq
lim f1x2 = q
xS-q
[–4, 3] by [–20, 90]
lim f1x2 = - q
xSq
lim f1x2 = - q
xS-q
For #25–28, the end behavior of a polynomial is governed by
the highest-degree term.
20.
25. lim f1x2 = q , lim f1x2 = q
xSq
xS-q
26. lim f1x2 = - q , lim f1x2 = q
xSq
xS-q
27. lim f1x2 = - q , lim f1x2 = q
xSq
[–10, 10] by [–100, 130]
lim f1x2 = q
xSq
lim f1x2 = - q
xS-q
xS-q
28. lim f1x2 = - q , lim f1x2 = - q
xSq
xS-q
29. (a); There are 3 zeros: they are –2.5, 1, and 1.1.
30. (b); There are 3 zeros: they are 0.4, approximately
0.429 (actually 3/7), and 3.
31. (c); There are 3 zeros: approximately –0.273 (actually
–3/11), –0.25, and 1.
21.
32. (d); There are 3 zeros: –2, 0.5, and 3.
For #33–35, factor or apply the quadratic formula.
33. –4 and 2
[–5, 5] by [–14, 6]
lim f1x2 = q
xSq
lim f1x2 = q
xS-q
34. –2 and 2/3
35. 2/3 and –1/3
For #36–38, factor out x, then factor or apply the quadratic
formula.
36. 0, –5, and 5
22.
37. 0, –2/3, and 1
38. 0, –1, and 2
39. Degree 3; zeros: x=0 (multiplicity 1, graph crosses
x-axis), x=3 (multiplicity 2, graph is tangent)
y
[–2, 6] by [–100, 25]
10
lim f1x2 = q
xSq
lim f1x2 = q
xS-q
23.
6
[–3, 5] by [–50, 50]
lim f1x2 = q
xSq
lim f1x2 = q
xS-q
x
84
Chapter 2
Polynomial, Power, and Rational Functions
40. Degree 4; zeros: x=0 (multiplicity 3, graph crosses
x-axis), x=2 (multiplicity 1, graph crosses x-axis)
45. Zeros: –2.47, –1.46, 1.94
y
5
4
x
[–3, 3] by [–10, 10]
46. Zeros: –4.53, 2
41. Degree 5; zeros: x=1 (multiplicity 3, graph crosses
x-axis), x=–2 (multiplicity 2, graph is tangent)
y
10
[–5, 3] by [–20, 90]
47. Zeros: –4.90, –0.45, 1, 1.35
–5
5
x
–10
42. Degree 6; zeros: x=3 (multiplicity 2, graph is tangent),
x=–5 (multiplicity 4, graph is tangent)
y
[–6, 2] by [–5, 5]
48. Zeros: –1.98, –0.16, 1.25, 2.77, 3.62
100,000
10
x
[–3, 4] by [–100, 100]
49. 0, –6, and 6. Algebraically — factor out x first.
43. Zeros: –2.43, –0.74, 1.67
50. –11, –1, and 10. Graphically. Cubic equations can be
solved algebraically, but methods of doing so are more
complicated than the quadratic formula.
[–3, 2] by [–10, 10]
44. Zeros: –1.73, 0.26, 4.47
[–15, 15] by [–800, 800]
51. –5, 1, and 11. Graphically.
[–2, 5] by [–8, 22]
[–10, 15] by [–300, 150]
Section 2.3
Polynomial Functions of Higher Degree with Modeling
85
must cross the x-axis at least once. That is to say, f(x) takes
on both positive and negative values, and thus by the
Intermediate Value Theorem, f(x)=0 for some x.
52. –6, 2, and 8. Graphically.
62. f(x)=x9-x+50 has an odd-degree leading term,
which means that in its end behavior it will go toward
–q at one end and toward q at the other. Thus the graph
must cross the x-axis at least once. That is to say, f(x) takes
on both positive and negative values, and thus by the
Intermediate Value Theorem, f(x)=0 for some x.
[–10, 15] by [–500, 500]
For #53–56, the “minimal” polynomials are given; any constant
(or any other polynomial) can be multiplied by the answer
given to give another answer.
63. (a)
53. f(x)=(x-3)(x+4)(x-6)
=x3-5x2-18x+72
54. f(x)=(x+2)(x-3)(x+5)=x3+4x2-11x-30
55. f(x)= 1 x - 13 2 1 x + 132 (x-4)
=(x2-3)(x-4)=x3-4x2-3x+12
56. f(x)=(x-1) 1x - 1 - 122 1x - 1 + 122
2
3
2
=(x-1)[(x-1) -2]=x -3x +x+1
[0, 60] by [–10, 210]
(b) y=0.051x2+0.97x+0.26
(c)
57. y=0.25x3-1.25x2-6.75x+19.75
[0, 60] by [–10, 210]
(d) y(25)≠56.39 ft
[–4, 4] by [–10, 30]
58. y=0.074x3-0.167x2+0.611x+4.48
(e) Using the quadratic equation to solve
0=0.051x2+0.97x+(0.26-300), we find two
answers: x=67.74 mph and x=–86.76 mph.
Clearly the negative value is extraneous.
64. (a) P(x)=R(x)-C(x) is positive if 29.73<x<541.74
(approx.), so they need between 30 and 541 customers.
(b) P(x)=60000 when x=200.49 or x=429.73. Either
201 or 429 customers gives a profit slightly over
$60,000; 200 or 430 customers both yield slightly less
than $60,000.
[–3, 8] by [–2, 30]
59. y=–2.21x4+45.75x3-339.79x2+1075.25x-1231
65. (a)
[0, 0.8] by [0, 1.20]
[0, 10] by [–25, 45]
4
(b) 0.3391 cm from the center of the artery
3
2
60. y=–0.017x +0.226x +0.289x -3.202x-21
66. (a) The height of the box will be x, the width will be
15-2x, and the length 60-2x.
(b) Any value of x between approximately 0.550 and
6.786 inches.
[–1, 14] by [–25, 35]
61. f(x)=x7+x+100 has an odd-degree leading term,
which means that in its end behavior it will go toward
–q at one end and toward q at the other. Thus the graph
[0, 8] by [0, 1500]
86
Chapter 2
Polynomial, Power, and Rational Functions
67. The volume is V(x)=x(10-2x)(25-2x); use any x
with 0<x 0.929 or 3.644 x<5.
[0.9999, 1.0001] by [–1*10–7, 1*10–7]
[0, 5] by [0, 300]
68. Determine where the function is positive: 0<x<21.5.
(The side lengths of the rectangle are 43 and 62 units.)
[0, 25] by [0, 12,000]
69. True. Because f is continuous and
f(1)=(1)3-(1)2-2=–2<0
while f(2)=(2)3-(2)2-2=2>0,
the Intermediate Value Theorem assures us that the graph
of f crosses the x-axis (f(x)=0) somewhere between
x=1 and x=2.
78. This also has a maximum near x=1 — but this time a
window such as [0.6, 1.4] by [–0.1, 0.1] reveals that the
graph actually rises above the x-axis and has a maximum
at (0.999, 0.025).
[0.6, 1.4] by [–0.1, 0.1]
79. A maximum and minimum are not visible in the standard
window, but can be seen on the window
[0.2, 0.4] by [5.29, 5.3].
70. False. If a>0, the graph of f(x)=(x+a)2 is obtained
by translating the graph of f(x)=x2 to the left by a
units. Translation to the right corresponds to a<0.
71. When x=0, f(x)=2(x-1)3+5=2(–1)3+5=3.
The answer is C.
72. In f(x)=(x-2)2(x+2)3(x+3)7, the factor x-2
occurs twice. So x=2 is a zero of multiplicity 2, and the
answer is B.
[0.2, 0.4] by [5.29, 5.30]
80. A maximum and minimum are not visible in the
standard window, but can be seen on the window
[0.95, 1.05] by [–6.0005, –5.9995].
73. The graph indicates three zeros, each of multiplicity 1:
x=–2, x=0, and x=2. The end behavior indicates a
negative leading coefficient. So f(x)=–x(x+2)(x-2),
and the answer is B.
74. The graph indicates four zeros: x=–2 (multiplicity 2),
x=0 (multiplicity 1) and x=2 (multiplicity 2). The end
behavior indicates a positive leading coefficient. So
f(x)=x(x+2)2(x-2), and the answer is A.
75. The first view shows the end behavior of the function, but
obscures the fact that there are two local maxima and a
local minimum (and 4 x-axis intersections) between –3
and 4. These are visible in the second view, but missing is
the minimum near x=7 and the x-axis intersection near
x=9. The second view suggests a degree 4 polynomial
rather than degree 5.
[0.95, 1.05] by [–6.0005, –5.9995]
81. The graph of y=3(x3-x) (shown on the window
[–2, 2] by [–5, 5]) increases, then decreases, then increases;
the graph of y=x3 only increases. Therefore, this graph
cannot be obtained from the graph of y=x3 by the transformations studied in Chapter 1 (translations, reflections,
and stretching/shrinking). Since the right side includes only
these transformations, there can be no solution.
76. The end behavior is visible in the first window, but not the
details of the behavior near x=1. The second view
shows those details, but at the loss of the end behavior.
77. The exact behavior near x=1 is hard to see. A
zoomed–in view around the point (1, 0) suggests that the
graph just touches the x-axis at 0 without actually crossing
it — that is, (1, 0) is a local maximum. One possible window is [0.9999, 1.0001] by [–1 10–7, 1 10–7].
[–2, 2] by [–5, 5]
Section 2.3
Polynomial Functions of Higher Degree with Modeling
87
(in the geometric sense) to HGB, and also ABC is
similar to AFH. Therefore:
HG
BG
8 D - u
FH
AF
, or =
, and also
,
=
=
EC
BC
x
D
AB
BC
y - 8
y - 8
8
D - u
or
. Then =
.
=
y
D
x
y
82. The graph of y=x4 has a “flat bottom,” while the graph
of y=x4+3x3-2x-3 (shown on [–4, 2] by [–8, 5])
is “bumpy.” Therefore, this graph cannot be obtained from
the graph of y=x4 through the transformations of
Chapter 1. Since the right side includes only these transformations, there can be no solution.
A
E
y
F
D– u
H
x
[–4, 2] by [–8, 5]
83. (a) Substituting x=2, y=7, we find that
7=5(2-2)+7, so Q is on line L, and also
f(2)=–8+8+18-11=7, so Q is on the graph
of f(x).
(b) Window [1.8, 2.2] by [6, 8]. Calculator output will not
show the detail seen here.
[1.8, 2.2] by [6, 8]
(c) The line L also crosses the graph of f(x) at
(–2, –13).
[–5, 5] by [–25, 25]
y2 - y1
84. (a) Note that f(a)=an and f(–a)=–an; m=
x2 - x1
n
n
n
-a - a
-2a
=
=
=an-1.
-a - a
-2a
1/(n-1) n
n/(n-1)
) =a
(b) First observe that f(x0)=(a
Using point–slope form:
y-an/(n-1)=an-1(x-a1/(n-1)).
.
8
D– u
B
G
C
D
8
8
= 1 - . Multiply both sides by
x
y
xy: 8y = xy - 8x. Subtract xy from both sides and
factor: y 18 - x2 = -8. Divide both sides by 8 - x:
-8
y =
. Factor out -1 from numerator and
8 - x
8
denominator: y =
.
x - 8
(c) Applying the Pythagorean Theorem to EBC and
ABC, we have x2+D2=202 and y2+D2=302,
which combine to give D2=400-x2=900-y2,
or y2-x2=500. Substituting y=8x/(x-8),
2
8x
b -x2=500, so that
we get a
x - 8
64x2
-x2=500, or 64x2-x2(x-8)2
1x - 8 2 2
=500(x-8)2. Expanding this gives 500x2 - 8000x +
32,000 = 64x2 - x4 + 16x3 - 64x2. This is equivalent to x4 - 16x3 + 500x2 - 8000x + 32,000 = 0.
(b) Equation (a) says
(d) The two solutions are x≠5.9446 and x≠11.7118.
Based on the figure, x must be between 8 and 20 for
this problem, so x≠11.7118. Then D= 2202 - x2
≠16.2121 ft.
86. (a) Regardless of the value of b, f1 - b 2 = 1 - b,
lim f1x2 = +q, lim f1x2 = - q, and the graph of
f has a y-intercept of 1. If 0 b 0 13, the graph of f is
strictly increasing. If 0 b 0 7 13, f has one local maximum and one local minimum. If 0 b 0 is large, the graph
of f appears to have a double root at 0 and a single
root at -b, because f1 x2 = x3 + bx2 = x2 1x + b2
for large x.
xSq
(c) With n=3 and a=3, this equation becomes
y-33/2=32(x-31/2), or y=9 1 x - 132 +3 13
=9x-613. So y=x3.
xS-q
(b) Answers will vary.
(c)
[–5, 5] by [–30, 30]
85. (a) Label the points of the
_____diagram as shown, adding the
horizontal segment FH. Therefore, ECB is similar
[–18.8, 18.8] by [–1000, 1000]
88
Chapter 2
Polynomial, Power, and Rational Functions
■ Section 2.4 Real Zeros of Polynomial
Functions
4.
7
2x2-5x +2
2x+1R 4x3-8x2+2x-1
4x3+2x2
Quick Review 2.4
–10x2+2x
1. x2 - 4x + 7
–10x2-5x
5
2. x - x - 3
2
3. 7x3 + x2 - 3
2
7x-1
7
7x+2
9
-2
2
7
x +
3
3
5. x1 x2 - 4 2 = x1x2 - 22 2 = x1 x + 22 1 x - 2 2
4. 2x2 -
f1 x2 = a 2x2 - 5x +
7
9
b 12x + 12 - ;
2
2
9>2
f1 x2
7
2
= 2x - 5x + 2x + 1
2
2x + 1
6. 6 1x2 - 9 2 = 61x2 - 32 2 = 61 x + 3 2 1 x - 3 2
7. 4 1x2 + 2x - 152 = 4 1x + 5 2 1 x - 3 2
8. x1 15x2 - 22x + 82 = x13x - 2 2 15x - 4 2
9. 1 x3 + 2x2 2 - 1x + 2 2 = x2 1x + 22 - 1 1 x + 2 2
= 1x + 2 2 1x2 - 1 2 = 1 x + 2 2 1 x + 12 1 x - 1 2
5.
x2-4x+ 12
x2+2x-1R x4-2x3+3x2-4x+6
10. x1 x3 + x2 - 9x - 9 2 = x3 1x3 + x2 2 - 19x + 92 4
= x1 3 x2 1 x + 1 2 - 9 1x + 1 2 4
= x1x + 1 2 1x2 - 9 2 = x1x + 1 2 1x2 - 32 2
= x1x + 1 2 1x + 3 2 1x - 3 2
x4+2x3- x2
–4x3+4x2-4x
–4x3-8x2+4x
12x2- 8x+6
12x2+24x-12
Section 2.4 Exercises
1.
–32x+18
x- 1
x-1R x -2x+3
f1 x2 = 1 x2 - 4x + 122 1x2 + 2x - 12 - 32x + 18;
f1x2
- 32x + 18
= x2 - 4x + 12 + 2
2
x + 2x - 1
x + 2x - 1
2
2
x- x
–x+3
–x+1
6.
2
f(x)=(x-1)2+2;
2.
f1 x2
x - 1
= x - 1 +
x2-3x+ 5
x2+1R x4-3x3+6x2-3x+5
2
x - 1
x4
+ x2
–3x3+5x2-3x
x2- x+ 1
x+1R x3+0x2+0x-1
–3x3
x3+ x2
2
–x +0x
-3x
2
5x
+5
5x2
+5
0
–x2- x
f1 x2 = 1 x2 - 3x + 52 1x2 + 12 ;
x-1
x+1
–2
f(x)=(x2-x+1)(x+1)-2;
f1 x2
2
= x2 - x + 1 x + 1
x + 1
3.
7.
–1|
1
1
x+3R x +4x +7x- 9
4
2
x3+3x2
8.
3
–5
3
–2
–1
6
–9
–6
9
–11
2
2x - 5x + 7x - 3x + 1
x - 3
= 2x3 + x2 + 10x + 27 +
x2+7x
x2+3x
4x- 9
4x+12
-21
f1 x2 = 1 x2 + x + 42 1 x + 3 2 - 21;
f1 x2
21
= x2 + x + 4 x + 3
x + 3
x2 + 1
= x2 - 3x + 5
- 11
x3 - 5x2 + 3x - 2
= x2 - 6x + 9 +
x + 1
x + 1
x2+ x+ 4
3
f1x2
3|
2
2
–5
82
x - 3
7
–3
1
6
3
30
81
1
10
27
82
Section 2.4
9.
9670
9x3 + 7x2 - 3x
= 9x2 + 97x + 967 +
x - 10
x - 10
10|
9
4
10.
7
–3
90
970
9670
97
967
9670
3
2
–5|
3
3
11.
–2|
3|
3x + x - 4x + 9x - 3
x + 5
1602
x + 5
1
–4
9
–3
–15
70
–330
1605
–14
66
–321
1602
= -5x3 - 20x2 - 80x - 317 +
–5
–5
0
3
–20
–80
–320 –1268
–20
–80
–317 –1269
1
0
–22
21
0
15
–21
–7
0
1
3
b ax - b
2
2
1
= 1x - 2 2 12x - 12 12x - 32
2
19
3
2
=2x -8x + 2 x-3
5
b
2
=x(x+3)(x+1)(2x-5)
30. 2(x+3)(x+1)(x) a x -
–1
0
–42
29. 2(x-2) a x -
=2x4+3x3-14x2-15x
31. Since f(–4)=f(3)=f(5)=0, it must be that (x+4),
(x-3), and (x-5) are factors of f. So
f(x)=k(x+4)(x-3)(x-5) for some constant k.
= x7 - 2x6 + 4x5 - 8x4 + 16x3 - 32x2 + 64x - 128
255
+
x + 2
1
42
44
28. 2(x+1)(x-3)(x+5)=2x3+6x2-26x-30
-1269
4 - x
0
–23
27. 2(x+2)(x-1)(x-4)=2x3-6x2-12x+16
x8 - 1
12.
x + 2
–2|
5
–12
–10
f(x)=(x+2)(x-3)(5x-7)
5x4 - 3x + 1
4 - x
4|
5
5
= 3x3 - 14x2 + 66x - 321 +
0
0
0
0
–1
–2
4
–8
16 –32
64 –128
0
256
–2
4
–8
16-32
64 –128
255
13. The remainder is f(2)=3.
Since f(0)=180, we must have k=3. So
f(x)=3(x+4)(x-3)(x-5)
32. Since f(–2)=f(1)=f(5)=0, it must be that (x+2),
(x-1), and (x-5) are factors of f. So
f(x)=k(x+2)(x-1)(x-5) for some constant k.
Since f(–1)=24, we must have k=2, so
f(x)=2(x+2)(x-1)(x-5)
33. Possible rational zeros:
14. The remainder is f(1)=–4.
15. The remainder is f(–3)=–43.
16. The remainder is f(–2)=2.
17. The remainder is f(2)=5.
19. Yes: 1 is a zero of the second polynomial.
1
— ; 1 is a zero.
6
;1, ;2, ;7, ; 14
, or —1, —2, —7,
;1, ;3
1 2 7 14 7
—14, ; , ; , ; , ; ; is a zero.
3 3 3
3 3
20. Yes: 3 is a zero of the second polynomial.
21. No: when x=2, the second polynomial evaluates to 10.
22. Yes: 2 is a zero of the second polynomial.
23. Yes: -2 is a zero of the second polynomial.
24. No: when x = -1, the second polynomial evaluates to 2.
25. From the graph it appears that (x+3) and (x-1) are
factors.
1|
5
5
5
;1
1 1
, or —1, — , — ,
;1, ;2, ;3, ; 6
2 3
34. Possible rational zeros:
18. The remainder is f(–1)=23.
–3|
89
26. From the graph it appears that (x+2) and (x-3) are
factors.
0
9
Real Zeros of Polynomial Functions
–7
–49
51
–15
66
–51
–22
–17
0
5
–17
–17
0
f(x)=(x+3)(x-1)(5x-17)
35. Possible rational zeros:
;1, ;3, ;9
1
, or —1, —3, —9, — ,
;1, ;2
2
3
9 3
; , ; ; is a zero.
2
2 2
;1, ;2, ;3, ; 4, ;6, ;12
, or —1,
;1, ;2, ;3, ;6
1
3
1
2 4 1
4
—2, —3, —4, —6, —12, ; , ; , ; , ; , ; , ; ; - and
2
2
3
3 3 6
3
3
are zeros.
2
36. Possible rational zeros:
37. 3|
2
2
–4
1
–2
6
6
21
2
7
19
Since all numbers in the last line are 0, 3 is an upper
bound for the zeros of f.
Chapter 2
90
38. 5|
2
2
Polynomial, Power, and Rational Functions
–5
–5
–1
10
25
100
5
20
99
Since all values in the last line are 0, 5 is an upper
bound for the zeros of f(x).
39. 2|
1
1
–1
1
1
4
4
2
2
6
14
1
3
7
2
–6
–7
9
2
12
18
33
126
6
11
42
128
Since all values in the last line are 0, 3 is an upper
bound for the zeros of f(x).
41. –1|
3
3
–4
1
3
–3
7
–8
–7
8
–5
Since the values in the last line alternate signs, -1 is a
lower bound for the zeros of f(x).
42. –3|
1
2
–3
2
3
–15
1
–4
7
0
0
0
1
–4
7
–2
3
–2
3
–1
–5
–3
–12
52
–188
–13
47
–191
5|
–11
–7
8
–34
–30
205
–990
4910
6
–41
198
–982
4876
6
–11
–7
8
–34
6
5|
1
–84
1
–4
–129
–5|
5|
2
–8
3
420 –4080
20,440
816 –4088 –20,443
396
–8
3
5
5
–620 –1120
–5640
1
–124
–224 –1128
–5637
–5
–141
–10
75
330 –2730 14,105
2
–15
–66
546 –2821 14,130
2
–5
–141
10
25
–580 –1820 –9555
216
5
–116
–364 –1911 –9530
216
–91
–91
25
25
For #49–56, determine the rational zeros using a grapher (and
the Rational Zeros Test as necessary). Use synthetic division
to reduce the function to a quadratic polynomial, which can
be solved with the quadratic formula (or otherwise). The first
two are done in detail; for the rest, we show only the synthetic
division step(s).
;1, ;2, ;3, ; 6
, or
;1, ;2
1
3
3
;1, ;2, ;3, ;6, ; , ; . The only rational zero is .
2
2
2
Synthetic division (below) leaves 2x2-4, so the irra-
30
95
440
2240
19
88
448
2206
19
3/2 |
2
2
–3
–4
6
3
0
–6
0
–4
0
50. Possible rational zeros: ;1, ;3, ;9. The only rational
zero is -3. Synthetic division (below) leaves x2 - 3, so
the irrational zeros are ; 23.
1
1
3
–3
–9
–3
0
9
0
–3
0
51. Rational: -3; irrational: 1 ; 23
–1
0
21
–5
30
–150
645 –3320
1
–6
30
–129
664 –3323
1
–1
0
21
19
–3
5
20
100
605
3120
4
20
121
624 –3117
1
45
–9
396
48. Synthetic division shows that the lower/upper bounds
tests were not met. There are zeros not shown (approx.
–8.036 and 9.038), because –5 and 5 are not bounds for
zeros of f(x).
–3|
46. By the Upper and Lower Bound Tests, –5 is a lower
bound and 5 is an upper bound. No zeros outside window.
–5|
–5
1
1
–129
tional zeros are ; 22.
45. By the Upper and Lower Bound Tests, –5 is a lower
bound and 5 is an upper bound. No zeros outside window.
6
–4
49. Possible rational zeros:
Since the values in the last line alternate signs, –4 is a
lower bound for the zeros of f(x).
–5|
1
2
Since the values in the last line alternate signs, 0 is a lower
bound for the zeros of f(x).
44. –4|
5|
5
1
–1
5
–10
Since the values in the last line alternate signs, -3 is a
lower bound for the zeros of f(x).
43. 0|
–5|
–12
Since all values in the last line are 0, 2 is an upper
bound for the zeros of f(x).
40. 3|
47. Synthetic division shows that the Upper and Lower
Bound Tests were not met. There are zeros not shown
(approx. –11.002 and 12.003), because –5 and 5 are not
bounds for zeros of f(x).
–3
–3|
1
1
–8
–6
–3
6
6
1
–2
–2
0
52. Rational: 4; irrational: 1 ; 22
4|
1
–6
7
4
4
–8
–4
1
–2
–1
0
Section 2.4
53. Rational: –1 and 4; irrational: ; 22
–1|
4|
1
–3
–6
6
8
–1
4
2
–8
1
–4
–2
8
0
1
–4
–2
8
4
0
–8
1
0
–2
2|
1
–7
5
10
–1
2
5
–10
–5
10
1
–2
–5
10
2
0
–10
0
–5
0
2
–2
–7
–4
4
8
4
2
1
–1/2 | 2
1
2
1
–1
0
–1
0
2
0
56. Rational:
2/3 |
3
3
–5
15
–20
165
1
–3
4
–33
203
2
–11
–13
38
4
24
52
156
6
13
39
194
1
1
–13
38
A graph shows that 2 is most promising, so we verify
with synthetic division:
2|
1
0
1
2
–11
(b) Potential rational zeros:
Factors of 38 ;1, ;2, ;19, ;38
:
Factors of 1
;1
8
1
2
The Upper and Lower Bound Tests are met, so all real
zeros of f lie on the interval [–5, 4].
–7
2
4|
0
1
55. Rational: - and 4; irrational: none
2
4|
1
Upper bound:
–1
–2
1
61. (a) Lower bound:
0
1
91
59. Using the Remainder Theorem, the remainder is
1 -12 40 - 3 = -2.
60. Using the Remainder Theorem, the remainder is
163 - 17 = - 16.
–5|
54. Rational: –1 and 2; irrational: ; 25
–1|
Real Zeros of Polynomial Functions
2
–11
–13
38
2
8
–6
–38
4
–3
–19
0
Use the Remainder Theorem:
f1 -2 2 = 20 Z 0
2
; irrational: about –0.6823
3
–2
3
1
–2
2
0
2
2
0
3
3
0
f1 -1 2 = 39 Z 0
f11 2 = 17 Z 0
f(–38)=1,960,040
f(38)=2,178,540
f(–19)=112,917
f(19)=139,859
Since all possible rational roots besides 2 yield nonzero function values, there are no other rational roots.
57. The supply and demand graphs are shown on the window
[0, 50] by [0, 100]. They intersect when p=$36.27, at
which point the supply and demand equal 53.7.
[–5, 4] by [–1, 49]
(c) f1 x2 = 1 x - 2 2 1 x3 + 4x2 - 3x - 192
(d) From our graph, we find that one irrational zero of x
is x L 2.04.
(e) f1 x2 L 1 x - 2 2 1 x - 2.042 1x2 + 6.04x + 9.32162
[0, 50] by [0, 100]
62. (a) D L 0.0669t3 - 0.7420t2 + 2.1759t + 0.8250
58. The supply and demand graphs, shown on the window
[0, 150] by [0, 1600], intersect when p=$106.99. There
S(p)=D(p)=1010.15.
[1, 8.25] by [0, 5]
(b) When t=0, D L 0.8250 m.
[0, 150] by [0, 1600]
(c) The graph changes direction at t L 2.02 and at
t L 5.38. Lewis is approximately 2.74 m from the
motion detector at t = 2.02 and 1.47 m from the
motion detector at t = 5.38.
92
Chapter 2
Polynomial, Power, and Rational Functions
4pd
pd
=
13r2 + x2 2 x
15
6
24
= 13r2 + x2 2x
15
63. False. x-a is a factor if and only if f(a)=0. So (x+2)
is a factor if and only if f(–2)=0.
64. True. By the Remainder Theorem, the remainder when
f(x) is divided by x-1 is f(1), which equals 3.
0 = 16x - 3x2 + x2 2 x -
65. The statement f(3)=0 means that x=3 is a zero of f(x)
and that 3 is an x-intercept of the graph of f(x). And it follows that x-3 is a factor of f(x) and thus that the
remainder when f(x) is divided by x-3 is zero. So the
answer is A.
66. By the Rational Zeros Theorem, every rational root of
1
3
f(x) must be among the numbers —1, —3, ; , ; .
2
2
The answer is E.
2
67. f(x)=(x+2)(x +x-1)-3 yields a remainder
of –3 when divided by either x+2 or x2+x-1, from
which it follows that x+2 is not a factor of f(x) and that
f(x) is not evenly divisible by x+2. The answer is B.
0 = -2x3 + 6x2 -
8
5
8
5
4
5
Solving graphically, we find that x L 0.57 m, the depth
the buoy would sink.
0 = x3 - 3x2 +
71. (a) Shown is one possible view, on the window
[0, 600] by [0, 500].
68. Answers A through D can be verified to be true. And
because f(x) is a polynomial function of odd degree, its
graph must cross the x-axis somewhere. The answer is E.
[0, 600] by [0, 500]
4 3
pr . In this case, the
3
4
radius of the buoy is 1, so the buoy’s volume is p.
3
69. (a) The volume of a sphere is V =
(b) Total weight = volume
#
(b) The maximum population, after 300 days, is 460 turkeys.
(c) P=0 when t L 523.22 — about 523 days after
release.
weight
(d) Answers will vary. One possibility: After the population increases to a certain point, they begin to compete for food and eventually die of starvation.
unit volume
density. In this case, the density of the
=volume #
1
buoy is d, so, the weight Wb of the buoy is
4
4p 1
# d = dp .
Wb =
3
4
3
(c) The weight of the displaced water is
WH2O = volume # density. We know from geometry
that the volume of a spherical cap is
p
1 3r2 + h2 2 h, so,
V =
6
pd
p
WH2O =
13r2 + x2 2 x # d =
x13r2 + x2 2
6
6
72. (a) d is the independent variable.
(b) A good choice is [0, 172] by [0, 5].
(c) s = 1.25 when d L 95.777 ft. (found graphically)
73. (a) 2 sign changes in f(x), 1 sign change in
f(–x)=–x3+x2+x+1; 0 or 2 positive zeros, 1
negative zero.
(b) No positive zeros, 1 or 3 negative zeros.
(c) 1 positive zero, no negative zeros.
(d) 1 positive zero, 1 negative zero.
74.
(d) Setting the two weights equal, we have:
Wb = WH2O
pd
pd
=
1 3r2 + x2 2x
3
6
2 = 13r2 + x2 2x
0 = 16x - 3x2 + x2 2x - 2
0 = -2x3 + 6x2 - 2
0 = x3 - 3x2 + 1
Solving graphically, we find that x L 0.6527 m,
the depth that the buoy will sink.
70. The weight of the buoy, Wb, with density
4p 1
# d = 4pd . So,
3
5
15
Wb = WH2O
Wb =
1
d, is
5
[–5, 5] by [–1, 9]
The functions are not exactly the same, when x 3, we
1x - 32 12x2 + 3x + 42
have f1x2 =
1x - 32
= 2x2 + 3x + 4 = g1x2
The domain of f is 1 - q, 32 ´ 13, q 2 while the domain of
g is 1 - q, q 2 . f is discontinuous at x=3. g is continuous.
Section 2.5
75.
Complex Zeros and the Fundamental Theorem of Algebra
78. Let f1 x2 = x2 - 2 = 1x + 122 1x - 122 . Notice that
12 is a zero of f. By the Rational Zeros Theorem, the
only possible rational zeros of f are ;1 and ; 2. Because
12 is none of these, it must be irrational.
4x3 - 5x2 + 3x + 1
2x - 1
5 2
3
1
x + x +
2
2
2
1
x 2
2x3 =
zero of divisor
1
2
|
2 -
line for products
line for sums
2 -
Divide numerator and
denominator by 2.
5
2
3
2
1 Write coefficients of
2 dividend.
1 -
3
4
3
8
3
2
3
4
7
Quotient, remainder
8
Copy 2 into the first quotient position. Multiply 2 #
1
= 1
2
3 1
5
3
and add this to - . Multiply - # = - and add this to
2
2 2
4
3
3 1
3
1
. Multiply # = and add this to . The last line tells
2
4 2
8
2
us a x -
93
1
3
3
7
b a 2x2 - x + b +
2
2
4
8
1
5
3
=2x3- x2+ x+ .
4
2
2
76. Graph both functions in the same viewing window to see
if they differ significantly. If the graphs lie on top of each
other, then they are approximately equal in that viewng
window.
77. (a) g(x)=3f(x), so the zeros of f and the zeros of g are
identical. If the coefficients of a polynomial are
rational, we may multiply that polynomial by the least
common multiple (LCM) of the denominators of the
coefficients to obtain a polynomial, with integer
coefficients, that has the same zeros as the original.
(b) The zeros of f(x) are the same as the zeros of
6f1 x2 = 6x3 - 7x2 - 40x + 21. Possible rational
; 1, ; 3, ;7, ;21
zeros:
, or
; 1, ;2, ;3, ; 6
1 3
; 1, ;3, ; 7, ; 21, ; , ; ,
2
2
7 21 1
7 1 7
; , ; , ; , ; , ; , ; . The actual zeros are
2
2
3
3
6 6
–7/3, 1/2, and 3.
(c) The zeros of f(x) are the same as the zeros of
12f1 x2 = 12x3 - 30x2 - 37x + 30.
Possible rational zeros:
; 1, ;2, ; 3, ; 5, ;6, ;10, ;15, ;30
, or
;1, ;2, ; 3, ;4, ;6, ;12
5
1 3
; 1, ;2, ; 3, ; 5, ;6, ; 10, ; 15, ;30, ; , ; , ; ,
2 2
2
15 1 2
5
10 1
3 5 15 1
; ,; ,; ,; ,; ,; ,; ,; ,; ,; ,
2
3 3
3
3
4
4 4
4
6
1
5
5
; ,; ,; .
6 12 12
There are no rational zeros.
79. (a) Approximate zeros: –3.126, –1.075, 0.910, 2.291
(b) f1 x2 L g(x)
= 1x + 3.126 2 1x + 1.0752 1x - 0.910 2 1x - 2.291 2
(c) Graphically: Graph the original function and the
approximate factorization on a variety of windows
and observe their similarity. Numerically: Compute
f1 c2 and g1c2 for several values of c.
■ Section 2.5 Complex Zeros and the
Fundamental Theorem of Algebra
Exploration 1
1. f1 2i2 = 1 2i2 2 - i12i2 + 2 = -4 + 2 + 2 = 0;
f1 -i2 = 1 -i2 2 - i1 -i2 + 2 = -1 - 1 + 2 = 0; no.
2. g1 i2 = i2 - i + 1 1 + i2 = - 1 - i + 1 + i = 0;
g1 1 - i2 = 11 - i2 2 - 11 - i2 + 11 + i2 =
- 2i + 2i = 0; no.
3. The Complex Conjugate Zeros Theorem does not necessarily hold true for a polynomial function with complex
coefficients.
Quick Review 2.5
1. 13 - 2i2 + 1 -2 + 5i2 = 13 - 2 2 + 1 - 2 + 52i
= 1 + 3i
2. 1 5 - 7i2 - 1 3 - 2i2 = 15 - 32 + 1 -7 - 1 - 22 2 i
= 2 - 5i
3. 11 + 2i2 13 - 2i2 = 113 - 2i2 + 2i13 - 2i2
= 3 - 2i + 6i - 4i2
= 7 + 4i
4.
2 + 3i
2 + 3i # 1 + 5i
=
1 - 5i
1 - 5i 1 + 5i
2 + 10i + 3i + 15i2
12 + 52
- 13 + 13i
=
26
1
1
= - + i
2
2
=
5. (2x-3)(x+1)
6. (3x+1)(2x-5)
5 ; 125 - 4112 1 112
5 ; 1–19
=
7. x=
2
2
5
119
i
= ;
2
2
–3 ; 19 - 4 12 2 172
–3 ; 1–47
=
8. x=
4
4
3
147
i
=- ;
4
4
9.
10.
; 1, ;2
1
2
, or ; 1, ; 2, ; , ;
; 1, ; 3
3
3
; 1, ; 3
1
3
1
3
, or ; 1, ; 3, ; , ; , ; , ;
; 1, ; 2, ; 4
2
2
4
4
94
Chapter 2
Polynomial, Power, and Rational Functions
Section 2.5 Exercises
2
2
2
1. (x-3i)(x+3i)=x -(3i) =x +9. The factored
form shows the zeros to be x= ; 3i. The absence of real
zeros means that the graph has no x-intercepts.
2. (x+2)(x- 1 3 i)(x+ 1 3 i)=(x+2)(x2+3)
=x3+2x2+3x+6. The factored form shows the
zeros to be x=–2 and x= ; 1 3i. The real zero
x=–2 is the x-intercept of the graph.
3. (x-1)(x-1)(x+2i)(x-2i)
=(x2-2x+1)(x2+4)
=x4-2x3+5x2-8x+4. The factored form shows
the zeros to be x=1 (multiplicity 2) and x = ;2i.
The real zero x=1 is the x-intercept of the graph.
4. x(x-1)(x-1-i)(x-1+i)
=(x2-x)[x-(1+i)][x-(1-i)]
=(x2-x)[x2-(1-i+1+i) x+(1+1)]
=(x2-x)(x2-2x+2)=x4-3x3+4x2-2x.
The factored form shows the zeros to be x=0, x=1,
and x=1_i. The real zeros x=0 and x=1 are the
x-intercepts of the graph.
In #21–26, the number of complex zeros is the same as the
degree of the polynomial; the number of real zeros can be
determined from a graph. The latter always differs from the
former by an even number (when the coefficients of the
polynomial are real).
21. 2 complex zeros; none real.
22. 3 complex zeros; all 3 real.
23. 3 complex zeros; 1 real.
24. 4 complex zeros; 2 real.
25. 4 complex zeros; 2 real.
26. 5 complex zeros; 1 real.
In #27–32, look for real zeros using a graph (and perhaps the
Rational Zeros Test). Use synthetic division to factor the
polynomial into one or more linear factors and a quadratic
factor. Then use the quadratic formula to find complex zeros.
27. Inspection of the graph reveals that x=1 is the only real
zero. Dividing f(x) by x-1 leaves x2+x+5 (below).
The quadratic formula gives the remaining zeros of f(x).
1|
In #5–16, any constant multiple of the given polynomial is also
an answer.
5. (x-i)(x+i)=x2+1
4
0
1
1
5
1
1
5
0
6. (x-1+2i)(x-1-2i)=x2-2x+5
Zeros: x=1, x=–
7. (x-1)(x-3i)(x+3i)=(x-1)(x2+9)
=x3-x2+9x-9
f(x)
8. (x+4)(x-1+i)(x-1-i)
=(x+4)(x2-2x+2)=x3+2x2-6x+8
9. (x-2)(x-3)(x-i)(x+i)
=(x-2)(x-3)(x2+1)
=x4-5x3+7x2-5x+6
10. (x+1)(x-2)(x-1+i)(x-1-i)
=(x+1)(x-2)(x2-2x+2)
=x4-3x3+2x2+2x-4
13. (x-1)2(x+2)3=x5+4x4+x3-10x2-4x+8
14. (x+1)3(x-3)=x4-6x2-8x-3
2
15. (x-2) (x-3-i)(x-3+i)
=(x-2)2(x2-6x+10)
=(x2-4x+4)(x2-6x+10)
=x4-10x3+38x2-64x+40
1
119
;
i
2
2
1
119
119
1
ib d cx - a– +
ib d
2
2
2
2
= 1x - 12 c x - a -
1
= 1x - 1 2 12x + 1 + 119 i2 12x + 1 - 119 i2
4
7
143
;
i (applying
2
2
the quadratic formula to x2-7x+23).
28. Zeros: x=3 (graphically) and x=
3|
1 –10
4 –69
3 –21
11. (x-5)(x-3-2i)(x-3+2i)
=(x-5)(x2-6x+13)=x3-11x2+43x-65
12. (x+2)(x-1-2i)(x-1+2i)
=(x+2)(x2-2x+5)=x3+x+10
–5
1
1
–7
69
23
0
f(x)
= 1x - 32 c x - a
7
143
143
7
ib d cx - a +
ib d
2
2
2
2
1
= 1x - 3 2 12x - 7 + 143 i2 12x - 7 - 143 i2
4
123
1
;
i
2
2
(applying the quadratic formula to x2+x+6).
29. Zeros: x=_1(graphically) and x= -
16. (x+1)2(x+2+i)(x+2-i)
=(x+1)2(x2+4x+5)
=(x2+2x+1)(x2+4x+5)
=x4+6x3+14x2+14x+5
1|
In #17–20, note that the graph crosses the x-axis at oddmultiplicity zeros, and “kisses” (touches but does not cross)
the x-axis where the multiplicity is even.
–1|
17. (b)
f(x)=(x-1)(x+1) c x - a -
18. (c)
19. (d)
20. (a)
–6
1
1
2
7
6
1
2
7
6
0
1
2
7
6
–1
–1
–6
1
6
0
1
cx - a -
5
–1
1
123
1
ib d
2
2
123
1
+
ib d
2
2
1
= 1x - 1 2 1x + 12 12x + 1 + 123 i2 12x + 1 - 123 i2
4
Section 2.5
Complex Zeros and the Fundamental Theorem of Algebra
1
30. Zeros: x=–2 and x= (graphically) and
3
13
1
x=– ;
i (applying the quadratic formula to
2
2
2
3x +3x+3=3(x2+x+1)).
–2|
1/3|
3
8
6
3
–4
–4
2
3
2
2
–1
0
3
2
2
–1
3
1
1
1
3
3
0
1-i|
1
13
ib d
2
2
1
13
+
ib d
2
2
12x + 1 - 13 i2
3
7
31. Zeros: x= - and x= (graphically) and x=1 ; 2i
3
2
(applying the quadratic formula to 6x2-12x+30
=6(x2-2x+5)).
3/2|
–7
–1
67 –105
–14
49 –112
105
6 –21
48 –45
0
6 –21
48 –45
9 –18
6 –12
30
45
0
f(x)=(3x+7)(2x-3)[x-(1-2i)]
3
3
32. Zeros: x= - and x=5 (graphically) and x= ; i
5
2
=(5x+3)(x-5)(2x-3+2i)(2x-3-2i)
(applying the quadratic formula to 20x2-60x+65
=5(4x2-12x+13)).
269 –106 –195
100 –240
145
195
0
20 –48
29
39
–3/5| 20 –48
29
39
–12
20 –60
1
1
–1+i
36 –39
65
–1
6
–6
1+i
–2
–3-3i
6
–1+i
–3
3-3i
0
–3
3-3i
1-i
0
–3+3i
0
–3
0
34. First divide f(x) by x-4i. Then divide the result,
x3+4ix2-3x-12i, by x+4i. This leaves the
polynomial x2-3. Zeros: x= ;13, x = ; 4i
4i|
–4i|
1
0
13
0
–48
4i
–16
–12i
48
1
4i
–3
–12i
0
1
4i
–3
–12i
–4i
0
12i
0
–3
0
1
f(x)= 1 x - 13 2 1x + 132 1x - 4i2 1x + 4i2
35. First divide f(x) by x-(3-2i).Then divide the result,
x3+(–3-2i)x2-2x+6+4i, by x-(3+2i). This
leaves x2-2. Zeros: x= ; 12, x = 3 ; 2i
3 - 2i 1
-6
11
12 - 26
3 - 2i -13 -6 + 4i
26
1 -3 - 2i
-2
6 + 4i
0
3 + 2i
1
1
=(3x+7)(2x-3)(x-1+2i)(x-1-2i)
20 –148
–2
= 1x - 132 1x + 13 2 1x - 1 + i2 1x - 1 - i2
[x-(1+2i)]
5|
1
f(x)= 1x - 132 1x + 132 3x - 1 1 - i2 4 3x - 1 1 + i2 4
1
= 1x + 22 13x - 1 2 12x + 1 + 13 i2
4
–7/3| 6
1+i|
1
f(x)= 1 x + 2 2 13x - 1 2 c x - a –
cx - a–
33. First divide f(x) by x-(1+i) (synthetically). Then
divide the result, x3+(–1+i)x2-3x+(3-3i), by
x-(1-i). This leaves the polynominal x2-3.
Zeros: x = ;13, x = 1 ; i
–2
–6
95
0
f(x)=(5x+3)(x-5)[2x-(3-2i)]
[2x-(3+2i)]
=(5x+3)(x-5)(2x-3+2i)(2x-3-2i)
In #33–36, since the polynomials’ coefficients are real, for the
given zero z=a+bi, the complex conjugate z=a-bi
must also be a zero. Divide f(x) by x-z and x-z to
reduce to a quadratic.
-3 - 2i
3 + 2i
0
-2
0
-2
6 + 4i
-6 - 4i
0
f(x)= 1 x - 12 2 1x + 122 [x-(3-2i)]
[x-(3+2i)]
= 1x - 122 1x + 12 2 (x-3+2i)(x-3-2i)
36. First divide f(x) by x-(1+3i). Then divide the result,
x3+(–1+3i)x2-5x+5-15i, by x-(1-3i).
This leaves x2-5. Zeros: x= ; 15, x = 1 ; 3i
1 + 3i
1
1
1 - 3i
1
1
-2
1 + 3i
- 1 + 3i
5
-10
-5
-1 + 3i
1 - 3i
0
-5
0
-5
10
-5 - 15i
5 - 15i
5 - 15i
-5 + 15i
0
-50
50
0
f(x)= 1 x - 15 2 1x + 152 [x-(1-3i)]
[x-(1+3i)]
= 1x - 15 2 1x + 15 2 (x-1+3i)(x-1-3i)
96
Chapter 2
Polynomial, Power, and Rational Functions
For #37–42, find real zeros graphically, then use synthetic division to find the quadratic factors. Only the synthetic divison
step is shown.
37. f(x)=(x-2)(x2+x+1)
2 1 -1 -1 - 2
2
2
2
1
1
1
0
49. f(x) must have the form
a(x-3)(x+1)(x-2+i) (x-2-i); since
f(0)=a(–3)(1)(–2+i)(–2-i) =–15a=30, we
know that a=–2. Multiplied out, this gives
f(x)=–2x4+12x3-20x2-4x+30.
50. f(x) must have the form
a(x-1-2i)(x-1+2i) (x-1-i)(x-1+i);
since f(0)=a(–1-2i)(–1+2i)(–1-i)(–1+i)
=a(5)(2) =10a=20, we know that a=2. Multiplied
out, this gives f(x)=2x4-8x3+22x2-28x+20.
38. f(x)=(x-2)(x2+x+3)
2 1 -1 1 - 6
2 2
6
1
1 3
0
51. (a) The model is D≠–0.0820t3+0.9162t2-2.5126t
+3.3779.
39. f(x)=(x-1)(2x2+x+3)
1 2 - 1 2 -3
2 1
3
2
1 3
0
40. f(x)=(x-1)(3x2+x+2)
1 3 - 2 1 -2
3 1
2
3
1 2
0
[–1, 9] by [0, 5]
(b) Sally walks toward the detector, turns and walks away
(or walks backward), then walks toward the detector
again.
41. f(x)=(x-1)(x+4)(x2+1)
1 1 3 -3 3 - 4
1
4 1
4
1 4
1 4
0
-4
1
1
4
-4
0
1
0
1
(c) The model “changes direction” at t≠1.81 sec
(D≠1.35 m) and t≠5.64 sec (when D≠3.65 m).
52. (a) D≠0.2434t2-1.7159t+4.4241
4
-4
0
42. f(x)=(x-3)(x+1)(x2+4)
3 1 -2 1 -8 -12
3 3
12
12
1
1 4
4
0
-1
1
1
1
-1
0
4
0
4
[–1, 9] by [0, 6]
(b) Jacob walks toward the detector, then turns and walks
away (or walks backward).
4
-4
0
(c) The model “changes direction” at t≠3.52 (when
D≠1.40 m).
4
p
(15h2-h3)(62.5)= p (125)(20), so that
3
3
15h2-h3=160. Of the three solutions (found graphically),
only h≠3.776 ft makes sense in this setting.
43. Solve for h:
p
4
(15h2-h3)(62.5)= p (125)(45), so that
3
3
15h2-h3=360. Of the three solutions (found graphically),
only h≠6.513 ft makes sense in this setting.
44. Solve for h:
2
3
2
45. Yes: (x+2)(x +1)=x +2x +x+2 is one such
polynomial. Other examples can be obtained by multiplying any other quadratic with no real zeros by (x+2).
46. No; by the Complex Conjugate Zeros Theorem, for such a
polynomial, if 2i is a zero, so is –2i.
47. No: if all coefficients are real, 1-2i and 1+i must also
be zeros, giving 5 zeros for a degree 4 polynomial.
48. Yes: f(x)=(x-1-3i)(x-1+3i)(x-1-i)
(x-1+i)=x4-4x3+16x2-24x+20 is one such
polynomial; all other examples would be multiples of this
polynomial.
53. False. Complex, nonreal solutions always come in conjugate pairs, so that if 1-2i is a zero, then 1+2i must
also be a zero.
54. False. All three zeros could be real. For instance, the polynomial f(x)=x(x-1)(x-2)=x3-3x2+2x has
degree 3, real coefficients, and no non-real zeros.
(The zeros are 0, 1, and 2.)
55. Both the sum and the product of two complex conjugates
are real numbers, and the absolute value of a complex
number is always real. The square of a complex number,
on the other hand, need not be real. The answer is E.
56. Allowing for multiplicities other than 1, then, the polynomial can have anywhere from 1 to 5 distinct real zeros.
But it cannot have no real zeros at all. The answer is A.
57. Because the complex, non-real zeros of a real-coefficient
polynomial always come in conjugate pairs, a polynomial
of degree 5 can have either 0, 2, or 4 non-real zeros. The
answer is C.
58. A polynomial with real coefficients can never have an
odd number of non-real complex zeros. The answer is E.
Section 2.6
97
Real Part
Imaginary Part
62. f(–2i)=(–2i)3-(2-i)(–2i)2+(2-2i)(–2i)-4
7
8
–8
8
16
0
9
16
16
=8i+(2-i)(4)-(2-2i)(2i)-4=
8i+8-4i-4i-4-4=0.
One can also take the last number of the bottom row
from synthetic division.
0
32
59. (a) Power
10
(b) 1 1
11
11
11
+
+
+
+
63. Synthetic division shows that f(i)=0 (the remainder),
and at the same time gives
f(x)÷(x-i) = x2+3x -i=h(x), so
f(x)=(x-i)(x2+3x-i).
7
i2 = 8 - 8i
i2 8 = 16
i2 9 = 16 + 16i
i2 10 = 32i
i
(c) Reconcile as needed.
(b) a2-b2=0
2abi=i, so 2ab=1
(1)
(2)
(c) From (1), we have:
a2-b2=0
(a+b)(a-b)=0
a=–b, a=b
Substituting into (2), we find:
a=b: 2ab=1
a=–b: 2ab=1
2b2=1
–2b2=1
1
1
b2=
b2=–
2
2
1
b=—
C2
Since a and b must be real, we have
(a, b)= b ¢
(d) Checking ¢
1
1
1
C2
12
12 12
12
≤ r.
≤, ¢ –
,
,–
2
2
2
2
12 12
≤ first.
,
2
2
12
12 ¤
12 ¤
¢
≤ 1 1 + i2 2
+
i≤ =¢
2
2
2
1
2i
1
= i
= 11 + 2i + i2 2 = 1 1 - 1 + 2i2 =
2
2
2
12
12
≤
Checking ¢ –
, –
2
2
12
12 ¤
12 ¤
≤ 1 1 + i2 2
i≤ =¢–
2
2
2
1
2i
1
= i
= 11 + 2i + i2 2 = 1 1 - 1 + 2i2 =
2
2
2
¢–
(e) The two square roots of i are:
12
12
12
12
+
i ≤ and ¢ –
i≤
2
2
2
2
-4i
3i
-i
-1
1
0
64. Synthetic division shows that f(1+i)=0 (the
remainder), and at the same time gives
f (x)÷(x-1-i)=x2+1=h(x), so
f(x)=(x-1-i)(x2+1).
1 + i
1
b2=
C2
b=— i
3 - i
i
3
1
1
60. (a) (a+bi)(a+bi)=a2+2abi+bi2
=a2+2abi-b2
¢
Graphs of Rational Functions
-1 - i
1 + i
0
1
0
1
-1 - i
1 + i
0
65. From graphing (or the Rational Zeros Test), we expect
x=2 to be a zero of f(x)=x3-8. Indeed,
f(2)=8-8=0. So, x=2 is a zero of f(x). Using
synthetic division we obtain:
2 1 0 0 -8
2 4
8
1 2 4
0
f(x)=(x-2)(x2+2x+4). We then apply the quadratic formula to find that the cube roots of x3-8 are
2, - 1 + 23i, and -1 - 13i.
66. From graphing (or the Rational Zeros Test), we expect
x=–4 to be a zero of f(x)=x3+64. Indeed
f(–4)=–64+64=0, so x=–4 is a zero of f(x).
Using synthetic division, we obtain:
-4
1
0
-4
-4
1
0
16
16
64
-64
0
f(x)=(x+4)(x2-4x+16). We then apply the quadratic formula to find that the cube roots of x3+64 are
–4, 2 + 2 13i, and 2 - 2 13i.
■ Section 2.6 Graphs of Rational Functions
Exploration 1
1. g1 x2 =
1
x - 2
61. f(i)=i3-i(i)2+2i(i)+2=–i+i-2+2=0.
One can also take the last number of the bottom row
from synthetic division.
[–3, 7] by [–5, 5]
98
Chapter 2
Polynomial, Power, and Rational Functions
1
x - 5
2. h1 x2 = -
Section 2.6 Exercises
1. The domain of f(x)=1/(x+3) is all real numbers
x Z -3. The graph suggests that f(x) has a vertical
asymptote at x=–3.
[–1, 9] by [–5, 5]
3. k1 x2 =
3
- 2
x + 4
[–4.7, 4.7] by [–4, 4]
As x approaches –3 from the left, the values of f(x)
decrease without bound. As x approaches –3 from the
right, the values of f(x) increase without bound. That is,
lim - f1 x2 = - q and lim + f1 x2 = q .
xS-3
xS-3
2. The domain of f(x)=–3/(x-1) is all real numbers
x Z 1. The graph suggests that f(x) has a vertical
asymptote at x=1.
[–8, 2] by [–5, 5]
Quick Review 2.6
1. f(x)=(2x-1)(x+3) ⇒ x=–3 or x =
1
2
4
2. f1 x2 = 1 3x + 42 1 x - 22 ⇒ x = - or x = 2
3
3. g1x2 = 1x + 22 1x - 22 ⇒ x = ;2
4. g1 x2 = 1 x + 12 1x - 12 ⇒ x = ;1
5. h1 x2 = 1 x - 12 1 x2 + x + 1 2 ⇒ x = 1
6. h1x2 = 1 x - i2 1 x + i2 ⇒ no real zeros
7.
2
x-3R 2x+1
[–4.7, 4.7] by [–12, 12]
As x approaches 1 from the left, the values of f(x)
increase without bound. As x approaches 1 from the right,
the values of f(x) decrease without bound. That is,
lim- f1x2 = q and lim+ f1x2 = - q .
xS1
xS1
3. The domain of f(x)=–1/(x2-4) is all real numbers
x Z -2, 2. The graph suggests that f(x) has vertical
asymptotes at x=–2 and x=2.
2x-6
7
Quotient: 2, Remainder: 7
8.
2
2x-1R 4x+3
4x-2
5
9.
Quotient: 2, Remainder: 5
3
xR 3x-5
3x
–5
10.
Quotient: 3, Remainder: –5
5
2
2xR 5x-1
5x
[–4.7, 4.7] by [–3, 3]
As x approaches –2 from the left, the values of f(x)
decrease without bound, and as x approaches –2 from the
right, the values of f(x) increase without bound. As x
approaches 2 from the left, the values of f(x) increase
without bound, and as x approaches 2 from the right, the
values of f(x) decrease without bound. That is,
lim - f1 x2 = - q , lim + f1x2 = q , lim- f1x2 = q , and
xS-2
lim+ f1x2 = - q .
xS2
–1
Quotient:
5
, Remainder: –1
2
xS-2
xS2
Section 2.6
4. The domain of f(x)=2/(x2-1) is all real numbers
x Z -1, 1. The graph suggests that f(x) has vertical
asymptotes at x=–1 and x=1.
Graphs of Rational Functions
99
8. Translate right 1 unit, translate up 3 units.
Asymptotes: x=1, y=3.
y
6
6
x
[–4.7, 4.7] by [–6, 6]
As x approaches –1 from the left, the values of f(x)
increase without bound, and as x approaches –1 from the
right, the values of f(x) decrease without bound. As x
approaches 1 from the left, the values of f(x) decrease
without bound, and as x approaches 1 from the right, the
values of f(x) increase without bound. That is,
lim - f1x2 = q , lim + f1x2 = - q , lim- f1x2 = - q ,
xS-1
xS-1
9. Translate left 4 units, vertically stretch by 13, translate
down 2 units. Asymptotes: x=–4, y=–2.
y
8
xS1
and lim+ f1x2 = q .
xS1
6
5. Translate right 3 units. Asymptotes: x=3, y=0.
x
y
5
5
x
10. Translate right 5 units, vertically stretch by 11, reflect across
x-axis, translate down 3 units. Asymptotes: x=5,
y=–3.
y
20
6. Translate left 5 units, reflect across x-axis, vertically
stretch by 2. Asymptotes: x=–5, y=0.
y
10
5
x
4
11. lim- f1x2 = q
xS3
12. lim+ f1x2 = - q
xS3
13. lim f1x2 = 0
7. Translate left 3 units, reflect across x-axis, vertically
stretch by 7, translate up 2 units.
Asymptotes: x=–3, y=2.
y
xSq
14. lim f1x2 = 0
xS-q
15. lim + f1 x2 = q
xS-3
16. lim - f1 x2 = - q
10
xS-3
17. lim f1x2 = 5
xS-q
6
x
18. lim f1x2 = 5
xSq
x
Chapter 2
100
Polynomial, Power, and Rational Functions
19. The graph of f(x)=(2x2-1)/(x2+3) suggests that
there are no vertical asymptotes and that the horizontal
asymptote is y=2.
[–4.7, 4.7] by [–4, 4]
The domain of f(x) is all real numbers, so there are
indeed no vertical asymptotes. Using polynomial long
division, we find that
7
2x2 - 1
= 2 - 2
x2 + 3
x + 3
When the value of 0 x 0 is large, the denominator x2+3 is
a large positive number, and 7/(x2+3) is a small positive
number, getting closer to zero as 0 x 0 increases. Therefore,
f(x)=
The domain of f(x)=(2x+1)/(x2-x)=
(2x+1)/[x(x-1)] is all real numbers x Z 0, 1, so
there are indeed vertical asymptotes at x=0 and x=1.
Rewriting one rational expression as two, we find that
2x
1
2x + 1
= 2
+ 2
f1x2 = 2
x - x
x - x
x - x
2
1
=
+ 2
x - 1
x - x
When the value of 0 x 0 is large, both terms get close to
zero. Therefore,
lim f1x2 = lim f1 x2 = 0,
xSq
xS-q
so y=0 is indeed a horizontal asymptote.
22. The graph of f(x)=(x-3)/(x2+3x) suggests that
there are vertical asymptotes at x=–3 and x=0, with
lim - f1 x2 = - q , lim + f1x2 = q , lim- f1x2 = q , and
xS-3
xS-3
xS0
lim+ f1x2 = - q , and that the horizontal asymptote
xS0
is y=0.
lim f1x2 = lim f1 x2 = 2,
xSq
xS-q
so y=2 is indeed a horizontal asymptote.
20. The graph of f(x)=3x2/(x2+1) suggests that there are
no vertical asymptotes and that the horizontal asymptote
is y=3.
[–4.7, 4.7] by [–4, 4]
The domain of f(x) is all real numbers, so there are
indeed no vertical asymptotes. Using polynomial long
division, we find that
f1 x2 =
3
3x2
= 3 - 2
x + 1
x + 1
The domain of f(x)=(x-3)/(x2+3x)=
(x-3)/[x(x+3)] is all real numbers x Z -3, 0, so
there are indeed vertical asymptotes at x=–3 and x=0.
Rewriting one rational expression as two, we find that
x
3
x - 3
= 2
- 2
f1 x2 = 2
x + 3x
x + 3x
x + 3x
1
3
=
- 2
x + 3
x + 3x
When the value of 0 x 0 is large, both terms get close to
zero. Therefore,
lim f1x2 = lim f1 x2 = 0,
xSq
2
When the value of 0 x 0 is large, the denominator x2+1 is
a large positive number, and 3/(x2+1) is a small positive
number, getting closer to zero as 0 x 0 increases. Therefore,
lim f1x2 = lim f1 x2 = 3,
xSq
[–4.7, 4.7] by [–4, 4]
xS-q
so y=0 is indeed a horizontal asymptote.
23. Intercepts: a 0,
2
b and (2, 0). Asymptotes: x=–1,
3
x=3, and y=0.
xS-q
so y 3 is indeed a horizontal asymptote.
21. The graph of f(x)=(2x+1)/(x2-x) suggests that
there are vertical asymptotes at x=0 and x=1, with
lim- f1x2 = q , lim+ f1 x2 = - q , lim- f1 x2 = - q , and
xS0
xS0
xS1
lim+ f1x2 = q , and that the horizontal asymptote is
xS1
y=0.
[–4, 6] by [–5, 5]
2
24. Intercepts: a 0, - b and (–2, 0). Asymptotes: x=–3,
3
x=1, and y=0.
[–4.7, 4.7] by [–12, 12]
[–6, 4] by [–5, 5]
Section 2.6
25. No intercepts. Asymptotes: x=–1, x=0, x=1, and
y=0.
Graphs of Rational Functions
101
31. (d); Xmin=–2, Xmax=8, Xscl=1, and Ymin=–3,
Ymax=3, Yscl=1.
32. (b); Xmin=–6, Xmax=2, Xscl=1, and Ymin=–3,
Ymax=3, Yscl=1.
33. (a); Xmin=–3, Xmax=5, Xscl=1, and Ymin=–5,
Ymax=10, Yscl=1.
34. (f); Xmin=–6, Xmax=2, Xscl=1, and Ymin=–5,
Ymax=5, Yscl=1.
[–4.7, 4.7] by [–10, 10]
26. No intercepts. Asymptotes: x=–2, x=0, x=2, and
y=0.
[–4, 4] by [–5, 5]
27. Intercepts: (0, 2), (–1.28, 0), and (0.78, 0). Asymptotes:
x=1, x=–1, and y=2.
35. (e); Xmin=–2, Xmax=8, Xscl=1, and Ymin=–3,
Ymax=3, Yscl=1.
36. (c); Xmin=–3, Xmax=5, Xscl=1, and Ymin=–3,
Ymax=8, Yscl=1.
37. For f(x)=2/(2x2-x-3), the numerator is never
zero, and so f(x) never equals zero and the graph has no
x-intercepts. Because f(0)=–2/3, the y-intercept is
–2/3. The denominator factors as 2x2-x-3
=(2x-3)(x+1), so there are vertical asymptotes at
x=–1 and x=3/2. And because the degree of the
numerator is less than the degree of the denominator, the
horizontal asymptote is y=0. The graph supports this
information and allows us to conclude that
lim f1 x2 = q, lim + f1 x2 = - q , lim - f1x2 = - q ,
xS-1 -
xS 13>22
xS-1
and lim + f1 x2 = q.
xS 13>22
The graph also shows a local maximum of –16/25 at
x=1/4.
[–5, 5] by [–4, 6]
28. Intercepts: (0, –3), (–1.84, 0), and (2.17, 0). Asymptotes:
x=–2, x=2, and y=–3.
[–4.7, 4.7] by [–3.1, 3.1]
2
Intercept: ¢0, - ≤
3
[–5, 5] by [–8, 2]
3
29. Intercept: a 0, b . Asymptotes: x=–2, y=x-4.
2
[–20, 20] by [–20, 20]
7
30. Intercepts: a 0, - b , (–1.54, 0), and (4.54, 0).
3
Asymptotes: x=–3, y=x-6.
3
3
Domain: (–q, –1) ª ¢ -1, ≤ ª ¢ , q ≤
2
2
16
Range: ¢ - q, - ≤ ª (0, q)
25
3
Continuity: All x Z -1,
2
1
Increasing on (–q, –1) and ¢ -1, ≤
4
3
1 3
Decreasing on c , b and a , q b
4 2
2
Not symmetric.
Unbounded.
1 16
Local maximum at a , - b
4
25
Horizontal asymptote: y=0
Vertical asymptotes: x=–1 and x=3/2
End behavior: lim f1 x2 = lim f1x2 = 0
xS-q
[–30, 30] by [–40, 20]
xSq
102
Chapter 2
Polynomial, Power, and Rational Functions
38. For g(x)=2/(x2+4x+3), the numerator is never
zero, and so g(x) never equals zero and the graph has no
x-intercepts. Because g(0)=2/3, the y-intercept is 2/3.
The denominator factors as
x2+4x+3=(x+1)(x+3),
so there are vertical asymptotes at x=–3 and x=–1.
And because the degree of the numerator is less than
the degree of the denominator, the horizontal asymptote
is y=0. The graph supports this information and allows
us to conclude that
lim g1x2 = q, lim + g1x2 = - q , lim - g1x2 = - q ,
xS-3 -
xS-3
xS-1
1
b , 11, 0 2
12
Domain: (–q, –3) ª (–3, 4) ª (4, q)
Range: (–q, q)
Continuity: All x Z -3, 4
Decreasing on (–q, –3), (–3, 4), and (4, q)
Not symmetric.
Unbounded.
No local extrema.
Horizontal asymptote: y=0
Vertical asymptotes: x=–3 and x=4
End behavior: lim h1x2 = lim h 1x2 = 0
xS-q
and lim + g1x2 = q.
xS-1
The graph also shows a local maximum of –2 at x=–2.
[–6.7, 2.7] by [–5, 5]
2
Intercept: ¢0, ≤
3
Domain: (–q, –3) ª (–3, –1) ª (–1, q)
Range: (–q, –2] ª (0, q)
Continuity: All x Z -3, -1
Increasing on (–q, –3) and (–3, –2]
Decreasing on [–2, –1) and (–1, q)
Not symmetric.
Unbounded.
Local maximum at (–2, –2)
Horizontal asymptote: y=0
Vertical asymptotes: x = -3 and x = -1
End behavior: lim g1 x2 = lim g1x2 = 0
xS-q
xSq
2
39. For h(x)=(x-1)/(x -x-12), the numerator is
zero when x=1, so the x-intercept of the graph is 1.
Because h(0)=1/12, the y-intercept is 1/12.
The denominator factors as
x2-x-12=(x+3)(x-4),
so there are vertical asymptotes at x=–3 and x=4.
And because the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote is
y=0. The graph supports this information and allows us
to conclude that
lim - h1 x2 = - q, lim + h1 x2 = q , lim- h1x2 = - q ,
xS-3
Intercepts: a 0,
xS-3
and lim+ h1x2 = q.
xS4
The graph shows no local extrema.
[–5.875, 5.875] by [–3.1, 3.1]
xS4
xSq
40. For k(x)=(x+1)/(x2-3x-10), the numerator is
zero when x=–1, so the x-intercept of the graph is –1.
Because k(0)=–1/10, the y-intercept is –1/10. The
denominator factors as
x2-3x-10=(x+2)(x-5),
so there are vertical asymptotes at x=–2 and x=5.
And because the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote is
y=0. The graph supports this information and allows us
to conclude that
lim k1x2 = - q, lim+ k 1x2 = q,
lim k 1x2 = - q , and lim+ k1x2 = q.
xS5 xS-2 -
xS2
xS5
The graph shows no local extrema.
[–9.4, 9.4] by [–1, 1]
Intercepts: 1 -1, 0 2, 10, -0.12
Domain: (–q, –2) ª (–2, 5) ª (5, q)
Range: (–q, q)
Continuity: All x Z -2, 5
Decreasing on (–q, –2), (–2, 5), and (5, q)
Not symmetric.
Unbounded.
No local extrema.
Horizontal asymptote: y=0
Vertical asymptotes: x=–2 and x=5
End behavior: lim k1 x2 = lim k1 x2 = 0
xS-q
xSq
41. For f(x)=(x2+x-2)/(x2-9), the numerator
factors as
x2+x-2=(x+2)(x-1),
so the x-intercepts of the graph are –2 and 1. Because
f(0)=2/9, the y-intercept is 2/9. The denominator
factors as
x2-9=(x+3)(x-3),
so there are vertical asymptotes at x=–3 and x=3.
And because the degree of the numerator equals the
degree of the denominator with a ratio of leading terms
that equals 1, the horizontal asymptote is y=1. The graph
supports this information and allows us to conclude that
lim - f1 x2 = q, lim + f1 x2 = - q, lim- f1x2 = - q ,
xS-3
xS-3
and lim+ f1x2 = q.
xS3
xS3
Section 2.6
The graph also shows a local maximum of about 0.260 at
about x=–0.675.
2
Intercepts: 1 -2, 02 , 1 1, 0 2, a 0, b
9
Domain: (–q, –3) ª (–3, 3) ª (3, q)
Range: (–q, 0.260] ª (1, q)
Continuity: All x Z -3, 3
Increasing on (–q, –3) and (–3, –0.675]
Decreasing on [–0.675, 3) and (3, q)
Not symmetric.
Unbounded.
Local maximum at about (–0.675, 0.260)
Horizontal asymptote: y=1
Vertical asymptotes: x=–3 and x=3
End behavior: lim f1 x2 = lim f1x2 = 1.
xS-q
xSq
42. For g(x)=(x2-x-2)/(x2-2x-8), the numerator
factors as
x2-x-2=(x+1)(x-2),
so the x-intercepts of the graph are –1 and 2. Because
g(0)=1/4, the y-intercept is 1/4. The denominator
factors as
x2-2x-8=(x+2)(x-4),
so there are vertical asymptotes at x=–2 and x=4.
And because the degree of the numerator equals the
degree of the denominator with a ratio of leading terms
that equals 1, the horizontal asymptote is y=1. The graph
supports this information and allows us to conclude that
lim - g1x2 = q, lim + g1x2 = - q, lim- g1x2 = - q ,
xS-2
xS-2
xS4
x2 + 2x - 2
2
= x ,
x + 2
x + 2
so the end-behavior asymptote of h(x) is y=x. The graph
supports this information and allows us to conclude that
lim - h1 x2 = q and lim + h1 x2 = - q.
xS-2
xS-2
The graph shows no local extrema.
[–9.4, 9.4] by [–15, 15]
3
Intercepts: (–3, 0), (1, 0), a 0, - b
2
Domain: (–q, –2) ª (–2, q)
Range: (–q, q)
Continuity: All x Z -2
Increasing on (–q, –2) and (–2, q)
Not symmetric.
Unbounded.
No local extrema.
Horizontal asymptote: None
Vertical asymptote: x=–2
Slant asymptote: y=x
End behavior: lim h1x2 = - q and lim h1x2 = q.
xS-q
xSq
2
and lim+ g1 x2 = q.
xS4
The graph also shows a local maximum of about 0.260 at
about x=0.324.
[–9.4, 9.4] by [–3, 3]
1
Intercepts: (–1, 0), (2, 0), a 0, b
4
Domain: (–q, –2) ª (–2, 4) ª (4, q)
Range: (–q, 0.260] ª (1, q)
Continuity: All x Z -2, 4
Increasing on (–q, –2) and (–2, 0.324]
Decreasing on [0.324, 4) and (4, q)
Not symmetric.
Unbounded.
Local maximum at about (0.324, 0.260)
Horizontal asymptote: y=1
Vertical asymptotes: x=–2 and x=4
End behavior: lim g1 x2 = lim g1x2 = 1
xS-q
103
43. For h(x)=(x2+2x-3)/(x+2), the numerator
factors as
x2+2x-3=(x+3)(x-1),
so the x-intercepts of the graph are –3 and 1. Because
h(0)=–3/2, the y-intercept is –3/2. The denominator is
zero when x=–2, so there is a vertical asymptote at
x=–2. Using long division, we rewrite h(x) as
h1x2 =
[–9.4, 9.4] by [–3, 3]
Graphs of Rational Functions
xSq
44. For k(x)=(x -x-2)/(x-3), the numerator
factors as
x2-x-2=(x+1)(x-2),
so the x-intercepts of the graph are –1 and 2. Because
k(0)=2/3, the y-intercept is 2/3. The denominator is
zero when x=3, so there is a vertical asymptote at
x=3. Using long division, we rewrite k(x) as
x2 - x - 2
4
h1x2 =
= x + 2 +
,
x - 3
x - 3
so the end-behavior asymptote of k(x) is y=x+2.
The graph supports this information and allows us to
conclude that lim- k(x)=–q and lim+ k(x)=q.
xS3
xS3
The graph shows a local maximum of 1 at x=1 and a
local minimum of 9 at x=5.
[–9.4, 9.4] by [–10, 20]
104
Chapter 2
Polynomial, Power, and Rational Functions
2
Intercepts: (–1, 0), (2, 0), a 0, b
3
Domain: (–q, 3) ª (3, q)
Range: (–q, 1] ª [9, q)
Continuity: All x Z 3
Increasing on (–q, 1] and [5, q)
Decreasing on [1, 3) and (3, 5]
Not symmetric.
Unbounded.
Local maximum at (1, 1); local minimum at (5, 9)
Horizontal asymptote: None
Vertical asymptote: x=3
Slant asymptote: y=x+2
End behavior: lim k(x)=–q and lim k(x)=q.
xS-q
(a)
[–10, 10] by [–30, 60]
(b)
xSq
45. Divide x2-2x-3 by x-5 to show that
x2 - 2x + 3
18
f1 x2 =
= x + 3 +
.
x - 5
x - 5
The end-behavior asymptote of f(x) is y=x+3.
(a)
[–50, 50] by [–1500, 2500]
48. Divide x3+1 by x-1 to show that
x3 + 1
2
f(x)=
=x2+x+1+
.
x - 1
x - 1
The end-behavior asymptote of f(x) is y=x2+x+1.
(a)
[–10, 20] by [–10, 30]
[–8, 8] by [–20, 40]
(b)
(b)
[–40, 40] by [–40, 40]
46. Divide 2x2+2x-3 by x+3 to show that
2x2 + 2x - 3
9
f(x)=
=2x-4+
.
x + 3
x + 3
The end-behavior asymptote of f(x) is y=2x-4.
[–50, 50] by [–1500, 2500]
49. Divide x4-2x+1 by x-2 to show that
f(x)=
(a)
x4 - 2x + 1
13
=x3+2x2+4x+6+
.
x - 2
x - 2
The end-behavior asymptote of f(x) is
y=x3+2x2+4x+6.
(a)
[–15, 10] by [–30, 20]
(b)
[–5, 5] by [–100, 200]
(b)
[–40, 40] by [–40, 40]
47. Divide x3-x2+1 by x+2 to show that
x3 - x2 + 1
11
f(x)=
=x2-3x+6.
x + 2
x + 2
The end-behavior asymptote of f(x) is y=x2-3x+6.
[–20, 20] by [–5000, 5000]
Section 2.6
Graphs of Rational Functions
105
50. Divide x5+1 by x2+1 to show that
x5 + 1
x + 1
f(x)= 2
=x3-x+ 2
.
x + 1
x + 1
The end-behavior asymptote of f(x) is y=x3-x.
(a) There are no vertical asymptotes, since the denominator x2+1 is never zero.
[–10, 15] by [–5, 10]
(b)
[–20, 20] by [–5000, 5000]
51. For f(x)=(3x2-2x+4)/(x2-4x+5), the numerator
is never zero, and so f(x) never equals zero and the graph
has no x-intercepts. Because f(0)=4/5, the y-intercept is
4/5. The denominator is never zero, and so there are no vertical asymptotes. And because the degree of the numerator
equals the degree of the denominator with a ratio of leading
terms that equals 3, the horizontal asymptote is y=3. The
graph supports this information. The graph also shows a
local maximum of about 14.227 at about x=2.445 and a
local minimum of about 0.773 at about x=–0.245.
[–15, 15] by [–5, 15]
4
Intercept: ¢0, ≤
5
Domain: (–q, q)
Range: [0.773, 14.227]
Continuity: (–q, q)
Increasing on [–0.245, 2.445]
Decreasing on (–q, –0.245], [2.445, q)
Not symmetric.
Bounded.
Local maximum at (2.445, 14.227); local minimum at
(–0.245, 0.773)
Horizontal asymptote: y=3
No vertical asymptotes.
End behavior: lim f(x)= lim f(x)=3
xS-q
xSq
52. For g(x)=(4x2+2x)/(x2-4x+8), the numerator
factors as
4x2+2x=2x(2x+1),
so the x-intercepts of the graph are –1/2 and 0. Because
g(0)=0, the y-intercept is 0. The denominator is never
zero, and so there are no vertical asymptotes. And
because the degree of the numerator equals the degree of
the denominator with a ratio of leading terms that equals
4, the horizontal asymptote is y=4. The graph supports
this information. The graph also shows a local maximum
of about 9.028 at about x=3.790 and a local minimum of
about –0.028 at about x=–0.235.
1
Intercepts: ¢ - , 0 ≤ , (0, 0)
2
Domain: (–q, q)
Range: [–0.028, 9.028]
Continuity: (–q, q)
Increasing on [–0.235, 3.790]
Decreasing on (–q, –0.235], [3.790, q).
Not symmetric.
Bounded.
Local maximum at (3.790, 9.028); local minimum
at (–0.235, –0.028)
Horizontal asymptote: y=4
No vertical asymptotes.
End behavior: lim g(x)= lim g(x)=4
xS-q
xSq
53. For h(x)=(x3-1)/(x-2), the numerator factors as
x3-1=(x-1)(x2+x+1),
so the x-intercept of the graph is 1. The y-intercept is
h(0)=1/2. The denominator is zero when x=2, so the
vertical asymptote is x=2. Because we can rewrite h(x) as
x3 - 1
7
h(x)=
=x2+2x+4+
,
x - 2
x - 2
we know that the end-behavior asymptote is
y=x2+2x+4. The graph supports this information
and allows us to conclude that
lim- h(x)=–q, lim+ h(x)=q.
xS2
xS2
The graph also shows a local maximum of about 0.586 at
about x=0.442, a local minimum of about 0.443 at about
x=–0.384, and another local minimum of about 25.970
at about x=2.942.
[–10, 10] by [–20, 50]
1
Intercepts: (1, 0), ¢ 0, ≤
2
Domain: (–q, 2) ª (2, q)
Range: (–q, q)
Continuity: All real x Z 2
Increasing on [–0.384, 0.442], [2.942, q)
Decreasing on (–q, –0.384], [0.442, 2), (2, 2.942]
Not symmetric.
Unbounded.
Local maximum at (0.442, 0.586); local minimum at
(–0.384, 0.443) and (2.942, 25.970)
No horizontal asymptote. End-behavior asymptote:
y=x2+2x+4
Vertical asymptote: x=2
End behavior: lim h(x)= lim h(x)=q
xS -q
xSq
106
Chapter 2
Polynomial, Power, and Rational Functions
54. For k(x)=(x3-2)/(x+2), the numerator is zero
3
3
when x= 12
, so the x-intercept of the graph is 12
.
The y-intercept is k(0)=–1. The denominator is zero
when x=–2, so the vertical asymptote is x=–2.
Because we can rewrite k(x) as
x3 - 2
10
k(x)=
=x2-2x+4,
x + 2
x - 2
we know that the end-behavior asymptote is
y=x2-2x+4. The graph supports this information
and allows us to conclude that
lim - k(x)=q, lim + k(x)=–q.
xS-2
xS-2
The graph also shows a local minimum of about 28.901 at
about x=–3.104.
Range: (–q, q)
1
2
Increasing on [–0.184, 0.5), (0.5, q)
Decreasing on (–q, –0.184]
Not symmetric.
Unbounded.
Local minimum at (–0.184, 0.920)
No horizontal asymptote. End-behavior
1
1
3
asymptote: y= x2- x+
2
4
8
1
Vertical asymptote: x =
2
End behavior: lim f(x)= lim f(x)=q
Continuity: All x Z
xS-q
[–10, 10] by [–20, 50]
3
Intercepts: ( 12
, 0), (0, –1)
Domain: (–q, –2) ª (–2, q)
Range: (–q, q)
Continuity: All real x Z -2
Increasing on [–3.104, –2), (–2, q)
Decreasing on (–q, –3.104]
Not symmetric.
Unbounded.
Local minimum at (–3.104, 28.901)
No horizontal asymptote. End-behavior asymptote:
y=x2-2x+4
Vertical asymptote: x=–2
End behavior: lim k(x)= lim k(x)=q
xS- q
xS1>2
The graph also shows a local minimum of about 0.920 at
about x=–0.184.
[–5, 5] by [–10, 10]
Intercepts: (1.755, 0), (0, 1)
1
Domain: All x Z
2
xS2
xS2
The graph also shows a local minimum of about 37.842 at
about x=2.899.
xSq
55. f(x)=(x3-2x2+x-1)/(2x-1) has only one
x-intercept, and we can use the graph to show that it is
about 1.755. The y-intercept is f(0)=1. The denominator is zero when x=1/2, so the vertical asymptote is
x=1/2. Because we can rewrite f(x) as
x3 - 2x2 + x - 1
f(x)=
2x - 1
1 2
3
1
7
= x - x+ ,
2
4
8
16 12x - 1 2
we know that the end-behavior asymptote is
1
3
1
y= x2- x+ . The graph supports this information
2
4
8
and allows us to conclude that
lim - f(x)=q, lim + f(x)=–q.
xS1>2
xSq
56. g(x)=(2x3-2x2-x+5)/(x-2) has only one
x-intercept, and we can use the graph to show that it is
about –1.189. The y-intercept is g(0)=–5/2.
The denominator is zero when x=2, so the vertical
asymptote is x=2. Because we can rewrite g(x) as
2x3 - 2x2 - x + 5
11
g(x)=
=2x2+2x+3+
,
x - 2
x - 2
we know that the end-behavior asymptote is
y=2x2+2x+3. The graph supports this information
and allows us to conclude that
lim- g(x)=–q, lim+ g(x)=q.
[–10, 10] by [–20, 60]
Intercepts: (–1.189, 0), (0, –2.5)
Domain: All x Z 2
Range: (–q, q)
Continuity: All x Z 2
Increasing on [2.899, q)
Decreasing on (–q, 2), (2, 2.899]
Not symmetric.
Unbounded.
Local minimum at (2.899, 37.842)
No horizontal asymptote. End-behavior asymptote:
y=2x2+2x+3
Vertical asymptote: x=2
End behavior: lim g(x)= lim g(x)=q
xS-q
xSq
57. For h(x)=(x4+1)/(x+1), the numerator is never
zero, and so h(x) never equals zero and the graph has no
x-intercepts. Because h(0)=1, the y-intercept is 1. So the
one intercept is the point (0, 1). The denominator is zero
when x=–1, so x=–1 is a vertical asymptote. Divide
x4+1 by x+1 to show that
2
x4 + 1
h(x)=
=x3-x2+x-1+
.
x + 1
x + 1
The end-behavior asymptote of h(x) is
y=x3-x2+x-1.
Section 2.6
[–5, 5] by [–30, 30]
58. k(x)=(2x5+x2-x+1)/(x2-1) has only one
x-intercept, and we can use the graph to show that it is
about –1.108. Because k(0)=–1, the y-intercept is –1.
So the intercepts are (–1.108, 0) and (0, –1). The denominator is zero when x=—1, so x=–1 and x=1 are
vertical asymptotes. Divide 2x5+x2-x+1 by x2-1
to show that
2x5 + x2 - x + 1
x + 2
k(x)=
=2x3+2x+1+ 2
.
x2 - 1
x - 1
The end-behavior asymptote of k(x) is
y=2x3+2x+1.
Graphs of Rational Functions
107
[–5, 5] by [–25, 50]
61. h(x)=(2x3-3x+2)/(x3-1) has only one
x-intercept, and we can use the graph to show that it is
about –1.476. Because h(0)=–2, the y-intercept is –2.
So the intercepts are (–1.476, 0) and (0, –2). The denominator is zero when x=1, so x=1 is a vertical asymptote. Divide 2x3-3x+2 by x3-1 to show that
2x3 - 3x + 2
3x - 4
h(x)=
=2- 3
.
x3 - 1
x - 1
The end-behavior asymptote of h(x) is y=2, a horizontal
line.
[–5, 5] by [–5, 5]
[–3, 3] by [–20, 40]
59. For f(x)=(x5-1)/(x+2), the numerator factors as
x5-1=(x-1)(x4+x3+x2+x+1), and since the
second factor is never zero (as can be verified by
Descartes’ Rule of Signs or by graphing), the x-intercept
of the graph is 1. Because f(0)=–1/2, the y-intercept is
–1/2. So the intercepts are (1, 0) and (0, –1/2).
The denominator is zero when x=–2, so x=–2 is a
vertical asymptote. Divide x5-1 by x+2 to show that
x5 - 1
33
f(x)=
=x4-2x3+4x2-8x+16x + 2
x + 2
The end-behavior asymptote of f(x) is
y=x4-2x3+4x2-8x+16.
62. For k(x)=(3x3+x-4)/(x3+1), the numerator
factors as 3x3+x-4=(x-1)(3x2+3x+4),
and since the second factor is never zero (as can be verified
by Descartes’ Rule of Signs or by graphing), the x-intercept
of the graph is 1. Because k(0)=–4, the y-intercept is –4.
So the intercepts are (1, 0) and (0, –4). The denominator is
zero when x=–1, so x=–1 is a vertical asymptote. Divide
3x3+x-4 by x3+1 to show that
3x3 + x - 4
x - 7
k(x)=
=3+ 3
.
x3 + 1
x + 1
The end-behavior asymptote of k(x) is y=3, a horizontal
line.
[–5, 5] by [–10, 10]
[–10, 10] by [–200, 400]
60. For g(x)=(x5+1)/(x-1), the numerator factors as
x5+1=(x+1)(x4-x3+x2-x+1), and since the
second factor is never zero (as can be verified by graphing), the x-intercept of the graph is –1. Because
g(0)=–1, the y-intercept is –1. So the intercepts are
(–1, 0) and (0, –1). The denominator is zero when x=1,
so x=1 is a vertical asymptote. Divide x5+1 by x-1
to show that
x5 + 1
2
g(x)=
=x4+x3+x2+x+1+
.
x - 1
x - 1
The end-behavior asymptote of g(x) is
y=x4+x3+x2+x+1.
63. False. If the denominator is never zero, there will be no
vertical asymptote. For example, f(x)=1/(x2+1) is a
rational function and has no vertical asymptotes.
64. False. A rational function is the quotient of two polynomials, and 2x2 + 4 is not a polynomial.
65. The excluded values are those for which x3+3x=0,
namely 0 and –3. The answer is E.
66. g(x) results from f(x) by replacing x with x+3, which
represents a shift of 3 units to the left. The answer is A.
67. Since x+5=0 when x=–5, there is a vertical asymptote. And because x2/(x+5)=x-5+25/(x+5),
the end behavior is characterized by the slant asymptote
y=x-5. The answer is D.
108
Chapter 2
Polynomial, Power, and Rational Functions
68. The quotient of the leading terms is x4, so the answer is E.
The functions are identical except at x=–1, where
f(x) has a discontinuity.
69. (a) No: the domain of f is 1 - q, 3 2 ´ 13, q 2 ; the
domain of g is all real numbers.
(b) No: while it is not defined at 3, it does not tend
toward ; q on either side.
(c) Most grapher viewing windows do not reveal that f is
undefined at 3.
(d) Almost—but not quite; they are equal for all x 3.
1x + 2 2 1x - 12
x2 + x - 2
70. (a) f1 x2 =
=
= x + 2
x - 1
x - 1
= g1 x2 when x Z 1
`
Asymptotes `
f
x=1
`
`
` (0, 2) (–2, 0)
`
` (–q, 1) ª (1, q) `
Intercepts
Domain
[–5, 5] by [–5, 5]
Asymptotes
g
none
(0, 2) (–2, 0)
(–q, q)
The functions are identical at all points except x=1,
where f has a discontinuity.
Intercepts
f
g
x=1, x=–2
x=–2
a 0,
1
b
2
1
a 0, b
2
(–q, –2) ª
(–q, –2) ª (–2, q)
(–2, 1) ª (1, q)
Domain
x - 1
x - 1
1
=
=
1x + 2 2 1x - 12
x + 2
x2 + x - 2
=g(x) when, x Z -2
Except at x=1, where f has a discontinuity, the
functions are identical.
(d) f1 x2 =
[–5, 5] by [–5, 5]
1x + 1 2 1x - 1 2
x2 - 1
=
= x - 1
x + 1
x + 1
= g1 x2 when x Z - 1
(b) f1 x2 =
`
Asymptotes `
f
x=–1
`
`
` (0, –1) (1, 0)
`
` (–q, –1) ª (–1, q) `
Intercepts
Domain
g
none
(0, –1) (1, 0)
(–q, q)
The functions are identical except at x=–1, where f
has a discontinuity.
[–5.7, 3.7] by [–3.1, 3.1]
71. (a) The volume is f1x2 = k>x, where x is pressure and k
is a constant. f(x) is a quotient of polynomials and
hence is rational, but f1x2 = k # x -1, so is a power
function with constant of variation k and pressure -1.
(b) If f(x)=kxa, where a is a negative integer, then the
power function f is also a rational function.
k
, so k=(2.59)(0.866)=2.24294.
P
2.24294
If P=0.532, then V=
≠4.22 L.
0.532
(c) V =
72. (a)
[–5, 5] by [–5, 5]
(c) f1 x2 =
=
x2 - 1
x2 - 1
=
x3 - x2 - x + 1
1 x2 - 1 2 1 x - 12
1
= g1x2 when x Z -1
x - 1
`
Asymptotes `
Intercepts
Domain
`
f
x=1, x=–1
(0, –1)
`
`
`
g
x=1
(0, –1)
` (–q, –1) ª (–1, 1) ` (–q, 1) ª (1, q)
`
`
ª (1, q)
[0.5, 3.5] by [0, 7]
(b) One method for determining k is to find the power
regression for the data points using a calculator,
discussed in previous sections. By this method, we
find that a good approximation of the data points is
given by the curve y L 5.81 # x-1.88. Since –1.88 is
very close to –2, we graph the curve to see if
k=5.81 is reasonable.
Section 2.7
(c)
Solving Equations in One Variable
109
76. Horizontal asymptotes: y=—2.
Intercepts: (0, 2), (1, 0)
2 - 2x
x 0
f1 x2 = • x + 1
2
x 6 0
[0.5, 3.5] by [0, 7]
(d) At 2.2 m, the light intensity is approximately
1.20 W/m2.
At 4.4 m, the light intensity is approximately
0.30 W/m2.
73. Horizontal asymptotes: y=–2 and y=2.
3
3
Intercepts: a 0, - b , a , 0 b
2
2
2x - 3
x 0
x + 2
h1 x2 = μ
2x - 3
x 6 0
-x + 2
[–7, 13] by [–3, 3]
1
shifted horizontally
x
-d>c units, stretched vertically by a factor of
0 bc - ad 0 >c2, reflected across the x-axis if and only if
bc-ad<0, and then shifted vertically by a/c.
77. The graph of f is the graph of y =
78. Yes, domain = 1 - q, -12 ´ 1 -1, 0 2 ´ 10, 12 ´ 11, q 2;
range = 1 - q, 12 ´ 11, q 2; continuous and decreasing
on each interval within their common domain;
x-intercepts = 1 -1 ; 152>2; no y-intercepts; hole at
(0, 1); horizontal asymptote of y = 1; vertical asymptotes
of x = ;1; neither even nor odd; unbounded; no local
extrema; end behavior: lim f1x2 = lim g1x2 = 1.
0 x 0 Sq
[–5, 5] by [–5, 5]
74. Horizontal asymptotes: y=—3.
5
5
Intercepts: a 0, b , a - , 0 b
3
3
3x + 5
x 0
x + 3
h1 x2 = μ
3x + 5
x 6 0
-x + 3
0 x 0 Sq
■ Section 2.7 Solving Equations
in One Variable
Quick Review 2.7
1. The denominator is x2+x-12=(x-3)(x+4), so
the new numerator is 2x(x+4)=2x2+8x.
2. The numerator is x2-1=(x-1)(x+1), so the new
denominator is (x+1)(x+1)=x2+2x+1.
3. The LCD is the LCM of 12, 18, and 6, namely 36.
7
5
15
14
30
5
+
- =
+
12
18
6
36
36
36
1
= 36
[–5, 5] by [–5, 5]
75. Horizontal asymptotes: y=—3.
5
5
Intercepts: a 0, b , a , 0 b
4
3
5 - 3x
x 0
x + 4
f1 x2 = μ
5 - 3x
x 6 0
–x + 4
4. The LCD is x(x-1).
1
3x
x - 1
3
=
x - 1
x
x1x - 1 2
x1x - 1 2
3x - x + 1
=
x1x - 12
2x + 1
= 2
x - x
5. The LCD is (2x+1)(x-3).
x1x - 32
212x + 1 2
x
2
=
2x + 1
x - 3
1 2x + 12 1 x - 3 2
12x + 12 1x - 32
=
=
[–10, 10] by [–5, 5]
x2 - 3x - 4x - 2
12x + 12 1x - 3 2
x2 - 7x - 2
12x + 12 1x - 3 2
Chapter 2
110
Polynomial, Power, and Rational Functions
6. x2-5x+6=(x-2)(x-3) and
x2-x-6=(x+2)(x-3), so the LCD is
(x-2)(x-3)(x+2).
x + 1
3x + 11
- 2
x2 - 5x + 6
x - x - 6
1x + 1 2 1x + 2 2
1 3x + 11 2 1x - 2 2
=
1 x - 2 2 1x - 3 2 1x + 2 2
1x - 22 1x - 32 1x + 22
x2 + 3x + 2 - 3x2 - 5x + 22
1x - 2 2 1x - 3 2 1x + 2 2
- 2x2 - 2x + 24
=
1 x - 2 2 1x - 3 2 1x + 2 2
-21 x - 32 1 x + 42
=
1 x - 2 2 1x - 3 2 1x + 2 2
-2x - 8
=
,x Z 3
1 x - 2 2 1x + 2 2
=
7. For 2x2-3x-1=0: a=2, b=–3, and c=–1.
x =
-b ; 3b2 - 4ac
2a
=
- 1 - 32 ; 31 - 32 2 - 4 1 2 2 1 -1 2
=
3 ; 39 - 1 -8 2
2 12 2
4
=
3 ; 117
4
8. For 2x2-5x-1=0: a=2, b=–5, and c=–1.
x =
-b ; 3b2 - 4ac
2a
=
- 1 - 52 ; 31 - 52 2 - 4 1 2 2 1 -1 2
=
5 ; 325 - 1 -8 2
2 12 2
4
=
5 ; 133
4
9. For 3x2+2x-2=0: a=3, b=2, and c=–2.
x =
-b ; 3b2 - 4ac
2a
=
-2 ; 31 22 2 - 4 13 2 1 -2 2
=
-2 ; 34 - 1 -24 2
2132
6
=
-2 ; 128
6
-2 ; 2 27
- 1 ; 27
=
=
6
3
10. For x2-3x-9=0: a=1, b=–3, and c=–9.
x =
-b ; 3b2 - 4ac
2a
x + 5
1
x - 2
+
=
3
3
3
(x-2)+(x+5)=1
2x+3=1
2x=–2
x=–1
Numerically: For x=–1,
x + 5
-1 - 2
-1 + 5
x - 2
+
=
+
3
3
3
3
4
-3
+
=
3
3
1
=
3
1. Algebraically:
15
x
x2+2x=15 (x Z 0)
x2+2x-15=0
(x-3)(x+5)=0
x-3=0 or x+5=0
x=3 or
x=–5
Numerically: For x=3,
x+2=3+2=5 and
15
15
=
= 5.
x
3
For x=–5,
x+2=–5+2=–3 and
15
15
=
= -3.
x
-5
2. Algebraically: x+2=
3. Algebraically: x + 5 =
=
=
3 ; 39 - 1 -36 2
2 11 2
=
3 ; 145
2
14
x
(x Z 0)
x2+5x=14
x2+5x-14=0
(x-2)(x+7)=0
x-2=0 or x+7=0
x=2 or
x=–7
Numerically: For x=2,
x+5=2+5=7 and
14
14
=
= 7.
x
2
For x=–7,
x+5=–7+5=–2 and
14
14
=
= -2.
x
-7
1
2
= 4
x
x - 3
(x-3)-2x=4x(x-3) (x Z 0, 3)
–x-3=4x2-12x
–4x2+11x-3=0
4. Algebraically:
x =
- 1 - 32 ; 31 - 32 2 - 4 1 1 2 1 -9 2
2
3 ; 3 15
=
2
Section 2.7 Exercises
=
-11 ; 3112 - 41 -4 2 1 -3 2
2 1 -42
-11 ; 173
-8
11 - 173
11 + 173
L 2.443 or x =
L 0.307
8
8
Numerically: Use a graphing calculator to support your
answers numerically.
x =
Section 2.7
4x
12
=
x - 3
x - 3
x(x-3)+4x=12 (x Z 3)
x2-3x+4x=12
x2+x-12=0
(x+4)(x-3)=0
x+4=0
or x-3=0
x=–4 or
x=3 — but x=3 is extraneous.
Numerically: For x=–4,
4 1 -4 2
16
12
4x
= -4 +
= -4 +
= - and
x +
x - 3
-4 - 3
7
7
12
12
12
=
= - .
x - 3
-4 - 3
7
Solving Equations in One Variable
111
5. Algebraically: x +
3
2
+
= 8
x - 1
x
3x+2(x-1)=8x(x-1)
5x-2=8x2-8x
–8x2+13x-2=0
6. Algebraically:
x =
=
(x Z 0, 1)
-13 ; 3132 - 4 1 -8 2 1 -2 2
[–9.4, 9.4] by [–15, 15]
Then the solutions are x=–5 and x=3.
12
9. Algebraically: x+ =7
x
x2+12=7x
1x Z 0 2
x2-7x+12=0
(x-3)(x-4)=0
x-3=0 or x-4=0
x=3 or
x=4
12
Graphically: The graph of f(x)=x+ -7 suggests
x
that the x-intercepts are 3 and 4.
2 1 -8 2
-13 ; 3105
- 16
13 + 3105
13 - 3105
L 1.453 or x =
L 0.172
16
16
Numerically: Use a graphing calculator to support your
answers numerically.
10
7. Algebraically: x +
= 7
x
2
x +10=7x
(x Z 0)
x2-7x+10=0
(x-2)(x-5)=0
x-2=0 or x-5=0
x=2 or
x=5
x =
Graphically: The graph of f1x2 = x +
10
- 7 suggests
x
[–9.4, 9.4] by [–15, 5]
Then the solutions are x=3 and x=4.
6
10. Algebraically: x+ =–7
x
x2+6=–7x
1x Z 02
2
x +7x+6=0
(x+6)(x+1)=0
x+6=0
or x+1=0
x=–6 or
x=–1
6
Graphically: The graph of f(x)=x+ +7 suggests
x
that the x-intercepts are –6 and –1.
that the x-intercepts are 2 and 5.
[–9.4, 9.4] by [–5, 15]
[–9.4, 9.4] by [–15, 5]
Then the solutions are x=2 and x=5.
15
8. Algebraically: x+2=
x
x2+2x=15
1 x Z 02
x2+2x-15=0
(x+5)(x-3)=0
x+5=0
or x-3=0
x=–5 or
x=3
Graphically: The graph of f(x)=x+2that the x-intercepts are –5 and 3.
15
suggests
x
Then the solutions are x=–6 and x=–1.
1
1
11. Algebraically: 2= 2
x + 1
x + x
[and x2+x=x(x+1)]
2(x2+x)-x=1
(x Z 0, –1)
2x2+x-1=0
(2x-1)(x+1)=0
2x-1=0 or x+1=0
1
x=
or
x=–1
2
— but x=–1 is extraneous.
1
1
Graphically: The graph of f(x)=2- 2
x + 1
x + x
1
suggests that the x-intercept is . There is a hole at
2
x=–1.
112
Chapter 2
Polynomial, Power, and Rational Functions
4x
3
15
+
= 2
x + 4
x - 1
x + 3x - 4
[and x2+3x-4=(x+4)(x-1)]
14. Algebraically:
4x(x-1)+3(x+4)=15 (x Z-4, 1)
4x2-x-3=0
(4x+3)(x-1)=0
4x+3=0
or x-1=0
3
x= or
x=1
4
— but x=1 is extraneous.
[–4.7, 4.7] by [–4, 4]
1
Then the solution is x= .
2
3
12
12. Algebraically: 2= 2
x + 4
x + 4x
[and x2+4x=x(x+4)]
Graphically: The graph of
4x
3
15
f(x)=
suggests that
+
- 2
x + 4
x - 1
x + 3x - 4
3
the x-intercept is - . There is a hole at x=1.
4
2(x2+4x)-3x=12
(x Z 0, –4)
2x2+5x-12=0
(2x-3)(x+4)=0
2x-3=0 or x+4=0
3
x=
or
x=–4
2
— but x=–4 is extraneous.
Graphically: The graph of f(x)=2-
12
3
- 2
x + 4
x + 4x
3
suggests that the x-intercept is . There is a hole at
2
x=–4.
[–12.4, 6.4] by [–5, 10]
3
Then the solution is x= - .
4
3
3
x - 3
+ 2
= 0
x
x + 1
x + x
[and x2+x=x(x+1)]
(x-3)(x+1)-3x+3=0
(x Z 0, –1)
x2-5x=0
x(x-5)=0
x=0 or x-5=0
x=0 or
x=5 — but x=0 is extraneous.
15. Algebraically:
[–4.7, 4.7] by [–4, 4]
3
Then the solution is x= .
2
1
7
3x
13. Algebraically:
+
= 2
x + 5
x - 2
x + 3x - 10
[and x2+3x-10=(x+5)(x-2)]
3x(x-2)+(x+5)=7
(x Z -5, 2)
3x2-5x-2=0
(3x+1)(x-2)=0
3x+1=0
or x-2=0
1
x=–
or
x=2
3
— but x=2 is extraneous.
Graphically: The graph of
3x
1
7
f(x)=
suggests that
+
- 2
x + 5
x - 2
x + 3x - 10
1
the x-intercept is - . There is a hole at x=2.
3
[–14.4, 4.4] by [–3, 9]
1
Then the solution is x= - .
3
Graphically: The graph of
3
3
x - 3
f(x)=
suggests that the
+ 2
x
x + 1
x + x
x-intercept is 5. The x-axis hides a hole at x=0.
[–6.4, 12.4] by [–10, 5]
Then the solution is x=5.
x + 2
4
2
= 0
+ 2
x
x - 1
x - x
2
[and x -x=x(x-1)]
(x+2)(x-1)-4x+2=0
(x Z 0, 1)
x2-3x=0
x(x-3)=0
x=0 or x-3=0
x=0 or
x=3 — but x=0 is extraneous.
16. Algebraically:
Section 2.7
Graphically: The graph of
x + 2
4
2
f(x)=
+ 2
x
x - 1
x - x
suggests that the x-intercept is 3. The x-axis hides a hole
at x=0.
Solving Equations in One Variable
113
19. There is no x-intercept at x=–2. That is the extraneous
solution.
20. There is no x-intercept at x=3. That is the extraneous
solution.
21. Neither possible solution corresponds to an x-intercept of
the graph. Both are extraneous.
22. There is no x-intercept at x=3. That is the extraneous
solution.
23.
[–4.7, 4.7] by [–5, 5]
Then the solution is x=3.
3 - x
6
3
=
+ 2
x + 2
x
x + 2x
[and x2+2x=x(x+2)]
3x+6=(3-x)(x+2)
(x Z - 2, 0)
3x+6=–x2+x+6
x2+2x=0
x(x+2)=0
x=0 or x+2=0
x=0 or
x=–2
— but both solutions are extraneous.
No real solutions.
2
+x=5
x - 1
2+x(x-1)=5(x-1)
x2-x+2=5x-5
x2-6x+7=0
17. Algebraically:
Then there are no real solutions.
2
6
x + 3
= 2
x
x + 3
x + 3x
[and x2+3x=x(x+3)]
(x+3)2-2x=6
(x Z -3, 0)
x2+4x+3=0
(x+1)(x+3)=0
x+1=0
or x+3=0
x=–1 or
x=–3
— but x=–3 is extraneous.
x = 3 + 12 L 4.414 or
x = 3 - 12 L 1.586
[–4.7, 4.7] by [–3, 3]
Then the solution is x=–3.
x2 - 6x + 5
= 3
x2 - 2
x2-6x+5=3(x2-2)
24.
–2x2-6x+11=0
1x Z ; 122
- 1 - 62 ; 31 -62 2 - 4 1 -2 2 111 2
21 -2 2
6 ; 1124
-3 ; 131
x =
=
-4
2
x =
25.
18. Algebraically:
Graphically: The graph of
x + 3
2
6
f(x)=
- 2
x
x + 3
x + 3x
suggests that the x-intercept is –1. There is a hole at
x=–3.
- 1 -62 ; 31 -6 2 2 - 4 11 2 172
2 11 2
6 ; 18
x =
= 3 ; 12
2
x =
Graphically: The graph of
3 - x
3
6
suggests that there
f1 x2 =
+ 2
x + 2
x
x + 2x
are no x-intercepts. There is a hole at x=–2, and the
x-axis hides a “hole” at x=0.
[–4.7, 4.7] by [–3, 3]
(x Z 1)
26.
x =
-3 + 131
L 1.284 or
2
x =
-3 - 131
L -4.284
2
x2 - 2x + 1
= 0
x + 5
x2-2x+1=0
(x-1)2=0
x-1=0
x=1
(x Z 5)
2
5
3x
+
= 2
x + 2
x - 1
x + x - 2
[and x2+x-2=(x+2)(x-1)]
3x(x-1)+2(x+2)=5
(x Z -2, 1)
3x2-x-1=0
x =
- 1 -12 ; 31 -1 2 2 - 4 13 2 1 -12
1 ; 113
x =
6
2 132
x =
1 + 113
L 0.768 or
6
x =
1 - 113
L -0.434
6
114
27.
Chapter 2
Polynomial, Power, and Rational Functions
5
15
4x
+
= 2
x + 4
x - 1
x + 3x - 4
[and x2+3x-4=(x+4)(x-1)]
4x(x-1)+5(x+4)=15
(x Z - 4, 1)
4x2+x+5=0
The discriminant is b2-4ac=12-4(4)(5)=–79<0.
There are no real solutions.
28.
5
15
3x
+
= 2
x + 1
x - 2
x - x - 2
[and x2-x-2=(x+1)(x-2)]
3x(x-2)+5(x+1)=15
(x Z -1, 2)
3x2-x-10=0
(3x+5)(x-2)=0
3x+5=0
or x-2=0
5
or
x=2
x = 3
— but x=2 is extraneous.
5
The solution is x = - .
3
5
= 8
x
x3+5=8x
(x Z 0)
Using a graphing calculator to find the x-intercepts of
f(x)=x3-8x+5 yields the solutions
x L -3.100, x L 0.661, and x L 2.439.
29. x2 +
3
= 7
x
3
x -3=7x
(x Z 0)
Using a graphing calculator to find the x-intercepts of
f(x)=x3-7x-3 yields the solutions
x≠–2.398, x≠–0.441, and x≠2.838.
33. (a) C1x2 =
3000 + 2.12x
x
(b) A profit is realized if C1x2 6 2.75, or
3000 + 2.12x 6 2.75x. Then 3000 6 0.63x,
so that x 7 4761.9—4762 hats per week.
(c) They must have 2.75x-(3000+2.12x)>1000 or
0.63x>4000: 6350 hats per week.
34. (a) P(10)=200, P(40)=350, P(100)=425
(b) As t S q, P1 t2 S 500. So, yes. The horizontal asymptote is y=500.
(c) lim P1t2 = lim a 500 -
9000
b = 500, so the bear
t + 20
population will never exceed 500.
tSq
tSq
35. (a) If x is the length, then 182/x is the width.
182
364
P1 x2 = 2x + 2 a
b = 2x +
x
x
(b) The graph of P(x)=2x+364/x has a minimum
when x≠13.49, so that the rectangle is square. Then
P(13.49)=2(13.49)+364/13.49≠53.96 ft.
36. (a)
1.5 in.
30. x2 -
31. (a) The total amount of solution is (125+x) mL; of this,
the amount of acid is x plus 60% of the original
amount, or x+0.6(125).
(b) y=0.83
x + 75
= 0.83. Multiply both sides by
x + 125
x+125, then rearrange to get 0.17x = 28.75, so that
x L 169.12 mL.
x + 0.35 1100 2
x + 35
=
32. (a) C1x2 =
x + 100
x + 100
(c) C(x)=
(b) Graph C(x) along with y=0.75; observe where
the first graph intersects the second.
0.75 in.
2
40 in
1 in.
1 in.
The height of the print material is 40/x. The total
area is
40
A1x2 = 1x + 0.75 + 1 2 a
+ 1.5 + 1 b
x
40
+ 2.5 b
= 1x + 1.75 2 a
x
(b) The graph of A1x2 = 1x + 1.75 2 a
40
+ 2.5 b
x
has a minimum when x≠5.29, so the dimensions
are about 5.29+1.75=7.04 in. wide by
40/5.29+2.5≠10.06 in. high. And
A(5.29)≠70.8325 in.2
37. (a) Since V=∏r2h, the height here is V/(∏r2). And
since in general, S=2∏r2+2∏rh=2∏r2+2V/r,
here S(x)=2∏x2+1000/x
(0.5 L=500 cm3).
[0, 250] by [0, 1]
For x=160, C(x)=0.75. Use 160 mL.
x + 35
= 0.75, multiply by x+100
(c) Starting from
x + 100
and rearrange to get 0.25x = 40, so that x = 160 mL
That is how much pure acid must be added.
(b) Solving 2∏x2+1000/x=900 graphically by finding
the zeros of f(x)=2∏x2+1000/x-900 yields two
solutions: either x≠1.12 cm, in which case
h≠126.88 cm, or x≠11.37 cm, in which case
h≠1.23 cm.
Section 2.7
38. (a) If x is the length, then 1000/x is the width. The total
area is A(x)=(x+4)(1000/x+4).
(b) The least area comes when the pool is square, so
that x = 11000 L 31.62 ft. With dimensions
35.62 ft*35.62 ft for the plot of land,
A(31.62)≠1268.98 ft2.
1
1
1
=
+
39. (a)
R
R1
R2
1
1
1
=
+
R
2.3
x
2.3x=xR+2.3R
2.3x
R1 x2 =
x + 2.3
(b) Graphically: The function f1x2 =
17
53
5
3
x
x + 43
has a zero at x L 20.45.
[0, 25] by [–1, 1]
For R=1.7, x≠6.52 ohms.
Algebraically: 1 h 40 min=1
40. (a) If x is the length, then 200/x is the width.
200
400
P1x2 = 2x + 2 a
b = 2x +
x
x
2
5
h= h
3
3
5
17
53
=
+
3
x
x + 43
5x(x+43)=51(x+43)+159x
5x2+215x=51x+2193+159x
2
5x +5x-2193=0
Using the quadratic formula and selecting the positive
solution yields
-5 + 143,885
x =
L 20.45.
10
The rate of the bike was about 20.45 mph.
(b) 70 = 2x +
43. (a)
41. (a) Drain A can drain 1/4.75 of the pool per hour, while
drain B can drain 1/t of the pool per hour. Together,
they can drain a fraction
1
1
t + 4.75
D1t 2 =
+ =
4.75
t
4.75t
of the pool in 1 hour.
(b) The information implies that D(t)=1/2.6, so we solve
1
1
1
=
+ .
2.6
4.75
t
1
1
1
Graphically: The function f1t2 =
- has
2.6
4.75
t
a zero at t≠5.74 h, so that is the solution.
115
42. (a) With x as the bike speed, x+43 is the car speed.
Biking time=17/x and driving time=53/(x+43),
so
53
17
.
+
T =
x
x + 43
(b) 2.3x=xR+2.3R
2.3R
x =
2.3 - R
400
x
70x=2x2+400
2x2-70x+400=0
The quadratic formula gives
x≠7.1922 or x≠27.8078.
When one of those values is considered as the
length, the other is the width. The dimensions are
7.1922 m* 27.8078 m.
Solving Equations in One Variable
[0, 15] by [0, 120]
(b) When x=15, y=120-500/(15+8)≠98.3.
In 2005, sales are estimated at about $98.3 billion.
44. (a)
[0, 35] by [0, 3000]
(b) When x=35, y=3000-39,500/(35+9)≠2102.
In 2005, the number of wineries is estimated to be 2102.
[0, 10] by [–0.25, 0.25]
1
1
1
=
+
2.6
4.75
t
4.75t=2.6t+2.6(4.75)
2.6 14.752
t =
4.75 - 2.6
≠5.74
Algebraically:
45. False. An extraneous solution is a value that, though
generated by the solution-finding process, does not work
in the original equation. In an equation containing
rational expressions, an extraneous solution is typically a
solution to the version of the equation that has been
cleared of fractions but not to the original version.
46. True. For a fraction to equal zero, the numerator has to be
zero, and 1 is not zero.
116
47.
Chapter 2
Polynomial, Power, and Rational Functions
6
3x
=
x + 2
x + 2
x(x+2)-3x=6
(x Z - 2)
x2-x-6=0
(x-3)(x+2)=0
x=3 or x=–2 — but x=–2 is extraneous.
The answer is D.
x -
6
3
[and x2+2x=x(x+2)]
= 2
x
x + 2x
x2+2x-3(x+2)=6
(x Z -2, 0)
x2-x-12=0
(x+3)(x-4)=0
x=–3 or x=4
The answer is C.
48. 1 -
49.
x
2
14
+
= 2
x + 2
x - 5
x - 3x - 10
[and x2-3x-10=(x+2)(x-5)]
x(x-5)+2(x+2)=14
(x Z -2, 5)
x2-3x-10=0
(x+2)(x-5)=0
x=–2 or x=5 — but both solutions are extraneous.
The answer is E.
50. 0.2*10 or 2=liters of pure acid in 20% solution
0.30*30 or 9=liters of pure acid in 30% solution
L of pure acid
=concentration of acid
L of mixture
2 + 9
11
=0.275=27.5%
=
10 + 30
40
The answer is D.
51. (a) The LCD is x2+2x=x(x+2).
x - 3
3
6
f(x)=
+
+ 2
x
x + 2
x + 2x
1x - 32 1x + 2 2
3x
6
+ 2
+ 2
=
2
x + 2x
x + 2x
x + 2x
x2 - x - 6 + 3x + 6
=
x2 + 2x
2
x + 2x
= 2
x + 2x
(b) All x Z 0, –2
(c) f1 x2 = b
1
undefined
x Z - 2, 0
x = - 2 or x = 0
(d) The graph appears to be the horizontal line y=1
with holes at x=–2 and x=0.
1
1 + x
y(1+x)=(1+x)+1
y+xy=x+2
xy-x=2-y
2 - y
x =
y - 1
52. y = 1 +
1
1 - x
y(1-x)=(1-x)-1
y-xy=–x
y=xy-x
y
x =
y - 1
53. y = 1 -
54. y = 1 +
1
1
x
1 +
x
1 + x
y(1+x)=(1+x)+x
y+xy=2x+1
xy-2x=1-y
1 - y
x =
y - 2
y = 1 +
1
55. y = 1 +
1
1 - x
1 - x
y = 1 +
1 - x + 1
1 - x
y = 1 +
2 - x
y(2-x)=(2-x)+(1-x)
2y-xy=3-2x
2y-3=xy-2x
2y - 3
x =
y - 2
1 +
■ Section 2.8 Solving Inequalities
in One Variable
Exploration 1
1. (a)
(+)(–)(+)
Negative
(+)(–)(+)
Negative
–3
(+)(+)(+)
Positive x
2
(b)
[–4.7, 4.7] by [–3.1, 3.1]
This matches the definition in part (c).
[–5, 5] by [–250, 50]
2. (a)
(–) (+) (–)(+) (–) (+) (–)(+) (–) (+)(+)(+)
Positive
Positive
Negative x
–2
–1
Section 2.8
Solving Inequalities in One Variable
117
;1, ; 2, ; 4, ;8
; 1, ; 3
4
1
2
8
or ; 1, ; , ; 2, ; , ; 4, ; , ;8, ;
3
3
3
3
(b) A graph suggests that –2 and 1 are good candidates
for zeros.
(b)
10. (a)
[–3, 1] by [–30, 20]
3. (a)
(+)(+) (–)(–) (+)(+)(+)(–) (+)(+)(+)(–)
Positive
Negative
Negative x
–4
2
(b)
–2 |
3
–1
–10
8
–6
14
–8
1|
3
–7
4
0
3
–4
3
–4
0
2
2
3x -x -10x+8=(x+2)(x-1)(3x-4)
Section 2.8 Exercises
1. (a) f(x)=0
[–5, 5] by [–3000, 2000]
(b) f(x) 7 0
when –2 6 x 6 –1 or x 7 5
(c) f(x) 6 0
when x 6 –2 or –1 6 x 6 5
(–)(–) (–) (+)(–)(–) (+)(+)(–) (+)(+)(+)
Negative Positive Negative Positive
–2
–1
5
Quick Review 2.8
1. lim f(x)=q, lim f(x)=–q
xSq
xS –q
2. (a) f(x)=0
2. lim f(x)=–q, lim f(x)=–q
xSq
xS –q
3. lim g(x)=q, lim g(x)=q
xSq
xS –q
1
when –4 6 x 6 - or x 7 7
3
(c) f(x) 6 0
when x 6 –4 or -
x2 - 3x - 4x - 2
=
7.
1 2x + 1 2 1 x - 3 2
1 2x + 1 2 1x - 3 2
x2 - 7x - 2
x2 - 7x - 2
=
=
12x + 12 1x - 3 2
2x2 - 5x - 3
x1 3x - 42 + 1x + 1 2 1x - 1 2
2
3. (a) f(x)=0
2
3x - 4x + x - 1
1x - 1 2 13x - 4 2
1x - 1 2 13x - 42
4x2 - 4x - 1
4x2 - 4x - 1
=
=
1x - 12 13x - 4 2
3x2 - 7x + 4
=
; 1, ;3
1
3
or ;1, ; , ; 3, ;
; 1, ;2
2
2
(b) A graph suggests that –1 and
for zeros.
-1
2
3>2
2
2
1
-2
-1
3
2
x
3
x1 x - 32 - 212x + 1 2
9. (a)
1
6 x 6 7
3
(–)(–) (–) (–)(–)(+) (–)(+)(+) (+)(+)(+)
Negative Positive Negative Positive
–4
7
–1
x3 - 3
6.
x
8.
1
when x=7, - , –4
3
xS –q
x + 5
5.
x
x
(b) f(x) 7 0
4. lim g(x)=q, lim g(x)=–q
xSq
3
when x=–2, –1, 5
-4
1
-3
3
0
3
are good candidates
2
-3
3
0
3
2x3+x2-4x-3=(x+1) a x - b (2x+2)
2
=(x+1)(2x-3)(x+1)
when x=–7, –4, 6
(b) f(x) 7 0
when x 6 –7 or –4 6 x 6 6 or x 7 6
(c) f(x) 6 0
when –7 6 x 6 –4
(–)(–) (–)2 (+)(–)(–)2 (+)(+)(–)2 (+)(+)(+)2
Positive Negative Positive Positive x
–7
–4
6
4. (a) f(x)=0
3
when x= - , 1
5
(b) f(x) 7 0
3
when x 6 - or x 7 1
5
(c) f(x) 6 0
when
(–)(+)(–)
Positive
–3
5
5. (a) f(x)=0
-
3
6 x 6 1
5
(+)(+)(–)
Negative
(+)(+)(+)
Positive
x
1
when x=8, –1
(b) f(x) 7 0
when –1 6 x 6 8 or x 7 8
(c) f(x) 6 0
when x 6 –1
(+)(–)2(–)3 (+)(–)2(+)3 (+)(+)2(+)3
Negative
Positive
Positive
–1
8
x
118
Chapter 2
6. (a) f(x)=0
Polynomial, Power, and Rational Functions
when x=–2, 9
(b) f(x) 7 0
when –2 6 x 6 9 or x 7 9
(c) f(x) 6 0
when x 6 –2
3
(–) (+) (–)4 (+)3(+) (–)4 (+)3(+)(+)4
Negative
Positive
Positive
–2
9
x
7. (x+1)(x-3)2=0 when x=–1, 3
(–)(–)2
Negative
(+)(–)2
Positive
–1
(+)(+)2
Positive
x
3
By the sign chart, the solution of (x+1)(x-3)2 7 0 is
(–1, 3) ª (3, q).
1
4
8. (2x+1)(x-2)(3x-4)=0 when x= - , 2,
2
3
(–)(–) (–) (+)(–)(–) (+)(–)(+) (+)(+)(+)
Negative Positive Negative Positive
4
2
–1
2
x
3
By the sign chart, the solution of
1
4
(2x+1)(x-2)(3x-4) 0 is a - q, - d ª c , 2 d .
2
3
12. By the Rational Zeros Theorem, the possible rational
zeros are ; 1, ; 2, ; 3, ; 6. A graph suggests that –1,
2, and 3 are good candidates to be zeros.
-1
1 -4
1
6
-1
5 -6
2
1 -5
6
0
2 -6
1 -3
0
x3-4x2+x+6=(x+1)(x-2)(x-3)=0
when x=–1, 2, 3.
(–)(–) (–) (+)(–)(–) (+)(+)(–) (+)(+)(+)
Negative Positive Negative Positive
–1
2
3
x
By the sign chart, the solution of
(x+1)(x-2)(x-3) 0 is (–q, –1] ª [2, 3].
13. The zeros of f(x)=x3-x2-2x appear to be –1, 0,
and 2. Substituting these values into f confirms this. The
graph shows that the solution of x3-x2-2x 0 is
[–1, 0] ª [2, q).
9. (x+1)(x2-3x+2)=(x+1)(x-1)(x-2)=0
when x=–1, 1, 2
(–)(–) (–) (+)(–)(–) (+)(+)(–) (+)(+)(+)
Negative Positive Negative Positive
–1
1
2
x
By the sign chart, the solution of
(x+1)(x-1)(x-2) 6 0 is (–q, –1) ª (1, 2).
10. (2x-7)(x2-4x+4)=(2x-7)(x-2)2=0 when
7
x= , 2
2
(–)(–)2
Negative
(–)(+)2
Negative
(+)(+)2
Positive
x
[–5, 5] by [–5, 5]
14. The zeros of f(x)=2x3-5x2+3x appear to be 0, 1,
3
and . Substituting these values into f confirms this.
2
The graph shows that the solution of 2x3-5x2+3x 6 0
3
is (–q, 0) ª a 1, b .
2
7
2
2
By the sign chart, the solution of (2x-7)(x-2)2 7 0
7
is a , qb .
2
11. By the Rational Zeros Theorem, the possible rational
1
3
zeros are ; 1, ; , ; 2, ; 3, ; , ; 6. A graph
2
2
1
suggests that –2, , and 3 are good candidates to be zeros.
2
-2
2 - 3 -11
6
-4
14 -6
3
2 -7
3
0
6
-3
2 -1
0
[–3, 3] by [–5, 5]
15. The zeros of f(x)=2x3-5x2-x+6 appear to be
3
–1, and 2. Substituting these values into f confirms this.
2
The graph shows that the solution of
3
2x3-5x2-x+6 7 0 is a -1, b ª (2, q).
2
2x3-3x2-11x+6=(x+2)(x-3)(2x-1)=0
1
when x=–2, 3,
2
(–)(–) (–) (+)(–)(–) (+)(–)(+) (+)(+)(+)
Negative Positive Negative Positive
1
–2
3
x
2
By the sign chart, the solution of
1
(x+2)(x-3)(2x-1) 0 is c - 2, d ª [3, q).
2
[–3, 3] by [–7, 7]
Section 2.8
Solving Inequalities in One Variable
119
16. The zeros of f(x)=x3-4x2-x+4 appear to be –1,
1, and 4. Substituting these values into f confirms this. The
graph shows that the solution of x3-4x2-x+4 0
is (–q, –1] ª [1, 4].
[–3, 3] by [–3, 23]
21. f(x)=(x2+4)(2x2+3)
(a) The solution is (–q, q), because both factors of f(x)
are always positive.
[–3, 7] by [–10, 10]
17. The only zero of f(x)=3x3-2x2-x+6 is found
graphically to be x≠–1.15. The graph shows that the
solution of 3x3-2x2-x+6 0 is approximately
[–1.15, q).
(b) (–q, q), for the same reason as in part (a).
(c) There are no solutions, because both factors of f(x)
are always positive.
(d) There are no solutions, for the same reason as in
part (c).
22. f(x)=(x2+1)(–2-3x2)
(a) There are no solutions, because x2+1 is always positive and –2-3x2 is always negative.
(b) There are no solutions, for the same reason as in part
(a).
[–3, 3] by [–10, 10]
18. The only zero of f(x)=–x3-3x2-9x+4 is found
graphically to be x≠0.39. The graph shows that the solution of –x3-3x2-9x+4 6 0 is approximately
(0.39, q).
(c) (–q, q), because x2+1 is always positive and
–2-3x2 is always negative.
(d) (–q, q), for the same reason as in part (c).
23. f(x)=(2x2-2x+5)(3x-4)2
The first factor is always positive because the leading
term has a positive coefficient and the discriminant
(–2)2-4(2)(5)=–36 is negative. The only zero is
x=4/3, with multiplicity two, since that is the solution
for 3x-4=0.
(a) True for all x Z
[–5, 5] by [–10, 10]
4
3
(b) (–q, q)
19. The zeros of f(x)=2x4-3x3-6x2+5x+6 appear
3
to be –1, , and 2. Substituting these into f confirms this.
2
The graph shows that the solution of
3
2x4-3x3-6x2+ 5x+6 6 0 is a , 2 b .
2
(c) There are no solutions.
(d) x =
4
3
24. f(x)=(x2+4)(3-2x)2
The first factor is always positive. The only zero is
x=3/2, with multiplicity two, since that is the solution
for 3-2x=0.
(a) True for all x Z
3
2
(b) (–q, q)
(c) There are no solutions.
[–5, 5] by [–10, 10]
(d) x =
4
3
2
20. The zero of f(x)=3x -5x -12x +12x+16
4
appear to be - ,–1, and 2. Substituting these into f
3
confirms this. The graph shows that the solution of
3x4-5x3-12x2+12x+16 0 is
4
a - q, - d ª [–1, q).
3
3
2
25. (a) f(x)=0 when x=1
3
(b) f(x) is undefined when x= - , 4
2
(c) f(x) 7 0 when -
3
6 x 6 1 or x 7 4
2
3
(d) f(x) 6 0 when x 6 - or 1 6 x 6 4
2
Negative
1
(b) f(x) is undefined when x = 4, x -2
(+)
(+)(+)
Positive
(c) f(x) 7 0 when -2 6 x 6 1 or x 7 4
(d) f(x) 6 0 when 1 6 x 6 4
x
4
7
26. (a) f(x)=0 when x= , –1
2
(c) f(x) 7 0 when - 5 6 x 6 -1 or x 7
7
2
(d) f(x) 6 0 when x 6 -5 or - 1 6 x 6
7
2
undefined
(+)
(–)(+)
x
31. (a) f(x)=0 when x=3
(b) f(x) is undefined when x = 4, x 6 3
(c) f(x) 7 0 when 3 6 x 6 4 or x 7 4
(d) None. f(x) is never negative
(–)(–)
(+)
0 (–)(+) 0 (+)(+)
(+)
(+)
Negative Positive Negative Positive
7
–5
–1
(+)(+)
(–)2
0
x
Undefined
2
Positive
3
27. (a) f(x)=0 when x=0, –3
(+)(+)
(+)2
32. (a) None. f(x)is never 0
(b) f(x) is undefined when x 5
(c) f(x) 7 0 when x 7 0
(c) f(x)>0 when 5 6 x 6 q
(d) f(x) 6 0 when - 3 6 x 6 0
(d) None. f(x) is never negative
(–)(+)
(+)(+)
Undefined Negative
Positive
–3
0
x
9
2
Undefined
(b) None. f(x) is never undefined
x - 1
x - 1
=
has points of
2
1x + 22 1x - 2 2
x - 4
potential sign change at x=–2, 1, 2.
x
29. (a) f(x)=0 when x=–5
1
(b) f(x) is undefined when x = - , x = 1, x 6 -5
2
1
(c) f(x) 7 0 when - 5 6 x 6 - or x 7 1
2
undefined
undefined
1
6 x 6 1
2
2
(+)
(+)(–)
(+)
(+)(+)
By the sign chart, the solution of
x
x
x - 1
6 0 is
x2 - 4
(–q, –2) ª (1, 2).
x + 2
x + 2
=
has points of
1x + 32 1x - 3 2
x2 - 9
potential sign change at x=–3, –2, 3.
34. f(x)=
(–)
(–)(–)
(+)
(+)(+)
Undefined Positive Negative Positive
–5
1
–1
(–)
0
(+)(–)
Negative Positive Negative Positive
–2
1
2
2
(+)
(+)(–)
(–)
(–)(–)
(–)
0
(+)(–)
(+)
(+)(–)
undefined
(–)2 |–| 0 (–)2 |+| 0 (+)2 |+|
Positive
Positive
Positive
0
–9
undefined
(d) None. f(x) is never negative
(+)
(–)(–)
x
Positive
33. f(x)=
9
(c) f(x) 7 0 when x Z - , 0
2
(d) f(x) 6 0 when -
(+)
(+)(+)
5
undefined
28. (a) f(x)=0 when x=0, -
undefined
0
undefined
0
x
Positive
4
(b) f(x) is undefined when x 6 –3
0
(+)
(+)(+)
Undefined Positive Negative Positive
–2
1
4
(b) f(x) is undefined when x=–5
(–)(–)
(–)
(–)
0
(–)(+)
undefined
Positive
–3
2
(+)
(+)(–)
undefined
Negative
(–)
0
(+)(–)
30. (a) f(x)=0 when x=1
undefined
(–)
(–)(–)
Polynomial, Power, and Rational Functions
undefined
Chapter 2
undefined
120
(+)
(+)(+)
Negative Positive Negative Positive
–3
–2
3
By the sign chart, the solution of
(–q, –3) ª (–2, 3).
x
x + 2
6 0 is
x2 - 9
Section 2.8
1x + 12 1x - 1 2
x2 - 1
=
has points of
2
x + 1
1x2 + 1 2
potential sign change at x=–1, 1.
35. f(x)=
(+)(–)
(+)
0
Positive
0
Negative
–1
(+)(+)
(+)
x
Positive
Negative Positive Negative Positive
–2
0
2
x2 - 1
0 is [–1, 1].
x2 + 1
1x + 22 1x - 2 2
x2 - 4
=
has points of
2
x + 4
x2 + 4
potential sign change at x=–2, 2.
36. f(x)=
(–)(–)
(+)
(+)(–)
(+)
0
Positive
0
Negative
–2
x
Positive
2
1x + 42 1x - 3 2
x + x - 12
=
x2 - 4x + 4
1x - 2 2 2
potential sign change at x=–4, 2, 3.
(–)(–) 0
(–)2
(+)(–)
(–)2
undefined
37. f(x)=
(+)(–) 0
(+)2
has points of
x
undefined
(+)(+)
(–)2
x + x - 12
7 0 is
x2 - 4x + 4
x
x2 + 3x - 10
6 0 is
x2 - 6x + 9
(–5, 2).
x 1 x + 1 2 1x - 1 2
x3 - x
has points of
=
2
x + 1
x2 + 1
potential sign change at x=–1, 0, 1.
39. f(x)=
By the sign chart, the solution of
Positive
x3 - x
0 is
x2 + 1
x - 3
6 0 is
0x + 20
0
(–)(+)
Negative
(+)(+)
Positive
x
1
2
By the sign chart, the solution of 12x - 1 2 1x + 4 6 0 is
1
a -4, b .
2
44. f(x)= 1 3x - 4 2 12x + 1 has a point of potential sign
1
4
change at x= . Note that the domain of f is c - , q b .
3
2
0
0
(–)(+)
Negative
–1
x
x
43. f(x)= 1 2x - 1 2 1x + 4 has a point of potential sign
1
change at x= . Note that the domain of f is [–4, q).
2
Undefined
(–)(–)(–) 0 (–)(+)(–) 0 (+)(+)(–) 0 (+)(+)(+)
(+)
(+)
(+)
(+)
Negative Positive Negative Positive
–1
0
1
(+)
|+|
3
(–q, –2) ª (–2, 3).
Undefined
–4
(+)(+)
(+)2
By the sign chart, the solution of
0
Negative
–2
0
Positive Negative Positive Positive
–5
2
3
[–1, 0] ª [1, q).
(–)
|+|
By the sign chart, the solution of
1x + 52 1x - 2 2
x2 + 3x - 10
38. f(x)= 2
has points of
=
x - 6x + 9
1x - 3 2 2
potential sign change at x=–5, 2, 3.
(+)(–) 0
(–)2
x
2
x - 3
has points of potential sign change at
0x + 20
x=–2, 3.
Negative
(–q, –4) ª (3, q).
(–)(–) 0
(–)2
(+) |+|
Positive
By the sign chart, the solution of x 0 x - 2 0 7 0 is
(0, 2) ª (2, q).
(–)
|–|
2
By the sign chart, the solution of
0
(+) |–|
Positive
42. f(x)=
(+)(+)
(+)2
Positive Negative Negative Positive
–4
2
3
x3 - 4x
0 is
x2 + 2
41. f(x)=x 0 x - 2 0 has points of potential sign change at
x=0, 2.
0
x2 - 4
7 0 is
x2 + 4
x
(–q, –2] ª [0, 2].
(–) |–|
Negative
2
(–q, –2) ª (2, q).
By the sign chart, the solution of
0
(+)(+)
(+)
By the sign chart, the solution of
x1 x + 2 2 1x - 22
x3 - 4x
has points of
=
2
x + 2
x2 + 2
potential sign change at x=–2, 0, 2.
(–)(–)(–) 0 (–)(+)(–) 0 (+)(+)(–) 0 (+)(+)(+)
(+)
(+)
(+)
(+)
1
By the sign chart, the solution of
121
40. f(x)=
undefined
(–)(–)
(+)
Solving Inequalities in One Variable
2
(+)(+)
Positive
x
4
3
By the sign chart, the solution of 13x - 4 2 12x + 1 0
4
is c , q b .
3
Polynomial, Power, and Rational Functions
By the sign chart, the solution of
1x - 5 2 4
(–)4
(–)(–)
undefined
x1x + 32
at x=–3, 0, 5.
(–)4
(–)(+)
x3 1x - 2 2
1x + 32
(–)4 0
(+)(+)
2
6 0 is (0, 2).
By the sign chart, the solution of
1x - 5 2 4
x 1x + 32
0 is
2
x3 - 2
47. f(x)=x has points of potential sign
=
x
x
3
change at x=0, 12.
undefined
Positive
Negative
0
x
2
By the sign chart, the solution of x2 -
2
7 0 is
x
3
(–q, 0) ª ( 22
, q).
4
x + 4
has points of potential sign
=
x
x
3
change at x=0, - 14.
Positive
(+)
(–)
undefined
48. f(x)=x2 +
0
(–)
(–)(–)
(–)
(+)(–)
Negative
3
– 4
(–q, –5) ª (–2, 1).
51. f(x)= 1 x + 3 2 0 x - 1 0 has points of potential sign
change at x=–3, 1.
0
(–) |–|
Negative
0
(+) |–|
Positive
3
(–q, - 14
] ª (0, q).
(+) |+|
Positive
x
1
52. f(x)= 1 3x + 5 2 2 0 x - 2 0 is always 0 or positive since
1 3x + 5 2 2 0 for all real x and 0 x - 2 0 0 for all
real x. Thus the inequality 13x + 5 2 2 0 x - 2 0 6 0 has no
solution.
1x - 5 2 0 x - 2 0
has points of potential sign
12x - 2
change at x=2, 5. Note that the domain of f is 1 1, q 2 .
(–)|–|
(+)
0
(–)|+|
(+)
0
(+)|+|
(+)
x
By the sign chart, the solution of
0
By the sign chart, the solution of x2 +
x
1
2
7 0 is
x + 2
x - 1
Undefined Negative Negative Positive
1
2
5
(+)
(+)
Positive
(–)
(+)(+)
By the sign chart, the solution of
53. f(x)=
3
(–)
(–)
1
2
-x - 5
has points of
=
x + 2
x - 1
1x + 22 1x - 12
potential sign change at x=–5, –2, 1.
By the sign chart, the solution of 1x + 3 2 0 x - 1 0 0 is
[–3, q).
(+)
(+)
Positive
3
1
1
+
0 is
x + 1
x - 3
50. f(x)=
–3
0
x
Positive Negative Positive Negative
–5
–2
1
2
(–)
(+)
(+)
(+)(+)
By the sign chart, the solution of
(+) 0
(–)(–)
x
(–q, –3) ª (0, q).
(–)
(–)
(+)
(+)(–)
(–q, –1) ª [1, 3).
(+)4
(+)(+)
Positive Negative Positive Positive
–3
0
5
(–)
0
(+)(–)
Negative Positive Negative Positive
–1
1
3
x
0 has points of potential sign change
undefined
46. f(x)=
(–)
(–)(–)
(–)3(–) 0 (+)3(–) 0 (+)3(+)
(+)2
(+)2
(+)2
Positive Positive Negative Positive
–3
0
2
2 1x - 12
1
1
has points of
+
=
x + 1
x - 3
1x + 12 1x - 32
potential sign change at x=–1, 1, 3.
49. f(x)=
undefined
has points of potential sign change at
undefined
undefined
(–)3(–)
(–)2
2
undefined
x3 1x - 2 2
1x + 32
x=–3, 0, 2.
45. f(x)=
undefined
Chapter 2
undefined
122
4
0 is
x
is [5, q).
x
1x - 5 2 0 x - 2 0
12x - 2
0
Section 2.8
54. f(x)=
x2 1x - 4 2 3
has points of potential sign change at
undefined
1x + 1
x=0, 4. Note that the domain of f is (–1, q).
By the sign chart, the solution of
x
x2 1x - 4 2 3
1x + 1
123
60. The circumference of the base of the cone is 8p-x,
r=4-
x
x 2
, and h= 16 - a 4 b . The volume
2p
B
2p
1
x 2
x 2
is v= p a 4 b
b .
16 - a 4 3
2p B
2p
To solve v 21, graph v-21 and find the zeros:
x≠1.68 in. or x≠9.10.
(–)2(–)3 0 (+)2(–)3 0 (+)2(+)3
(+)
(+)
(+)
Undefined Negative Negative Positive
–1
0
4
Solving Inequalities in One Variable
6 0 is
(–1, 0) ª (0, 4).
55. One way to solve the inequality is to graph
y=3(x-1)+2 and y=5x+6 together, then find
the interval along the x-axis where the first graph is below
or intersects the second graph. Another way is to solve for
x algebraically.
56. Let x be the number of hours worked. The repair charge
is 25+18x; this must be less than $100. Starting with
25+18x 6 100, we have 18x 6 75, so x 6 4.166»
Therefore, the electrician worked no more than 4 hours
7.5 minutes (which rounds to 4 hours).
57. Let x 7 0 be the width of a rectangle; then the length is
2x-2 and the perimeter is P=2[x+(2x-2)].
Solving P 6 200 and 2x-2 7 0 (below) gives
1 in. 6 x 6 34 in.
2[x+(2x-2)] 6 200 and 2x-2 7 0
2(3x-2) 6 200
2x 7 2
6x-4 6 200
x 7 1
6x 6 204
x 6 34
[0, 26] by [–25, 25]
From the graph, the solution of v-21 0 is approximately [1.68, 9.10]. The arc length should be in the range
of 1.68 in. x 9.10 in.
1
61. (a) L = 500 cm3
2
500
px2
500
S = 2pxh + 2px2 = 2px a 2 b + 2px2
px
1000
1000 + 2px3
+ 2px2 =
=
x
x
V = px2h = 500 1 h =
(b) Solve S 6 900 by graphing
finding its zeros:
x≠1.12 and x≠11.37
58. Let x be the number of candy bars made. Then the costs
are C=0.13x+2000, and the income is I=0.35x.
Solving C 6 I (below) gives x 7 9090.91. The company
will need to make and sell 9091 candy bars to make a
profit.
0.13x+2000 6 0.35x
2000 6 0.22x
x 7 9090.91
59. The lengths of the sides of the box are x, 12-2x, and
15-2x, so the volume is x(12-2x)(15-2x). To solve
x(12-2x)(15-2x) 100, graph
f(x)=x(12-2x)(15-2x)-100 and find the zeros:
x≠0.69 and x≠4.20.
[0, 15] by [–1000, 1000]
From the graph, the solution of S-900 6 0 is
approximately (1.12, 11.37). So the radius is between
1.12 cm and 11.37 cm. The corresponding height must
be between 1.23 cm and 126.88 cm.
(c) Graph S and find the minimum graphically.
[0, 15] by [0, 1000]
[0, 6] by [–100, 100]
From the graph, the solution of f(x) 0 is approximately [0, 0.69] ª [4.20, 6]. The squares should be such that
either 0 in. x 0.69 in. or 4.20 in. x 6 in.
1000 + 2px3
- 900 and
x
The minimum surface area is about 348.73 cm2.
62. (a)
1
1
1
=
+
R
2.3
x
2.3x=Rx+2.3R=R(x+2.3)
2.3x
R =
x + 2.3
(b) R 1.7 1
Polynomial, Power, and Rational Functions
2.3x
1.7
x + 2.3
2.3x
- 1.7 0
x + 2.3
2.3x - 1.7 1x + 2.3 2
70. The expression (x2-1)2 cannot be negative for any real x,
and it can equal zero only for x=—1. The answer is A.
undefined
0
x + 2.3
0.6x - 3.91
0
x + 2.3
0.6x - 3.91
The function f(x)=
has a point of
x + 2.3
391
potential sign change at x=
L 6.5. Note that the
60
domain of f is (–2.3, q).
1x - 1 2 1x + 22 2
1x - 32 1x + 12
Vertical asymptotes: x=–1, x=3
x-intercepts: (–2, 0), (1, 0)
4
y-intercept: a 0, b
3
71. f(x)=
(–)(–)2 0 (–)(+)2
(–)(–)
(–)(–)
(–)(+)2 0 (+)(+)2
(–)(+)
(–)(+)
undefined
Chapter 2
undefined
124
(+)(+)2
(+)(+)
Negative Negative Positive Negative Positive x
–2
–1
1
3
By hand:
(–)
(+)
0
(+)
(–)
Undefined Negative
Positive
391 6.5
–2.3
y
30
x
60
By the sign chart, the solution of f(x) 0 is about
[6.5, q). The resistance in the second resistor is at
least 6.5 ohms.
–10
5
63. (a) y≠993.870x+19,025.768
10
x
–30
Grapher:
5
[0 25] b [0 400 000]
(b) From the graph of y L 7.883x3 - 214.972x2 +
6479.797x + 62,862.278, we find that y = 250,000
when x L 28. The per capita income will exceed
$250,000 in the year 2008.
65. False. Because the factor x4 has an even power, it does not
change sign at x=0.
72. g(x)=
1x - 32 4
1x - 32 4
(–)4
(–)(–)
(–)4
(–)(+)
(–)4 0
(+)(+)
Sketch:
y
600
67. x must be positive but less than 1. The answer is C.
69. The statement is true so long as the denominator is negative and the numerator is nonzero. Thus x must be less
than three but nonzero. The answer is D.
(+)4
(+)(+)
Positive Negative Positive Positive
–4
0
3
66. True. Because the denominator factor (x+2) has an odd
power (namely 1), it changes sign at x=–2.
68. The statement is true so long as the numerator does not
equal zero. The answer is B.
[0, 10] by [–40, 40]
=
x1x + 42
x2 + 4x
Vertical asymptotes: x=–4, x=0
x-intercept: (3, 0)
y-intercept: None
undefined
64. (a) y L 7.883x3 - 214.972x2 + 6479.797x + 62,862.278
[–5, 5] by [–5, 5]
undefined
5 b
(b) From the graph of y = 993.870x + 19,025.768, we
find that y = 40,000 when x L 21.2. The per capita
income will exceed $40,000 in the year 2011.
10
x
x
Chapter 2
Grapher:
Review
125
3. Starting from y=x2, translate right 2 units and vertically
stretch by 3 (either order), then translate up 4 units.
y
10
[–10, 10] by [–10, 10]
[–20, 0] by [–1000, 1000]
73. (a) 0 x - 3 0 6 1>3 1 0 3x - 9 0 6 1 1
0 3x - 5 - 4 0 6 1 1 0 f1 x2 - 4 0 6 1.
For example:
0 f1 x2 - 4 0 = 0 1 3x - 52 - 4 0 = 0 3x - 9 0
1
=3 0 x - 3 0 6 3 a b = 1
3
(b) If x stays within the dashed vertical lines, f(x) will stay
within the dashed horizontal lines. For the example in
part (a), the graph shows that for
1
8
10
a that is, 0 x - 3 0 6 b , we have
6 x 6
3
3
3
3 6 f1x2 6 5 (that is, 0 f1x2 - 4 0 6 1).
(c) 0 x - 3 0 6 0.01 1 0 3x - 9 0 6 0.03 1
0 3x - 5 - 4 0 6 0.03 1 0 f1 x2 - 4 0 6 0.03. The
dashed lines would be closer to x = 3 and y = 4.
6
x
4. Starting from y=x2, translate left 3 units and reflect
across x-axis (either order), then translate up 1 unit.
y
10
10
x
74. When x2 - 4 0, y = 1, and when x2-4 0, y=0.
75. One possible answer: Given 0 6 a 6 b, multiplying both
sides of a 6 b by a gives a2 6 ab; multiplying by b gives
ab 6 b2. Then, by the transitive property of inequality, we
have a2 6 b2.
76. One possible answer: Given 0 6 a 6 b, multiplying both
1
1
1
sides of a 6 b by
gives 6 , which is equivalent to
ab
b
a
1
1
7 .
a
b
■ Chapter 2 Review
For #1 and 2, first find the slope of the line. Then use algebra
to put into y=mx+b format.
-9 - 1 - 2 2
-7
1. m=
=
= -1 , (y+9)=–1(x-4),
4 - 1 -3 2
7
y=–x-5
[–15, 5] by [–15, 5]
-2 - 6
-8
=
= -2 , (y+2)=–2(x-1),
1 - 1 -32
4
y=–2x
2. m=
[–5, 5] by [–5, 5]
5. Vertex: (–3, 5); axis: x=–3
6. Vertex: (5, –7); axis: x=5
7. f(x)=–2(x2+8x)-31
=–2(x2+8x+16)+32-31=–2(x+4)2+1;
Vertex: (–4, 1); axis: x=–4
8. g(x)=3(x2-2x)+2=3(x2-2x+1)-3+2=
3(x-1)2 -1; Vertex: (1, –1); axis: x=1
For #9–12, use the form y=a(x-h)2+k, where (h, k), the
vertex, is given.
9. h=–2 and k=–3 are given, so y=a(x+2)2-3.
5
Using the point (1, 2), we have 2=9a-3, so a= :
9
5
y= (x+2)2-3.
9
10. h=–1 and k=1 are given, so y=a(x+1)2+1.
Using the point (3, –2), we have –2=16a+1, so
3
3
a=– : y=– (x+1)2+1.
16
16
11. h=3 and k=–2 are given, so y=a(x-3)2-2.
1
Using the point (5, 0), we have 0=4a-2, so a= :
2
1
y= (x-3)2-2.
2
12. h=–4 and k=5 are given, so y=a(x+4)2+5.
Using the point (0, –3), we have –3=16a+5,
1
1
so a=– ; y=– (x+4)–2+5.
2
2
126
Chapter 2
Polynomial, Power, and Rational Functions
3
22. k=–2, a= . In Quadrant IV, f(x) is decreasing and
4
concave up since 0<a<1 . f is not defined for x<0.
13.
[–10, 7] by [–50, 10]
14.
[–1, 9] by [–10, 10]
23. k=–2, a=–3. In Quadrant IV, f is increasing
-2
-2
and concave down. f(–x)=–2(–x)–3=
=
1 -x2 3
- x3
2
= 3 = 2x–3=–f(x), so f is odd.
x
[–2, 4] by [–50, 10]
15.
[–4, 3] by [–30, 30]
[–5, 5] by [–5, 5]
16.
2
24. k= , a=–4. In Quadrant I, f(x) is decreasing and
3
2 #
1
2
concave up. f(–x)= 1 -x2 -4 =
3
3 1 -x2 4
2
2
= 4 = x-4 = f(x), so f is even.
3
3x
[–6, 7] by [–50, 30]
17. S=kr2
(k=4∏)
k
(k=gravitational constant)
d2
19. The force F needed varies directly with the distance x
from its resting position, with constant of variation k.
18. F=
20. The area of a circle A varies directly with the square of its
radius.
1
21. k=4, a= . In Quadrant I, f(x) is increasing and
3
concave down since 0<a<1.
[–3, 3] by [–1, 4]
25.
2
2x3 - 7x2 + 4x - 5
= 2x2 - x + 1 x - 3
x - 3
2x2- x +1
x-3R 2x3-7x2+4x-5
2x3-6x2
-x2+4x
-x2+3x
x-5
x-3
–2
[–10, 10] by [–10, 10]
1/3
1/3
f(–x)=4(–x) =–4x =–f(x), so f is odd.
Chapter 2
26.
5
x4 + 3x3 + x2 - 3x + 3
= x3 + x2 - x - 1 +
x + 2
x + 2
x3+ x2- x -1
x+2R x4+3x3+ x2-3x+3
x4+2x3
35. –3|
4
2
x3+2x2
2
–x -3x
–x2-2x
–x+3
–x-2
5
27.
2x4 - 3x3 + 9x2 - 14x + 7
x2 + 4
-2x + 3
= 2x2 - 3x + 1 +
x2 + 4
x +4R 2x -3x +9x -14x+7
4
3
2x4
2
+8x2
–3x3+ x2-14x+7
–3x3
-12x
x2-2x +7
x2
+4
–2x +3
-7
3x4 - 5x3 - 2x2 + 3x - 6
28.
= x3 - 2x2 + 1 +
3x + 1
3x + 1
x3-2x2
+1
3x+1R 3x4-5x3-2x2+3x-6
3x4+ x3
–6x3-2x2+3x-6
–6x3-2x2
3x-6
3x+1
-7
29. Remainder: f(–2)=–39
31. Yes: 2 is a zero of the second polynomial.
32. No: x=–3 yields 1 from the second polynomial.
1
1
–5
3
4
4
–12
24
–27
132
6
1
–6
–6
0
–3
2
0
1
–9
Yes, x=–3 is a lower bound for the zeros of f(x)
because all entries on the bottom row alternate signs
(remember that 0=–0).
;1, ;2, ;3, ; 6
,
;1, ;2
1 3 3
or —1, —2, —3, —6, — , — ; – and 2 are zeros.
2 2 2
37. Possible rational zeros:
;1, ;7
,
;1, ;2, ;3, ; 6
1 7 1 7 1 7 7
or —1, —7, — , — , — , — , — , — ; is a zero.
2 2 3 3 6 6 3
39. (1+i)3=(1+2i+i2)(1+i)=(2i)(1+i)
=–2+2i
40. (1+2i)2(1-2i)2=[(1+2i)(1-2i)]2=(1+22)2
=25
41. i29=i
42. 1-16 = 4i
For #43–44, use the quadratic formula.
43. x =
6 ; 136 - 52
6 ; 4i
=
= 3 ; 2i
2
2
44. x =
2 ; 14 - 16
2 ; 213 i
=
= 1 ; 13 i
2
2
45. (c) f(x)=(x-2)2 is a quadratic polynomial that has
vertex (2, 0) and y-intercept (0, 4), so its graph must be
graph (c).
46. (d) f(x)=(x-2)3 is a cubic polynomial that passes
through (2, 0) and (0, –8), so its graph must be graph (d).
47. (b) f(x)=(x-2)4 is a quartic polynomial that passes
through (2, 0) and (0, 16), so its graph must be graph (b).
5
0
15
0
3
19
In #49–52, use a graph and the Rational Zeros Test to determine zeros.
49. Rational: 0 (multiplicity 2) — easily seen by inspection.
Irrational: 5 — 12 (using the quadratic formula, after taking out a factor of x2). No non-real zeros.
Yes, x=5 is an upper bound for the zeros of f(x)
because all entries on the bottom row are
nonnegative.
34. 4 |
–2
48. (a) f(x)=(x-2)5 is a quintic polynomial that passes
through (2, 0) and (0, –32), so its graph must be graph (a).
30. Remainder: f(3)=–2
33. 5 |
–17
38. Possible rational zeros:
2x2-3x +1
2
–15
127
4
–8
9
–44
130
Yes, x=–3 is a lower bound for the zeros of f(x)
because all entries on the bottom row alternate signs.
36. –3|
x3+ x2
4
Review
–16
8
16
–12
16
0
32
192
4
0
8
48
180
Yes, x=4 is an upper bound for the zeros of f(x)
because all entries on the bottom row are nonnegative.
50. Rational: —2. Irrational: — 13 . No nonreal zeros. These
zeros may be estimated from a graph, or by dividing k(t)
by t-2 and t+2 then applying the quadratic formula,
or by using the quadratic formula on k(t) to determine
that t2 =
7 ; 149 - 48
, i.e., t2 is 3 or 4.
2
51. Rational: none. Irrational: approximately –2.34, 0.57, 3.77.
No non-real zeros.
128
Chapter 2
Polynomial, Power, and Rational Functions
52. Rational: none. Irrational: approximately –3.97, –0.19.
Two non-real zeros.
2
1
12 ; 1144 + 36
= ; 15 . Then
18
3
3
f(x)=(x+1)(9x2-12x-1).
–1 | 9 –3 –13 –1
3
3
53. The only rational zero is - . Dividing by x +
2
2
(below) leaves 2x2-12x+20, which has zeros
12 ; 1144 - 160
= 3 ; i . Therefore
4
f(x)=(2x+3)[x-(3-i)][x-(3+i)]
=(2x+3)(x-3+i)(x-3-i).
–3/2 | 2
2
–9
2
30
–3
18
–30
–12
20
0
12
4
–16
–12
5
–20
–15
0
9 –12
–1
0
2 –7
23 –31
–7
15
3/2 | 2 –7
16 –15
16 –15
3 –6
16 –15
2 –4
0
10
15
0
2
60. The two real zeros are –1 and - ; dividing by x+1 and
3
2
x+ leaves the quadratic factor 3x2-12x+15, so
3
f(x)=(3x+2)(x+1)(x2-4x+5).
=(5x-4)Qx-2- 27R Qx-2+ 27R .
1
1
2
f(x)=(5x-4)Sx-Q2+ 27RT Sx-Q2- 27RT
–24
12
1 | 2 –9
20 ; 1400 + 300
= 2 ; 17 . Therefore
10
5
–9
3
59. The two real zeros are 1 and ; dividing by x-1 and
2
3
x- leaves the quadratic factor 2x2-4x+10, so
2
f(x)=(2x-3)(x-1)(x2-2x+5).
4
4
54. The only rational zero is . Dividing by x5
5
(below) leaves 5x2-20x-15, which has zeros
4/5 |
58. The only rational zero is –1; dividing by x+1
leaves the quadratic factor 9x2-12x-1, which has zeros
–1| 3
–7 –3
–3
17
10
–2/3| 3 –10
–2
10 –7 –10
7
10
8 –10
2
5
55. All zeros are rational: 1, –1, , and – . Therefore
3
2
f(x)=(3x-2)(2x+5)(x-1)(x+1); this can be
confirmed by multiplying out the terms or graphing the
original function and the factored form of the function.
61. (x- 15)(x+ 15)(x-3)=x -3x -5x+15.
Other answers may be found by multiplying this
polynomial by any real number.
56. Since all coefficients are real, 1-2i is also a zero.
Dividing synthetically twice leaves the quadratic
x2-6x+10, which has zeros 3 ; i.
f(x)=[x-(1+2i)][x-(1-2i)][x-(3+i)]
[x-(3-i)]=(x-1-2i)(x-1+2i)
(x-3-i)(x-3+i)
63. (x-3)(x+2)(3x-1)(2x+1)
=6x4-5x3-38x2-5x+6 (This may be multiplied
by any real number.)
1+2i| 1
–8
27
–50
50
1+2i
1
–7+2i
–11-12i
40+20i
–50
16-12i
–10+20i
0
1-2i| 1
–7+2i
16-12i
–10+20i
1-2i
–6+12i
10-20i
–6
10
0
1
In #57–60, determine rational zeros (graphically or otherwise)
and divide synthetically until a quadratic remains. If more real
zeros remain, use the quadratic formula.
57. The only real zero is 2; dividing by x-2 leaves
the quadratic factor x2+x+1, so
f(x)=(x-2)(x2+x+1).
2 | 1 –1 –1 –2
1
2
2
2
1
1
0
3 –10
7
10
3 –12 15
0
3
0
2
62. (x+3)2=x2+6x+9 (This may be multiplied by
any real number.)
64. The third zero must be 1-i:
(x-2)(x-1-i)(x-1+i)=x3-4x2+6x-4
(This may be multiplied by any real number.)
65. (x+2)2(x-4)2=x4-4x3-12x2+32x+64
(This may be multiplied by any real number.)
66. The third zero must be 2+i, so
f(x)=a(x+1)(x-2-i)(x-2+i).
Since f(2)=6, a=2:
f(x)=2(x+1)(x-2-i)(x-2+i)
=2x3-6x2+2x+10.
2
; translate right 5 units and vertically
x - 5
stretch by 2 (either order), then translate down 1 unit.
Horizontal asymptote: y=–1; vertical asymptote: x=5.
67. f(x)=–1+
1
; translate left 2 units and reflect
x - 2
across x-axis (either order), then translate up 3 units.
Horizontal asymptote: y=3; vertical asymptote: x=–2.
68. f(x)=3-
Chapter 2
Review
129
69. Asymptotes: y=1, x=–1, and x=1.
Intercept: (0, –1).
[–10, 10] by [–10, 20]
[–5, 5] by [–5, 5]
70. Asymptotes: y=2, x=–3, and x=2.
7
Intercept: a 0, - b .
6
[–10, 10] by [–10, 10]
71. End-behavior asymptote: y=x-7.
5
Vertical asymptote: x=–3. Intercept: a 0, b .
3
5
y-intercept: a 0, b
2
x-intercept: (–2.55, 0)
Domain: All x Z -2
Range: (–q, q)
Continuity: All x Z -2
Increasing on [0.82, q)
Decreasing on (–q, –2), (–2, 0.82]
Not symmetric.
Unbounded.
Local minimum: (0.82, 1.63)
No horizontal asymptote. End-behavior
asymptote: y=x2-x
Vertical asymptote: x=–2.
End behavior: lim f1 x2 = lim f1x2 = q
xS - q
xSq
-x4 + x2 + 1
74. f1 x2 =
has two x-intercepts, and we
x - 1
can use the graph to show that they are about –1.27 and
1.27. The y-intercept is f(0)=–1. The denominator is
zero when x=1, so the vertical asymptote is x=1.
Because we can rewrite f(x) as
[–7, 3] by [–50, 30]
72. End-behavior asymptote: y=x-6.
Vertical asymptote: x=–3. Intercepts: approx.
7
(–1.54, 0), (4.54, 0), and a 0, - b .
3
[–15, 10] by [–30, 10]
x3 + x2 - 2x + 5
has only one x-intercept,
x + 2
and we can use the graph to show that it is about –2.552.
The y-intercept is f(0)=5/2. The denominator is zero
when x=–2, so the vertical asymptote is x=–2.
Because we can rewrite f(x) as
5
x3 + x2 - 2x + 5
= x2 - x +
,
f1 x2 =
x + 2
x + 2
2
we know that the end-behavior asymptote is y=x -x.
The graph supports this information and allows us to
conclude that
lim - = - q, lim + = q.
73. f1 x2 =
xS - 2
xS - 2
The graph also shows a local minimum of about 1.63 at
about x=0.82.
f1 x2 =
-x4 + x2 + 1
1
= -x3 - x2 +
,
x - 1
x - 1
we know that the end-behavior asymptote is
y=–x3-x¤. The graph supports this information and
allows us to conclude that lim- = - q and lim+ = q.
xS1
xS1
The graph shows no local extrema.
[–4.7, 4.7] by [–10, 10]
y-intercept: (0, 1)
x-intercepts: (–1.27, 0), (1.27, 0)
Domain: All x Z 1
Range: (–q, q)
Continuity: All x Z 1
Never increasing
Decreasing on (–q, 1), (1, q)
No symmetry.
Unbounded.
No local extrema.
No horizontal asymptote. End-behavior asymptote:
y=–x‹-x¤
Vertical asymptote: x=1
End behavior: lim f1 x2 = q; lim f1x2 = - q
xS - q
xSq
Chapter 2
130
Polynomial, Power, and Rational Functions
3
75. Multiply by x: 2x2-11x+12=0, so x= or x=4.
2
76. Multiply by (x+2)(x-3)=x2-x-6:
x(x-3)+5(x+2)=25, or x2+2x-15=0,
so x=–5 or x=3. The latter is extraneous; the only
solution is x=–5.
For #77–78, find the zeros of f(x) and then determine where
the function is positive or negative by creating a sign chart.
77. f(x)=(x-3)(2x+5)(x+2), so the zeros of f(x)
5
are x= e - , -2, 3 f .
2
lower bound (because all numbers on the bottom row
alternate signs). Yes, there is another zero (at x L 10.0002).
5|
1
–10
–3
28
20
–2
–5 |
5
–25
–140
–560
–2700
1
–5
–28
–112
–540
–2702
1
–10
–3
28
20
–2
–5
75
–360
1660
–8400
1
–15
72
–332
1680
–8402
84. (a) h=–16t2+170t+6
(–)(–) (–) (–)(+)(–) (–)(+)(+) (+)(+)(+)
x
Negative Positive Negative Positive
–2
3
–5
2
As our sign chart indicates, f(x)<0 on the interval
5
a - q, - b ª (–2, 3).
2
78. f(x)=(x-2)2(x+4)(3x+1), so the zeros of f(x)
1
are x= e -4, - , 2 f .
3
(+)(–) (–) (+)(+)(–) (+)(+)(+) (+)(+)(+)
x
Positive Negative Positive Positive
–4
2
–1
3
[0, 11] by [0, 500]
(b) When t≠5.3125, h≠457.5625.
(c) The rock will hit the ground after about 10.66 sec.
85. (a) V=(height)(width)(length)
=x(30-2x)(70-2x) in.3
(b) Either x≠4.57 or x≠8.63 in.
86. (a) & (b)
As our sign chart indicates, f(x) 0 on the interval
1
(–q, –4] ª c - , q b .
3
79. Zeros of numerator and denominator: –3, –2, and 2.
x + 3
Choose –4, –2.5, 0, and 3; 2
is positive at –2.5 and
x - 4
3, and equals 0 at –3, so the solution is [–3, –2) ª (2, q).
80.
x - 1
x2 - 7
- 1 = 2
. Zeros of numerator and
x - x - 6
x - x - 6
denominator: –2, 1, and 3. Choose –3, 0, 2, and 4;
x - 1
is negative at –3 and 2, so the solution
x2 - x - 6
is (–q, –2) ª (1, 3).
2
81. Since the function is always positive, we need only worry
about the equality (2x-1)2|x+3|=0. By inspection,
1
we see this holds true only when x= e -3, f .
2
82. 1x + 3 exists only when x –3, so we are concerned
only with the interval (–3, q). Further |x-4| is always 0
or positive, so the only possible value for a sign change is
1x - 1 2 |x - 4|
x=1. For –3<x<1,
is negative, and
1x + 3
1x - 12 |x + 4|
for 1<x<4 or 4<x<q,
is positive.
1x + 3
So the solution is (1, 4) (4, q).
83. Synthetic division reveals that we cannot conclude that
5 is an upper bound (since there are both positive and
negative numbers on the bottom row), while –5 is a
[0, 255] by [0, 2.5]
(c) When d≠170 ft, s≠2.088 ft.
(d) One possibility: The beam may taper off (become thinner) from west to east — e.g., perhaps it measures 8 in.
by 8 in. at the west end, but only 7 in. by 7 in. on the
east end. Then we would expect the beam to bend more
easily closer to the east end (though not at the extreme
east end, since it is anchored to the piling there).
Another possibility: The two pilings are made of
different materials.
87. (a) The tank is made up of a cylinder, with volume
4
∏x2(140-2x), and a sphere, with volume ∏x3.
3
4
Thus, V= ∏x3+∏x2(140-2x).
3
(b)
[0, 70] by [0, 1,500,000]
Chapter 2
(c) The largest volume occurs when x=70 (so it is
actually a sphere). This volume is
4
∏(70)3≠1,436,755 ft3.
3
88. (a) y=18.694x2-88.144x+2393.022
(b) y=
640
=800
0.8
(c) The deer population approaches (but never equals)
800.
1
1
1
1
1
1
x - 1.2
=
+
=
=
, so
.
1.2
x
R2
R2
1.2
x
1.2x
Then, R2 =
(b) y=–0.291x4+7.100x3-35.865x2
+48.971x+2336.634
131
91. (a) P(15)=325, P(70)=600, P(100)=648
92. (a)
[0, 15] by [0, 4500]
Review
1.2x
.
x - 1.2
(b) When x=3, R2 =
93. (a) C(x)=
3.6
3.6
=
= 2 ohms.
3 - 1.2
1.8
50
50 + x
(b) Shown is the window [0, 50] by [0, 1], with the graphs
of y=C(x) and y=0.6. The two graphs cross when
x≠33.33 ounces of distilled water.
[0, 15] by [0, 4500]
(c) For x=16, the quadratic model yields y≠$5768
and the quartic model yields y≠$3949.
(d) The quadratic model, which has a positive leading
coefficient, predicts that the amount of the Pell Grant
will always increase. The quartic model, which has a
negative leading coefficient, predicts that eventually
the amount of the Pell Grant will decrease over time.
89. (a) y=1.401x+4.331
[0, 50] by [0, 1]
50
=0.6 leads to
50 + x
50=0.6(50+x), so that 0.6x=20, or
100
x=
≠33.33.
3
(c) Algebraic solution of
94. (a) Let h be the height (in cm) of the can; we know the
1000
.
volume is 1 L=1000 cm3=∏x2h, so h=
px2
Then S=2∏x2+2∏xh=2∏x2+2000/x.
[0, 15] by [0, 30]
(b) y=0.188x2-1.411x+13.331
(b) Solve 2∏x2+2000/x=900, or equivalently,
2∏x3-900x+2000=0. Graphically we find that
either x≠2.31 cm and h≠59.75 cm, or x≠10.65
and h≠2.81 cm.
(c) Approximately 2.31<x<10.65 (graphically) and
2.81<h<59.75.
[0, 15] by [0, 30]
(c) Linear: Solving 1.401x+4.331=30 graphically, we
find that y=30 when x≠18.32. The spending will
exceed $30 million in the year 2008.
Quadratic: Solving 0.188x2-1.411x+13.331=30
graphically, we find that y=30 when x≠13.89.
The spending will exceed $30 million in the year 2003.
90. (a) Each shinguard costs $4.32 plus a fraction of the overhead: C=4.32+4000/x.
(b) Solve x(5.25-4.32-4000/x)=8000:
0.93x=12,000, so x≠12,903.23 — round up to 12,904.
95. (a) Let y be the height of the tank; 1000=x2y, so
y=1000/x2. The surface area equals the area of the
base plus 4 times the area of one side. Each side is a
rectangle with dimensions x × y, so S=x2+4xy
=x2+4000/x.
(b) Solve x2+4000/x=600, or x3-600x+4000=0
(a graphical solution is easiest): Either x=20, giving
dimensions 20 ft by 20 ft by 2.5 ft or x≠7.32, giving
approximate dimensions 7.32 by 7.32 by 18.66.
(c) 7.32<x<20 (lower bound approximate), so y must
be between 2.5 ft and about 18.66 ft.
132
Chapter 2
Polynomial, Power, and Rational Functions
Chapter 2 Project
Answers are based on the sample data shown in the table.
1.
3. The sign of a affects the direction the parabola opens. The
magnitude of a affects the vertical stretch of the graph.
Changes to h cause horizontal shifts to the graph, while
changes to k cause vertical shifts.
4. y≠–4.962x2-10.887x-5.141
5. y≠–4.968x2+10.913x-5.160
6. y≠–4.968x2+10.913x-5.160
≠–4.968 (x2-2.1967x+1.0386)
[0, 1.6] by [–0.1, 1]
2. We estimate the vertex to lie halfway between the two
data points with the greatest height, so that h is the
average of 1.075 and 1.118, or about 1.097. We estimate
k to be 0.830, which is slightly greater than the greatest
height in the data, 0.828.
Noting that y=0 when x=0.688, we solve
0=a(0.688-1.097)2+0.830 to find a≠–4.962. So
the estimated quadratic model is
y=–4.962(x-1.097)2+0.830.
= -4.968 c x2 - 2.1967x + a
- a
= -4.968 c a x -
2.1967 2
b
2
2.1967 2
b + 1.0386 R
2
2.1967 2
b - 0.1678 d
2
≠–4.968 (x-1.098)2+0.833
Section 3.1
Exponential and Logistic Functions
133
Chapter 3
Exponential, Logistic, and Logarithmic Functions
■ Section 3.1 Exponential and Logistic
Functions
Exploration 1
1. The point (0, 1) is common to all four graphs, and all four
functions can be described as follows:
Domain: 1 - q, q 2
Range: 1 0, q 2
Continuous
Always increasing
Not symmetric
No local extrema
Bounded below by y = 0, which is also the only
asymptote
lim f1x2 = q, lim f1 x2 = 0
xSq
2. The point (0, 1) is common to all four graphs, and all four
functions can be described as follows:
Domain: 1 - q, q 2
Range: 10, q 2
Continuous
Always decreasing
Not symmetric
No local extrema
Bounded below by y = 0, which is also the only
asymptote
lim g1 x2 = 0, lim g1x2 = q
xSq
xS–q
1 x
y1 = a b
2
xS–q
[–2, 2] by [–1, 6]
y1=2 x
[–2, 2] by [–1, 6]
1 x
y2 = a b
3
y2=3 x
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
1 x
y3 = a b
4
y3=4 x
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
1 x
y4 = a b
5
y4=5 x
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
134
Chapter 3
Exponential, Logistic, and Logarithmic Functions
Exploration 2
1
38
1
7. 6
a
8. b15
6.
1.
f1 x2 = 2 x
9. –1.4 since (–1.4)5=–5.37824
10. 3.1, since (3.1)4=92.3521
[–4, 4] by [–2, 8]
Section 3.1 Exercises
2.
1. Not an exponential function because the base is variable
and the exponent is constant. It is a power function.
f1x2 = 2 x
g1x2 = e0.4x
2. Exponential function, with an initial value of 1 and base of 3.
3. Exponential function, with an initial value of 1 and base
of 5.
4. Not an exponential function because the exponent is constant. It is a constant function.
[–4, 4] by [–2, 8]
5. Not an exponential function because the base is variable.
f1 x2 = 2 x
g1x2 = e0.5x
6. Not an exponential function because the base is variable.
It is a power function.
7. f1 02 = 3 # 50 = 3 # 1 = 3
8. f1 -2 2 = 6 # 3-2 =
[–4, 4] by [–2, 8]
6
2
=
9
3
1
3
9. f a b = -2 # 31>3 = -2 13
3
f1 x2 = 2 x
g1x2 = e0.6x
3
8
8
8
10. f a - b = 8 # 4-3>2 = 2 3>2 = 3 = = 1
2
8
2
12 2
11. f1 x2 =
[–4, 4] by [–2, 8]
3# 1 x
a b
2
2
1 x
12. g1 x2 = 12 # a b
3
f1 x2 = 2 x
g1x2 = e0.7x
[–4, 4] by [–2, 8]
f1 x2 = 2 x
g1x2 = e0.8x
13. f1 x2 = 3 # 1 122 x = 3 # 2x>2
1 x
14. g1 x2 = 2 # a b = 2e-x
e
15. Translate f1x2 = 2x by 3 units to the right. Alternatively,
1
1
g1 x2 = 2x-3 = 2 - 3 # 2x = # 2x = # f1x2, so it can be
8
8
1
obtained from f(x) using a vertical shrink by a factor of .
8
[–4, 4] by [–2, 8]
k = 0.7 most closely matches the graph of f(x).
3. k L 0.693
Quick Review 3.1
3
1. 2216 = -6 since 1 -6 2 3 = -216
2.
[–3, 7] by [–2, 8]
16. Translate f(x)=3x by 4 units to the left. Alternatively,
g(x)=3x±4=34 # 3x=81 # 3x=81 # f(x), so it can be
obtained by vertically stretching f(x) by a factor of 81.
5
3 125
= since 53 = 125 and 23 = 8
B 8
2
3. 272/3=(33)2/3=32=9
4. 45/2=(22)5/2=25=32
5.
1
212
[–7, 3] by [–2, 8]
Section 3.1
17. Reflect f1x2 = 4x over the y-axis.
Exponential and Logistic Functions
135
23. Reflect f1x2 = ex across the y-axis, horizontally shrink by
a factor of 3, translate 1 unit to the right, and vertically
stretch by a factor of 2.
[–2, 2] by [–1, 9]
18. Reflect f1x2 = 2x over the y-axis and then shift by 5 units
to the right.
[–2, 3] by [–1, 4]
24. Horizontally shrink f1 x2 = ex by a factor of 2, vertically
stretch by a factor of 3 and shift down one unit.
[–3, 7] by [–5, 45]
19. Vertically stretch f1x2 = 0.5x by a factor of 3 and then
shift 4 units up.
[–3, 3] by [–2, 8]
25. Graph (a) is the only graph shaped and positioned like
the graph of y = bx, b 7 1.
26. Graph (d) is the reflection of y = 2x across the y-axis.
27. Graph (c) is the reflection of y = 2x across the x-axis.
28. Graph (e) is the reflection of y = 0.5x across the x-axis.
[–5, 5] by [–2, 18]
20. Vertically stretch f1x2 = 0.6x by a factor of 2 and then
horizontally shrink by a factor of 3.
29. Graph (b) is the graph of y = 3-x translated down 2 units.
30. Graph (f) is the graph of y = 1.5x translated down 2 units.
31. Exponential decay; lim f1x2 = 0; lim f1x2 = q
xSq
xS–q
32. Exponential decay; lim f1x2 = 0; lim f1x2 = q
xSq
xS–q
33. Exponential decay: lim f1x2 = 0; lim f1x2 = q
xSq
xS–q
34. Exponential growth: lim f1x2 = q; lim f1x2 = 0
xSq
35. x 6 0
[–2, 3] by [–1, 4]
21. Reflect f1x2 = ex across the y-axis and horizontally
shrink by a factor of 2.
[–2, 2] by [–0.2, 3]
36. x 7 0
[–2, 2] by [–1, 5]
22. Reflect f1x2 = ex across the x-axis and y-axis. Then,
horizontally shrink by a factor of 3.
[–0.25, 0.25] by [0.5, 1.5]
[–3, 3] by [–5, 5]
xS–q
136
Chapter 3
Exponential, Logistic, and Logarithmic Functions
37. x 6 0
45.
[–3, 3] by [–2, 8]
Domain: 1 - q, q 2
Range: 10, q 2
Continuous
Always increasing
Not symmetric
Bounded below by y = 0, which is also the only asymptote
No local extrema
lim f1x2 = q, lim f1x2 = 0
[–0.25, 0.25] by [0.75, 1.25]
38. x 7 0
xSq
xS–q
46.
[–0.25, 0.25] by [0.75, 1.25]
39. y1 = y3, since 32x + 4 = 321x + 22 = 1 32 2 x + 2 = 9x + 2.
40. y2 = y3, since 2 # 23x–2 = 21 23x–2 = 21 + 3x–2 = 23x–1.
41. y-intercept: (0, 4). Horizontal asymptotes: y=0, y = 12.
[–3, 3] by [–2, 18]
Domain: 1 - q, q 2
Range: 10, q 2
Continuous
Always decreasing
Not symmetric
Bounded below by y = 0, which is the only asymptote
No local extrema
lim f1x2 = 0, lim f1x2 = q
[–10, 20] by [–5, 15]
42. y-intercept: (0, 3). Horizontal asymptotes: y = 0, y = 18.
xSq
xS–q
47.
[–5, 10] by [–5, 20]
43. y-intercept: (0, 4). Horizontal asymptotes: y = 0, y = 16.
[–5, 10] by [–5, 20]
44. y-intercept: (0, 3). Horizontal asymptotes: y = 0, y = 9.
[–5, 10] by [–5, 10]
[–2, 2] by [–1, 9]
Domain: 1 - q, q 2
Range: 10, q 2
Continuous
Always increasing
Not symmetric
Bounded below by y = 0, which is the only asymptote
No local extrema
lim f1x2 = q, lim f1x2 = 0
xSq
xS–q
Section 3.1
48.
Exponential and Logistic Functions
137
52. Let P(t) be the Columbus’s population t years after 1990.
Then with exponential growth, P1 t2 = P0 bt where
P0=632,910. From Table 3.7,
P(10)=632,910 b10=711,470. So,
b =
[–2, 2] by [–1, 9]
Domain: 1 - q, q 2
Range: 1 0, q 2
Continuous
Always decreasing
Not symmetric
Bounded below by y = 0, which is also the only asymptote
No local extrema
lim f1x2 = 0, lim f1x2 = q
xSq
xS–q
10 711,470
L 1.0118.
B 632,910
Solving graphically, we find that the curve
y = 632,91011.0118 2 t intersects the line y=800,000 at
t L 20.02. Columbus’s population will pass 800,000 in
2010.
53. Using the results from Exercises 51 and 52, we represent
Austin’s population as y=465,622(1.0350)t and
Columbus’s population as y=632,910(1.0118)t. Solving
graphically, we find that the curves intersect at t L 13.54.
The two populations will be equal, at 741,862, in 2003.
54. From the results in Exercise 53, the populations are equal
at 741,862. Austin has the faster growth after that, because
b is bigger (1.0350>1.0118). So Austin will reach 1 million first. Solving graphically, we find that the curve
y=465,622(1.0350)t intersects the line y=1,000,000 at
t L 22.22. Austin’s population will reach 1 million in 2012.
49.
[–3, 4] by [–1, 7]
Domain: 1 - q, q 2
Range: (0, 5)
Continuous
Always increasing
Symmetric about (0.69, 2.5)
Bounded below by y = 0 and above by y = 5; both are
asymptotes
No local extrema
lim f1x2 = 5, lim f1x2 = 0
xSq
xS–q
50.
55. Solving graphically, we find that the curve
12.79
y =
intersects the line y=10 when
11 + 2.402e-0.0309x 2
t L 69.67. Ohio’s population stood at 10 million in 1969.
19.875
L 1.794558
1 + 57.993e-0.0350051502
or 1,794,558 people
56. (a) P1 502 =
19.875
L 19.161673 or
1 + 57.993e-0.03500512102
19,161,673 people
(b) P1 2102 =
(c) lim P1t 2 = 19.875 or 19,875,000 people.
xSq
57. (a) When t = 0, B = 100.
(b) When t = 6, B L 6394.
58. (a) When t = 0, C = 20 grams.
[–3, 7] by [–2, 8]
Domain: 1 - q, q 2
Range: (0, 6)
Continuous
Always increasing
Symmetric about (0.69, 3)
Bounded below by y = 0 and above by y = 6; both
are asymptotes
No local extrema
lim f1x2 = 6, lim f1x2 = 0
xSq
xS – q
For #51–52, refer to Example 7 on page 285 in the text.
51. Let P(t) be Austin’s population t years after 1990. Then with
exponential growth, P1t2 = P0 bt where P0 = 465,622.
From Table 3.7, P1 10 2 = 465,622 b10 = 656,562. So,
b =
656,562
L 1.0350.
B 465,622
10
Solving graphically, we find that the curve
y = 465,622 11.0350 2 t intersects the line y=800,000 at
t L 15.75. Austin’s population will pass 800,000 in 2006.
(b) When t = 10,400, C L 5.647. After about 5700.22
years, 10 grams remain.
59. False. If a>0 and 0<b<1, or if a<0 and b>1, then
f1 x2 = a # bx is decreasing.
c
the horizontal asymptotes
1 + a # bx
are y=0 and y=c, where c is the limit of growth.
60. True. For f1x2 =
61. Only 8x has the form a # bx with a nonzero and b positive
but not equal to 1. The answer is E.
62. For b>0, f(0)=b0=1. The answer is C.
63. The growth factor of f1x2 = a # bx is the base b. The
answer is A.
64. With x>0, ax>bx requires a>b (regardless of
whether x<1 or x>1). The answer is B.
138
Chapter 3
Exponential, Logistic, and Logarithmic Functions
■ Section 3.2 Exponential and Logistic
Modeling
65. (a)
Quick Review 3.2
1. 0.15
2. 4%
[–5, 5] by [–2, 5]
3. (1.07)(23)
Domain: 1 - q, q 2
4. (0.96)(52)
1
Range: B - , q b
e
5. b2 =
Intercept: (0, 0)
Decreasing on 1 - q, -1 4 : Increasing on 3 - 1, q 2
1
Bounded below by y = e
1
Local minimum at a -1, - b
e
Asymptote y = 0.
lim f1x2 = q, lim f1 x2 = 0
xSq
6. b3 =
9
9
1
1
, so b = 3
= 3
= .
243
B 243
B 27
3
7. b=
6 838
≠1.01
B 782
8. b=
5 521
≠1.41
B 93
9. b=
4 91
≠0.61
B 672
10. b=
7 56
≠0.89
B 127
xS-q
(b)
160
= 4, so b = ; 14 = ;2.
40
Section 3.2 Exercises
For #1–20, use the model P1t2 = P0 11 + r2 t.
[–3, 3] by [–7, 5]
Domain: 1 - q, 0 2 ´ 10, q 2
Range: 1 - q, -e 4 ´ 1 0, q 2
No intercepts
Increasing on 1 - q, -1 4 ;
Decreasing on 3 -1, 0 2 ´ 10, q 2
Not bounded
Local maxima at 1 - 1, -e2
Asymptotes: x=0, y = 0.
lim g1 x2 = 0, lim g1x2 = - q
xSq
xS-q
66. (a) 2x = 122 2 2 = 24, so x = 4
3x
= 2x + 2,
2
3x = 4x + 4, x = -4
x
, 13 2 = 3
x+1
(d) 9 = 3
2 x
2. r=0.018, so P(t) is an exponential growth function
of 1.8%.
3. r=–0.032, so f(x) is an exponential decay function
of 3.2%.
4. r=–0.0032, so f(x) is an exponential decay function
of 0.32%.
5. r=1, so g(t) is an exponential growth function of 100%.
6. r=–0.95, so g(t) is an exponential decay function of 95%.
7. f1 x2 = 5 # 11 + 0.172 x = 5 # 1.17x 1x = years2
8. f1 x2 = 52 # 11 + 0.0232 x = 52 # 1.023x 1x = days2
(b) 3x = 33, so x = 3
(c) 8x>2 = 4x + 1, 122 2 x>2 = 122 2 x + 1 #
1. r=0.09, so P(t) is an exponential growth function of 9%.
x+1
, 2x = x + 1, x = 1
67. (a) y1—f(x) decreases less rapidly as x decreases.
(b) y3—as x increases, g(x) decreases ever more rapidly.
68. c = 2a: to the graph of 12a 2 x apply a vertical stretch by 2b,
since f1ax + b2 = 2ax + b = 2ax2b = 1 2b 2 1 2a 2 x.
69. a Z 0, c = 2.
70. a 6 0, c = 1.
71. a 7 0 and b 7 1, or a 6 0 and 0 6 b 6 1.
72. a 7 0 and 0 6 b 6 1, or a 6 0 and b 7 1.
73. Since 0<b<1, lim 11 + a # bx 2 = q and
xS-q
c
lim 11 + a # bx 2 = 1. Thus, lim
= 0 and
xSq
xS - q 1 + a # bx
c
lim
= c.
xSq 1 + a # bx
9. f1 x2 = 16 # 11 - 0.52 x = 16 # 0.5x 1x = months2
10. f1 x2 = 5 # 11 - 0.00592 = 5 # 0.9941x 1x = weeks2
11. f1 x2 = 28,900 # 11 - 0.0262 x = 28,900 # 0.974x
(x=years)
12. f1 x2 = 502,000 # 11 + 0.0172 x = 502,000 # 1.017x
(x=years)
13. f1 x2 = 18 # 11 + 0.0522 x = 18 # 1.052x 1x = weeks2
14. f1 x2 = 15 # 11 - 0.0462 x = 15 # 0.954x 1x = days2
15. f(x)=0.6 # 2x>3 (x=days)
16. f(x)=250 # 2x>7.5 = 250 # 22x>15 1x = hours 2
17. f(x)=592 # 2 - x>6 (x=years)
18. f(x)=17 # 2 - x>32 (x=hours)
2.875
= 1.25 = r + 1, so
2.3
f(x)=2.3 # 1.25x 1Growth Model2
19. f0 = 2.3,
Section 3.2
(a) In 1915: about P(25)≠12,315. In 1940: about
P(50)≠24,265.
For #21–22, use f1 x2 = f0 # bx
21. f0 = 4, so f1x2 = 4 # b . Since f1 52 = 4 # b = 8.05,
5
8.05
8.05
bfi=
, b= 5
L 1.15. f1x2 L 4 # 1.15x
4
B 4
22. f0 = 3, so f1x2 = 3 # bx. Since f1 42 = 3 # b4 = 1.49
1.49
4 1.49
L 0.84. f1x2 L 3 # 0.84x
, b=
b›=
3
B 3
For #23–28, use the model f1x2 =
139
31. The model is P1t2 = 62501 1.02752 t.
-4.64
= 0.8 = r + 1, so
-5.8
g(x)= -5.8 # 1 0.8 2 x 1Decay Model2
20. g0 = - 5.8,
x
Exponential and Logistic Modeling
c
.
1 + a # bx
40
= 20, 20 + 60b = 40,
1 + 3b
40
1
60b=20, b= , thus f(x)=
.
3
1 x
1 + 3# a b
3
23. c=40, a=3, so f(1)=
(b) P(t)=50,000 when t≠76.65 years after 1890 — in
1966.
32. The model is P1t2 = 42001 1.02252 t.
(a) In 1930: about P(20)≠6554. In 1945: about
P(35)≠9151.
(b) P(t)=20,000 when t≠70.14 years after 1910 —
about 1980.
1 t>14
33. (a) y = 6.6 a b , where t is time in days.
2
(b) After 38.11 days.
1 t>65
34. (a) y = 3.5 a b , where t is time in days.
2
(b) After 117.48 days.
60
= 24, 60 = 24 + 96b,
1 + 4b
60
3
96b=36, b= , thus f(x)=
.
8
3 x
1 + 4a b
8
35. One possible answer: Exponential and linear functions
are similar in that they are always increasing or always
decreasing. However, the two functions vary in how
quickly they increase or decrease. While a linear function
will increase or decrease at a steady rate over a given
interval, the rate at which exponential functions increase
or decrease over a given interval will vary.
128
= 32,
1 + 7b5
96
128=32+224bfi, 224bfi=96, bfi=
,
224
128
5 96
L 0.844, thus f1x2 L
b=
.
B 224
1 + 7 # 0.844x
36. One possible answer: Exponential functions and logistic
functions are similar in the sense that they are always
increasing or always decreasing. They differ, however, in
the sense that logistic functions have both an upper and
lower limit to their growth (or decay), while exponential
functions generally have only a lower limit. (Exponential
functions just keep growing.)
30
= 15, 30 = 15 + 75b3,
1 + 5b3
15
1
1
= , b= 3 L 0.585,
75b‹=15, b‹=
75
5
B5
30
thus f1x2 L
.
1 + 5 # 0.585x
37. One possible answer: From the graph we see that the
doubling time for this model is 4 years. This is the time
required to double from 50,000 to 100,000, from 100,000 to
200,000, or from any population size to twice that size.
Regardless of the population size, it takes 4 years for it to
double.
24. c=60, a=4, so f(1)=
25. c=128, a=7, so f(5)=
26. c=30, a=5, so f(3)=
20
= 10, 20 = 10 + 30b2,
1 + 3b2
1
1
L 0.58,
30b¤=10, b¤= , b=
3
B3
20
thus f(x)=
.
1 + 3 # 0.58x
27. c=20, a=3, so f(2)=
60
= 30, 60 = 30 + 90b8,
1 + 3b8
1
8 1
L 0.87,
90b°=30, b°= , b=
3
B3
60
thus f(x)=
.
1 + 3 # 0.87x
28. c=60, a=3, so f(8)=
29. P1t2 = 736,000 1 1.0149 2 t; P(t)=1,000,000 when
t≠20.73 years, or the year 2020.
38. One possible answer: The number of atoms of a radioactive substance that change to a nonradioactive state in a
given time is a fixed percentage of the number of radioactive atoms initially present. So the time it takes for half of
the atoms to change state (the half-life) does not depend
on the initial amount.
39. When t=1, B≠200—the population doubles every hour.
40. The half-life is about 5700 years.
For #41–42, use the formula P(h)=14.7 # 0.5h/3.6, where h is
miles above sea level.
41. P(10)=14.7 # 0.510/3.6=2.14 lb/in2
42. P1 h2 = 14.7 # 0.5h>3.6 intersects y = 2.5 when h≠9.20
miles above sea level.
30. P1t2 = 478,000 1 1.0628 2 t; P(t)=1,000,000 when
t≠12.12 years, or the year 2012.
[–1, 19] by [–1, 9]
Chapter 3
140
Exponential, Logistic, and Logarithmic Functions
43. The exponential regression model is
P1t2 = 1149.61904 1 1.012133 2 t, where P1 t 2 is measured in
thousands of people and t is years since 1900. The predicted
population for Los Angeles for 2003 is P1103 2 L 3981, or
3,981,000 people. This is an overestimate of 161,000 people,
161,000
an error of
L 0.04 = 4% .
3,820,000
44. The exponential regression model using 1950–2000 data is
P1t2 = 20.84002 11.04465 2 t, where P1t2 is measured in
thousands of people and t is years since 1900. The predicted
population for Phoenix for 2003 is P1103 2 L 1874, or
1,874,000 people. This is an overestimate of 486,000 people,
486,000
an error of
L 0.35 = 35% .
1,388,000
The exponential regression model using 1960–2000 data is
P1t2 = 86.70393 11.02760 2 t, where P1t2 is measured in
thousands of people and t is years since 1900. The predicted population for Phoenix for 2003 is P1 103 2 L 1432, or
1,432,000 people. This is an overestimate of 44,000 people,
44,000
an error of
L 0.03 = 3% .
1,388,000
The equations in #45–46 can be solved either algebraically or
graphically; the latter approach is generally faster.
45. (a)
(b)
(c)
46. (a)
(b)
(c)
P(0)=16 students.
P(t)=200 when t≠13.97 — about 14 days.
P(t)=300 when t≠16.90 — about 17 days.
P(0)=11.
P(t)=600 when t≠24.51 — after 24 or 25 years.
As t S q , P1t2 S 1001—the population never rises
above this level.
47. The logistic regression model is
837.7707752
, where x is the numP1x2 =
1 + 9.668309563e -.015855579x
ber of years since 1900 and P1 x2 is measured in millions
of people. In the year 2010, x = 110, so the model predicts
a population of
837.7707752
=
P1110 2 =
1 + 9.668309563e 1 - .015855579211102
837.7707752
837.7707752
=
L 311.4
1 - 1.744113692
2.690019034
1 + 9.668309563e
L 311,400,000 people.
19.875
where x is the number of
1 + 57.993e-0.035005x
years after 1800 and P is measured in millions. Our model
is the same as the model in Exercise 56 of Section 3.1.
49. P1 x2 L
[0, 200] by [0, 20]
15.64
, where x is the number of
1 + 11799.36e-0.043241x
years since 1800 and P is measured in millions.
50. P1 x2 L
As x S q, P1x2 S 15.64, or nearly 16 million, which is
significantly less than New York’s population limit of
20 million. The population of Arizona, according to our
models, will not surpass the population of New York. Our
graph confirms this.
[0, 500] by [0, 25]
51. False. This is true for logistic growth, not for exponential
growth.
52. False. When r<0, the base of the function, 1+r, is
merely less than 1.
53. The base is 1.049=1+0.049, so the constant percentage
growth rate is 0.049=4.9%. The answer is C.
54. The base is 0.834=1-0.166, so the constant percentage
decay rate is 0.166=16.6%. The answer is B.
55. The growth can be modeled as P(t)=1 # 2t/4. Solve
P(t)=1000 to find t≠39.86. The answer is D.
56. Check S(0), S(2), S(4), S(6), and S(8). The answer is E.
694.27
, where x is the number of
1 + 7.90e-0.017x
years since 1900 and P is measured in millions.
P(100)≠277.9, or 277,900,000 people.
57. (a) P1 x2 L
(b) The logistic model underestimates the 2000 population by about 3.5 million, an error of around 1.2%.
[–1, 109] by [0, 310]
1,301,614
, which is the same model as
1 + 21.603 # e-0.05055t
the solution in Example 8 of Section 3.1. Note that t
represents the number of years since 1900.
48. P1t2 L
[0, 120] by [–500,000, 1,500,000]
(c) The logistic model predicted a value closer to the
actual value than the exponential model, perhaps
indicating a better fit.
58. (a) Using the exponential growth model and the data
from 1900–2050, Mexico’s population can be represented by M(x)≠13.62 # 1.018x where x is the number of years since 1900 and M is measured in millions.
Using 1900–2000 data for the U.S., and the exponential growth model, the population of the United States
can be represented by P(x)≠80.55 # 1.013x, where x
is the number of years since 1900 and P is measured
in millions. Since Mexico’s rate of growth outpaces the
United States’ rate of growth, the model predicts that
Section 3.3
Mexico will eventually have a larger population. Our
graph indicates this will occur at x L 349, or 2249.
Logarithmic Functions and Their Graphs
■ Section 3.3 Logarithmic Functions
and Their Graphs
Exploration 1
1.
[0, 500] by [0, 10000]
(b) Using logistic growth models and the same data,
165.38
M1x2 L
while
1 + 39.65e-0.041x
798.80
P1x2 L
1 + 9.19e-0.016x
Using this model, Mexico’s population will not exceed
that of the United States, confirmed by our graph.
[–6, 6] by [–4, 4]
2. Same graph as part 1.
Quick Review 3.3
[0, 500] by [0, 800]
(c) According to the logistic growth models, the maximum sustainable populations are:
Mexico—165 million people.
U.S.—799 million people
(d) Answers will vary. However, a logistic model acknowledges that there is a limit to how much a country’s
population can grow.
e-x - ex
e-x - e-1-x2
=
2
2
ex - e-x
= -a
b = -sinh1 x2 , so the function is odd.
2
59. sinh1 -x2 =
e-x + e-1-x2
e-x + ex
ex + e-x
=
=
2
2
2
=cosh(x), so the function is even.
60. cosh1 -x2 =
ex - e-x
sinh1 x2
2
61. (a)
= x
cosh 1x2
e + e-x
2
ex - e-x #
2
ex - e-x
=
= tanh 1x2.
x
-x = x
2
e + e
e + e-x
e-x - e-1-x2
e-x - ex
(b) tanh(–x)= -x
= -x
-1-x2
e + ex
e + e
x
-x
e - e
=- x
= - tanh1 x2 , so the function is odd.
e + e-x
ex - e-x
ex + e-x
2ex
ex + e-x + ex - e-x
=
= x
x
-x
e + e
e + e-x
x
2
2
e
= x a
b =
,
e
1 + e-x e-x
1 + e-2x
which is a logistic function of c=2, a=1, and k=2.
(c) f(x)=1+tanh(x)=1+
1.
1
= 0.04
25
2.
1
= 0.001
1000
3.
1
= 0.2
5
4.
1
= 0.5
2
233
= 25 = 32
228
326
6. 24 = 32 = 9
3
5.
7. 51/2
8. 101/3
1 1>2
9. a b
= e-1>2
e
10. a
1 1>3
b
= e-2>3
e2
Section 3.3 Exercises
1. log4 4=1 because 41=4
2. log6 1=0 because 60=1
3. log2 32=5 because 25=32
4. log3 81=4 because 34=81
3
5. log5 125 =
6. log6
1
2
3
because 52/3= 125
3
2
1
1
= - because 6-2>5 = 2>5 = 5
5
6
136
136
5
141
Chapter 3
142
Exponential, Logistic, and Logarithmic Functions
7. log 103=3
42. Starting from y=ln x: translate up 2 units.
4
8. log 10,000=log 10 =4
9. log 100,000=log 105=5
10. log 10–4=–4
3
11. log 110=log 101/3=
1
3
-3
1
=log 10–3/2=
2
11000
3
13. ln e =3
12. log
[–5, 5] by [–3, 4]
43. Starting from y=ln x: reflect across the y-axis and
translate up 3 units.
14. ln e–4=–4
1
15. ln =ln e–1=–1
e
16. ln 1=ln e0=0
4
17. ln 1e
=ln e1/4=
18. ln
1
2e7
= ln e
-7>2
1
4
-7
=
2
[–4, 1] by [–3, 5]
44. Starting from y=ln x: reflect across the y-axis and
translate down 2 units.
19. 3, because blogb3 = 3 for any b>0.
20. 8, because blogb8 = 8 for any b>0.
21. 10log 10.52 = 10log10 10.52 = 0.5
22. 10log 14 = 10log10 14 = 14
23. eln 6=eloge 6=6
24. eln 11>52=eloge11>52 = 1>5
25. log 9.43≠0.9745≠0.975 and 100.9745≠9.43
26. log 0.908≠–0.042 and 10–0.042≠0.908
[–4, 1] by [–5, 1]
45. Starting from y=ln x: reflect across the y-axis and
translate right 2 units.
27. log (–14) is undefined because –14<0.
28. log (–5.14) is undefined because –5.14<0.
29. ln 4.05≠1.399 and e1.399≠4.05
30. ln 0.733≠–0.311 and e–0.311≠0.733
31. ln (–0.49) is undefined because –0.49<0.
32. ln (–3.3) is undefined because –3.3<0.
33. x=102=100
34. x=104=10,000
35. x=10–1=
1
=0.1
10
36. x=10–3=
1
=0.001
1000
[–7, 3] by [–3, 3]
46. Starting from y=ln x: reflect across the y-axis and
translate right 5 units.
37. f(x) is undefined for x>1. The answer is (d).
38. f(x) is undefined for x<–1. The answer is (b).
39. f(x) is undefined for x<3. The answer is (a).
40. f(x) is undefined for x>4. The answer is (c).
[–6, 6] by [–4, 4]
47. Starting from y=log x: translate down 1 unit.
41. Starting from y=ln x: translate left 3 units.
[–5, 15] by [–3, 3]
[–5, 5] by [–3, 3]
Section 3.3
48. Starting from y=log x: translate right 3 units.
Logarithmic Functions and Their Graphs
53.
[–1, 9] by [–3, 3]
[–5, 15] by [–3, 3]
Domain: (2, q)
Range: (–q, q)
Continuous
Always increasing
Not symmetric
Not bounded
No local extrema
Asymptote at x=2
lim f(x)=q
49. Starting from y=log x: reflect across both axes and
vertically stretch by 2.
xSq
[–8, 1] by [–2, 3]
54.
50. Starting from y=log x: reflect across both axes and
vertically stretch by 3.
[–2, 8] by [–3, 3]
Domain: (–1, q)
Range: (–q, q)
Continuous
Always increasing
Not symmetric
Not bounded
No local extrema
Asymptote: x=–1
lim f(x)=q
[–8, 7] by [–3, 3]
51. Starting from y=log x: reflect across the y-axis,
translate right 3 units, vertically stretch by 2, translate
down 1 unit.
xSq
55.
[–5, 5] by [–4, 2]
52. Starting from y=log x: reflect across both axes,
translate right 1 unit, vertically stretch by 3, translate up
1 unit.
[–6, 2] by [–2, 3]
[–2, 8] by [–3, 3]
Domain: (1, q)
Range: (–q, q)
Continuous
Always decreasing
Not symmetric
Not bounded
No local extrema
Asymptotes: x=1
lim f(x)=–q
xSq
143
144
Chapter 3
Exponential, Logistic, and Logarithmic Functions
60. I=12 # 10–0.0705≠10.2019 lumens.
56.
[–3, 7] by [–2, 2]
61. The logarithmic regression model is
y = -246461780.3 + 32573678.51 ln x, where x is the
year and y is the population. Graph the function and use
TRACE to find that x L 2023 when y L 150,000,000.
The population of San Antonio will reach 1,500,000 people in the year 2023.
Domain: (–2, q)
Range: (–q, q)
Continuous
Always decreasing
Not symmetric
Not bounded
No local extrema
Asymptotes: x=–2
lim f(x)=–q
xSq
57.
[1970, 2030] by [600,000, 1,700,000]
[–3, 7] by [–3, 3]
Domain: (0, q)
Range: (–q, q)
Continuous
Increasing on its domain
No symmetry
Not bounded
No local extrema
Asymptote at x=0
lim f1x2 = q
62. The logarithmic regression model is
y = 56360880.18 - 7337490.871 ln x, where x is the year
and y is the population. Graph the function and use
TRACE to find that x L 2024 when y L 500,000. The
population of Milwaukee will reach 500,000 people in the
year 2024.
xSq
58.
[–7, 3, 1] by [–10, 10, 2]
Domain: (–q, 2)
Range: (–q, q)
Continuous
Decreasing on its domain
No symmetry
Not bounded
No local extrema
Asymptote at x=2.
lim f1x2 = q
xS-q
59. (a) ı=10 log a
(b) ı=10 log a
(c) ı=10 log a
10-11
b = 10 log 10 = 10 1 1 2 =10 dB
10-12
10-5
b = 10 log 107 = 10 1 7 2 =70 dB
10-12
103
b = 10 log 1015 = 10 1 15 2 =150 dB
10-12
[1970, 2030] by [400,000, 800,000]
63. True, by the definition of a logarithmic function.
64. True, by the definition of common logarithm.
65. log 2≠0.30103. The answer is C.
66. log 5≠0.699 but 2.5 log 2≠0.753. The answer is A.
67. The graph of f(x)=ln x lies entirely to the right of the
origin. The answer is B.
68. For f1x2 = 2 # 3x, f-1 1x2 = log3 1x>22
because f-1 1f1x2 2 = log3 12 # 3x>22
=log3 3x
=x
The answer is A.
Section 3.4
69.
f(x)
Domain
Range
Intercepts
Asymptotes
`
`
3x
(–q, q)
`
(0, q)
`
(0, 1)
`
y=0
`
`
`
`
`
log3 x
(0, q)
Properties of Logarithmic Functions
3. log 23≠0.90309, 3 log 2≠3(0.30103)≠0.90309
4. log 5=log a
10
b =log 10-log 2≠1-0.30103
2
=0.69897
(–q, q)
(1, 0)
x=0
5. log 16=log 24=4 log 2≠1.20412
log 32=log 25=5 log 2≠1.50515
log 64=log 26=6 log 2≠1.80618
6. log 25=log 52=2 log 5=2 log a
10
b
2
=2(log 10-log 2)≠1.39794
log 40 = log 14 # 102 = log 4 + log 10 L 1.60206
100
log 50 = log a
b = log 100 - log 2 L 1.69897
2
The list consists of 1, 2, 4, 5, 8, 16, 20, 25, 32, 40, 50, 64,
and 80.
[–6, 6] by [–4, 4]
70.
f(x)
Domain
Range
Intercepts
Asymptotes
`
`
5
x
(–q, q)
`
`
`
(0, q)
(0, 1)
y=0
`
`
`
`
`
Exploration 2
log5 x
1. False
2. False; log£ (7x)=log3 7+log3 x
(0, q)
3. True
(–q, q)
4. True
(1, 0)
5. False; log
x=0
6. True
x
=log x-log 4
4
7. False; log5 x2=log5 x+log5 x=2 log5 x
8. True
Quick Review 3.4
1. log 102=2
2. ln e3=3
[–6, 6] by [–4, 4]
e
71. b=1e
. The point that is common to both graphs
is (e, e).
3. ln e–2=–2
4. log 10–3=–3
5.
6.
x5y - 2
x2y-4
= x5–2y–2–(–4)=x3y2
u-3v7
v7-2
v5
=
=
u
u-2v2
u-2 - 1-32
7. (x6y–2)1/2=(x6)1/2(y–2)1/2=
[–3.7, 5.7] by [–2.1, 4.1]
72. 0 is not in the domain of the logarithm functions because
0 is not in the range of exponential functions; that is, ax is
never equal to 0.
0x03
0y0
8. (x–8y12)3/4=(x–8)3/4(y12)3/4=
1u v 2
2 -4 1>2
9.
73. Reflect across the x-axis.
127 u v
2
6 - 6 1>3
1x-2y3 2 -2
1x3y-2 2 -3
=
0u0 0v0
3u2v-2
x4y-6
74. Reflect across the x-axis.
10.
■ Section 3.4 Properties of Logarithmic
Functions
Section 3.4 Exercises
=
-2
x-9y6
=
=
0y09
x6
1
30u0
x13
y12
1. ln 8x=ln 8+ln x=3 ln 2+ln x
2. ln 9y=ln 9+ln y=2 ln 3+ln y
Exploration 1
1. log (2 # 4)≠0.90309,
log 2+log 4≠0.30103+0.60206≠0.90309
8
2. log a b ≠0.60206, log 8-log 2≠0.90309-0.30103
2
≠0.60206
3
3. log =log 3-log x
x
2
4. log =log 2-log y
y
5. log2 y5=5 log2 y
145
Chapter 3
146
Exponential, Logistic, and Logarithmic Functions
6. log2 x–2=–2 log2 x
3 2
7. log x y =log x‹+log y¤=3 log x+2 log y
8. log xy3=log x+log y‹=log x+3 log y
2
x
= ln x¤-ln y‹=2 ln x-3 ln y
y3
10. log 1000x4=log 1000+log x4=3+4 log x
34. log4 x=
log x
log 4
35. log1/2(x+y)=
9. ln
11. log
1
1
1
4 x
= (log x-log y)= log x - log y
Ay
4
4
4
3
1x
1
1
1
= (ln x-ln y)= ln x - ln y
3
3
3
3
1y
13. log x+log y=log xy
12. ln
14. log x+log 5=log 5x
36. log1/3(x-y)=
log 11>22
= -
log1 x + y 2
log 1 1>3 2
= -
log1 x - y 2
log1x + y2
log1x - y2
log 2
log 3
37. Let x=logb R and y=logb S.
Then bx=R and by=S, so that
bx
R
= y = bx - y
S
b
R
logb a b = logb bx - y = x - y = logb R - logb S
S
38. Let x=logb R. Then bx=R, so that
Rc=(bx)c=bc # x
logb Rc = logb bc # x = c # x = c logb R
15. ln y-ln 3=ln(y/3)
16. ln x-ln y=ln(x/y)
17.
1
3
log x=log x1/3=log 1x
3
18.
1
5
log z=log z1/5=log 1z
5
39. Starting from g(x)=ln x: vertically shrink by a factor
1/ln 4≠0.72.
19. 2 ln x+3 ln y=ln x2+ln y3=ln (x2y3)
20. 4 log y-log z=log y4-log z=log a
y4
b
z
21. 4 log (xy)-3 log (yz)=log (x4y4)-log (y3z3)
=log a
4 4
xy
y3z3
b = log a
4
xy
z3
b
[–1, 10] by [–2, 2]
40. Starting from g(x)=ln x: vertically shrink by a factor
1/ln 7≠0.51.
22. 3 ln (x3y)+2 ln (yz2)=ln (x9y3)+ln (y2z4)
=ln (x9y5z4)
In #23–28, natural logarithms are shown, but common (base-10)
logarithms would produce the same results.
23.
ln 7
L 2.8074
ln 2
[–1, 10] by [–2, 2]
ln 19
L 1.8295
24.
ln 5
41. Starting from g(x)=ln x: reflect across the x-axis, then
vertically shrink by a factor 1/ln 3≠0.91.
25.
ln 175
L 2.4837
ln 8
26.
ln 259
L 2.2362
ln 12
27.
ln 12
ln 12
= L - 3.5850
ln 0.5
ln 2
28.
ln 29
ln 29
= L - 2.0922
ln 0.2
ln 5
29. log3 x=
ln x
ln 3
30. log7 x=
ln x
ln 7
31. log2(a+b)=
32. log5(c-d)=
33. log2 x=
log x
log 2
[–1, 10] by [–2, 2]
42. Starting from g(x)=ln x: reflect across the x-axis, then
shrink vertically by a factor of 1>ln 5 L 0.62.
ln1a + b 2
ln 2
ln1 c - d2
ln 5
[–1, 10] by [–2, 2]
43. (b): [–5, 5] by [–3, 3], with Xscl=1 and Yscl=1
(graph y=ln(2-x)/ln 4).
44. (c): [–2, 8] by [–3, 3], with Xscl=1 and Yscl=1
(graph y=ln(x-3)/ln 6).
Section 3.4
45. (d): [–2, 8] by [–3, 3], with Xscl=1 and Yscl=1
(graph y=ln(x-2)/ln 0.5).
Properties of Logarithmic Functions
147
50.
46. (a): [–8, 4] by [–8, 8], with Xscl=1 and Yscl=1
(graph y=ln(3-x)/ln 0.7).
47.
[–1, 9] by [–2, 8]
Domain: (0, q)
Range: (–q, q)
Continuous
Always increasing
Asymptote: x=0
lim f(x)=q
[–1, 9] by [–1, 7]
Domain: (0, q)
Range: (–q, q)
Continuous
Always increasing
Asymptote: x=0
lim f(x)=q
xSq
f(x)=log2 (8x)=
xSq
51. In each case, take the exponent of 10, add 12, and
multiply the result by 10.
(a) 0
ln 18x2
ln 1 22
48.
(b) 10
(c) 60
(d) 80
(e) 100
(f) 120
1 1 = 100 2
52. (a) R=log
(b) R=log
[–1, 9] by [–5, 2]
Domain: (0, q)
Range: (–q, q)
Continuous
Always decreasing
Asymptote: x=0
lim f(x)=–q
xSq
f(x)=log1>3 (9x)=
I
= –0.00235(40)=–0.094, so
12
I=12 # 10-0.094 L 9.6645 lumens.
I
=–0.0125(10)=–0.125, so
12
I=12 # 10-0.125 L 8.9987 lumens.
54. log
ln 19x2
1
ln a b
3
[–10, 10] by [–2, 3]
Domain: (–q, 0) ª (0, q)
Range: (–q, q)
Discontinuous at x=0
Decreasing on interval (–q, 0); increasing on
interval (0, q)
Asymptote: x=0
lim f(x)=q, lim f(x)=q,
xS –q
300
+ 3.5 = log 75+3.5≠5.3751
4
53. log
49.
xSq
250
+ 4.25 = log 125+4.25≠6.3469.
2
55. From the change-of-base formula, we know that
ln x
1 #
=
ln x L 0.9102 ln x.
f1 x2 = log3x =
ln 3
ln 3
f(x) can be obtained from g1x2 = ln x by vertically
stretching by a factor of approximately 0.9102.
56. From the change-of-base formula, we know that
log x
1 #
f(x)=log0.8x=
log x L -10.32 log x.
=
log 0.8
log 0.8
f(x) can be obtained from g(x)=log x by reflecting
across the x-axis and vertically stretching by a factor of
approximately 10.32.
57. True. This is the product rule for logarithms.
58. False. The logarithm of a positive number less than 1 is
negative. For example, log 0.01=–2.
59. log 12=log (3 # 4)=log 3+log 4 by the product rule.
The answer is B.
60. log9 64=(ln 64)/(ln 9) by the change-of-base formula.
The answer is C.
61. ln x 5=5 ln x by the power rule. The answer is A.
148
Chapter 3
Exponential, Logistic, and Logarithmic Functions
62. log1>2 x2 = 2 log1>2 0 x 0
ln 0 x 0
= 2
ln11>2 2
ln 0 x 0
= 2
ln 1 - ln 2
ln 0 x 0
= -2
ln 2
= - 2 log2 0 x 0
(d) log r=(–0.30) log (450)+2.36≠1.58, r≠37.69,
very close
(e) One possible answer: Consider the power function
y=a # xb then:
log y=log (a # xb)
=log a+log xb
=log a+b log x
=b(log x)+log a
which is clearly a linear function of the form
f(t)=mt+c where m=b, c=log a, f(t)=log y
and t=log x. As a result, there is a linear relationship between log y and log x.
The answer is E.
63. (a) f(x)=2.75 # x5.0
(b) f(7.1)≠49,616
(c) ln(x) 1.39
1.87
2.14
2.30
ln(y) 7.94
10.37
11.71
12.53
p
[0, 3] by [0, 15]
(d) ln(y)=5.00 ln x+1.01
(e) a≠5, b≠1 so f(x)=e1x5=ex5≠2.72x5. The two
equations are the same.
64. (a) f(x)=8.095 # x–0.113
(b) f(9.2)=8.095 # (9.2)–0.113≠6.30
For #67–68, solve graphically.
(c) ln(x) 0.69
1.10
1.57
2.04
ln(y) 2.01
1.97
1.92
1.86
p
66. log 4=log 22=2 log 2
log 6=log 2+log 3
log 8=log 23=3 log 2
log 9=log 32=2 log 3
log 12=log 3+log 4=log 3+2 log 2
log 16=log 24=4 log 2
log 18=log 2+log 9=log 2+2 log 3
log 24=log 2+log 12=3 log 2+log 3
log 27=log 33=3 log 3
log 32=log 25=5 log 2
log 36=log 6+log 6=2 log 2+2 log 3
log 48=log 4+log 12=4 log 2+log 3
log 54=log 2+log 27=log 2+3 log 3
log 72=log 8+log 9=3 log 2+2 log 3
log 81=log 34=4 log 3
log 96=log (3 # 32)=log 3+log 32=log 3+5 log 2
67. ≠6.41 6 x 6 93.35
68. ≠1.26 x 14.77
69. (a)
[0.5, 2.5] by [1.8, 2.1]
[–1, 9] by [–2, 8]
(d) ln(y)=–0.113 ln (x)+2.09
(e) a≠–0.113, b≠2.1 so f(x)=e2.1 # x–0.113
≠8.09x–0.113.
Domain of f and g: (3, q)
(b)
65. (a) log(w) –0.70 –0.52 0.30 0.70 1.48 1.70 1.85
log(r)
p
2.62 2.48 2.31 2.08 1.93 1.85 1.86
[0, 20] by [–2, 8]
Domain of f and g: (5, q)
(c)
[–1, 2] by [1.6, 2.8]
(b) log r=(–0.30) log w+2.36
(c)
[–7, 3] by [–5, 5]
[–1, 2] by [1.6, 2.8]
Domain of f: (–q, –3) ª (–3, q)
Domain of g: (–3, q)
Answers will vary.
Section 3.5
70. Recall that y=loga x can be written as x=ay.
Let y=loga b
ay=b
log ay=log b
y log a =log b
log b
y=
= loga b
log a
71. Let y=
y=
log x
ln x
. By the change-of-base formula,
log x
log e
= log x #
= log e L 0.43
log x
log x
log e
Thus, y is a constant function.
3. f1 g1x2 2 =
1
1
ln1e3x 2 = 13x2 = x and
3
3
g(f(x))=e311>3 ln x2=eln x = x.
4. f1 g1x2 2 = 3 log 110x>6 2 2 = 6 log 110x>6 2 = 61x>6 2 = x
2
and g1f1x2 2 = 1013 log x 2>6 = 1016 log x2>6 = 10log x = x.
5. 7.783 * 108 km
6. 1 * 10-15 m
7. 602,000,000,000,000,000,000,000
8. 0.000 000 000 000 000 000 000 000 001 66 (26 zeros
between the decimal point and the 1)
9. 11.86 * 105 2 13.1 * 107 2 = 11.86 2 13.1 2 * 105 + 7
=5.766 * 1012
10.
72.
Equation Solving and Modeling
8
8 * 10-7
=
* 10-7 - 1-62 = 1.6 * 10-1
-6
5
5 * 10
Section 3.5 Exercises
For #1–18, take a logarithm of both sides of the equation,
when appropriate.
1 x>5
1. 36 a b
3
1 x>5
a b
3
1 x>5
a b
3
x
5
x
[–1, 9] by [–3, 2]
Domain: (1, q)
Range: (–q, q)
Continuous
Increasing
Not symmetric
Vertical asymptote: x=1
lim f(x)=q
xSq
–1
ex
One-to one, hence invertible (f (x)=e )
■ Section 3.5 Equation Solving and Modeling
Exploration 1
1. log
log
log
log
log
log
log
log
log
log
14 # 102 L 1.60206
14 # 102 2 L 2.60206
14 # 103 2 L 3.60206
14 # 104 2 L 4.60206
14 # 105 2 L 5.60206
14 # 106 2 L 6.60206
14 # 107 2 L 7.60206
14 # 108 2 L 8.60206
14 # 109 2 L 9.60206
14 # 1010 2 L 10.60206
2. The integers increase by 1 for every increase in a power
of 10.
3. The decimal parts are exactly equal.
4. 4 # 1010 is nine orders of magnitude greater than 4 # 10.
Quick Review 3.5
In #1–4, graphical support (i.e., graphing both functions on a
square window) is also useful.
1. f1 g1 x2 2 = e2 ln1x
= ln1 ex 2 = x.
1>2
2
= eln x = x and g1 f1x2 2 =ln(e2x)1/2
2. f1 g1 x2 2 = 101log x 2>2 = 10log x = x and
g(f(x))=log 110x>2 2 2 = log 110x 2 = x.
2
1 x>3
2. 32 a b
4
1 x>3
a b
4
1 x>3
a b
4
x
3
x
3. 2 # 5x>4
5x>4
5x>4
x
4
x
4. 3 # 4x>2
4x>2
4x>2
x
2
x
= 4
=
1
9
1 2
= a b
3
= 2
= 10
= 2
1
16
1 2
= a b
4
=
= 2
= 6
= 250
= 125
= 53
= 3
= 12
= 96
= 32
= 45>2
5
=
2
= 5
5. 10-x>3 = 10, so -x>3 = 1, and therefore x = - 3.
6. 5-x>4 = 5, so -x>4 = 1, and therefore x = - 4.
7. x = 104 = 10,000
8. x=25=32
9. x - 5 = 4-1, so x = 5 + 4-1 = 5.25.
10. 1 - x = 41, so x = -3.
149
150
Chapter 3
Exponential, Logistic, and Logarithmic Functions
11. x =
ln 4.1
= log1.06 4.1 L 24.2151
ln 1.06
12. x =
ln 1.6
= log0.98 1.6 L -23.2644
ln 0.98
30. Multiply both sides by 2 # 2x, leaving 1 2x 2 2 + 1 = 6 # 2x, or
12x 2 2 - 6 # 2x + 1 = 0. This is quadratic in 2x, leading to
6 ; 136 - 4
2x =
= 3 ; 212. Then
2
ln13 ; 2 122
x =
= log2 1 3 ; 2122 L ;2.5431.
ln 2
13. e0.035x = 4, so 0.035x = ln 4, and therefore
x =
1
ln 4 L 39.6084.
0.035
31. Multiply both sides by 2ex, leaving 1ex 2 2 + 1 = 8ex, or
1ex 2 2 - 8ex + 1 = 0. This is quadratic in ex, leading to
14. e0.045x = 3, so 0.045x = ln 3, and therefore
x =
1
ln 3 L 24.4136.
0.045
15. e-x =
32. This is quadratic in ex, leading to
-5 ; 7
-5 ; 125 + 24
=
. Of these two
4
4
1
-5 + 7
1
numbers, only
= is positive, so x = ln
4
2
2
≠–0.6931.
3
L -0.4055.
2
ex =
5
5
, so -x = ln , and therefore
3
3
x = -ln
8 ; 164 - 4
= 4 ; 115 . Then
2
x=ln 14 ; 1152 L ;2.0634.
3
3
, so -x = ln , and therefore
2
2
x = -ln
16. e-x =
ex =
5
L -0.5108.
3
33.
1
17. ln(x-3)= , so x - 3 = e1>3, and therefore
3
x = 3 + e1>3 L 4.3956.
18. log 1x + 22 = -2, so x + 2 = 10-2, and therefore
x = -2 + 10-2 = -1.99.
19. We must have x1 x + 1 2 7 0, so x 6 -1 or x 7 0.
Domain: 1 - q, -1 2 ´ 10, q 2 ; graph (e).
20. We must have x 7 0 and x + 1 7 0, so x 7 0.
Domain: 1 0, q 2; graph (f).
x
7 0, so x 6 -1 or x 7 0.
x + 1
Domain: 1 - q, - 1 2 ´ 10, q 2 ; graph (d).
21. We must have
22. We must have x 7 0 and x + 1 7 0, so x 7 0.
Domain: 1 0, q 2; graph (c).
23. We must have x 7 0. Domain: 10, q 2 ; graph (a).
24. We must have x2 7 0, so x Z 0.
Domain: 1 - q, 0 2 ´ 10, q 2 ; graph (b).
34.
3
500
= 1 + 25e0.3x, so e0.3x =
= 0.06, and therefore
200
50
1
x =
ln 0.06 L -9.3780.
0.3
400
1
= 1 + 95e-0.6x, so e-0.6x =
, and therefore
150
57
1
1
x =
ln
L 6.7384.
-0.6 57
35. Multiply by 2, then combine the logarithms to obtain
x + 3
x + 3
ln
= 0. Then
= e0 = 1, so x + 3 = x2.
x2
x2
The solutions to this quadratic equation are
1 ; 11 + 12
1
1
x =
= ; 113 L 2.3028.
2
2
2
36. Multiply by 2, then combine the logarithms to obtain
x2
x2
log
= 2. Then
= 102 = 100, so
x + 4
x + 4
x2 = 1001x + 4 2. The solutions to this quadratic equation
100 ; 110000 + 1600
= 50 ; 10129. The
2
original equation requires that x 7 0, so 50 - 10129 is
extraneous; the only actual solution is
x = 50 + 10129 L 103.852.
are x =
For #25–38, algebraic solutions are shown (and are generally
the only way to get exact answers). In many cases solving
graphically would be faster; graphical support is also useful.
2
25. Write both sides as powers of 10, leaving 10log x = 106, or
x2 = 1,000,000. Then x = 1000 or x = -1000.
2
26. Write both sides as powers of e, leaving eln x = e4, or
x2 = e4. Then x = e2 L 7.389 or x = -e2 L -7.389.
37. ln[(x-3)(x+4)]=3 ln 2, so (x-3)(x+4)=8, or
x2 + x - 20 = 0. This factors to (x-4)(x+5)=0, so
x=4 (an actual solution) or x=–5 (extraneous, since
x-3 and x+4 must be positive).
4
27. Write both sides as powers of 10, leaving 10log x = 102, or
x4 = 100. Then x2 = 10, and x = ; 110.
6
28. Write both sides as powers of e, leaving eln x = e12, or
x6 = e12. Then x2 = e4, and x = ; e2.
29. Multiply both sides by 3 # 2x, leaving 12x 2 2 - 1=12 # 2x,
or 1 2x 2 2 - 12 # 2x - 1 = 0. This is quadratic in 2x,
12 ; 1144 + 4
leading to 2x =
= 6 ; 137. Only
2
6 + 137 is positive, so the only answer is
ln 16 + 1372
= log2 16 + 1372 L 3.5949.
x =
ln 2
38. log[(x-2)(x+5)]=2 log 3, so (x-2)(x+5)=9,
or x2 + 3x - 19 = 0.
3
1
- 3 ; 19 + 76
= - ; 185. The actual
2
2
2
1
3
solution is x = - + 185 L 3.1098; since x-2 must
2
2
be positive, the other algebraic solution,
3
1
x= - - 185, is extraneous.
2
2
Then x =
39. A $100 bill has the value of 1000, or 10‹, dimes so they
differ by an order of magnitude of 3.
Section 3.5
40. A 2 kg hen weighs 2000, or 2 # 103, grams while a 20 g
canary weighs 2 # 10 grams. They differ by an order of
magnitude of 2.
41. 7-5.5=1.5. They differ by an order of magnitude of 1.5.
42. 4.1-2.3=1.8. They differ by an order of magnitude of
1.8.
43. Given
I1
= 95
I0
I2
b2 = 10 log
= 65,
I0
we seek the logarithm of the ratio I1/I2.
b1 = 10 log
10 log
I1
I2
- 10 log
= b1 - b2
I0
I0
I1
I2
10 a log
- log b = 95 - 65
I0
I0
I1
= 30
10 log
I2
I1
= 3
log
I2
The two intensities differ by 3 orders of magnitude.
44. Given
I1
= 70
I0
I2
b2 = 10 log
= 10,
I0
b1 = 10 log
(b)
Equation Solving and Modeling
3H + 4 of carbonated water
3H 4 of household ammonia
+
=
10-3.9
= 108
10-11.9
(c) They differ by an order of magnitude of 8.
48. (a) Stomach acid: –log [H+]=2.0
log [H+]=–2.0
[H+]=10–2.0=1*10–2
Blood: –log [H+]=7.4
log [H+]=–7.4
[H+]=10–7.4≠3.98*10–8
(b)
3H + 4 of stomach acid
3 H 4 of blood
+
=
10-2
L 2.51 * 105
10-7.4
(c) They differ by an order of magnitude of 5.4.
The equations in #49–50 can be solved either algebraically or
graphically; the latter approach is generally faster.
49. Substituting known information into
T(t)=Tm + 1T0 - Tm 2e-kt leaves T(t)=22+70e–kt.
2
Using T(12)=50=22+70e–12k, we have e–12k= , so
5
1
2
k= - ln L 0.0764. Solving T(t)=30 yields
12 5
t≠28.41 minutes.
50. Substituting known information into T(t)
=Tm + 1T0 - Tm 2 e-kt leaves T(t)=65+285e–kt.
Using T(20)=120=65+285e–20k, we have
11
1
11
e–20k= , so k=– ln ≠0.0823. Solving
57
20 57
T(t)=90 yields t≠29.59 minutes.
51. (a)
we seek the logarithm of the ratio I1/I2.
10 log
I1
I2
- 10 log
= b1 - b2
I0
I0
I2
I1
- log b = 70 - 10
I0
I0
I1
= 60
10 log
I2
I1
= 6
log
I2
The two intensities differ by 6 orders of magnitude.
10 a log
45. Assuming that T and B are the same for the two quakes,
we have 7.9 = log a1 - log T + B and
6.6=log a2 - log T + B, so 7.9 - 6.6 = 1.3
=log(a1/a2). Then a1>a2 = 101.3, so a1 L 19.95a2—the
Mexico City amplitude was about 20 times greater.
[0, 40] by [0, 80]
(b)
[0, 40] by [0, 80]
T1 x2 L 79.47 # 0.93x
(c) T(0)+10=79.47+10≠89.47°C
52. (a)
46. If T and B were the same, we have
7.2 = log a1 - log T + B and 6.6 = log a2 - log T + B,
so 7.2-6.6=0.6=log 1a1>a2 2. Then a1>a2 = 100.6, so
a1 L 3.98a2—Kobe’s amplitude was about 4 times greater.
47. (a) Carbonated water: –log [H+]=3.9
log [H+]=–3.9
[H+]=10–3.9≠1.26*10–4
Household ammonia: –log [H+]=11.9
log [H+]=–11.9
[H+]=10–11.9≠1.26*10–12
151
[0, 35] by [0, 90]
152
Chapter 3
Exponential, Logistic, and Logarithmic Functions
(b)
58. Linear — the scatterplot of (x, y) is exactly linear
(y=2x+3.)
[0, 35] by [0, 90]
T1 x2 L 79.96 # 0.93x
[0, 5] by [0, 13]
(c) T(0)+0≠79.96=79.96°C.
59. False. The order of magnitude of a positive number is its
common logarithm.
53. (a)
60. True. In the formula T1t 2 = Tm + 1T0 - Tm 2e-kt, the
term 1T0 - Tm 2e-kt goes to zero at t gets large, so that
T(t) approaches Tm.
[0, 20] by [0, 15]
(b) The scatter plot is better because it accurately represents the times between the measurements. The equal
spacing on the bar graph suggests that the measurements were taken at equally spaced intervals, which
distorts our perceptions of how the consumption has
changed over time.
54. Answers will vary.
55. Logarithmic seems best — the scatterplot of (x, y) looks
most logarithmic. (The data can be modeled by
y=3+2 ln x.)
61. 23x–1=32
23x–1=25
3x-1=5
x=2
The answer is B.
62. ln x=–1
eln x=e–1
1
x =
e
The answer is B.
63. Given
a1
+ B = 8.1
T
a2
R2 = log
+ B = 6.1,
T
we seek the ratio of amplitudes (severities) a1/a2.
R1 = log
a log
a1
a2
+ B b - a log
+ B b = R1 - R2
T
T
[0, 5] by [0, 7]
log
56. Exponential — the scatterplot of (x, y) is exactly
exponential. (The data can be modeled by y=2 # 3x.)
a1
a2
- log
= 8.1 - 6.1
T
T
a1
log
= 2
a2
a1
= 102=100
a2
The answer is E.
[0, 5] by [0, 200]
57. Exponential — the scatterplot of (x, y) is exactly
3
exponential. (The data can be modeled by y= # 2x.)
2
[0, 5] by [0, 30]
64. As the second term on the right side of the formula
T1 t2 = Tm + 1T0 - Tm 2 e-kt indicates, and as the graph
confirms, the model is exponential.
The answer is A.
65. A logistic regression a f1x2 =
4443
b most
1 + 169.96e-0.0354x
closely matches the data, and would provide a natural
“cap” to the population growth at approx. 4.4 million
people. (Note: x=number of years since 1900.)
Section 3.5
66. The logistic regression model a f1x2 =
2930
b
1 + 17.29e-0.0263x
matches the data well and provides a natural cap of 2.9
million people. (Note: x=number of years since 1900.)
67. (a)
[–3, 3] by [0, 10]
As k increases, the bell curve stretches vertically. Its
height increases and the slope of the curve seems to
steepen.
(b)
[–3, 3] by [0, 1]
As c increases, the bell curve compresses horizontally.
Its slope seems to steepen, increasing more rapidly to
(0, 1) and decreasing more rapidly from (0, 1).
u
=10n, u, v 7 0
v
u
log =log 10n
v
log u-log v=n log 10
log u-log v=n(1)=n
For the initial expression to be true, either u and v are
both powers of ten, or they are the same constant k multiplied by powers of 10 (i.e., either u=10k and v=10m
or u = a # 10k and v = a # 10m, where a, k, and m are
constants). As a result, u and v vary by an order of
magnitude n. That is, u is n orders of magnitude greater
than v.
68. Let
Equation Solving and Modeling
70. Since T0 L 66.156 and Tm = 4.5, we have
166.156 - 4.52 e-kt=61.656 * 1 0.927702 t
61.656e-kt=61.656 * 10.92770 2 t
61.656
e-kt=
* 10.92770 2 t
61.656
e-kt=1 * 10.927702 t
ln e-kt=ln 11 # 10.92770 2 t 2
-kt=ln (1) + ln 10.927702 t
-kt=0 + t ln 10.927702
k= - ln 10.927702
≠0.075
71. One possible answer: We “map” our data so that all points
(x, y) are plotted as (ln x, y). If these “new” points are
linear—and thus can be represented by some standard
linear regression y=ax+b—we make the same substitution 1x S ln x2 and find y=a ln x+b, a logarithmic
regression.
72. One possible answer: We “map” our data so that all points
(x, y) are plotted as (ln x, ln y). If these “new” points are
linear—and thus can be represented by some standard
linear regression y=ax+b—we make the same “mapping” 1x S ln x, y S ln y 2 and find ln y=a ln x+b.
Using algebra and the properties of algorithms, we have:
ln y=a ln x+b
eln y=ea ln x+b
y=ea ln x # eb
a
=eln x # eb
b# a
=e x
=c xa, where c = eb, exactly the power regression
The equations and inequalities in #73–76 must be solved
graphically—they cannot be solved algebraically. For #77–78,
algebraic solution is possible, although a graphical approach
may be easier.
73. x≠1.3066
69. (a) r cannot be negative since it is a distance.
[–1, 5] by [–1, 6]
74. x≠0.4073 or x≠0.9333
[–10, 10] by [–10, 30]
(b) 30, 10 4 by 3 - 5, 3 4 is a good choice. The maximum
energy, approximately 2.3807, occurs when r L 1.729.
153
[0, 2] by [–1, 1]
75. 0 6 x 6 1.7115 1 approx. 2
[0, 10] by [–5, 3]
[–1, 2] by [–2, 8]
154
Chapter 3
Exponential, Logistic, and Logarithmic Functions
76. x - 20.0855 1approx. 2
Section 3.6 Exercises
1. A=1500(1+0.07)6≠$2251.10
2. A=3200(1+0.08)4≠$4353.56
3. A=12,000(1+0.075)7≠$19,908.59
4. A=15,500(1+0.095)12≠$46,057.58
[–40, 10] by [–1, 4]
x
77. log x - 2 log 3 7 0, so log 1x>92 7 0. Then 7 100 = 1,
9
so x 7 9.
78. log 1x + 12 - log 6 6 0, so log
x + 1
6 0.
6
x + 1
6 100 = 1, so x + 1 6 6, or x 6 5. The
6
original equation also requires that x + 1 7 0, so the
solution is -1 6 x 6 5.
Then
■ Section 3.6 Mathematics of Finance
k
10
20
30
40
50
60
70
80
90
100
0.07 20
b ≠$2122.17
4
6. A=3500 a 1 +
0.05 40
b ≠$5752.67
4
7. A=40,500 a 1 +
0.038 240
b ≠$86,496.26
12
8. A=25,300 a 1 +
0.045 300
b ≠$77,765.69
12
9. A=1250e(0.054)(6)≠$1728.31
10. A=3350e(0.062)(8)≠$5501.17
11. A=21,000e(0.037)(10)≠$30,402.43
12. A=8875e(0.044)(25)≠$26,661.97
13. FV=500 #
Exploration 1
1.
5. A=1500 a 1 +
`
`
`
`
`
`
`
`
`
`
`
a1 +
0.07 24
b - 1
4
≠$14,755.51
0.07
4
a1 +
0.06 48
b - 1
4
≠$20,869.57
0.06
4
a1 +
0.0525 120
b
- 1
12
≠$70,819.63
0.0525
12
a1 +
0.065 300
b
- 1
12
≠$456,790.28
0.065
12
A
1104.6
1104.9
14. FV=300 #
1105
1105
1105.1
1105.1
15. FV=450 #
1105.1
1105.1
1105.1
1105.1
A approaches a limit of about 1105.1.
2. y=1000e0.1≠1105.171 is an upper bound (and asymptote) for A(x). A(x) approaches, but never equals, this
bound.
16. FV=610 #
17. PV=815.37 #
1 - a1 +
Quick Review 3.6
1. 200 # 0.035 = 7
2. 150 # 0.025 = 3.75
3.
18. PV=1856.82 #
0.047 - 60
b
12
≠$43,523.31
0.047
12
1 - a1 +
1
# 7.25% = 1.8125%
4
1
# 6.5% L 0.5417%
4.
12
19. R=
78
= 0.65 = 65%
5.
120
28
= 0.35 = 35%
6.
80
7. 0.32x=48 gives x=150
8. 0.84x=176.4 gives x=210
9. 300 (1+0.05)=315 dollars
10. 500 (1+0.45)=522.50 dollars
20. R=
PV # i
1 - 11 + i2 -n
PV # i
1 - 11 + i2 -n
0.065 -360
b
12
≠$293,769.01
0.065
12
0.054
b
12
=
≠$293.24
0.054 -72
1 - a1 +
b
12
1 18,0002 a
0.072
b
12
=
≠$1401.47
0.072 -180
1 - a1 +
b
12
1154,000 2 a
Section 3.6
In #21–24, the time must be rounded up to the end of the next
compounding period.
21. Solve 2300 a 1 +
83
0.09 4t
b = 4150 : (1.0225)4t= , so
4
46
1 ln1 83>46 2
L 6.63 years — round to 6 years
4 ln 1.0225
9 months (the next full compounding period).
t=
22. Solve 8000 a 1 +
0.09 12t
b = 16,000 : (1.0075)12t=2, so
12
ln 2
1
t=
L 7.73 years — round to 7 years
12 ln 1.0075
9 months (the next full compounding period).
23. Solve 15,000 a 1 +
0.08 12t
b = 45,000 : (1.0067)12t=3, so
12
ln 3
1
L 13.71 years — round to 13 years
12 ln 1.0067
9 months (the next full compounding period). Note: A
graphical solution provides t L 13.78 years—round to 13
years 10 months.
t=
0.08 4t
24. Solve 1.5 a 1 +
b = 3.75 : (1.02)4t=2.5, so
4
1 ln 2.5
t=
L 11.57 years — round to 11 years
4 ln 1.02
9 months (the next full compounding period).
r 13652152
25. Solve 22,000 a 1 +
b
= 36,500 :
365
1 +
73 1>1825
r
, so r≠10.13%
= a b
365
44
26. Solve 8500 a 1 +
r 1122152
b
= 3 # 8500 :
12
r
1 +
= 31>60, so r≠22.17%
12
6
27. Solve 14.6 (1+r) =22:
110 1>6
1+r= a
b , so
73
r≠7.07%.
28. Solve 18 (1+r)8=25: 1+r= a
25 1>8
b , so r≠4.19%.
18
In #29–30, the time must be rounded up to the end of the next
compounding period.
0.0575 4t
ln 2
1
29. Solve a 1 +
b = 2 : t=
L 12.14
4
4 ln 1.014375
— round to 12 years 3 months.
0.0625 12t
b = 3:
12
ln 3
1
L 17.62 — round to 17 years
t=
12 ln 1 1 + 0.0625>12 2
8 months.
30. Solve a 1 +
rt
For #31–34, use the formula S=Pe .
31. Time to double: solve 2=e0.09t, leading to
1
t=
ln 2≠7.7016 years. After 15 years:
0.09
S=12,500e(0.09)(15)≠ $48,217.82
Mathematics of Finance
155
32. Time to double: solve 2=e0.08t, leading to
1
t=
ln 2≠8.6643 years. After 15 years:
0.08
S=32,500e(0.08)(15)≠ $107,903.80.
1
33. APR: solve 2=e4r, leading to r= ln 2≠17.33%.
4
After 15 years: S=9500e(0.1733)(15)≠$127,816.26 (using
the “exact” value of r).
1
34. APR: solve 2=e6r, leading to r= ln 2≠11.55%.
6
After 15 years: S=16,800e(0.1155)(15)≠$95,035.15 (using
the “exact” value of r).
In #35–40, the time must be rounded up to the end of the
next compounding period (except in the case of continuous
compounding).
1 ln 2
0.04 4t
b = 2: t =
L 17.42 — round
4
4 ln 1.01
to 17 years 6 months.
35. Solve a 1 +
1 ln 2
0.08 4t
b = 2: t =
L 8.751 — round
4
4 ln 1.02
to 9 years (almost by 8 years 9 months).
36. Solve a 1 +
37. Solve 1.07t=2: t=
ln 2
L 10.24 — round to 11 years.
ln 1.07
1 ln 2
0.07 4t
b = 2: t =
L 9.99 —
4
4 ln 1.0175
round to 10 years.
38. Solve a 1 +
39. Solve a 1 +
0.07 12t
1
ln 2
b = 2: t =
12
12 ln1 1 + 0.07>122
≠9.93 — round to 10 years.
40. Solve e0.07t=2: t=
1
ln 2≠9.90 years.
0.07
For #41–44, observe that the initial balance has no effect on
the APY.
41. APY= a 1 +
0.06 4
b - 1 L 6.14%
4
42. APY= a 1 +
0.0575 365
b
- 1 L 5.92%
365
43. APY=e0.063-1≠6.50%
44. APY= a 1 +
0.047 12
b - 1 L 4.80%
12
45. The APYs are a 1 +
0.05 12
b - 1 L 5.1162% and
12
a1 +
0.051 4
b - 1 L 5.1984% . So, the better investment
4
is 5.1% compounded quarterly.
1
46. The APYs are 5 % = 5.125% and e0.05-1≠5.1271%.
8
So, the better investment is 5% compounded continuously.
n
1 1 + i2 - 1
For #47–50, use the formula S=R
.
i
0.0726
= 0.00605 and R=50, so
12
1 1.006052 11221252 - 1
L $42,211.46.
S=50
0.00605
47. i=
156
Chapter 3
Exponential, Logistic, and Logarithmic Functions
0.155
= 0.0129» and R=50, so
12
11.0129 2 11221202 - 1
S=50
L $80,367.73.
0.0129
48. i=
0.124
= 0.0103; solve
12
11.0103 2 11221202 - 1
250,000=R
to obtain R≠$239.42
0.0103
per month (round up, since $239.41 will not be adequate).
49. i=
0.045
= 0.00375; solve
12
11.00375 2 11221302 - 1
120,000=R
to obtain R≠$158.03
0.00375
per month (round up, since $158.02 will not be adequate).
50. i=
For #51–54, use the formula A=R
1 - 11 + i2 -n
i
.
0.0795
= 0.006625; solve
12
1 - 11.0066252 -1122142
9000=R
to obtain R≠$219.51
0.006625
per month.
51. i=
0.1025
= 0.0085417; solve
12
1 - 11.00854172 -1122132
4500=R
to obtain R≠$145.74
0.0085417
per month (roundup, since $145.73 will not be adequate).
52. i=
0.0875
= 0.0072917; solve
12
1 - 11.0072917 2 -11221302
86,000=R
to obtain R≠$676.57
0.0072917
per month (roundup, since $676.56 will not be adequate).
53. i=
0.0925
= 0.0077083; solve 100,000=
12
1 - 11.0077083 2 -11221252
to obtain R≠$856.39 per
R
0.0077083
month (roundup, since $856.38 will not be adequate).
54. i=
0.12
= 0.01, solve
12
1 - 1 1.012 -n
86,000=1050
; this leads to
0.01
860
19
(1.01)–n =1 =
, so n≠171.81
1050
105
months, or about 14.32 years. The mortgage will be
paid off after 172 months (14 years, 4 months). The
last payment will be less than $1050. A reasonable
estimate of the final payment can be found by taking
the fractional part of the computed value of n above,
0.81, and multiplying by $1050, giving about $850.50.
To figure the exact amount of the final payment,
1 - 11.01 2 -171
solve 86,000=1050
+R (1.01)–172
0.01
(the present value of the first 171 payments, plus the
present value of a payment of R dollars 172 months
from now). This gives a final payment of R≠$846.57.
55. (a) With i=
(b) The total amount of the payments under the original
plan is 360 # $884.61 = $318,459.60. The total using the
higher payments is 172 # $1050 = $180,660 (or
171 # $1050 + $846.57 = $180,396.57 if we use the
correct amount of the final payment)—a difference of
$137,859.60 (or $138,063.03 using the correct final
payment).
56. (a) After 10 years, the remaining loan balance is
11.01 2 120 - 1
86,000(1.01)120-884.61
L $80,338.75
0.01
(this is the future value of the initial loan balance,
minus the future value of the loan payments). With
$1050 payments, the time required is found by solving
1 - 1 1.012 -n
80,338.75=1050
; this leads to
0.01
(1.01)–n≠0.23487, so n≠145.6 months, or about
12.13 (additional) years. The mortgage will be paid off
after a total of 22 years 2 months, with the final payment being less than $1050. A reasonable estimate of
the final payment is (0.6)($1050)≠$630.00 (see the
previous problem); to figure the exact amount, solve
1 - 11.01 2 -145
+R(1.01)–146,
80,338.75 = 1050
0.01
which gives a final payment of R≠$626.93.
(b) The original plan calls for a total of $318,459.60 in
payments; this plan calls for 120 # $884.61+
146 # $1050=$259,453.20 (or 120 # $884.61+
145 # $1050+$626.93=$259,030.13)— a savings
of $59,006.40 (or $59,429.47).
57. One possible answer: The APY is the percentage increase
from the initial balance S(0) to the end-of-year balance
S(1); specifically, it is S(1)/S(0)-1. Multiplying the initial balance by P results in the end-of-year balance being
multiplied by the same amount, so that the ratio remains
unchanged. Whether we start with a $1 investment, or a
r k
$1000 investment, APY= a 1 + b - 1.
k
58. One possible answer: The APR will be lower than the
APY (except under annual compounding), so the bank’s
offer looks more attractive when the APR is given.
Assuming monthly compounding, the APY is about
4.594%; quarterly and daily compounding give approximately 4.577% and 4.602%, respectively.
59. One possible answer: Some of these situations involve
counting things (e.g., populations), so that they can only
take on whole number values — exponential models which
predict, e.g., 439.72 fish, have to be interpreted in light of
this fact.
Technically, bacterial growth, radioactive decay, and compounding interest also are “counting problems” — for example, we cannot have fractional bacteria, or fractional atoms
of radioactive material, or fractions of pennies. However,
because these are generally very large numbers, it is easier to
ignore the fractional parts. (This might also apply when one
is talking about, e.g., the population of the whole world.)
Another distinction: while we often use an exponential
model for all these situations, it generally fits better (over
long periods of time) for radioactive decay than for most
Chapter 3
of the others. Rates of growth in populations (esp. human
populations) tend to fluctuate more than exponential
models suggest. Of course, an exponential model also fits
well in compound interest situations where the interest
rate is held constant, but there are many cases where
interest rates change over time.
60. (a) Steve’s balance will always remain $1000, since
interest is not added to it. Every year he receives 6%
of that $1000 in interest: 6% in the first year, then
another 6% in the second year (for a total of
2 # 6%=12%), then another 6% (totaling
3 # 6%=18%), etc. After t years, he has earned 6t% of
the $1000 investment, meaning that altogether he has
1000+1000 # 0.06t=1000(1+0.06t).
(b) The table is shown below; the second column gives
values of 1000(1.06)t. The effects of annual compounding show up beginning in year 2.
Years
0
1
2
3
4
5
6
7
8
9
10
`
`
`
`
`
`
`
`
`
`
`
`
`
Not
Compounded
1000.00
1060.00
1120.00
1180.00
1240.00
1300.00
1360.00
1420.00
1480.00
1540.00
1600.00
`
` Compounded
`
`
`
`
`
`
`
`
`
`
`
1000.00
1060.00
1123.60
1191.02
1262.48
1338.23
1418.52
1503.63
1593.85
1689.48
1790.85
69. (a) Matching up with the formula S=R
Review
157
1 1 + i2 n - 1
,
i
where i=r/k, with r being the rate and k being the
number of payments per year, we find r=8%.
(b) k=12 payments per year.
(c) Each payment is R=$100.
70. (a) Matching up with the formula A=R
1 - 11 + i2 -n
i
where i=r/k, with r being the rate and k being
number of payments per year, we find r=8%.
,
(b) k=12 payments per year.
(c) Each payment is R=$200.
■ Chapter 3 Review
1
3
1. f a b = -3 # 41>3 = -3 24
3
3
6
2
=
2. f a - b = 6 # 3-3>2 =
2
127
13
For #3–4, recall that exponential functions have the form
f(x)=a # bx.
3. a = 3, so f1 22 = 3 # b2 = 6, b2 = 2, b = 12,
f(x)=3 # 2x>2
1
4. a = 2, so f1 32 = 2 # b3 = 1, b3 = , b = 2-1>3,
2
f1 x2 = 2 # 2-x>3
5. f(x)=2–2x+3 — starting from 2x, horizontally shrink
1
by , reflect across y-axis, and translate up 3 units.
2
61. False. The limit, with continuous compounding, is
A = Pert = 100 e0.05 L $105.13.
62. True. The calculation of interest paid involves compounding, and the compounding effect is greater for longer
repayment periods.
63. A = P1 1 + r>k 2 kt = 225011 + 0.07>42 4162 L $3412.00.
The answer is B.
12
64. Let x=APY. Then 1+x=(1+0.06/12) ≠1.0617.
So x≠0.0617. The answer is C.
[–4, 6] by [0, 10]
1
6. f(x)=2–2x — starting from 2x, horizontally shrink by ,
2
reflect across the y-axis, reflect across x-axis.
65. FV=R((1+i)n-1)/i=
300((1+0.00375)240-1)/0.00375≠$116,437.31.
The answer is E.
66. R=PV i/(1-(1+i)–n)
=120,000(0.0725/12)/(1-(1+0.0725/12)–180)
≠$1095.44.
The answer is A.
67. The last payment will be $364.38.
68. One possible answer:
The answer is (c). This graph shows the loan balance
decreasing at a fairly steady rate over time. By contrast,
the early payments on a 30-year mortgage go mostly
toward interest, while the late payments go mostly toward
paying down the debt. So the graph of loan balance
versus time for a 30-year mortgage at double the interest
rate would start off nearly horizontal and more steeply
decrease over time.
[–2, 3] by [9, –1]
7. f(x)=–2–3x-3 — starting from 2x, horizontally shrink
1
by , reflect across the y-axis, reflect across x-axis,
3
translate down 3 units.
[–2, 3] by [9, –1]
158
Chapter 3
Exponential, Logistic, and Logarithmic Functions
8. f(x)=2–3x+3 — starting from 2x, horizontally shrink
1
by , reflect across the y-axis, translate up 3 units.
3
14. Exponential growth function
lim f1x2 = q , lim f1x2 = 1
xSq
xS-q
[–5, 5] by [–5, 15]
[–2, 3] by [–1, 9]
15.
1
9. Starting from ex, horizontally shrink by , then
2
3
translate right units — or translate right 3 units,
2
1
then horizontally shrink by .
2
[–1, 4] by [–10, 30]
Domain: (–q, q)
Range: (1, q)
Continuous
Always decreasing
Not symmetric
Bounded below by y=1, which is also the only
asymptote
No local extrema
lim f1x2 = 1 , lim f1x2 = q
[–1, 3] by [0, 10]
1
10. Starting from ex, horizontally shrink by , then
3
4
translate right units — or translate right 4 units,
3
1
then horizontally shrink by .
3
xSq
xS-q
16.
[–5, 5] by [–10, 50]
Domain: (–q, q)
Range: (–2, q)
Continuous
Always increasing
Not symmetric
Bounded below by y=–2, which is also the only
asymptote
No local extrema
lim g1 x2 = q , lim g1x2 = -2
[–1, 3] by [0, 25]
11. f1 0 2 =
100
= 12.5, lim f1x2 = 0, lim f1x2 = 20
xS-q
xSq
5 + 3
y-intercept: (0, 12.5); Asymptotes: y=0 and y=20
12. f1 0 2 =
50
50
, lim f1x2 = 0, lim f1 x2 = 10
=
xSq
5 + 2
7 xS-q
50
y-intercept: a 0, b ≠(0, 7.14)
7
Asymptotes: y=0, y=10
xSq
17.
13. It is an exponential decay function.
lim f1x2 = 2, lim f1 x2 = q
xSq
xS-q
[–5, 10] by [–5, 15]
xS-q
[–5, 10] by [–2, 8]
Domain: (–q, q)
Range: (0, 6)
Continuous
Increasing
Symmetric about (1.20, 3)
Bounded above by y=6 and below by y=0, the two
asymptotes
No extrema
lim f1x2 = 6 , lim f1x2 = 0
xSq
xS-q
Chapter 3
18.
Review
159
44
=22,
1 + 3e-5b
22
1
= ,
44=22+66e–5b, 66e–5b=22, e–5b=
66
3
1
–5b ln e=ln L –1.0986, so b≠0.22.
3
44
Thus, f1 x2 L
1 + 3e-0.22x
26. c=44, a=3, so f(5)=
[–300, 500] by [0, 30]
Domain: (–q, q)
Range: (0, 25)
Continuous
Always increasing
Symmetric about (–69.31, 12.5)
Bounded above by y=25 and below by y=0, the
two asymptotes
No local extrema
lim g1 x2 = 25 , lim g1x2 = 0
xSq
xS-q
For #19–22, recall that exponential functions are of the form
f1x2 = a # 11 + r2 kx .
19. a=24, r=0.053, k=1; so f(x)=24 # 1.053x, where
x=days.
20. a=67,000, r=0.0167, k=1, so
f(x)=67,000 # 1.0167x, where x=years.
1
21. a=18, r=1, k= , so f(x)=18 # 2 x>21, where
21
x=days.
1
1
22. a=117, r= - , k=
, so
2
262
x>262
1
f(x)=117 # a b
= 117 # 2 -x>262, where x=hours.
2
27. log2 32=log2 25=5 log2 2=5
28. log3 81=log3 34=4 log3 3=4
1
1
1
3
29. log 110 = log 103= log 10=
3
3
30. ln
1
2e
7
=ln e
- 72
7
7
= - ln e = 2
2
31. x=35=243
32. x=2y
x
33. a b = e-2
y
y
x = 2
e
a
34. a b = 10-3
b
b
a =
1000
35. Translate left 4 units.
For #23–26, recall that logistic functions are expressed in
c
f(x)=
.
1 + ae - bx
30
=20,
1 + 1.5e-2b
1
30=20+30e–2b, 30e–2b=10, e–2b= ,
3
1
–2b ln e=ln L –1.0986, so b≠0.55.
3
30
Thus, f1 x2 =
1 + 1.5e-0.55x
23. c=30, a=1.5, so f(2)=
20
=15,
1 + 2.33e-3b
1
20=15+35e–3b, 35e–3b=5, e–3b= ,
7
1
–3b ln e=ln L –1.9459, so b≠0.65.
7
20
Thus, f1 x2 =
1 + 2.33e-0.65x
20
25. c=20, a=3, so f(3)=
=10,
1 + 3e-3b
10
1
= ,
20=10+30e–3b, 30e–3b=10, e–3b=
30
3
1
–3b ln e=ln L –1.0986, so b≠0.37.
3
20
Thus, f1 x2 L
1 + 3e-0.37x
24. c=20, a≠2.33, so f(3)=
[–6, 7] by [–6, 5]
36. Reflect across y-axis and translate right 4 units — or
translate left 4 units, then reflect across y-axis.
[–6, 7] by [–6, 5]
37. Translate right 1 unit, reflect across x-axis, and translate
up 2 units.
[0, 10] by [–5, 5]
160
Chapter 3
Exponential, Logistic, and Logarithmic Functions
Local minima at (–0.61, –0.18)
and (0.61, –0.18)
No asymptotes
lim f1x2 = q , lim f1 x2 = q
38. Translate left 1 unit, reflect across x-axis, and translate up
4 units.
xSq
xS-q
42.
[–1.4, 17.4] by [–4.2, 8.2]
39.
[0, 15] by [–4, 1]
Domain: (0, q)
1
Range: a–q, R L (–q, 0.37]
e
[–4.7, 4.7] by [–3.1, 3.1]
Domain: (0, q)
1
Range: B - , q b L [–0.37, q)
e
Continuous
Decreasing on (0, 0.37]; increasing on [0.37, q)
Not symmetric
Bounded below
1
1
Local minimum at a , - b
e
e
lim f1x2 = q
xSq
40.
Continuous
Increasing on (0, e] L (0, 2.72],
Decreasing [e, q) L [2.72, q)
Not symmetric
Bounded above
1
Local maximum at ae, b L (2.72, 0.37)
e
Asymptotes: y=0 and x=0
lim f1x2 = 0
xSq
43. x=log 4≠0.6021
44. x=ln 0.25=–1.3863
45. x=
ln 3
≠22.5171
ln 1.05
46. x=e5.4=221.4064
47. x=10–7=0.0000001
[–4.7, 4.7] by [–3.1, 3.1]
Domain: (0, q)
Range: [–0.18, q)
Continuous
Decreasing on (0, 0.61]; increasing on [0.61, q)
Not symmetric
Bounded below
Local minimum at (0.61, –0.18)
No asymptotes
lim f1x2 = q
xSq
41.
48. x=3+
ln 5
≠4.4650
ln 3
49. log2x=2, so x=22=4
7
50. log3x= , so x=37/2=27 13 ≠46.7654
2
51. Multiply both sides by 2 # 3x, leaving (3x)2-1=10 # 3x,
or (3x)2-10 # 3x-1=0. This is quadratic in 3x, leading
10 ; 1100 + 4
to 3x=
= 5 ; 126. Only 5+ 126
2
is positive, so the only answer is x=log3(5+ 1262
≠2.1049.
52. Multiply both sides by 4+e2x, leaving 50=44+11e2x,
1
6
so 11e2x=6. Then x= ln ≠–0.3031.
2 11
53. log[(x+2)(x-1)]=4, so (x+2)(x-1)=104.
[–4.7, 4.7] by [–3.1, 3.1]
Domain: (–q, 0) ª (0, q)
Range: [–0.18, q)
Discontinuous at x=0
Decreasing on (–q, –0.61], (0, 0.61];
Increasing on [–0.61, 0), [0.61, q)
Symmetric across y-axis
Bounded below
The solutions to this quadratic equation are
1
x= (–1 ; 140,009), but of these two numbers, only
2
1
the positive one, x = 1 140,009 - 12 L 99.5112,
2
works in the original equation.
Chapter 3
3x + 4
= 5 , so 3x+4=e5(2x+1).
2x + 1
4 - e5
Then x = 5
L -0.4915 .
2e - 3
54. ln
55. log2 x=
log x
73. (a) f(0)=90 units.
(b) f(2)≠32.8722 units.
(c)
[0, 4] by [0, 90]
74. (a) P(t)=123,000(1-0.024)t=123,000(0.976)t.
log 5
58. log 1>2 1 4x 2 = log1>2 4 + log1>2 x3 = - 2 + 3 log 1>2 x
3
= -2 - 3 log2x
3 log x
= -2 log 2
(b) P(t)=90,000 when t=
ln190>1232
ln 0.976
≠12.86 years.
75. (a) P(t)=89,000(1-0.018)t=89,000(0.982)t.
(b) P(t)=50,000 when t=
59. Increasing, intercept at (1, 0). The answer is (c).
60. Decreasing, intercept at (1, 0). The answer is (d).
61. Intercept at (–1, 0). The answer is (b).
ln150>892
ln 0.982
≠31.74 years.
76. (a) P(0)≠5.3959 — 5 or 6 students.
(b) P(3)≠80.6824 — 80 or 81 students.
62. Intercept at (0, 1). The answer is (a).
(c) P(t)=100 when 1+e4–t=3, or t=4-ln 2
≠3.3069 — sometime on the fourth day.
63. A=450(1+0.046)3≠$515.00
(d) As t S q , P(t) S 300.
0.062 1421172
b
64. A=4800 a 1 +
≠$13,660.81
4
65. A=Pert
r
66. i= , n=kt, so FV = R #
k
a1 +
kt
r
b - 1
k
r
a b
k
0.055 1-122152
b
b
550 a 1 - a 1 +
12
67. PV =
≠$28,794.06
0.055
a
b
12
0.0725 1-2621152
b
b
26
68. PV =
≠$226,396.22
0.0725
b
a
26
1 5
69. 20e–3k=50, so k=– ln L –0.3054.
3 2
953 a 1 - a 1 +
3
70. 20e =30, so k= -ln L –0.4055.
2
–k
71. P(t)=2.0956 # 1.01218t, where x is the number of years
since 1900. In 2005, P(105)=2.0956 # 1.01218105≠7.5
million.
14.3614
, where x is the number of
1 1 + 2.0083e-0.0249t 2
years since 1900. In 2010, P(110)≠12.7 million.
72. P(t)=
161
ln x
ln 2
56. log1>6 1 6x2 2 = log1>6 6 + log1>6 x2 = log1>6 6 + 2 log 1>6 0 x 0
2 ln 0 x 0
2 ln 0 x 0
2 ln 0 x 0
= -1 = -1 +
= -1 +
ln 1>6
ln 6
ln 6-1
57. log5 x=
Review
77. (a) P(t)=20 # 2t, where t is time in months. (Other
possible answers: 20 # 212t if t is in years, or 20 # 2t/30
if t is in days).
(b) P(12)=81,920 rabbits after 1 year.
P(60)≠2.3058 × 1019 rabbits after 5 years.
(c) Solve 20 # 2t=10,000 to find t=log2 500
≠8.9658 months — 8 months and about 29 days.
78. (a) P(t)=4 # 2t=2t+2, where t is time in days.
(b) P(4)=64 guppies after 4 days. P(7)=512 guppies
after 1 week.
(c) Solve 4 # 2t=2000 to find t=log2 500=8.9658 days
— 8 days and about 23 hours.
1 t>1.5
, where t is time in seconds.
79. (a) S(t)=S0 # a b
2
(b) S(1.5)=S0/2. S(3)=S0/4.
1 60>1.5
1 40
=S0 # a b , then
(c) If 1 g=S(60)=S0 # a b
2
2
S0=240≠1.0995*1012 g=1.0995*109 kg
=1,099,500 metric tons.
1 t>2.5
, where t is time in seconds.
80. (a) S(t)=S0 # a b
2
(b) S(2.5)=S0/2. S(7.5)=S0/8.
1 60>2.5
1 24
=S0 # a b , then
(c) If 1 g=S(60)=S0 # a b
2
2
S0=224=16,777,216 g=16,777.216 kg.
162
Chapter 3
Exponential, Logistic, and Logarithmic Functions
81. Let a1=the amplitude of the ground motion of the Feb 4
quake, and let a2=the amplitude of the ground motion
of the May 30 quake. Then:
a2
a1
+ B and 6.9 = log
+ B
6.1 = log
T
T
a2
a1
a log
+ B b - a log
+ B b = 6.9 - 6.1
T
T
a2
a1
log
- log
= 0.8
T
T
a2
log
= 0.8
a1
a2
= 100.8
a1
a2 L 6.31 a1
The ground amplitude of the deadlier quake was approximately 6.31 times stronger.
82. (a) Seawater:
–log [H+]=7.6
log [H+]=–7.6
[H+]=10–7.6≠2.51 × 10–8
Milk of Magnesia:
–log [H+]=10.5
log [H+]=–10.5
[H+]=10–10.5≠3.16 × 10–11
3 H+ 4 of Seawater
10-7.6
=
L 794.33
(b)
+
3 H 4 of Milk of Magnesia
10-10.5
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
(c) They differ by an order of magnitude of 2.9.
0.08 4t
Solve 1500 a 1 +
b =3750: (1.02)4t=2.5,
4
1 ln 2.5
L 11.5678 years — round to 11 years
so t=
4 ln 1.02
9 months (the next full compounding period).
Solve 12,500e0.09t=37,500: e0.09t=3,
1
so t=
ln 3=12.2068 years.
0.09
700
t=133.83 ln
L 137.7940 — about 11 years 6 months.
250
500
t=133.83 ln
L 308.1550 — about 25 years 9 months.
50
0.0825 12
r= a 1 +
b -1≠8.57%
12
r=e0.072-1≠7.47%
I=12 # 10(–0.0125)(25)=5.84 lumens
ln x
logb x=
. This is a vertical stretch if e–1<b<e
ln b
(so that |ln b|<1), and a shrink if 0<b<e–1 or b>e.
(There is also a reflection if 0<b<1.)
log x
1
logb x=
. This is a vertical stretch if <b<10
log b
10
1
(so that |log b|<1), and a shrink if 0<b< or
10
b>10. (There is also a reflection if 0<b<1.)
g(x)=ln[a # bx]=ln a+ln bx=ln a+x ln b. This
has a slope ln b and y-intercept ln a.
(a) P(0)=16 students.
(b) P(t)=800 when 1+99e–0.4t=2, or e0.4t=99,
1
so t=
ln 99≠11.4878 — about 1112 days.
0.4
(c) P(t)=400 when 1+99e–0.4t=4, or e0.4t=33,
1
so t=
ln 33≠8.7413 — about 8 or 9 days.
0.4
94. (a) P(0)=12 deer.
(b) P(t)=1000 when 1+99e–0.4t=1.2, so
1
1
0.2
t= ln
L 15.5114 — about 15 years.
0.4
99
2
(c) As t S q , P(t) S 1200 (and the population never
rises above that level).
95. The model is T=20+76e–kt, and T(8)=65
1 45
45
=20+76e–8k. Then e–8k= , so k= - ln
76
8 76
≠0.0655. Finally, T=25 when 25=20+76e–kt,
1
5
so t= - ln
L 41.54 minutes.
k 76
96. The model is T=75+145e–kt, and T(35)=150
75
15
1
=75+145e–35k. Then e–35k=
, so k= - ln
145
35 29
≠0.0188. Finally, T=95 when 95=75+145e–kt,
1
20
so t= - ln
L 105.17 minutes.
k 145
1 1 + i2 n - 1
97. (a) Matching up with the formula S=R
,
i
where i=r/k, with r being the rate and k being the
number of payments per year, we find r=9%.
(b) k=4 payments per year.
(c) Each payment is R=$100.
98. (a) Matching up with the formula A=R
1 - 11 + i2 - n
i
where i=r/k, with r being the rate and k being the
number of payments per year, we find r=11%.
,
(b) k=4 payments per year.
(c) Each payment is R=$200.
99. (a) Grace’s balance will always remain $1000, since interest is not added to it. Every year she receives 5% of
that $1000 in interest; after t years, she has been paid
5t% of the $1000 investment, meaning that altogether
she has 1000+1000 # 0.05t=1000(1+0.05t).
(b) The table is shown below; the second column gives
values of 1000e0.05t. The effects of compounding continuously show up immediately.
Years
0
1
2
3
4
5
6
7
8
9
10
`
`
`
`
`
`
`
`
`
`
`
`
`
Not
Compounded
1000.00
1050.00
1100.00
1150.00
1200.00
1250.00
1300.00
1350.00
1400.00
1450.00
1500.00
`
`
`
`
`
`
`
`
`
`
`
`
`
Compounded
1000.00
1051.27
1105.17
1161.83
1221.40
1284.03
1349.86
1419.07
1491.82
1568.31
1648.72
Chapter 3
Chapter 3 Project
Answers are based on the sample data shown in the table.
2. Writing each maximum height as a (rounded) percentage
of the previous maximum height produces the following
table.
Bounce Number
Percentage Return
0
1
N/A
79%
2
77%
3
76%
4
78%
5
The average is 77.8%
79%
3.
Review 163
8. H would be changed by varying the height from which the
ball was dropped. P would be changed by using a different
type of ball or a different bouncing surface.
9. y=HP x
=H(eln P)x
=He(ln P)x
=2.7188 e–0.251x
10. ln y=ln (HP x)
=ln H+x ln P
This is a linear equation.
11. Bounce Number
0
ln (Height)
1.0002
1
0.76202
2
0.50471
3
0.23428
4
–0.01705
5
–0.25125
[–1, 6] by [0, 3]
4. Each successive height will be predicted by multiplying
the previous height by the same percentage of rebound.
The rebound height can therefore be predicted by the
equation y=HP x where x is the bounce number. From
the sample data, H=2.7188 and P≠0.778.
5. y=HP x becomes y≠2.7188 # 0.778x.
6. The regression equation is y≠2.733 # 0.776x. Both H and
P are close to, though not identical with, the values in the
earlier equation.
7. A different ball could be dropped from the same original
height, but subsequent maximum heights would in general
change because the rebound percentage changed. So P
would change in the equation.
[–1, 6] by [–1.25, 1.25]
The linear regression produces
Y=ln y≠–0.253x+1.005. Since ln y≠(ln P)x+ln H,
the slope of the line is ln P and the Y-intercept (that is, the
ln y-intercept) is ln H.
164
Chapter 4
Trigonometric Functions
Chapter 4
Trigonometric Functions
■ Section 4.1 Angles and Their Measures
Exploration 1
1. 2∏r
2. 2∏ radians (2∏ lengths of thread)
3. No, not quite, since the distance ∏r would require a piece
of thread ∏ times as long, and ∏>3.
4. ∏ radians
Quick Review 4.1
For #9–16, use the formula s=r¨, and the equivalent forms
r=s/¨ and ¨=s/r.
9. 60° #
p
p
=
rad
180°
3
10. 90° #
p
p
=
rad
180°
2
11. 120° #
p
2p
=
rad
180°
3
12. 150° #
p
5p
=
rad
180°
6
1. C=2∏ # 2.5=5∏ in.
2. C=2∏ # 4.6=9.2∏ m
13. 71.72° #
p
≠1.2518 rad
180°
p
≠0.2065 rad
180°
3. r =
1
# 12 = 6 m
p
2p
14. 11.83° #
4. r =
1
2p
15. 61°24'= a 61 +
p
24 °
b =61.4° #
≠1.0716 rad
60
180°
16. 75°30'= a 75 +
p
30 °
b =75.5° #
≠1.3177 rad
60
180°
#
8 =
4
ft
p
5. (a) s=47.52 ft (b) s=39.77 km
6. (a) v=26.1 m/sec (b) v=8.06 ft/sec
7. 60
mi
hr
#
5280
ft
mi
#
1 hr
= 88 ft>sec
3600 sec
17.
p 180°
#
= 30°
6 p
8. 45
mi
hr
#
5280
ft
mi
#
1 hr
= 66 ft>sec
3600 sec
18.
∏ 180°
#
= 45°
4 ∏
9. 8.8
ft
sec
10. 132
ft
sec
1 mi
5280 ft
3600
sec
= 6 mph
hr
19.
p 180°
#
= 18°
10 p
3600
sec
= 90 mph
hr
20.
3p 180°
#
= 108°
p
5
21.
12 °
b =23.2°
60
24 °
b =35.4°
2. 35°24'= a 35 +
60
7p 180°
#
= 140°
p
9
22.
13p 180°
#
= 117°
p
20
44
15 °
+
b =118.7375°
60
3600
30
36 °
+
b =48.51°
4. 48°30'36"= a 48 +
60
3600
24. 1.3 #
#
#
1 mi
5280 ft
#
#
Section 4.1 Exercises
1. 23°12'= a 23 +
3. 118°44'15"= a 118 +
23. 2 #
180
L 114.59°
p
180
L 74.48°
p
25. s=50 in.
5. 21.2°=21°(60 # 0.2)'=21°12'
26. s=70 cm
6. 49.7°=49°(60 # 0.7)'=49°42'
27. r=6/∏ ft
7. 118.32°=118°(60 # 0.32)'=118°19.2'
=118°19'(60 # 0.2)"=118°19'12"
28. r=7.5/∏ cm
8. 99.37°=99°(60 # 0.37)'=99°22.2'
=99°22'(60 # 0.2)"=99°22'12"
4
30. ¨= radians
7
29. ¨=3 radians
31. r=
360
cm
∏
Section 4.1
32. s=(5 ft)(18°) a
33. ¨=s1>r1 =
2p
p
b =
ft
360°
2
34. ¨=s1>r1 =4.5 rad and r2 = s2>¨ = 16 km
∏
∏
=
rad, so the curved side
180 °
18
#
11∏
in. The two straight sides measures 11 in.
18
11∏
L 24 inches.
each, so the perimeter is 11+11+
18
measures
45. v=44 ft/sec and r=13 in., so
sec
1 ft #
rad
ft #
◊=v/r= a 44
60
b ÷ a 13 in. #
2∏
b
sec
min
12 in.
rev
≠387.85 rpm.
40. (a) SSW is 202.5°. (b) WNW is 292.5°. (c) NNW is 337.5°.
R
WR
S
=
1S =
mm.
W
100
100
25.4 mm=1 in., so
WR
WR # 1
=
in.
S =
100 25.4
2540
WR
WR
(b) D+2S=D+2 a
in.
b =D+
2540
1270
215 # 60
(c) Taurus: D = 16 +
L 26.16 in.
1270
225 # 60
Charger: D = 18 +
L 28.63 in.
1270
235 # 70
Mariner: D = 16 +
L 28.95 in.
1270
245 # 65
Ridgeline: D = 17 +
L 29.54 in.
1270
47. ◊=2000 rpm and r=5 in., so
teeth #
v=r◊= a 5 in. # 12
b
in.
rev #
rad # 1 min
a 2000
2p
b ≠12,566.37 teeth per
min
rev 60 sec
second.
41. ESE is closest at 112.5°.
48.
36. The angle is 100 ° #
Then
∏
5∏
=
rad, so
180 °
9
5∏
7 =
r.
9
r =
63
L 4 cm.
5∏
37. Five pieces of track form a semicircle, so each arc has
a central angle of ∏/5 radians. The inside arc length is
ri 1∏>52 and the outside arc length is ro 1∏>5 2. Since
ro 1 ∏>5 2 - ri 1∏>5 2 = 3.4 inches , we conclude that
ro - ri = 3.4 15>∏ 2 L 5.4 inches.
38. Let the diameter of the inner (red) circle be d. The inner
circle’s perimeter is 37.7 inches, which equals ∏d. Then
the next-largest (yellow) circle has a perimeter of
∏ 1d + 6 + 6 2 = ∏d + 12∏ = 37.7 + 12∏
L 75.4 inches.
39. (a) NE is 45°. (b) NNE is 22.5°. (c) WSW is 247.5°.
46. (a)
35°
c.
42. SW is closest at 225°.
43. The angle between them is ¨=9°42'=9.7°≠0.1693
radians, so the distance is about
s=r¨=(25)(0.1693)≠ 4.23 statute miles.
a
60 min
1 mi
800pd in.
ba
ba
b L 2.38d mi>hr.
1 min
1 hr
63,360 in.
Vehicle
d
Speed≠2.38d
Taurus
26.16
62.26 mph
Charger
28.63
68.14 mph
Mariner
28.95
68.90 mph
1 mi
pd
pd in.
ba
b =
mi>rev, so each mile
(b) a
1 rev
63,360 in.
63,360
63,360
20,168
L
requires
revolutions.
pd
d
20,168
L 770.95 revolutions
Taurus:
26.16
20,168
L 696.65 revolutions
Mariner:
28.95
The Taurus must make just over 74 more revolutions.
a.
128°
44. Since C = pd, a tire travels a distance pd with each revolution.
(a) Each tire travels at a speed of 800 pd in. per minute, or
165
(c) In each revolution, the tire would cover a distance of
∏dnew rather than ∏dold, so that the car would travel
1∏dnew 2> 1 ∏dold 2 = dnew>dold = 28>26.16 L 1.07
miles for every mile the car’s instruments would show.
Both the odometer and speedometer readings would
be low.
9
rad and s2 = r2¨ =36
11
35. The angle is 10°
Angles and Their Measures
310°
b.
49.
47° 4 mi
38°
2 mi
50. 257 naut mi #
51. 895 stat mi #
3956p stat mi
≠296 statute miles
10,800 naut mi
10,800 naut mi
≠778 nautical miles
3956p stat mi
52. (a) Lane 5 has inside radius 37 m, while the inside radius
of lane 6 is 38 m, so over the whole semicircle,
the difference is 38∏-37∏=∏≠3.142 m.
(This would be the answer for any two adjacent lanes.)
(b) 38∏-33∏=5∏≠15.708 m.
166
Chapter 4
Trigonometric Functions
4
53. (a) s=r¨=(4)(4∏)=16∏≠50.265 in., or ∏
3
≠4.189 ft.
(b) r¨=2∏≠6.283 ft.
54. s=r¨=(52) a
55. (a) ◊1=120
p
13
b = ∏≠0.908 ft
180
45
rev #
rad # 1 min
2p
=4∏ rad/sec
min
rev 60 sec
(b) v=R◊1=(7 cm) a 4p
(c) ◊2=v/r= a 28p
56. (a) ◊=135
rad
b =28∏ cm/sec
sec
cm
b ÷(4 cm)=7∏ rad/sec
sec
rev #
rad # 1 min
2p
=4.5∏ rad/sec
min
rev 60 sec
(b) v=r◊=(1.2 m) a 4.5p
69. The difference in latitude is 44°59'-29°57'=15°02'
=902 minutes of arc, which is 902 naut mi.
70. The difference in latitude is 42°20'-33°45'=8°35'
=515 minutes of arc, which is 515 naut mi.
71. The whole circle’s area is pr2; the sector with central
¨
# pr2 = 1 ¨r2.
angle ¨ makes up ¨/2∏ of that area, or
2p
2
p
1
72. (a) A= (5.9)2 a b ≠3.481∏=10.936 ft2.
2
5
1
(b) A= (1.6)2(3.7)=4.736 km2.
2
73.
rad
b =5.4∏ m/sec
sec
1
(c) The radius to this halfway point is r*= r=0.6 m,
2
rad
b =2.7∏ m/sec.
so v=r*◊=(0.6 m) a 4.5p
sec
57. True. In the amount of time it takes for the merry-goround to complete one revolution, horse B travels a
distance of 2∏r, where r is B’s distance from the center.
In the same time, horse A travels a distance of
2∏(2r)=2(2∏r) — twice as far as B.
58. False. If all three radian measures were integers, their sum
would be an integer. But the sum must equal ∏, which is
not an integer.
59. x° = x° a
68. The difference in latitude is 47°36'-37°47'=9°49'
=589 minutes of arc, which is 589 naut mi.
∏ rad
∏x
b =
. The answer is C.
180 °
180
60. If the perimeter is 4 times the radius, the arc is two radii
long, which implies an angle of 2 radians. The answer is A.
61. Let n be the number of revolutions per minute.
n rev
60 min
1 mi
26∏ in.
ba
ba
ba
b
a
1 rev
1 min
1 hr
63,360 in.
L 0.07735 n mph.
Solving 0.07735 n=10 yields n L 129.
The answer is B.
62. The size of the circle does not affect the size of the
angle. The radius and the subtended arc length both
double, so that their ratio stays the same.
The answer is C.
In #63–66, we need to “borrow” 1° and change it to 60' in
order to complete the subtraction.
63. 122°25'-84°23'=38°02'
64. 117°09'-74°0'=43°09'
65. 93°16'-87°39'=92°76'-87°39'=5°37'
66. 122°20'-80°12'=42°08'
37°
340°
B
60 mi
A
74. Bike wheels: ◊1=v1>r= 166 ft>sec # 12 in.>ft2 ÷(14 in.)
≠56.5714 rad/sec. The wheel sprocket must have
the same angular velocity: ◊2 = ◊1≠56.5714 rad/sec.
For the pedal sprocket, we first need the velocity of the
chain, using the wheel sprocket: v2 L 1 32 in.2 156.5714 rad>sec 2
≠84.8571 in./sec. Then the pedal sprocket’s angular
velocity is ◊3= 1 84.8571 in.>sec 2 ÷(4.5 in.)≠18.9
rad/sec.
■ Section 4.2 Trigonometric Functions of
Acute Angles
Exploration 1
1. sin and csc, cos and sec, and tan and cot.
2. tan ¨
3. sec ¨
4. 1
5. sin ¨ and cos ¨
Exploration 2
1. Let ¨=60°. Then
13
sin ¨=
≠0.866
2
1
cos ¨=
2
tan ¨= 13≠1.732
csc ¨=
2
≠1.155
13
sec ¨=2
cot ¨=
1
≠0.577
13
In #67–70, find the difference in the latitude. Convert this difference to minutes; this is the distance in nautical miles. The
Earth’s diameter is not needed.
2. The values are the same, but for different functions. For
example, sin 30° is the same as cos 60°, cot 30° is the same
as tan 60°, etc.
67. The difference in latitude is 34°03'-32°43'=1°20'
=80 minutes of arc, which is 80 naut mi.
3. The value of a trig function at ¨ is the same as the value
of its co-function at 90°-¨.
Section 4.2
Quick Review 4.2
1. x = 252 + 52= 150=5 12
2. x = 282 + 12 2= 1208=4 113
3. x = 2102 - 82 = 6
4. x = 24 2 - 2 2 = 212 = 2 23
5. 8.4 ft # 12
6. 940 ft #
in
= 100.8 in.
ft
1 mi
47
=
L 0.17803 mi
5280 ft
264
7. a=(0.388)(20.4)=7.9152 km
8. b =
23.9
L 13.895 ft
1.72
9. Å=13.3 #
10. ı=5.9 #
2.4
≠1.0101 (no units)
31.6
6.15
≠4.18995 (no units)
8.66
Section 4.2 Exercises
4
3
4
5
5
1. sin ¨= , cos ¨= , tan ¨= , csc ¨= , sec ¨= ,
5
5
3
4
3
3
cot ¨= .
4
8
7
8
1113
, cos ¨=
, tan ¨= ; csc ¨=
,
7
8
1113
1113
7
1113
sec ¨=
, cot ¨= .
7
8
2. sin ¨=
12
5
12
13
, cos ¨= , tan ¨= ; csc ¨= ,
13
13
5
12
5
13
sec ¨= , cot ¨= .
5
12
3. sin ¨=
8
15
8
17
4. sin ¨= , cos ¨= , tan ¨= ; csc ¨= ,
17
17
15
8
17
15
sec ¨= , cot ¨= .
15
8
5. The hypotenuse length is 272 + 112 = 2170 , so
7
11
7
1170
sin ¨=
, cos ¨=
, tan ¨= ; csc ¨=
,
11
7
1170
1170
11
1170
sec ¨=
, cot ¨= .
11
7
6. The adjacent side length is 282 - 62 = 228 = 227, so
3
17
3
4
sin ¨= , cos ¨=
, tan ¨=
; csc ¨= ,
4
4
3
17
4
17
sec ¨=
, cot ¨=
.
3
17
7. The opposite side length is 2112 - 82 = 257 , so
157
8
157
11
sin ¨=
, cos ¨= , tan ¨=
; csc ¨=
,
11
11
8
157
11
8
sec ¨= , cot ¨=
.
8
157
Trigonometric Functions of Acute Angles
167
8. The adjacent side length is 2132 - 92 = 288 = 2222,
2 122
13
9
9
so sin ¨= , cos ¨=
, tan ¨=
; csc ¨= ,
13
13
9
2122
13
2122
sec ¨=
, cot ¨=
.
9
2122
9. Using a right triangle with hypotenuse 7 and legs 3
(opposite) and 272 - 32 = 240 = 2210 (adjacent),
3
3
2 110
we have sin ¨= , cos ¨=
, tan ¨=
;
7
7
2110
7
7
2110
csc ¨= , sec ¨=
, cot ¨=
.
3
3
2 110
10. Using a right triangle with hypotenuse 3 and legs 2
(opposite) and 232 - 2 2 = 25 (adjacent), we have
2
15
2
3
sin ¨= , cos ¨=
, tan ¨=
; csc ¨= ,
3
3
2
15
15
3
sec ¨=
, cot ¨=
.
2
15
11. Using a right triangle with hypotenuse 11 and legs 5
(adjacent) and 2112 - 52 = 296 = 4 26 (opposite),
416
5
416
we have sin ¨=
, cos ¨= , tan ¨=
;
11
11
5
11
11
5
csc ¨=
, sec ¨= , cot ¨=
.
5
416
416
12. Using a right triangle with hypotenuse 8 and legs 5
(adjacent) and 282 - 52 = 239 (opposite), we have
139
5
139
8
sin ¨=
, cos ¨= , tan ¨=
; csc ¨=
,
8
8
5
139
8
5
sec ¨= , cot ¨=
.
5
139
13. Using a right triangle with legs 5 (opposite) and
9 (adjacent) and hypotenuse 252 + 92 = 2106, we have
5
9
5
1106
sin ¨=
, cos ¨=
, tan ¨= ; csc ¨=
,
9
5
1106
1106
9
1106
sec ¨=
, cot ¨= .
9
5
14. Using a right triangle with legs 12 (opposite) and
13 (adjacent) and hypotenuse 212 2 + 132 = 2313,
12
13
12
we have sin ¨=
, cos ¨=
, tan ¨= ;
13
1313
1313
1313
1313
13
csc ¨=
, sec ¨=
, cot ¨= .
12
13
12
15. Using a right triangle with legs 3 (opposite) and
11 (adjacent) and hypotenuse 232 + 112 = 2130 ,
11
3
3
we have sin ¨=
, cos ¨=
, tan ¨= ;
11
1130
1130
11
1130
1130
csc ¨=
, sec ¨=
, cot ¨= .
3
11
3
168
Chapter 4
Trigonometric Functions
16. Using a right triangle with hypotenuse 12 and legs
45. ¨=60°=
p
3
46. ¨=45°=
p
4
47. ¨=30°=
p
6
48. ¨=30°=
p
6
5 (opposite) and 212 2 - 52 = 2119 (adjacent),
5
5
1119
we have sin ¨= , cos ¨=
, tan ¨=
;
12
12
1119
12
12
1119
csc ¨= , sec ¨=
, cot ¨=
.
5
5
1119
17. Using a right triangle with hypotenuse 23 and legs
50. z =
51. y =
32
L 20.78
tan 57°
52. x = 14 sin 43° L 9.55
53. y = 6>sin 35° L 10.46
9
8 17
9
we have sin ¨= , cos ¨=
, tan ¨=
;
23
23
8 17
For #55–58, choose whichever of the following formulas is
appropriate:
23
23
8 17
, sec ¨=
, cot ¨=
.
9
9
8 17
5 (adjacent) and 2172 - 52 = 2264 = 2 266
5
2166
, cos ¨= ,
17
17
2 166
17
17
5
tan ¨=
; csc ¨=
, sec ¨= , cot ¨=
.
5
5
2166
2 166
(opposite), we have sin ¨=
13
2
20. 1
21. 13
22. 2
1
12
=
2
12
24.
54. x = 50 cos 66° L 20.34
b
tan ı
a
b = 2c2 - a2=c cos Å=c sin ı=a tan ı=
tan Å
a
a
b
b
c = 2a2 + b2 =
=
=
=
cos ı
sin Å
sin ı
cos Å
If one angle is given, subtract from 90° to find the other angle.
a = 2c2 - b2=c sin Å=c cos ı=b tan Å=
18. Using a right triangle with hypotenuse 17 and legs
23.
15
L 26.82
sin 34°
9 (opposite) and 2232 - 92 = 2448 = 8 27 (adjacent),
csc ¨=
19.
23
L 29.60
cos 39°
49. x =
2
2 13
=
3
13
25. sec 45° = 1>cos 45° L 1.4142. Squaring this result yields
2.0000, so sec 45° = 12.
26. sin 60° L 0.8660. Squaring this result yields
0.7500=3/4, so sin 60° = 13>4 = 13>2.
27. csc 1∏>32 = 1>sin 1 ∏>3 2 L 1.1547. Squaring this result
yields 1.3333 or essentially 4/3, so
csc 1∏>32 = 14>3 = 2> 13 = 213>3.
28. tan 1 ∏>32 L 1.73205. Squaring this result yields 3.0000, so
tan 1 ∏>32 = 13.
For #29–40, the answers marked with an asterisk (*) should be
found in DEGREE mode; the rest should be found in RADIAN
mode. Since most calculators do not have the secant, cosecant,
and cotangent functions built in, the reciprocal versions of
these functions are shown.
29. ≠0.961*
30. ≠0.141*
31. ≠0.943*
32. ≠0.439*
33. ≠0.268
34. ≠0.208
1
L 1.524*
35.
cos 49°
1
L 3.072*
36.
sin 19°
12.3
a
=
L 33.79,
tan Å
tan 20°
a
12.3
c =
=
L 35.96, ı = 90° - Å = 70°
sin Å
sin 20°
55. b =
56. a=c sin Å=10 sin 41°≠6.56,
b=c cos Å=10 cos 41°≠7.55, ı=90°-Å=49°
57. b=a tan ı=15.58 tan 55°≠22.25,
c =
a
15.58
=
L 27.16, Å = 90° - ı = 35°
cos b
cos 55°
58. b=a tan ı=5 tan 59°≠8.32,
a
5
c =
=
L 9.71, Å=90°-ı=31°
cos ı
cos 59°
59. 0. As ¨ gets smaller and smaller, the side opposite ¨ gets
smaller and smaller, so its ratio to the hypotenuse
approaches 0 as a limit.
60. 1. As ¨ gets smaller and smaller, the side adjacent to ¨
approaches the hypotenuse in length, so its ratio to the
hypotenuse approaches 1 as a limit.
61. h=55 tan 75°≠205.26 ft
62. h=5+120 tan 8°≠21.86 ft
63. A = 12 #
5
L 74.16 ft2
sin 54°
64. h=130 tan 82.9°≠1043.70 ft
65. AC=100 tan 75°1242≠378.80 ft
37.
1
L 0.810
tan 0.89
38.
1
L 3.079
cos 1.24
39.
1
L 2.414
tan1 p>8 2
40.
1
L 3.236
sin1 p>10 2
41. ¨=30°=
p
6
42. ¨=60°=
p
3
43. ¨=60°=
p
3
44. ¨=45°=
p
4
66. Connect the three points on the arc to the center of the
circle, forming three triangles, each with hypotenuse 10 ft.
The horizontal legs of the three triangles have lengths
10 cos 67.5°≠3.827, 10 cos 45°≠7.071, and
10 cos 22.5°≠9.239. The widths of the four strips
are therefore,
3.827-0=3.827 (strip A)
7.071-3.827=3.244 (strip B)
9.239-7.071=2.168 (strip C)
10-9.239=0.761 (strip D)
Allen needs to correct his data for strips B and C.
Section 4.3
67. False. This is only true if ¨ is an acute angle in a right
triangle. (Then it is true by definition.)
68. False. The larger the angle of a triangle, the smaller its
cosine.
1
1
= is undefined. The answer is E.
69. sec 90° =
cos 90°
0
70. sin ¨ =
opp
hyp
=
3
. The answer is A.
5
71. If the unknown slope is m, then m sin ¨ =–1, so
1
m = = - csc u. The answer is D.
sin u
Trigonometry Extended: The Circular Functions
■ Section 4.3 Trigonometry Extended: The
Circular Functions
Exploration 1
1. The side opposite ¨ in the triangle has length y and the
hypotenuse has length r. Therefore
opp
y
sin ¨ =
= .
hyp
r
2. cos ¨ =
adj
hyp
=
x
r
=
y
x
opp
72. For all ¨, –1 cos ¨ 1. The answer is B.
3. tan ¨ =
73. For angles in the first quadrant, sine values will be
increasing, cosine values will be decreasing and only
tangent values can be greater than 1. Therefore, the first
column is tangent, the second column is sine, and the third
column is cosine.
r
x
r
4. cot ¨= ; sec ¨= ; csc ¨=
y
x
y
74. For angles in the first quadrant, secant values will be
increasing, and cosecant and cotangent values will be
decreasing. We recognize that csc (30°)=2. Therefore,
the first column is secant, the second column is cotangent,
and the third column is cosecant.
75. The distance dA from A to the mirror is 5 cos 30°; the distance from B to the mirror is dB=dA-2. Then
dA - 2
dB
2
=
= 5 PB =
cos ı
cos 30°
cos 30°
4
L 2.69 m.
= 5 13
76. Let P be the point at which we should aim; let Å and ı be
the angles as labeled in #73. Since Å=ı, tan Å=tan ı.
P should be x inches to the right of C, where x is chosen
x
30 - x
so that tan Å=
. Then
= tan ı=
15
10
10x=15(30-x), so 25x=450, which gives x=18.
Aim 18 in. to the right of C (or 12 in. to the left of D).
77. One possible proof:
a 2
b 2
1sin ¨ 2 + 1 cos ¨ 2 = a b + a b
c
c
a2
b2
= 2 + 2
c
c
a 2 + b2
=
c2
2
c
= 2
(Pythagorean theorem: a2+b2=c2.)
c
= 1
2
2
78. Let h be the length of the altitude to base b and denote
the area of the triangle by A. Then
h
= sin ¨
a
‹ h=a sin ¨
1
Since A= bh, we can substitute h=a sin ¨ to get
2
1
A = ab sin ¨.
2
169
adj
Exploration 2
1. The x-coordinates on the unit circle lie between –1 and 1,
and cos t is always an x-coordinate on the unit circle.
2. The y-coordinates on the unit circle lie between –1 and 1,
and sin t is always a y-coordinate on the unit circle.
3. The points corresponding to t and –t on the number line
are wrapped to points above and below the x-axis with the
same x-coordinates. Therefore cos t and cos (–t) are equal.
4. The points corresponding to t and –t on the number line
are wrapped to points above and below the x-axis with
exactly opposite y-coordinates. Therefore sin t and
sin (–t) are opposites.
5. Since 2∏ is the distance around the unit circle, both t and
t+2∏ get wrapped to the same point.
6. The points corresponding to t and t+∏ get wrapped to
points on either end of a diameter on the unit circle.
These points are symmetric with respect to the origin and
therefore have coordinates (x, y) and (–x, –y).
Therefore sin t and sin (t+∏) are opposites, as are cos t
and cos (t+∏).
7. By the observation in (6), tan t and tan(t+∏) are ratios
-y
y
of the form and
, which are either equal to each
x
-x
other or both undefined.
8. The sum is always of the form x2 + y2 for some (x, y) on
the unit circle. Since the equation of the unit circle is
x2 + y2 = 1, the sum is always 1.
9. Answers will vary. For example, there are similar statements that can be made about the functions cot, sec,
and csc.
Quick Review 4.3
1. –30°
2. –150°
3. 45°
4. 240°
5. tan
p
13
1
=
=
6
3
13
6. cot
p
=1
4
170
Chapter 4
7. csc
p
= 12
4
8. sec
p
=2
3
Trigonometric Functions
9. Using a right triangle with hypotenuse 13 and legs
5 (opposite) and 2132 - 52 = 12 (adjacent), we have
5
12
5
13
sin ¨= , cos ¨= , tan ¨= ; csc ¨= ,
13
13
12
5
13
12
sec ¨= , cot ¨= .
12
5
5
2
2
, cos ¨=
, tan ¨=– ;
5
129
129
5
129
129
csc ¨=–
, sec ¨=
, cot ¨=– .
2
5
2
11. sin ¨=–
1
1
, cos ¨=
, tan ¨=–1;
12
12
csc ¨=– 12, sec ¨= 12, cot ¨=–1.
12. sin ¨=–
10. Using a right triangle with hypotenuse 17 and legs
15 (adjacent) and 2172 - 152 = 8 (opposite), we have
8
15
8
17
sin ¨= , cos ¨= , tan ¨= ; csc ¨= ,
17
17
15
8
15
17
sec ¨= , cot ¨= .
15
8
For #13–16, determine the quadrant(s) of angles with the given
measures, and then use the fact that sin t is positive when the
terminal side of the angle is above the x-axis (in Quadrants I
and II) and cos t is positive when the terminal side of the angle
is to the right of the y-axis (in quadrants I and IV). Note that
since tan t=sin t>cos t, the sign of tan t can be determined
from the signs of sin t and cos t: if sin t and cos t have the same
sign, the answer to (c) will be ‘ +’; otherwise it will be ‘–’. Thus
tan t is positive in Quadrants I and III.
Section 4.3 Exercises
13. These angles are in Quadrant I. (a)+(i.e., sin t 7 0).
(b)+(i.e., cos t 7 0). (c)+(i.e., tan t 7 0).
1. The 450° angle lies on the positive–y axis
(450°-360°=90°), while the others are all
coterminal in Quadrant II.
5p
5p
p
+ 2p = ≤ ,
angle lies in Quadrant I ¢ 3
3
3
while the others are all coterminal in Quadrant IV.
2. The -
In #3–12, recall that the distance from the origin is
r= 2x2 + y2.
2
1
15
, cos ¨=–
, tan ¨=–2; csc ¨=
,
2
15
15
1
sec ¨=– 15, cot ¨=– .
2
3. sin ¨=
3
4
3
5
4. sin ¨=– , cos ¨= , tan ¨=– ; csc ¨=– ,
5
5
4
3
5
4
sec ¨= , cot ¨=– .
4
3
1
1
, cos ¨=–
, tan ¨=1; csc ¨=– 12,
12
12
sec ¨=– 12, cot ¨=1.
5. sin ¨=–
5
3
5
, cos ¨=
, tan ¨=– ;
3
134
134
3
134
134
csc ¨=–
, sec ¨=
, cot ¨=– .
5
3
5
6. sin ¨=–
4
3
4
5
7. sin ¨= , cos ¨= , tan ¨= ; csc ¨= ,
5
5
3
4
3
5
sec ¨= , cot ¨= .
3
4
3
2
3
, cos ¨=–
, tan ¨= ;
2
113
113
113
113
2
csc ¨=–
, sec ¨=–
, cot ¨= .
3
2
3
8. sin ¨=–
9. sin ¨=1, cos ¨=0, tan ¨ undefined; csc ¨=1,
sec ¨ undefined, cot ¨=0.
10. sin ¨=0, cos ¨=–1, tan ¨=0; csc ¨ undefined,
sec ¨=–1, cot ¨ undefined.
14. These angles are in Quadrant II. (a) + . (b) –. (c) –.
15. These angles are in Quadrant III. (a) –. (b) –. (c) + .
16. These angles are in Quadrant IV. (a) –. (b) + . (c) –.
For #17–20, use strategies similar to those for the previous
problem set.
17. 143° is in Quadrant II, so cos 143° is negative.
18. 192° is in Quadrant III, so tan 192° is positive.
19.
20.
21.
22.
23.
24.
7p
7p
rad is in Quadrant II, so cos
is negative.
8
8
4p
4p
rad is in Quadrant II, so tan
is negative.
5
5
y
= 1 1 y = x.
A (2, 2); tan 45° =
x
y
2p
2p
=
= - 13.
B (–1, 13); tan
is in Quadrant II,
3
x
3
so x is negative.
7p
C (– 13, –1);
is in Quadrant III, so x and y are both
6
7p
1
=
.
negative. tan
6
13
D (1, – 13); –60º is in Quadrant IV, so x is positive
while y is negative. tan 1 - 60°2 = - 13.
For #25–36, recall that the reference angle is the acute angle
formed by the terminal side of the angle in standard position
and the x-axis.
25. The reference angle is 60°. A right triangle with a 60°
angle at the origin has the point P(–1, 13) as one vertex,
x
1
with hypotenuse length r=2, so cos 120°= =– .
r
2
26. The reference angle is 60°. A right triangle with a 60°
angle at the origin has the point P(1, – 13) as one vertex,
y
so tan 300°= =– 13.
x
Section 4.3
27. The reference angle is the given angle,
p
. A right triangle
3
p
radian angle at the origin has the point P(1, 13)
3
as one vertex, with hypotenuse length r=2, so
r
p
sec = =2.
3
x
with a
p
p
. A right triangle with a radian
4
4
angle at the origin has the point P(1, 1) as one vertex,
r
3p
with hypotenuse length r= 12, so csc
= = 12.
4
y
28. The reference angle is
p
(in fact, the given angle is
6
p
p
coterminal with ). A right triangle with a radian
6
6
angle at the origin has the point P( 13, 1) as one vertex,
y
1
13p
with hypotenuse length r=2, so sin
= = .
6
r
2
29. The reference angle is
p
(in fact, the given angle is
3
p
p
coterminal with ). A right triangle with a radian
3
3
angle at the origin has the point P(1, 13) as one vertex,
1
7p
x
with hypotenuse length r=2, so cos
= = .
3
r
2
30. The reference angle is
p
(in fact, the given angle is
4
p
p
coterminal with ). A right triangle with a radian
4
4
angle at the origin has the point P(1, 1) as one vertex,
y
-15p
so tan
= =1.
4
x
31. The reference angle is
p
p
. A right triangle with a radian
4
4
angle at the origin has the point P(–1, –1) as one vertex,
x
13p
so cot
= =1.
4
y
32. The reference angle is
33. cos
13
23p
11p
=cos
=
6
6
2
34. cos
12
17p
p
=cos =
4
4
2
35. sin
13
11p
5p
=sin
=–
3
3
2
36. cot
19p
7p
=cot
= 13
6
6
37. –450° is coterminal with 270°, on the negative y-axis.
(a) –1 (b) 0 (c) Undefined
38. –270° is coterminal with 90°, on the positive y-axis.
(a) 1 (b) 0 (c) Undefined
39. 7∏ radians is coterminal with ∏ radians, on the negative
x-axis. (a) 0 (b) –1 (c) 0
11p
3p
40.
radians is coterminal with
radians, on the
2
2
negative y-axis. (a) –1 (b) 0 (c) Undefined
Trigonometry Extended: The Circular Functions
41.
171
p
-7p
radians is coterminal with radians, on the positive
2
2
y-axis. (a) 1 (b) 0 (c) Undefined
42. –4∏ radians is coterminal with 0 radians, on the positive
x-axis. (a) 0 (b) 1 (c) 0
43. Since cot ¨ 7 0, sin ¨ and cos ¨ have the same sign, so
sin ¨
15
15
=
sin ¨=+ 21 - cos2 ¨ =
, and tan ¨=
.
3
cos ¨
2
44. Since tan ¨ 6 0, sin ¨ and cos ¨ have opposite signs,
115
so cos ¨=– 21 - sin2 ¨ =–
, and
4
cos u
cot ¨=
=– 115.
sin u
121
sin ¨
, so tan ¨=
5
cos ¨
5
2
1
=
.
=–
and sec ¨=
cos ¨
121
121
46. sec ¨ has the same sign as cos ¨, and since cot ¨ 7 0,
sin ¨ must also be negative. With x=–3, y=–7, and
7
r= 232 + 72 = 158, we have sin ¨=–
and
158
3
cos ¨=–
.
158
45. cos ¨=+ 21 - sin2 ¨ =
47. Since cos ¨ 6 0 and cot ¨ 6 0, sin ¨ must be positive.
With x=–4, y=3, and r= 24 2 + 32 = 5, we have
5
5
sec ¨=– and csc ¨= .
4
3
48. Since sin ¨ 7 0 and tan ¨ 6 0, cos ¨ must be negative.
With x=–3, y=4, and r= 24 2 + 32 = 5, we have
3
5
csc ¨= and cot ¨=– .
4
4
49. sin a
p
1
p
+ 49,000p b =sin a b =
6
6
2
50. tan (1,234,567∏)-tan (7,654,321∏)
=tan (∏)-tan (∏)=0
51. cos a
5,555,555p
p
b = cos a b = 0
2
2
52. tan a
3p - 70,000p
3p
b =tan a
b =undefined.
2
2
53. The calculator’s value of the irrational number ∏ is necessarily an approximation. When multiplied by a very large
number, the slight error of the original approximation is
magnified sufficiently to throw the trigonometric functions off.
54. sin t is the y-coordinate of the point on the unit circle
after measuring counterclockwise t units from (1, 0). This
will repeat every 2∏ units (and not before), since the
distance around the circle is 2∏.
55. Â=
sin 83°
≠1.69
sin 36°
56. sin ¨™=
sin 42°
≠0.44
1.52
57. (a) When t=0, d=0.4 in.
(b) When t=3, d=0.4e–0.6 cos 12≠0.1852 in.
172
Chapter 4
Trigonometric Functions
58. When t=0, ¨=0.25 (rad). When t=2.5,
¨=0.25 cos 2.5≠–0.2003 rad.
59. The difference in the elevations is 600 ft, so d=600/sin ¨.
Then:
(a) d=600 12≠848.53 ft.
72. These coordinates give the lengths of the legs of the triangles from #71, and these triangles are congruent. For
example, the length of the horizontal leg of the triangle
with vertex P is given by the (absolute value of the) xcoordinate of P; this must be the same as the (absolute
value of the) y-coordinate of Q.
(b) d=600 ft.
y
(c) d≠933.43 ft.
p
=103.25.
6
2p
April (t=4): 72.4+61.7 sin
≠125.83.
3
June (t=6): 72.4+61.7 sin ∏=72.4.
5p
October (t=10): 72.4+61.7 sin
≠18.97.
3
December (t=12): 72.4+61.7 sin 2∏=72.4. June and
December are the same; perhaps by June most people
have suits for the summer, and by December they are
beginning to purchase them for next summer (or as
Christmas presents, or for mid-winter vacations).
60. January (t=1): 72.4+61.7 sin
61. True. Any angle in a triangle measures between 0° and
180°. Acute angles <
( 90°) determine reference triangles
in Quadrant I, where the cosine is positive, while obtuse
angles (>90°) determine reference triangles in
Quadrant II, where the cosine is negative.
62. True. The point determines a reference triangle in
Quadrant IV, with r = 282 + 1 -6 2 2 = 10. Thus
sin ¨=y/x=–6/10 =–0.6
63. If sin ¨=0.4, then sin (–¨)+csc ¨=–sin ¨+
=–0.4+
1
sin u
1
=2.1. The answer is E.
0.4
64. If cos ¨=0.4, then cos (¨+∏)=–cos ¨=–0.4. The
answer is B.
65. (sin t)2+(cos t)2=1 for all t. The answer is A.
66. sin ¨= - 21 - cos2 ¨, because tan ¨
=(sin ¨)/(cos ¨)>0. So sin ¨ = -
B
1 -
25
12
=- .
169
13
Q(–b, a)
t+ π
2
t
P(a, b)
t
(1, 0)
x
73. One possible answer: Starting from the point (a, b) on the
unit circle—at an angle of t, so that cos t=a—then measuring a quarter of the way around the circle (which corresponds to adding ∏> 2 to the angle), we end at (–b, a),
so that sin 1t + p>2 2 = a. For (a, b) in Quadrant I, this is
shown in the figure above; similar illustrations can be
drawn for the other quadrants.
74. One possible answer: Starting from the point (a, b) on the
unit circle—at an angle of t, so that sin t=b—then measuring a quarter of the way around the circle (which corresponds to adding ∏> 2 to the angle), we end at (–b, a), so
that cos 1 t + p>22 = –b=–sin t. For (a, b) in Quadrant I,
this is shown in the figure above; similar illustrations can
be drawn for the other quadrants.
75. Starting from the point (a, b) on the unit circle—at an
angle of t, so that cos t=a—then measuring a quarter of
the way around the circle (which corresponds to adding
∏>2 to the angle), we end at (–b, a), so that sin 1t + p>22 =
a. This holds true when (a, b) is in Quadrant II, just as it
did for Quadrant I.
y
The answer is A.
67. Since sin ¨ 7 0 and tan ¨ 6 0, the terminal side
5p
must be in Quadrant II, so ¨= .
6
68. Since cos ¨ 7 0 and sin ¨ 6 0, the terminal side must be
11p
in Quadrant IV, so ¨=
.
6
69. Since tan ¨ 6 0 and sin ¨ 6 0, the terminal side must be
7p
in Quadrant IV, so ¨= .
4
70. Since sin ¨ 6 0 and tan ¨ 7 0, the terminal side must be
5p
in Quadrant III, so ¨= .
4
71. The two triangles are congruent: both have hypotenuse 1,
and the corresponding angles are congruent—the smaller
acute angle has measure t in both triangles, and the two
acute angles in a right triangle add up to ∏> 2.
P(a, b)
t
t+π
(1, 0)
x
2
Q(–b, a)
76. (a) Both triangles are right triangles with hypotenuse 1,
and the angles at the origin are both t (for the triangle
on the left, the angle is the supplement of ∏-t).
Therefore the vertical legs are also congruent; their
lengths correspond to the sines of t and ∏-t.
(b) The points P and Q are reflections of each other
across the y-axis, so they are the same distance (but
opposite directions) from the y-axis. Alternatively, use
the congruent triangles argument from part (a).
Section 4.4
77. Seven decimal places are shown so that the slight
differences can be seen. The magnitude of the relative
error is less than 1% when œ¨œ 6 0.2441 (approximately).
This can be seen by extending the table to larger values
sin ¨ - ¨
of ¨, or by graphing `
` -0.01.
sin ¨
78. Let (x, y) be the coordinates of the point that corresponds
to t under the wrapping. Then
x2 + y2
y 2
1
1+(tan t)2=1+ a b =
= 2=(sec t)2.
x
x2
x
(Note that x2+y2=1 because (x, y) is on the unit circle.)
79. This Taylor polynomial is generally a very
good approximation for sin ¨—in fact, the
relative error (see #77) is less than 1% for
œ¨œ 6 1 (approx.). It is better for ¨ close to 0;
it is slightly larger than sin ¨ when ¨ 6 0 and
slightly smaller when ¨ 7 0.
80. This Taylor polynomial is generally
a very good approximation for cos ¨
—in fact, the relative error (see #77)
is less than 1% for œ¨œ 6 1.2 (approx.).
It is better for ¨ close to 0; it is slightly
larger than cos ¨ when ¨ Z 0.
Graphs of Sine and Cosine: Sinusoids
77.
¨
sin ¨
–0.03
–0.0299955
0.0000045
0.0001500
–0.02
–0.0199987
0.0000013
0.0000667
–0.01
–0.0099998
0.0000002
0.0000167
0
0
0.01
0.0099998
–0.0000002
0.0000167
0.02
0.0199987
–0.0000013
0.0000667
0.03
0.0299955
–0.0000045
0.0001500
¨
¨
2 sin ¨ - ¨ 2
sin ¨
sin ¨-¨
0
79.
—
1
sin ¨- ¢ ¨- ¨3 ≤
6
1
¨- ¨3
6
sin ¨
–0.3
–0.2955202
–0.2955000
–0.0000202
–0.2
–0.1986693
–0.1986667
–0.0000027
–0.1
–0.0998334
–0.0998333
–0.0000001
cos ¨
0
0
0
0
0.1
0.0998334
0.0998333
0.0000001
0.2
0.1986693
0.1986667
0.0000027
0.3
0.2955202
0.2955000
0.0000202
1
1
1- ¨2+ ¨4
2
24
1
1
cos ¨-(1- ¨2+ ¨4)
2
24
–0.3
0.9553365
0.9553375
–0.0000010
–0.2
0.9800666
0.9800667
–0.0000001
–0.1
–0.0000000
0.9950042
0.9950042
0
1
1
0.1
0.9950042
0.9950042
–0.0000000
0.2
0.9800666
0.9800667
–0.0000001
0.3
0.9553365
0.9553375
–0.0000010
■ Section 4.4 Graphs of Sine and Cosine:
Sinusoids
173
0
Quick Review 4.4
1. In order: +,+,-,2. In order: +,-,-,+
Exploration 1
1. ∏/2 (at the point (0, 1))
3. In order: +,-,+,∏
3∏
=
180°
4
2. 3∏/2 (at the point (0, –1))
4. 135° #
3. Both graphs cross the x-axis when the y-coordinate on the
unit circle is 0.
5. - 150° #
4. (Calculator exploration)
5. The sine function tracks the y-coordinate of the point as it
moves around the unit circle. After the point has gone
completely around the unit circle (a distance of 2∏), the
same pattern of y-coordinates starts over again.
6. Leave all the settings as they are shown at the start of the
Exploration, except change Y2T to cos(T).
6. 450° #
∏
5∏
= 180°
6
5∏
∏
=
180°
2
7. Starting with the graph of y1, vertically stretch by 3 to
obtain the graph of y2.
8. Starting with the graph of y1, reflect across y-axis to
obtain the graph of y2.
9. Starting with the graph of y1, vertically shrink by 0.5 to
obtain the graph of y2.
174
Chapter 4
Trigonometric Functions
10. Starting with the graph of y1, translate down 2 units to
obtain the graph of y2.
15. For y=–(3/2) sin 2x, the amplitude is 3/2, the period is
2∏/2=∏, and the frequency is 1/∏.
Section 4.4 Exercises
In #1–6, for y=a sin x, the amplitude is | a |. If | a |>1, there
is a vertical stretch by a factor of | a |, and if | a |<1, there is
a vertical shrink by a factor of | a |. When a<0, there is also a
reflection across the x-axis.
1. Amplitude 2; vertical stretch by a factor of 2.
2. Amplitude 2/3; vertical shrink by a factor of 2/3.
3. Amplitude 4; vertical stretch by a factor of 4, reflection
across the x-axis.
[–3, 3] by [–4, 4]
16. For y=–4 sin (2x/3), the amplitude is 4, the period is
2∏/(2/3)=3∏, and the frequency is 1/(3∏).
4. Amplitude 7/4; vertical stretch by a factor of 7/4, reflection across the x-axis.
5. Amplitude 0.73; vertical shrink by a factor of 0.73.
6. Amplitude 2.34; vertical stretch by a factor of 2.34, reflection across the x-axis.
In #7–12, for y=cos bx, the period is 2∏/| b | . If | b |>1,
there is a horizontal shrink by a factor of 1/| b |, and if
| b |<1, there is a horizontal stretch by a factor of 1/| b |.
When b<0, there is also a reflection across the y-axis. For
y=a cos bx, a has the same effects as in #1–6.
[–3, 3] by [–4, 4]
Note: the frequency for each graph in #17–22 is 1/(2∏).
17. Period 2∏, amplitude=2
18. Period 2∏,
amplitude=2.5
y
7. Period 2∏/3; horizontal shrink by a factor of 1/3.
8. Period 2∏/(1/5)=10∏; horizontal stretch by a factor of
1/(1/5)=5.
9. Period 2∏/7; horizontal shrink by a factor of 1/7, reflection across the y-axis.
y
2
2.5
␲
–␲
x
␲
–␲
–2
x
–2.5
10. Period 2∏/0.4=5∏; horizontal stretch by a factor of
1/0.4=2.5, reflection across the y-axis.
11. Period 2∏/2=∏; horizontal shrink by a factor of 1/2.
Also a vertical stretch by a factor of 3.
19. Period 2∏, amplitude=3
y
y
12. Period 2∏/(2/3)=3∏; horizontal stretch by a factor of
1/(2/3)=3/2. Also a vertical shrink by a factor of 1/4.
In #13–16, the amplitudes of the graphs for y=a sin bx and
y=a cos bx are governed by a, while the period is governed
by b, just as in #1–12. The frequency is 1/period.
20. Period 2∏,
amplitude=2
3
2
␲
–␲
x
–3
13. For y=3 sin (x/2), the amplitude is 3, the period is
2∏/(1/2)=4∏, and the frequency is 1/(4∏).
␲
–␲
x
–2
21. Period 2∏, amplitude=0.5 22. Period 2∏,
amplitude=4
y
y
4
0.5
[–3, 3] by [–4, 4]
14. For y=2 cos (x/3), the amplitude is 2, the period is
2∏/(1/3)=6∏, and the frequency is 1/(6∏).
[–3, 3] by [–4, 4]
␲
–␲
–0.5
x
␲
–␲
–4
x
Section 4.4
23. Period ∏, amplitude=5,
frequency=1/∏
24. Period 4∏,
amplitude=3,
frequency = 1/(4∏)
y
y
5
x
1.5␲
–6␲
6␲
x
–3
–5
175
p
3∏
≤ ; minimum: –1 (at 0, —∏, — 2∏).
39. Maximum: 1 ¢at — , —
2
2
7p
p
3p
5p
Zeros: — , —
,—
,—
.
4
4
4
4
40. Maximum: 2 ¢at -
3
–1.5␲
Graphs of Sine and Cosine: Sinusoids
∏ 3∏
3∏ p
≤ ; minimum: –2 ¢at – , ≤ .
,
2 2
2 2
Zeros: 0, —∏, —2∏.
41. y=sin x has to be translated left or right by an odd multiple of ∏. One possibility is y=sin (x + ∏).
p
plus an even
2
multiple of ∏. One possibility is y=sin (x-∏/2).
42. y=sin x has to be translated right by
26. Period ∏/2,
amplitude=20,
frequency = 2/∏
25. Period 2∏/3,
amplitude=0.5,
frequency=3/2∏
y
y
␲
x
–0.75␲
–0.5
x
0.75␲
–20
27. Period 8∏,
amplitude=4,
frequency=1/(8∏)
28. Period 2∏/5,
amplitude=8,
frequency = 5/(2∏)
y
1
and
3
vertically shrink by 0.5. The period is 2∏/3.
3 3
2∏ 2∏
,
d by c - , d .
Possible window: c 3 3
4 4
20
0.5
–␲
43. Starting from y=sin x, horizontally shrink by
44. Starting from y=cos x, horizontally shrink
1
by and vertically stretch by 1.5. The period is ∏/2.
4
∏ ∏
Possible window: c - , d by [–2, 2].
2 2
y
8
4
–12␲
12␲
x
0.6␲
x
–0.6␲
–8
–4
[ –2∏ , 2∏ ] by [–0.75, 0.75]
3
3
For #43
29. Period ∏; amplitude 1.5; [–2∏, 2∏] by [–2, 2]
30. Period 2∏/3; amplitude 2; c -
2∏ 2∏
,
d by [–4, 4]
3 3
31. Period ∏; amplitude 3; [–2∏, 2∏] by [–4, 4]
32. Period 4∏; amplitude 5; [–4∏, 4∏] by [–10, 10]
33. Period 6; amplitude 4; [–3, 3] by [–5, 5]
[ –∏ , ∏ ] by [–0.75, 0.75]
2
2
For #44
45. Starting from y=cos x, horizontally stretch by 3,
2
vertically shrink by , reflect across x-axis. The period
3
is 6∏. Possible window: [–6∏, 6∏] by [–1, 1].
46. Starting from y=sin x, horizontally stretch by 5 and
3
vertically shrink by . The period is 10∏. Possible
4
window: [–10∏, 10∏] by [–1, 1].
34. Period 2; amplitude 3; [–4, 4] by [–5, 5]
3∏
∏
and ≤ ;
2
2
3∏
∏
minimum: –2 ¢at – and ≤ .
2
2
Zeros: 0, —∏, —2∏.
35. Maximum: 2 ¢at –
36. Maximum: 3 (at 0); minimum: –3 (at —2∏). Zeros: —∏.
37. Maximum: 1 (at 0, —∏, —2∏); minimum:
7p
p
3∏
p
3p
5p
≤ . Zeros: — , —
–1 ¢at — and —
,—
,—
.
2
2
4
4
4
4
1
3∏
∏
¢at –
and ≤ ;
2
2
2
1
∏
3∏
≤ . Zeros: 0, —∏, —2∏.
minimum: - ¢at - and
2
2
2
38. Maximum:
[–6∏, 6∏] by [–1, 1]
For #45
[–10∏, 10∏] by [–1, 1]
For #46
3
and
2∏
vertically stretch by 3. The period is 3. Possible window:
[–3, 3] by [–3.5, 3.5].
47. Starting from y=cos x, horizontally shrink by
176
Chapter 4
Trigonometric Functions
4
,
∏
vertically stretch by 2, and reflect across x-axis. The period
is 8. Possible window: [–8, 8] by [–3, 3].
48. Starting from y=sin x, horizontally stretch by
65. Amplitude 2, period 1, phase shift 0, vertical translation
1 unit up.
2
66. Amplitude 4, period , phase shift 0, vertical translation
3
2 units down.
5
7
67. Amplitude , period 2∏, phase shift - , vertical transla3
2
tion 1 unit down.
[–8, 8] by [–3, 3]
[–3, 3] by [–3.5, 3.5]
For #47
For #48
5
49. Starting with y1, vertically stretch by .
3
50. Starting with y1, translate right
∏
units and vertically
12
1
shrink by .
2
69. y=2 sin 2x (a=2, b=2, h=0, k=0).
70. y=3 sin[2(x+0.5)] (a=3, b=2, h=0.5, k=0).
71. (a) There are two points of intersection in that interval.
(b) The coordinates are (0, 1) and (2∏, 1.3–2∏)
≠(6.28, 0.19). In general, two functions intersect
where cos x=1, i.e., x=2n∏, n an integer.
72. a=4 and b =
1
51. Starting with y1, horizontally shrink by .
2
52. Starting with y1, horizontally stretch by 2 and vertically
2
shrink by .
3
For #53–56, graph the functions or use facts about sine and
cosine learned to this point.
53. (a) and (b)
54. (a) and (b)
55. (a) and (b) — both functions equal cos x
56. (a) and (c) — sin a 2x +
=sin c a 2x -
2
68. Amplitude , period 8∏, phase shift 3, vertical translation
3
1 unit up.
∏
b
4
∏
∏
∏
b + d =cos a 2x - b
4
2
4
In #57–60, for y=a sin (b(x-h)), the amplitude is | a |, the
period is 2∏/| b |, and the phase shift is h.
57. One possibility is y=3 sin 2x.
58. One possibility is y=2 sin (2x/3).
59. One possibility is y=1.5 sin 12(x-1).
60. One possibility is y=3.2 sin 14(x-5)
∏
61. Amplitude 2, period 2∏, phase shift , vertical translation
4
1 unit up.
∏
b d - 1.
4
∏
Amplitude 3.5, period ∏, phase shift , vertical translation
4
1 unit down.
2∏
4∏
.
=
3.5
7
73. The height of the rider is modeled by
2∏
h=30-25 cos a t b , where t=0 corresponds
40
to the time when the rider is at the low point. h=50
-4
2∏
2∏
when
=cos a t b . Then
t L 2.498, so t
5
40
40
≠15.90 sec.
74. The length L must be the distance traveled in 30 min by
an object traveling at 540 ft/sec:
ft
L=1800 sec # 540
= 972,000 ft, or about 184 miles.
sec
75. (a) A model of the depth of the tide is
∏
d=2 cos c
1t - 7.2 2 d + 9, where t is hours since
6.2
midnight. The first low tide is at 1:00 A.M. (t=1).
(b) At 4:00 A.M. (t=4): about 8.90 ft. At 9:00 P.M.
(t=21): about 10.52 ft.
(c) 4:06 A.M. (t=4.1 — halfway between 1:00 A.M. and
7:12 A.M.).
76. (a) 1 second.
(b) Each peak corresponds to a heartbeat —there are
60 per minute.
(c)
62. Rewrite as y=–3.5 sin c 2 a x -
∏
b d + 0.5.
63. Rewrite as y=5 cos c 3 a x 18
2∏
∏
Amplitude 5, period
, phase shift , vertical
3
18
1
translation units up.
2
64. Amplitude 3, period 2∏, phase shift –3, vertical translation 2 units down.
[0, 10] by [80, 160]
Section 4.4
Graphs of Sine and Cosine: Sinusoids
177
77. (a) The maximum d is approximately 21.4. The amplitude
is (21.4-7.2)/2=7.1.
Scatterplot:
[0, 13] by [10, 80]
[0, 2.1] by [7, 22]
(b) The period appears to be slightly greater than 0.8, say
0.83.
p
1x - 7 2 b + 44.
6
Start with the general form sinusoidal function
y = a cos1 b1x - h2 2 + k, and find the variables a, b, h,
and k as follows:
80. One possible answer is y = 24 cos a
(c) Since the function has a minimum at t=0, we use
an inverted cosine model:
d(t)=–7.1 cos (2∏t/0.83)+14.3.
The amplitude is 0 a 0 =
(d)
2p
p
2p
1 0b0 =
= .
0b0
12
6
Again, we can arbitrarily choose to use the positive value,
p
so b = .
6
The maximum is at month 7, so the phase shift h=7.
68 + 20
The vertical shift k =
= 44.
2
68 - 20
= 24. We can arbitrarily
2
choose to use the positive value, so a=24.
The period is 12 months. 12 =
[0, 2.1] by [7, 22]
78. (a) The amplitude is 12.7, half the diameter of the
turntable.
(b) The period is 1.8, as can be seen by measuring from
minimum to minimum.
(c) Since the function has a minimum at t=0, we use
an inverted cosine model:
d(t)=–12.7 cos (2∏t/1.8)+72.7.
[0, 13] by [10, 80]
(d)
81. False. Since y=sin 2x is a horizontal stretch of y=sin 4x
by a factor of 2, y=sin 2x has twice the period, not half.
Remember, the period of y=sin bx is 2∏/| b |.
[0, 4.1] by [59, 86]
p
1x - 7 2 b + 57.5.
6
Start with the general form sinusoidal function
y = a cos1b1 x - h2 2 + k, and find the variables a, b, h,
and k as follows:
79. One possible answer is T = 21.5 cos a
The amplitude is 0 a 0 =
79 - 36
= 21.5. We can arbitrari2
ly choose to use the positive value, so a=21.5.
2p
p
2p
1 0b0 =
= .
0b0
12
6
Again, we can arbitrarily choose to use the positive value,
p
so b = .
6
The maximum is at month 7, so the phase shift h=7.
79 + 36
The vertical shift k =
= 57.5.
2
The period is 12 months. 12 =
82. True. Any cosine curve can be converted to a sine curve of
the same amplitude and frequency by a phase shift, which
can be accomplished by an appropriate choice of C
(a multiple of ∏/2).
83. The minimum and maximum values differ by twice the
amplitude. The answer is D.
84. Because the graph passes through (6, 0), f(6)=0. And 6
plus exactly two periods equals 96, so f(96)=0 also. But
f(0) depends on phase and amplitude, which are
unknown. The answer is D.
85. For f(x)=a sin (bx+c), the period is 2∏/| b |, which
here equals 2∏/420=∏/210. The answer is C.
86. There are 2 solutions per cycle, and 2000 cycles in the
interval. The answer is C.
87. (a)
[–∏, ∏] by [–1.1, 1.1]
178
Chapter 4
Trigonometric Functions
(b) cos x≠0.0246x4+0x3-0.4410x2+0x+0.9703. The
coefficients given as “0” here may show up as very small
numbers (e.g., 1.44*10–14) on some calculators. Note
that cos x is an even function, and only the even powers
of x have nonzero (or a least “non-small”) coefficients.
(c) The Taylor polynomial is
1
1 4
1 - x2 +
x = 1 - 0.5000x2 + 0.04167x4; the
2
24
coefficients are fairly similar.
88. (a)
[–∏, ∏] by [–1.1, 1.1]
(b) sin x≠–0.0872x3+0x2+0.8263x+0. The
coefficients given as “0” here may show up as very
small numbers (e.g., 3.56*10 –15) on some calculators. Note that sin x is an odd function, and only the
odd powers of x have nonzero (or a least “non-small”)
coefficients.
1
(c) The Taylor polynomial is x - x3 = x - 0.16667x3;
6
the coefficients are somewhat similar.
2∏
1
89. (a) p =
sec
=
524∏
262
(b) f = 262
1
(“cycles per sec”), or 262 Hertz (Hz).
sec
(c)
[0, 0.025] by [–2, 2]
90. Since the cursor moves at a constant rate, its distance
from the center must be made up of linear pieces as
shown (the slope of the line is the rate of motion). A
graph of a sinusoid is included for comparison.
[0, 2∏] by [–1.5, 1.5]
91. (a) a-b must equal 1.
(b) a-b must equal 2.
(c) a-b must equal k.
92. (a) a-b must equal 1.
(b) a-b must equal 2.
(c) a-b must equal k.
For #93–96, note that A and C are one period apart.
Meanwhile, B is located one-fourth of a period to the right of
A, and the y-coordinate of B is the amplitude of the sinusoid.
93. The period of this function is ∏ and the amplitude is 3.
∏
B and C are located (respectively) units and ∏ units to
4
3∏
the right of A. Therefore, B=(0, 3) and C= a , 0 b .
4
94. The period of this function is 2∏ and the amplitude is 4.5.
∏
B and C are located (respectively) units and 2∏ units to
2
3∏
the right of A. Therefore, B= a , 4.5 b and
4
9∏
C = a , 0b
4
2∏
and the amplitude is 2.
3
2∏
∏
B and C are located respectively units and
units to
6
3
3∏
∏
the right of A. Therefore, B= a , 2 b and C= a , 0 b
4
4
95. The period of this function is
96. The first coordinate of A is the smallest positive x such
n + 1
∏ must
that 2x-∏=n∏, n and integer, so x=
2
∏
equal . The period of this function ∏ is and the ampli2
∏
tude is 3. B and C are located (respectively) units and
4
∏
∏ units to the right of A. Therefore, A= a , 0 b ,
2
3∏
3∏
B = a , 3 b and C = a , 0 b .
4
2
97. (a) Since sin (–¨)=–sin ¨ (because sine is an odd function) a sin [–B(x-h)]+k=–a sin[B(x-h)]+k.
Then any expression with a negative value of b can be
rewritten as an expression of the same general form
but with a positive coefficient in place of b.
(b) A sine graph can be translated a quarter of a period
to the left to become a cosine graph of the same sinu1 2p
soid. Thus y = a sin c b a 1x - h2 + #
bd + k
4 b
p
b b d + k has the same
= a sin c b a x - a h 2b
graph as y = a cos3 b1 x - h2 4 + k. We therefore
p
choose H = h .
2b
(c) The angles ¨+∏ and ¨ determine diametrically opposite points on the unit circle, so they have point symmetry with respect to the origin. The y-coordinates are
therefore opposites, so sin(¨+∏)=–sin ¨.
(d) By the identity in (c). y = a sin3b1x - h2 + p4 + k
= -a sin3b1x - h2 4 + k. We therefore choose
p
H = h - .
b
(e) Part (b) shows how to convert y = a cos3b1 x - h2 4 + k
to y = a sin3 b1 x - H2 4 + k, and parts (a) and (d)
show how to ensure that a and b are positive.
Section 4.5
Graphs of Tangent, Cotangent, Secant, and Cosecant
179
■ Section 4.5 Graphs of Tangent, Cotangent,
Secant, and Cosecant
Exploration 1
1. The graphs do not seem to intersect.
2. Set the expressions equal and solve for x:
–k cos x=sec x
–k cos x=1/cos x
–k(cos x)2=1
(cos x)2=–1/k
Since k 7 0, this requires that the square of cos x be
negative, which is impossible. This proves that there is
no value of x for which the two functions are equal, so
the graphs do not intersect.
[–
∏ ∏
, ] by [–6, 6]
2 2
6. The graph of y=–cot 3x results from shrinking the
1
graph of y=cot x horizontally by a factor of and
3
reflecting it across the x-axis. There are vertical asymptotes at
2∏
∏
∏
x=. . . ., , - , 0, , . . . .
3
3
3
Quick Review 4.5
1. Period ∏
2. Period
2p
3
3. Period 6∏
4. Period 4∏
For #5–8, recall that zeros of rational functions are zeros of
the numerator, and vertical asymptotes are found at zeros of
the denominator (provided the numerator and denominator
have no common zeros).
5. Zero: 3. Asymptote: x=–4
6. Zero: –5. Asymptote: x=1
⎡–␲ , ␲ ⎤ by [–6, 6]
⎣ 3 3⎦
7. The graph of y=sec 3x results from shrinking the graph
1
of y=sec x horizontally by a factor of . There are
3
p
vertical asymptotes at odd multiples of .
6
7. Zero: –1. Asymptotes: x=2 and x=–2
8. Zero: –2. Asymptotes: x=0 and x=3
For #9–10, examine graphs to suggest the answer. Confirm by
checking f(–x)=f(x) for even functions and
f(–x)=–f(x) for odd functions.
9. Even: (–x)2+4=x2+4
1
1
10. Odd:
=–
1–x2
x
Section 4.5 Exercises
1. The graph of y=2 csc x must be vertically stretched by
2 compared to y=csc x, so y1=2 csc x and y2=csc x.
⎡– 2␲ , 2␲ ⎤ by [–6, 6]
⎣ 3 3⎦
8. The graph of y=csc 2x results from shrinking the graph
1
of y=csc x horizontally by a factor of . There are
2
∏
∏
vertical asymptotes at x = . . . ., -∏, - , 0, , . . . .
2
2
2. The graph of y=5 tan x must be vertically stretched
by 10 compared to y=0.5 tan x, so y1=5 tan x and
y2=0.5 tan x.
3. The graph of y=3 csc 2x must be vertically stretched by
1
3 and horizontally shrunk by compared to y=csc x, so
2
y1 = 3 csc 2x and y2 =csc x.
4. The graph of y=cot(x-0.5)+3 must be translated
3 units up and 0.5 units right compared to y=cot x, so
y1 =cot(x-0.5)+3 and y2 =cot x.
5. The graph of y=tan 2x results from shrinking the graph
1
of y = tan x horizontally by a factor of . There are
2
∏ ∏ 3∏
vertical asymptotes at x=. . . , - , ,
,.. . .
4 4 4
[–∏, ∏] by [–6, 6]
180
Chapter 4
Trigonometric Functions
9. The graph of y=2 cot 2x results from shrinking the
1
graph of y=cot x horizontally by a factor of and
2
stretching it vertically by a factor of 2. There are vertical
∏
∏
asymptotes at x=. . . ., -∏, - , 0, , . . . .
2
2
17. Domain: All reals except integer multiples of ∏
Range: (– q , q )
Continuous on its domain
Decreasing on each interval in its domain
Symmetric with respect to the origin (odd)
Not bounded above or below
No local extrema
No horizontal asymptotes
Vertical asymptotes x = k∏ for all integers k
End behavior: limq cot x and limq cot x do not exist.
xS
xS-
18. Domain: All reals except odd multiples of
[–
∏ ∏
, ] by [–6, 6]
2 2
x
10. The graph of y=3 tan ¢ ≤ results from stretching the
2
graph of y=tan x horizontally by a factor of 2 and
stretching it vertically by a factor of 3. There are vertical
asymptotes at x=. . . ., -∏, ∏, 3∏, . . . .
[–2∏, 2∏] by [–15, 15]
x
11. The graph of y=csc ¢ ≤ results from horizontally
2
stretching the graph of y=csc x by a factor of 2.
There are vertical asymptotes at
x=. . . ., -4∏, -2∏, 0, 2∏, . . . .
[–4∏, 4∏] by [–6, 6]
12. The graph of y=3 sec 4x results from horizontally
1
shrinking the graph of y=sec x by a factor of and
4
stretching it vertically by a factor of 3. There are vertical
p
asymptotes at odd multiples of .
8
Range: (– q , –1] ´ [1, q )
Continuous on its domain
On each interval centered at an even multiple of ∏:
decreasing on the left half of the interval and
increasing on the right half
On each interval centered at an odd multiple of ∏:
increasing on the left half of the interval and
decreasing on the right half
Symmetric with respect to the y-axis (even)
Not bounded above or below
Local minimum 1 at each even multiple of ∏, local
maximum –1 at each odd multiple of ∏
No horizontal asymptotes
Vertical asymptotes x = k∏/2 for all odd integers k
End behavior: limq sec x and limq sec x do not exist.
xS
∏ ∏
, ] by [–15, 15]
2 2
13. Graph (a); Xmin=–∏ and Xmax=∏
14. Graph (d); Xmin=–∏ and Xmax=∏
15. Graph (c); Xmin=–∏ and Xmax=∏
16. Graph (b); Xmin=–∏ and Xmax=∏
xS-
19. Domain: All reals except integer multiples of ∏
Range: (– q , –1] ´ [1, q )
Continuous on its domain
p
On each interval centered at x =
+ 2kp (k an integer):
2
decreasing on the left half of the interval and increasing on the right half
3p
On each interval centered at
+ 2kp: increasing on the
2
left half of the interval and decreasing on the right half
Symmetric with respect to the origin (odd)
Not bounded above or below
p
Local minimum 1 at each x =
+ 2kp, local maximum –1
2
3p
at each x =
+ 2kp, where k is an even integer in
2
both cases
No horizontal asymptotes
Vertical asymptotes: x = k∏ for all integers k
End behavior: limq csc x and limq csc x do not exist.
xS
[–
∏
2
xS-
20. Domain: All reals except odd multiples of ∏
Range: (– q , q )
Continuous on its domain
Increasing on each interval in its domain
Symmetric with respect to the origin (odd)
Not bounded above or below
No local extrema
No horizontal asymptotes
Vertical asymptotes x=k∏ for all odd integers k
End behavior: limq tan1 x>22 and limq tan1 x>2 2
xS
xSdo not exist.
Section 4.5
21. Starting with y=tan x, vertically stretch by 3.
Graphs of Tangent, Cotangent, Secant, and Cosecant
38. csc x=–1.5
1
sin x= 1.5
x≠∏-(–0.73)
≠3.87
22. Starting with y=tan x, reflect across x-axis.
23. Starting with y=csc x, vertically stretch by 3.
24. Starting with y=tan x, vertically stretch by 2.
25. Starting with y=cot x, horizontally stretch by 2,
vertically stretch by 3, and reflect across x-axis.
2
and
p
39. csc x=2
1
sin x=
2
x≠0.52 or
x≠∏-0.52
≠2.62
1
28. Starting with y=tan x, horizontally shrink by and
p
vertically stretch by 2 and shift down by 2 units.
40. tan x=0.3
x≠0.29 or
x≠∏+0.29
≠3.43
26. Starting with y=sec x, horizontally stretch by 2,
vertically stretch by 2, and reflect across x-axis.
27. Starting with y=tan x, horizontally shrink by
reflect across x-axis and shift up by 2 units.
29. sec x = 2
1
cos x =
2
∏
x =
3
30. csc x = 2
1
sin x =
2
5∏
x =
6
31. cot x = - 13
13
tan x = 3
5∏
x =
6
32. sec x = - 12
12
cos x = 2
5∏
x =
4
33. csc x = 1
sin x = 1
5∏
x =
2
34. cot x = 1
tan x = 1
x = -
3∏
4
35. tan x = 1.3
x L 0.92
36. sec x = 2.4
1
cos x =
2.4
x L 1.14
37. cot x=–0.6
1
tan x= 0.6
x≠–1.03+2∏
≠5.25
181
41. (a) One explanation: If O is the origin, the right triangles
with hypotenuses OP1 and OP2 , and one leg (each) on
the x-axis, are congruent, so the legs have the same
lengths. These lengths give the magnitudes of the
coordinates of P1 and P2 ; therefore, these coordinates
differ only in sign. Another explanation: The reflection of point (a, b) across the origin is (–a, –b).
(b) tan t=
sin t
b
= .
cos t
a
(c) tan(t-∏)=
sin 1t - p2
cos1t - p2
=
b
-b
=
= tan t.
-a
a
(d) Since points on opposite sides of the unit circle determine the same tangent ratio, tan(t_∏)=tan t for all
numbers t in the domain. Other points on the unit circle yield triangles with different tangent ratios, so no
smaller period is possible.
(e) The tangent function repeats every ∏ units; therefore,
so does its reciprocal, the cotangent (see also #43).
42. The terminal side passes through (0, 0) and (cos x, sin x);
sin x
sin x - 0
the slope is therefore m=
=
=tan x.
cos x - 0
cos x
1
1
1
1
43. For any x, a b (x+p)=
=
= a b (x).
f
f1x + p2
f1x2
f
This is not true for any smaller value of p, since this is
the smallest value that works for f.
44. (a), (b) The angles t and t+∏ determine points (cos t, sin t)
and (cos(t+∏), sin(t+∏)), respectively. These points are
on opposite sides of the unit circle, so they are reflections of
each other about the origin. The reflection of any point (a, b)
about the origin is (a, b), so cos(t+∏)=cos t and
sin(t+∏)=sin t.
sin 1t + p2
- sin t
sin t
=
=
=tan t.
cos1t + p2
-cos t
cos t
In order to determine that the period of tan t is ∏, we
would need to show that no p 6 ∏ satisfies
tan(t+p)= tan t for all t.
(c) tan(t+∏)=
45. (a) d=350 sec x=
(b) d≠16,831 ft
350
ft
cos x
182
Chapter 4
Trigonometric Functions
46. (a) x=800 cot y=
800
ft
tan x
(b) x≠5,051 ft
p
20
(c)
#
180°
= 9°
p
For #47–50, the equations can be rewritten (as shown), but
generally are easiest to solve graphically.
47. sin2 x=cos x; x≠ ; 0.905
2
48. cos x=sin x; x≠0.666 or x≠2.475
1
49. cos2 x= ; x≠ ; 1.107 or x≠ ; 2.034
5
50. 4 cos2 x=sin x; x≠1.082 or x≠2.060
[–∏, ∏] by [–10, 10]
60. They look similar on this window, but they are noticeably
different at the edges (near 0 and ∏). Also, if f were equal
1
1
p
to g, then it would follow that =–cos x= =xf
g
2
on this interval, which we know to be false.
51. False. f(x)=tan x is increasing only over intervals on
which it is defined, that is, intervals bounded by consecutive asymptotes.
52. True. Asymptotes of the secant function, sec x=1/cos x,
occur at all odd multiples of ∏> 2 (where cos x=0), and
these are exactly the zeros of the cotangent function,
cot x=cos x/sin x.
53. The cotangent curves are shaped like the tangent curves,
but they are mirror images. The reflection of tan x in the
x-axis is –tan x. The answer is A.
54. sec x “just barely” intersects its inverse, cos x, and when
cos x is shifted to produce sin x, that curve and the curve
of sec x do not intersect at all. The answer is E.
55. y=k/sin x and the range of sin x is [–1, 1].
The answer is D.
56. y=csc x=1/sin x has the same asymptotes as
y=cos x/sin x=cot x. The answer is C.
57. On the interval [–∏, ∏], f 7 g on about
(–0.44, 0) ´ (0.44, ∏).
[0, ∏] by [–10, 10]
p
∏
b (or csc x=sec a x - a + np b b
2
2
for any integer n) This is a translation to the right of
61. csc x=sec a x -
p
p
a or + np b units.
2
2
62. cot x=–tan a x -
p
b (or cot x=
2
p
+ np b ≤ for any integer n).
2
p
p
This is a translation to the right of a or + np b units,
2
2
and a reflection in the x-axis, in either order.
–tan ¢ x- a
63. d=30 sec x=
30
cos x
[–∏, ∏] by [–10, 10]
58. On the interval [–∏, ∏], f 7 g on about
p
p
(–∏, –2.24) ´ a - , 0 b ´ a , 2.24 b .
2
2
[–∏, ∏] by [–10, 10]
59. cot x is not defined at 0; the definition of “increasing on
(a, b)” requires that the function be defined everywhere
in (a, b). Also, choosing a=–∏> 4 and b=∏> 4, we have
a 6 b but f(a)=1 7 f(b)=–1.
[–0.5∏, 0.5∏] by [0, 100]
64. (a) For any acute angle ¨, cos ¨=sin a
p
- ¨ b —the sine
2
of the complement of ¨. This can be seen from the
right-triangle definition of sine and cosine: if one of
the acute angles is ¨, then the other acute angle is
p
- ¨, since all three angles in a triangle must add
2
to ∏. The side opposite the angle ¨ is the side adjacent
to the other acute angle.
(b) (cos t, sin t)
Section 4.6
(c) Using ^ ODA ~ ^ OCB (recall “~” means “similar
BC
DA
BC
to”),
=tan t=
=
, so BC=tan t.
OD
OC
1
(d) Using ^ ODA ~ ^ OCB,
so OB=
Graphs of Composite Trigonometric Functions
y = 2 sin 3x - 4 cos 2x
y = 2 sin 15x + 12 - 5 cos 5x
OD
1
OC
=
=cot t=
OA
OB
OB ,
1
=sec t.
cos t
(e) BC is a tangent segment (part of the tangent line); OB
is a secant segment (part of a secant line, which crosses
the circle at two points). The names “cotangent” and
“cosecant” arise in the same way as “cosine”—they
are the tangent and secant (respectively) of the complement. That is, just as BC and OB go with j BOC
(which has measure t), they also go with j OBC (the
p
complement of j BOC, with measure
- t).
2
kg
N
1
m
= (1.5 m) a 1050 3 b a 9.8
b
m
2
m
sec2
kg
(4.7*10–6 m)sec Ï≠0.03627 sec Ï
, so
sec2
sec Ï≠1.5990, and Ï≠0.8952 radians≠51.29°.
[–2∏, 2∏] by [–6, 6]
[–2∏, 2∏] by [–6, 6]
Not a Sinusoid
7x - 2
7x
y = cos a
b + sin a b
5
5
Sinusoid
y = 3 cos 2x + 2 sin7x
65. 0.058
66. (a)
1
1
1
1
1
=
= #
= cos(bx)=
y
a sec1 bx2
a sec 1 bx2
a
1
sin(bx+∏/2)
a
p
1
(b) y=0.2 sin a x + b
6
2
[–2∏, 2∏] by [–6, 6]
[–2∏, 2∏] by [–6, 6]
Sinusoid
Not a Sinusoid
Quick Review 4.6
1. Domain: (–q, q); range: [–3, 3]
2. Domain: (– q , q ), range: [–2, 2]
3. Domain: [1, q ); range: [0, q )
4. Domain: [0, q ); range: [0, q )
5. Domain: (– q , q ); range: [–2, q )
(c) a=1/0.2=5 and b=1/6
6. Domain: (– q , q ); range: [1, q )
x
(d) y=5 sec a b . The scatter plot is shown below, and
6
the fit is very good—so good that you should realize
that we made the data up!
7. As x S - q, f1 x2 S q; as x S q, f1 x2 S 0.
8. As x S - q, f1 x2 S - q; as x S q, f1x2 S 0.
9. føg(x)= 1 1x2 2-4=x-4, domain: [0, q ).
gøf(x)= 2x2 - 4, domain: (– q , –2] ´ [2, q ).
10. føg(x)=(cos x)2=cos2 x, domain: (– q , q ).
gøf(x)=cos(x2), domain: (– q , q ).
Section 4.6 Exercises
1. Periodic.
[–0.3, 8.7] by [2.32, 24]
■ Section 4.6 Graphs of Composite
Trigonometric Functions
Exploration 1
y = 3 sin x + 2 cos x
[–2∏, 2∏] by [–6, 6]
Sinusoid
183
[–2, 2] by [–1.5, 1.5]
y = 2 sin x - 3 cos x
2. Periodic.
[–2, 2] by [–2.5, 2.5]
[–2∏, 2∏] by [–6, 6]
Sinusoid
184
Chapter 4
Trigonometric Functions
3. Not periodic.
9. Since the period of cos x is 2∏, we have
cos2 (x+2∏)=(cos(x+2∏))2=(cos x)2=cos2 x.
The period is therefore an exact divisor of 2∏, and we see
graphically that it is ∏. A graph for –∏ x ∏ is
shown:
[–2, 2] by [–5, 20]
4. Not periodic.
[–, ] by [–1, 2]
[–2, 2] by [–5, 20]
10. Since the period of cos x is 2∏, we have
cos3 (x+2∏)=(cos(x+2∏))3=(cos x)3=cos3 x.
The period is therefore an exact divisor of 2∏, and we see
graphically that it is 2∏. A graph for –2∏ x 2∏ is
shown:
5. Not periodic.
[–2, 2] by [–1.5, 1.5]
[–2, 2] by [–6, 6]
6. Not periodic.
11. Since the period of cos x is 2∏, we have
2cos2 1x + 2∏ 2 = 21cos 1x + 2∏ 2 2 2= 2 1 cos x2 2
= 2cos2 x . The period is therefore an exact divisor of
2∏, and we see graphically that it is ∏. A graph for
–∏ x ∏ is shown:
[–2, 2] by [–12, 12]
7. Periodic.
[–, ] by [–1, 2]
[–2, 2] by [–10, 10]
12. Since the period of cos x is 2∏, we have
|cos3(x+2∏)|=|(cos(x+2∏))3|=|(cos x)3|
=|cos3 x|. The period is therefore an exact divisor of 2∏,
and we see graphically that it is ∏. A graph for
–∏ x ∏ is shown:
8. Periodic.
[–, ] by [–1, 2]
[–2, 2] by [–40, 40]
13. Domain: (– q , q ). Range: [0, 1].
[–2∏, 2∏] by [–0.25, 1.25]
Section 4.6
14. Domain: (– q , q ). Range: [0, 1].
Graphs of Composite Trigonometric Functions
20. –0.5x y 2 - 0.5x
[–10, 10] by [–10, 10]
[–2∏, 2∏] by [–0.25, 1.25]
15. Domain: all x Z n∏, n an integer. Range: [0, q ).
21. 1 - 0.3x y 3 - 0.3x
[–2∏, 2∏] by [–0.5, 4]
16. Domain: (– q, q ). Range: [–1, 1].
[–10, 10] by [–4, 8]
22. x y x + 2
[–4∏, 4∏] by [–1.2, 1.2]
p
+ np, n an integer. Range: (– q , 0].
17. Domain: all x Z
2
[–2∏, 2∏] by [–10, 0.2]
18. Domain: (– q, q ). Range: [–1, 0].
[–10, 10] by [–10, 10]
For #23–28, the function y1 + y2 is a sinusoid if both y1 and y2
are sine or cosine functions with the same period.
23. Yes (period 2∏)
24. Yes (period 2∏)
25. Yes (period 2)
26. No
27. No
28. No
For #29–34, graph the function. Estimate a as the amplitude of
the graph (i.e., the height of the maximum). Notice that the
value of b is always the coefficient of x in the original functions. Finally, note that a sin[b(x-h)]=0 when x=h, so
estimate h using a zero of f(x) where f(x)changes from negative to positive.
29. A≠3.61, b=2, and h≠0.49, so f(x)
≠3.61 sin[2(x-0.49)].
30. A≠2.24, b=3, and h≠–0.15, so f(x)
≠2.24 sin[3(x+0.15)].
31. A≠2.24, b=∏, and h≠0.35, so f(x)
≠2.24 sin[∏(x-0.35)].
[–2∏, 2∏] by [–1.25, .25]
In #19–22, the linear equations are found by setting the cosine
term equal to ; 1.
19. 2x - 1 y 2x + 1
32. A≠3.16, b=2∏, and h≠–0.05, so f(x)
≠3.16 sin[2∏(x +0.05)].
33. A≠2.24, b=1, and h≠–1.11, so f(x)
≠2.24 sin(x+1.11).
34. A≠3.16, b=2, and h≠0.16, so f(x)
≠3.16 sin[2(x-0.16)].
[–10, 10] by [–20, 20]
185
186
Chapter 4
Trigonometric Functions
35. The period is 2∏.
[0, 4∏] by [–1, 1]
[–∏, ∏] by [–3.5, 3.5]
50. f oscillates up and down between 2–x and –2–x.
As x S q, f1x2 S 0.
36. The period is 2∏.
[0, 2∏] by [–1, 1]
[–∏, ∏] by [–3, 3]
51. f oscillates up and down between
37. The period is 2∏.
1
1
and - .
x
x
As x S q, f1x2 S 0.
[–∏, ∏] by [–5, 5]
38. The period is 2∏.
[0, 4∏] by [–1.5, 1.5]
52. f oscillates up and down between e–x and –e–x.
As x S q, f1x2 S 0.
[–∏, ∏] by [–5, 5]
39. (a)
40. (d)
41. (c)
42. (b)
[0, 1.5∏] by [–1, 1]
43. The damping factor is e–x, which goes to zero as x gets
large. So damping occurs as x S q.
44. The damping factor is x, which goes to zero as x goes to
zero (obviously). So damping occurs as x S 0.
53. Period 2∏: sin[3(x+2∏)]+2cos[2(x+2∏)]=
sin(3x+6∏)+2 cos(2x+4∏)=sin 3x+2 cos 2x.
The graph, shows that no p<2∏ could be the period.
45. The amplitude, 15, is constant. So there is no damping.
46. The amplitude, ∏2, is constant. So there is no damping.
47. The damping factor is x 3, which goes to zero as x goes to
zero. So damping occurs as x S 0.
48. The damping factor is (2/3)x, which goes to zero as x gets
large. So damping occurs as x S q.
49. f oscillates up and down between 1.2
As x S q, f1x2 S 0.
–x
–x
and –1.2 .
[–2∏, 2∏] by [–3.4, 2.8]
Section 4.6
54. Period 2∏: 4 cos[2(x+2∏)]-2 cos[3(x+2∏)-1]
=4 cos(2x+4∏)-2 cos(3x-1+6∏)
=4 cos 2x-2 cos(3x-1). The graph, shows that no
p<2∏ could be the period.
Graphs of Composite Trigonometric Functions
187
58. Not periodic
[–4∏, 4∏] by [–50, 50]
59. Not periodic
[–2∏, 2∏] by [–7, 6]
55. Period 2∏:
2 sin[3(x+2∏)+1]-cos[5(x+2∏)-1]
=2 sin(3x+1+6∏)-cos(5x-1+10∏)
=2 sin(3x+1)-cos(5x-1). The graph, shows that
no p<2∏ could be the period.
[–4∏, 4∏] by [–13, 13]
60. Not periodic
[–2∏, 2∏] by [–3, 3]
56. Period 2∏:
3 cos[2(x+2∏)-1]-4 sin[3(x+2∏)-2]
=3 cos(2x-1+4∏)-4 sin(3x-2+6∏)
=3 cos(2x-1)-4 sin(3x-2). The graph, shows that
no p<2∏ could be the period.
[–4∏, 4∏] by [–13, 13]
61. Not periodic
[–4∏, 4∏] by [–7, 7]
[–2∏, 2∏] by [–8, 7]
62. Not periodic
1
57. Period 2∏: 2 sinB 1x + 2p 2 R 2 +2=
2
2 sin a 1 x + p b 2 +2= 2 -sin 1 x 2 + 2 = 2 sin 1 x 2 + 2.
2
2
2
The graph, shows that no p<2∏ could be the period.
[–4∏, 4∏] by [–10, 15]
For #63–70, graphs may be useful to suggest the domain and
range.
63. There are no restrictions on the value of x, so the domain
is ( - q, q ). Range: ( - q, q ).
[–4∏, 4∏] by [–1, 4]
64. There are no restrictions on the value of x, so the domain
is ( - q, q ). Range: ( - q, q ).
65. There are no restrictions on the value of x, so the domain
is ( - q, q ). Range: [1, q ).
66. There are no restrictions on the value of x, so the domain
is ( - q, q ). Range: ( - q, q ).
188
Chapter 4
Trigonometric Functions
67. sin x must be nonnegative, so the domain is
p ´ [–2∏, –∏] ´ [0, ∏] ´ [2∏, 3∏] ´ p; that is, all x
with 2n∏ x 12n + 1 2 ∏, n an integer. Range: [0, 1].
68. There are no restrictions on the value of x, so the domain
is ( - q, q ). Range: [–1, 1].
69. There are no restrictions on the value of x, since
∑sin x∑ 0, so the domain is (–q, q ). Range: [0, 1].
70. cos x must be nonnegative, so the domain is
p ´ c - 5p , - 3p d ´ c - p , p d ´ c 3p , 5p d ´ p ; that is, all
2
2
2 2
2 2
14n - 1 2p
14n + 1 2 p
x , n an integer.
x with
2
2
Range: [0, 1].
71. (a)
[0, 12] by [–0.5, 0.5]
(b) For t>0.51 (approximately).
72. (a) Using S1t2 = 75 1 1.04 2 t + 4 sin a
pt
b to estimate sales
3
(in millions of dollars) t years after 2005, we have
p#0
b = 75 million dollars.
S102 = 75 1 1.042 0 + 4 sin a
3
(b) The approximate annual growth rate is 4%.
(c) In 2013, t=8 so the sales are predicted by
p#8
S182 = 75 1 1.042 8 + 4 sin a
b L 106.1 million
3
dollars.
(d) To find the number of years in each economic cycle,
px
b , the trigonometric part of
find the period of sin a
3
px
2p
b is
= 6, so there
the model. The period of sin a
3
p>3
are 6 years in each economic cycle for the company.
73. No. This is suggested by a graph of y=sin x3; there is no
other section of the graph that looks like the section
between –1 and 1. In particular, there is only one zero of
the function in that interval (at x=0); nowhere else can
we find an interval this long with only one zero.
74. One explanation: The ‘v’-shaped section around x=0 is
unique — it does not appear anywhere else on the graph.
75. (a) — this is obtained by adding x to all parts of the
inequality - 1 sin x 1. In the second, after subtracting x from both sides, we are left with -sin x sin x,
which is false when sin x is negative.
81. False. The behavior near zero, with a relative minimum
of 0 at x=0, is not repeated anywhere else.
82. False. If two sinusoids have different periods, the sum of
the sinusoids is not a sinusoid. Example: sin x+sin 3x.
83. The negative portions of the graph of y=sin x are
reflected in the x-axis for y = @ sin [email protected] . This halves the
period. The answer is B.
84. f(–x)=(–x) sin (–x)=x sin x=f(x).
The answer is C.
85. f(–x)=–x+sin (–x)=–x-sin x=–f(x).
The answer is D.
86. The sum of two sinusoids is a sinusoid only when the two
sinusoids have the same period. The answer is D.
87. (a) Answers will vary — for example,
p
on a TI-81:
= 0.0661… L 0.07;
47.5
p
on a TI-82:
= 0.0668… L 0.07;
47
p
on a TI-85:
= 0.0498… L 0.05;
63
p
on a TI-92:
= 0.0263… L 0.03.
119
(b) Period: p=∏/125=0.0251». For any of the
TI graphers, there are from 1 to 3 cycles between each
pair of pixels; the graphs produced are therefore inaccurate, since so much detail is lost.
88. The amplitude is 3 hr 6 min, or 3.1 hr. The period is
2p
365 days (one could also use 365.26 days), so b =
.
365
The phase shift h is 80 days, so an answer is
2p
3.1 sin c
1 x - 80 2 d + 12 hours of daylight, where x is
365
the day of the year.
89. Domain: ( - q, q ). Range: [–1, 1]. Horizontal asymptote:
p
y=1. Zeros at ln a + np b , n a non-negative integer.
2
[–3, 3] by [–1.2, 1.2]
p
. Range: (0, q). Vertical
2
p
asympotes at missing points of domain: x=n∏+ .
2
90. Period: ∏. Domain: x Z np +
76. (b) — the first is impossible (even ignoring the middle
part) if x<0, since then –x x.
77. Graph (d), shown on [–2∏, 2∏] by [–4, 4]
78. Graph (a), shown on [–2∏, 2∏] by [–4, 4]
79. Graph (b), shown on [–2∏, 2∏] by [–4, 4]
80. Graph (c), shown on [–2∏, 2∏] by [–4, 4]
[–2.5∏, 2.5∏] by [–0.2, 5]
Section 4.7
91. Domain: 3 0, q 2 . Range: 1 - q, q 2 . Zeros at n∏, n a nonnegative integer.
Inverse Trigonometric Functions
189
96. Domain: 1 - q, 0 2 ´ 10, q 2 . Range: 1 - q, q 2 . Zeros
1
at
, n a non-zero integer. Note: the graph also suggests
np
the end-behavior asymptote y=x.
[–0.5, 4∏] by [–4, 4]
92. Domain: [–2, 2]. Range: [0, 2.94] (approximately). Zeros
at –2 and 2.
[–1, 1] by [–1, 1]
■ Section 4.7 Inverse Trigonometric Functions
Exploration 1
x
1. tan ¨= =x
1
[–2.5, 2.5] by [–0.5, 3.5]
93. Domain: 1 - q, 0 2 ´ 1 0, q 2 . Range: approximately
[–0.22, 1). Horizontal asymptote: y=0. Zeros at n∏, n a
non-zero integer.
x
2. tan–1 x=tan–1 a b =¨
1
3. 21 + x2 (by the Pythagorean theorem)
4. sin(tan–1(x))=sin(¨)=
x
21 + x2
5. sec(tan–1(x))=sec(¨)= 21 + x2
[–5∏, 5∏] by [–0.5, 1.2]
94. Domain: 1 - q, 0 2 ´ 1 0, q 2 . Range: 1 - q, q 2 . Horizontal
asymptote: y=0. Vertical asymptote: x=0. Zeros at n∏,
n a non-zero integer.
6. The hypotenuse is positive in either quadrant. The ratios
in the six basic trig functions are the same in every quadrant, so the functions are still valid regardless of the sign
of x. (Also, the sign of the answer in (4) is negative, as it
should be, and the sign of the answer in (5) is negative, as
it should be.)
Quick Review 4.7
1. sin x: positive; cos x: positive; tan x: positive
2. sin x: positive; cos x: negative; tan x: negative
3. sin x: negative; cos x: negative; tan x: positive
4. sin x: negative; cos x: positive; tan x: negative
5. sin
p
1
=
6
2
6. tan
p
=1
4
7. cos
2p
1
=–
3
2
8. sin
13
2p
=
3
2
9. sin
-∏
1
=–
6
2
10. cos
-p
1
=
3
2
[–4∏, 4∏] by [–0.5, 0.5]
95. Domain: 1 - q, 0 2 ´ 1 0, q 2 . Range: approximately
[–0.22, 1). Horizontal asymptote: y=1. Zeros at
a non-zero integer.
[–∏, ∏] by [–0.3, 1.2]
1
,n
np
Section 4.7 Exercises
For #1–12, keep in mind that the inverse sine and inverse
p p
tangent functions return values in c - , d , and the inverse
2 2
cosine function gives values in [0, ∏]. A calculator may
also be useful to suggest the exact answer. (A useful trick
is to compute, e.g., sin–1( 13/2)∏ and observe that this is
≠0.333, suggesting the answer ∏/3.)
1. sin–1 a
13
∏
b=
2
3
3. tan–1(0)=0
∏
1
2. sin–1 a - b =–
2
6
4. cos–1(1)=0
Chapter 4
190
Trigonometric Functions
1
p
5. cos–1 a b =
2
3
7. tan–1(–1)=–
9. sin–1 a -
6. tan–1(1)=
8. cos–1 a -
p
4
p
1
b = 4
12
11. cos–1(0)=
p
4
13
5p
b =
2
6
10. tan–1 1 - 132 = -
p
2
12. sin–1(1)=
p
3
p
2
13. approx. 21.22°
14. approx. 42.07°
15. approx. –85.43°
16. approx. 103.30°
17. approx. 1.172
18. approx. 1.527
19. approx. –0.478
20. approx. 2.593
21. y=tan–1(x2) is equivalent to tan y=x2,
–∏/2<y<∏/2. For x2 to get very large, y has to
approach ∏/2. So limq tan-1 1 x2 2 = ∏>2 and
xS
limq tan-1 1 x2 2 = ∏>2.
xS-
22. y=(tan–1 x)2 is equivalent to x = tan 1 ; 1y2 ,
0 y<∏2/4. For x to get very large, in the positive or
negative direction, y has to approach ∏2/4. So
limq 1tan -1x2 2 = ∏2>4 and limq 1tan -1x2 2 = ∏2>4.
xS
xS-
1
p
13
23. cos a sin–1 b =cos
=
2
6
2
24. sin(tan–1 1)=sin
p
12
=
4
2
25. sin–1 a cos
p
p
12
b =sin–1 a
b=
4
2
4
26. cos–1 a cos
12
7p
p
b = cos–1 a
b =
4
2
4
27. cos a 2 sin–1
1
p
1
b =cos a 2 # b =
2
6
2
28. sin[tan–1(–1)] =sin a –
12
∏
b =–
4
2
29. arcsin a cos
1
∏
∏
b =arcsin =
3
2
6
30. arccos a tan
∏
b =arccos 1=0
4
31. cos 1 tan–1 132 = cos a
35. Domain: (–q, q)
Range: (–∏/2, ∏/2)
Continuous
Increasing
Symmetric with respect to the origin (odd)
Bounded
No local extrema
Horizontal asymptotes: y=∏/2 and y=–∏/2
End behavior: limq tan-1x = ∏>2 and
xS
limq tan-1x = -∏>2
xS-
36. Domain: (–q, q)
Range: (0, ∏)
Continuous
Decreasing
Neither odd nor even (but symmetric with respect to the
point (0, ∏/2))
Bounded
No local extrema
Horizontal asymptotes: y=∏ and y=0
End behavior: limq cot-1x = 0 and limqcot-1x = ∏
xS
xS-
1 1
p p
37. Domain: c - , d . Range: c - , d . Starting from
2 2
2 2
1
–1
y=sin x, horizontally shrink by .
2
1 1
38. Domain: c - , d . Range: [0, 3∏]. Starting from
2 2
1
–1
y=cos x, horizontally shrink by and vertically
2
stretch by 3 (either order).
39. Domain: (–q, q). Range: a -
5p 5p
,
b . Starting from
2 2
y= tan–1 x, horizontally stretch by 2 and vertically
stretch by 5 (either order).
1
∏
b =
3
2
32. tan–1(cos ∏)=tan–1(–1)= -
34. Domain: [–1, 1]
Range: [0, ∏]
Continuous
Decreasing
Neither odd nor even (but symmetric with respect to the
point (0, ∏/2))
Bounded
Absolute maximum of ∏, absolute minimum of 0
No asymptotes
No end behavior (bounded domain)
∏
4
33. Domain: [–1, 1]
Range: [–∏/2, ∏/2]
Continuous
Increasing
Symmetric with respect to the origin (odd)
Bounded
Absolute maximum of ∏/2, absolute minimum of –∏/2
No asymptotes
No end behavior (bounded domain)
40. Domain: [–2, 2]. Range: [0, 3∏]. Starting from
y=arccos x=cos–1 x, horizontally stretch by 2 and
vertically stretch by 3 (either order).
41. First set ¨=sin–1 x and solve sin ¨=1, yielding
p
¨= + 2np for integers n. Since ¨=sin–1 x must be in
2
p
p p
c - , d , we have sin–1 x= , so x=1.
2 2
2
42. First set y=cos x and solve cos–1 y=1, yielding
y=cos 1. Then solve cos x=cos 1, which gives
x=1+2n∏ or x=–1+2n∏, for all integers n.
43. Divide both sides of the equation by 2, leaving
1
1
sin–1 x= , so x=sin L 0.479.
2
2
Section 4.7
1
45. For any x in [0, ∏], cos(cos–1x),=x. Hence, x= .
3
p p
p
is in
46. For any x in c - , d , sin–1(sin x)=x. Since
2 2
10
p
p p
c - , d , x= .
2 2
10
47. Draw a right triangle with horizontal leg 1, vertical leg x
(if x 7 0, draw the vertical leg “up”; if x 6 0, draw it
down), and hypotenuse 21 + x2. The acute angle
adjacent to the leg of length 1 has measure ¨=tan–1 x
x
(take ¨ 6 0 if x 6 0), so sin ¨=sin(tan–1 x)=
.
21 + x2
x> 0
x< 0
x
θ
x
1 + x2
1
48. Use the same triangles as in #47: draw a triangle with
horizontal leg 1, vertical leg x (up or down as x 7 0 or
x 6 0), and hypotenuse 21 + x2. The acute angle adjacent
to the leg of length 1 has measure ¨=tan–1 x (take
1
¨ 6 0 if x 6 0), so cos ¨=cos(tan–1 x)=
.
21 + x2
49. Draw a right triangle with horizontal leg 21 - x2, vertical leg x (if x 7 0, draw the vertical leg “up”; if x 6 0,
draw it down), and hypotenuse 1. The acute angle adjacent to the horizontal leg has measure
¨=arcsin x (take ¨ 6 0 if x 6 0), so
x
tan ¨=tan(arcsin x)=
.
21 - x2
x> 0
1
1 + 4x2
2x
θ
θ
2x
1 + 4x2
1
52. Draw a right triangle with horizontal leg 3x (if x 7 0,
draw the horizontal leg right; if x 6 0, draw it left),
vertical leg 21 - 9x2, and hypotenuse 1. If x 7 0, let ¨
be the acute angle adjacent to the horizontal leg; if x 6 0,
let ¨ be the supplement of this angle. Then ¨=arccos 3x,
so sin ¨=sin(arccos 3x)= 21 - 9x2.
x> 0
x< 0
1
3x
θ
191
x< 0
θ
1
1 + x2
x> 0
1 – 9x2
44. If tan–1 x=–1, then x=tan(–1) L –1.557.
Inverse Trigonometric Functions
1
θ
3x
53. (a) Call the smaller (unlabeled) angle in the lower left Å ;
2
2
then tan Å= , or Å=tan–1 (since Å is acute).
x
x
Also, ¨+Å is the measure of one acute angle in the
right triangle formed by a line parallel to the floor
14
and the wall; for this triangle tan(¨+Å)= . Then
x
–1 14
¨+Å=tan
(since ¨+Å is acute), so
x
14
14
2
¨=tan–1 -Å=tan–1 -tan–1 .
x
x
x
(b) Graph is shown. The actual maximum occurs at
x L 5.29 ft, where ¨ L 48.59°.
x< 0
1
x
θ
1 – x2
θ
x
1
1 – x2
50. Draw a right triangle with horizontal leg x (if x 7 0, draw
the horizontal leg right; if x 6 0, draw it left), vertical leg
21 - x2, and hypotenuse 1. If x 7 0, let ¨ be the acute
angle adjacent to the horizontal leg; if x 6 0, let ¨ be the
supplement of this angle. Then ¨=arccos x, so
x
cot ¨=cot(arccos x)=
.
21 - x2
x< 0
1 – x2
x> 0
1
θ
x
1
θ
[0, 25] by [0, 55]
(c) Either x L 1.83 or x L 15.31—these round to 2 ft
or 15 ft.
54. (a) ¨ is one acute angle in the right triangle with leg
x
lengths x (opposite) and 3 (adjacent); thus tan ¨= ,
3
–1 x
and ¨=tan
(since ¨ is acute).
3
(b) Graph is shown (using DEGREE mode). Negative values of x correspond to the point Q being “upshore”
from P (“into” the picture) instead of downshore (as
shown in the illustration). Positive angles are angles
that point downshore; negative angles point upshore.
x
51. Draw a right triangle with horizontal leg 1, vertical leg 2x
(up or down as x 7 0 or x 6 0), and hypotenuse
21 + 4x2. The acute angle adjacent to the leg of length 1
has measure ¨=arctan 2x (take ¨ 6 0 if x 6 0), so
1
cos ¨=cos(arctan 2x)=
.
21 + 4x2
[–20, 20] [–90, 90]
(c) ¨=tan–1 5≠78.69°
192
Chapter 4
55. (a) ¨=tan–1
Trigonometric Functions
s
.
500
65. (a) Domain all reals, range [–∏/2, ∏/2], period 2∏.
(b) As s changes from 10 to 20 ft, ¨ changes from about
1.1458° to 2.2906°—it almost exactly doubles (a
99.92% increase). As s changes from 200 to 210 ft, ¨
changes from about 21.80° to 22.78°—an increase of
less than 1°, and a very small relative change (only
about 4.25%).
(c) The x-axis represents the height and the y-axis represents the angle: the angle cannot grow past 90° (in
fact, it approaches but never exactly equals 90°).
[–2∏, 2∏] by [–0.5∏, 0.5∏]
(b) Domain all reals, range [0, ∏], period 2∏.
56. (a) Sin(x) exists for all x, but sin–1(x) is restricted to
[–1, 1]. The domain of f(x) is [–1, 1]. The range is
[–1, 1].
(b) Since the domains of sin–1(x) and cos–1(x) are [–1, 1],
p
the domain of g(x) is [–1, 1]. The range is e f .
2
(c) Since |sin(x)| 1 for all x, h(x) exists for all x and its
p p
domain is (– q , q ). The range is c - , d .
2 2
(d) Sin(x) exists for all x, but cos–1(x) is restricted to
[–1, 1]. The domain of k(x) is [–1, 1]. The range is
[0, 1].
[–2∏, 2∏] by [–0, ∏]
(c) Domain all reals except ∏/2+n∏ (n an integer),
range (–∏/2, ∏/2), period ∏. Discontinuity is not
removable.
(e) Since 0sin(x)0 1 for all x, q(x) exists for all x and its
domain is (– q , q ). The range is [0, ∏].
57. False. This is only true for –1 x 1, the domain of
the sin–1 function. For x<–1 and for x>1, sin (sin–1 x)
is undefined.
58. True. The end behavior of y=arctan x determines two
horizontal asymptotes, since limq arctan x = -∏>2 and
xS-
lim arctan x = ∏>2.
xS q
59. cos (5∏/6)= - 13>2, so cos-1 1 - 13>2 2 = 5∏>6. The
answer is E.
[–2∏, 2∏] by [–∏, ∏]
66. (a) Let ¨=sin–1(2x). Then the adjacent side of the right
triangle is 212 - 12x2 2 = 21 - 4x2 .
cos(¨)=
1
60. sin–1 (sin ∏)=sin–1 0=0. The answer is C.
61. sec (tan–1 x) = 21 + tan2 1tan-1x2 = 21 + x2.
The answer is C.
62. The range of f(x)=arcsin x=sin
[–∏/2, ∏/2]. The answer is E.
–1
x is, by definition,
63. The cotangent function restricted to the interval (0, ∏) is
one-to-one and has an inverse. The unique angle y
between 0 and ∏ (non-inclusive) such that cot y=x is
called the inverse cotangent (or arccotangent) of x, denoted cot–1 x or arccot x. The domain of y=cot–1 x is
(– q , q ) and the range is (0, ∏).
64. In the triangle below, A=sin–1 x and B=cos–1 x. Since
A and B are complementary angles, A+B=∏/2. The
left-hand side of the equation is only defined for
–1 x 1.
1
A
B
2x
θ
1 – 4x2
(b) Let ¨=tan–1(x). Then the hypotenuse is 21 + x2.
sec2(¨)= a
1 + x2
21 + x2 2
b =1+x2
1
x
θ
1
(c) Let ¨=cos–1 1 1x2 . Then the opposite side of the
right triangle is 212 - 1 1x2 2 = 11 - x .
11 - x
sin(¨)=
= 11 - x
1
1
x
21 - 4x2
= 21 - 4x2
1
1–x
θ
x
Section 4.7
(d) Let ¨=cot–1(x). Then the hypotenuse is
212 + x2 = 21 + x2 . –csc2(¨)=– a
21 + x
b
1
2
2
=–(1+x2)=–x2-1.
1 + x2
1
θ
x
(e) Let ¨=sec–1(x2). Then the opposite leg of the right
triangle is 2 1x2 2 2 - 12 = 2x4 - 1 .
tan(¨)=
2x4 - 1
= 2x4 - 1.
1
x2
1
p
67. y= -tan–1 x.
2
1
(Note that y=tan–1 a b does not have the correct
x
range for negative values of x.)
68. (a) cos(sin
–1
x) or sin(cos
1
–1
x)
(b) sin(tan
–1
x) or cos(cot
1 + x2
42 - d
= tan-1x. We know that y=42 is the upper
a
p
horizontal asymptote and thus it corresponds to y = .
2
42 - d
p
p
42 - d
-1
So,
= tan x =
1
= . Solving this
a
2
a
2
p
for d in terms of a yields d = 42 - a b a.
2
If d = 24 + a
p
p
b a and d = 42 - a b a, then
2
2
p
18
p
.
24 + a b a=42 - a b a. So, 18=∏a, and a =
p
2
2
The arctangent function with horizontal asymptotes at
18
y=24 and y=42 will be y =
tan-1x + 33.
p
70. (a) As in Example 5, ¨ can only be in Quadrant I or
Quadrant IV, so the horizontal side of the triangle can
only be positive.
1
x
x
(b) tan a sin-1 a b b =
=
4
x
s
2x - x2
x
1 – x2
–1
1
s
2x4 - x2
(c) sin a cos-1 a b b = 2 =
x
x
x2
(See figure below.)
x)
x
x2
1
–1
(c) tan(sin
193
Substitute this value for a into either of the two equations
p
18
for d to get: d = 24 + a b a b =24+9=33 or
p
2
p
18
d = 42 - a b a b =42-9=33.
p
2
x4 – 1
θ
Inverse Trigonometric Functions
x) or cot(cos
–1
s
θ
x)
x
1
x
71. (a) The horizontal asymptote of the graph on the left is
p
y = .
2
1 – x2
69. In order to transform the arctangent function to a function that has horizontal asymptotes at y=24 and y=42,
we need to find a and d that will satisfy the equation
y=a tan–1 x+d. In other words, we are shifting the
p
horizontal asymptotes of y=tan–1 x from y = - and
2
p
to the new asymptotes y=24 and y=42.
y =
2
–1
–1
Solving y=a tan x+d and y=24 for tan x in
terms of a and d yields 24=a tan–1 x+d; so,
24 - d
= tan-1x. We know that y=24 is the lower
a
p
horizontal asymptote and thus it corresponds to y = - .
2
24 - d
p
p
24 - d
-1
So,
= tan x = - 1
= - . Solving
a
2
a
2
p
this for d in terms of a yields d = 24 + a b a.
2
Solving y=a tan–1 x+d and y=42 for tan–1 x in
terms of a and d yields 42=a tan–1 x+d; so,
(b) The two horizontal asymptotes of the graph on the
3p
p
right are y =
and y =
.
2
2
1
(c) The graph of y = cos-1 a b will look like the graph on
x
the left.
(d) The graph on the left is increasing on both connected
intervals.
y
y
(−1, π)
3
(−1, π)
1
1
–1
–2
3
x
(1, 0)
–1
–2
x
(1, 0)
194
Chapter 4
Trigonometric Functions
72. (a) The horizontal asymptote of the graph on the left is
y = 0.
2. tan 34°1312=
(b) The two horizontal asymptotes of the graph on the
right are y=0 and y=∏.
≠68.01
1
(c) The graph of y = sin-1 a b will look like the graph
x
on the left.
h
y
(−1, 3π/2)
4
(1, π/2)
1
–1
3. Let d be the length of the horizontal leg. Then tan 10°
120 ft
120
=
, so d=
=120 cot 10°≠680.55 ft.
d
tan 10°
4. tan 14°=
4
(1, π/2)
34°13′12′′
100 ft
(d) The graph on the left is decreasing on both connected
intervals.
y
1
(−1, −π/2) –2
–1
90 ft
90
, so d=
=90 cot 14°≠361 ft.
d
tan 14°
14°
90 ft
d
1
x
h
, so h=100 tan 34°1312
100 ft
1
x
–2
■ Section 4.8 Solving Problems with
Trigonometry
Exploration 1
1. The parametrization should produce the unit circle.
2. The grapher is actually graphing the unit circle, but the
y-window is so large that the point never seems to get
above or below the x-axis. It is flattened vertically.
3. Since the grapher is plotting points along the unit circle,
it covers the circle at a constant speed. Toward the
extremes its motion is mostly vertical, so not much horizontal progress (which is all that we see) occurs. Toward
the middle, the motion is mostly horizontal, so it moves
faster.
4. The directed distance of the point from the origin at any
T is exactly cos T, and d=cos t models simple harmonic
motion.
Quick Review 4.8
5. Let / be the wire length (the hypotenuse); then
5
5 ft
cos 80°=
, so /=
=5 sec 80°≠28.79 ft.
O
cos 80°
Let h be the tower height (the vertical leg); then
h
tan 80°=
, so h=5 tan 80°≠28.36 ft.
5 ft
16 ft
16
, so /=
=16 sec 62°≠34.08 ft.
O
cos 62°
h
tan 62°=
, so h=16 tan 62°≠30.09 ft.
16 ft
6. cos 62°=
l
h
62°
16 ft
7. tan 80°112=
h
, so h=185 tan 80°112≠
185 ft
1051 ft.
h
1. b=15 cot 31°≠24.964, c=15 csc 31°≠29.124
80°1′12′′
2. a=25 cos 68°≠9.365, b=25 sin 68°≠23.180
3. b=28 cot 28°-28 cot 44°≠23.665,
c=28 csc 28°≠59.642, a=28 csc 44°≠40.308
4. b=21 cot 31°-21 cot 48°≠16.041,
c=21 csc 31°≠40.774, a=21 csc 48°≠28.258
5. complement: 58°, supplement: 148°
6. complement: 17°, supplement: 107°
185 ft
8. Let h be the height of the smokestack; then
h
tan 38°=
, so h=1580 tan 38°≠1234.43 ft.
1580 ft
h
9. tan 83°12=
, so h=100 tan 83°12≠839 ft.
100 ft
7. 45°
8. 202.5°
9. Amplitude: 3; period: ∏
10. Amplitude: 4; period: ∏/2
Section 4.8 Exercises
h
83°12′
All triangles in the supplied figures are right triangles.
h
, so h=300 tan 60°=300 13
300 ft
≠519.62 ft.
1. tan 60°=
100 ft
Section 4.8
10. ¨=sin–1
32 ft
≠3.9°
470 ft
470 ft
32 ft
θ
h
11. tan 55°=
, so h=10 tan 55°≠14.3 m
10 m
Solving Problems with Trigonometry
195
19. The difference in elevations is 1097 ft. If the width of the
1097 ft
canyon is w, then tan 19°=
, so
w
w=1097 cot 19°≠3186 ft.
19°
1097 ft
w
20. The distance from the base of the tower d satisfies
73 ft
tan 1°20=
, so d=73 cot 1°20≠3136 ft.
d
h
d
12. tan 29°48=
h
, so h=125 tan 29°48≠71.6 ft
125 ft
29°48′
h
21. The acute angle in the triangle has measure
O
180°-117°=63°, so tan 63°=
. Then
325 ft
O=325 tan 63°≠638 ft.
22. tan 17°=
125 ft
13. tan 35°=
1°20′
73 ft
55°
10 m
LP
, so LP=4.25 tan 35°≠2.98 mi.
4.25 mi
h
h
14. tan 35°= and tan 30°=
, so x=h cot 35°
x
x + 1000
and x+1000=h cot 30°. Then h cot 35°=
1000
h cot 30°-1000, so h=
≠3290.5 ft.
cot 30° - cot 35°
h
, so h=12 tan 17°≠3.67 mi.
12 mi
23. If h is the height of the vertical span, tan 15°=
h
, so
36.5 ft
h=36.5 tan 15°≠9.8 ft.
24. The distance d satisfies tan 5.25°=
760 ft
, so
d
d=760 cot 5.25°≠8271 ft.
5.25°
760 ft
30°
d
h
35°
x
1000 ft
15. Let x be the elevation of the bottom of the deck, and h be
x
the height of the deck. Then tan 30°=
and
200 ft
x + h
tan 40°=
, so x=200 tan 30° ft and x+h=
200 ft
200 tan 40° ft. Therefore h=200(tan 40°-tan 30°)≠
52.35 ft.
16. Let d be the distance traveled, and let x be car’s ending
distance from the base of the building. Then
100 ft
100 ft
tan 15°=
and tan 33°=
, so d+x=
d + x
x
100 cot 15° ft and x=100 cot 33° ft. Therefore
d=100(cot 15°-cot 33°)≠219 ft.
17. The two legs of the right triangle are the same length
(30 knots # 2 hr=60 naut mi), so both acute angles are
45°. The length of the hypotenuse is the distance:
60 12 L 84.85 naut mi. The bearing is 95°+45°=140°.
18. The two legs of the right triangle are 40 knots # 2 hr=
80 naut mi and 40 knots # 4 hr=160 naut mi. The
distance can be found with the Pythagorean Theorem:
d=80 15 L 178.885 naut mi. The acute angle at Fort
Lauderdale has measure tan–1 2, so the bearing is
65°+tan–1 2≠128.435° (see figure below)
65° 155°
80
Fort
160
Lauderdale
d
25. Let d be the distance from the boat to the shore, and let x
be the short leg of the smaller triangle. For the two triangles, the larger acute angles are 70° and 80°. Then
d
d
tan 80°= and tan 70°=
, or x=d cot 80°
x
x + 550
and x+550=d cot 70°. Therefore
550
d=
≠2931 ft.
cot 70° - cot 80°
26. If t is the time until the boats collide, the law enforcement
boat travels 23t naut mi. During that same time, the smugglers’ craft travels xt naut mi, where x is that craft’s speed.
These two distances are the legs of a right triangle
x
xt
(shown); then tan 15°=
, so x=23 tan 15°
=
23t
23
≠6.2 knots.
15°
23t
75° xt
27. (a) Frequency:
◊
16∏
=8 cycles/sec.
=
2∏
2∏
(b) d=6 cos 16∏t inches.
(c) When t=2.85, d≠1.854; this is about 4.1 in. left of
the starting position (when t=0, d=6).
28. (a) Frequency:
∏
1
◊
=
= cycle/sec.
2∏
2∏
2
(b) d=18 cos ∏t cm.
(c) Since 1 cycle takes 2 sec, there are 30 cycles/min.
196
Chapter 4
Trigonometric Functions
29. The frequency is 2 cycles/sec, so ◊ = 2 # 2∏
=4∏ radians/sec. Assuming the initial position is
d=3 cm: d=3 cos 4∏t.
30.
◊
=528, so ◊=1056∏ radians/sec.
2∏
31. (a) The amplitude is a=25 ft, the radius of the wheel.
[0, 13] by [42, 88]
(b) k=33 ft, the height of the center of the wheel.
(c)
◊
1
=
rotations/sec, so ◊=∏/10 radians/sec.
2∏
20
◊
1
=3 rpm= rotations/sec, so ◊=∏/10.
2∏
20
∏t
One possibility is h=–8 cos +9 m.
10
(b)
32. (a)
p
Algebraic solution: Solve 17 sin a 1t - 42 b + 65 = 70
6
for t.
17 sin a
p
1t - 42 b + 65 = 70
6
5
p
sin a 1t - 42 b =
6
17
5
p
1t - 42 = sin - 1 a b
6
17
p
1t - 42 L 0.299, 2.842
6
Note: sin ¨=sin (∏-¨)
t≠4.57, 9.43
[0, 30] by [–1, 20]
(c) h(4)≠6.5 m; h(10)=17 m.
33. (a) Given a period of 12, we have 12 =
12 0 b 0 = 2p so 0 b 0 =
2p
.
0b0
p
2p
= . We select the positive
12
6
p
.
6
(b) Using the high temperature of 82 and a low tempera34
82 - 48
=
so 0 a 0 = 17
ture of 48, we find 0 a 0 =
2
2
and we will select the positive value.
82 + 48
= 65
k =
2
(c) h is halfway between the times of the minimum and
maximum. Using the maximum at time t=7 and the
7 - 1
= 3. So,
minimum at time t=1, we have
2
h = 1 + 3 = 4.
p
(d) The fit is very good for y = 17 sin a 1t - 4 2 b + 65.
6
value so b =
Using either method to find t, find the day of the year
9.43
4.57 #
# 365 L 287.
365 L 139 and
as follows:
12
12
These represent May 19 and October 14.
34. (d) All have the correct period, but the others are incorrect
in various ways. Equation (a) oscillates between –25
and ±25, while equation (b) oscillates between –17
and ±33. Equation (c) is the closest among the incorrect formulas: it has the right maximum and minimum
values, but it does not have the property that h(0)=8.
This is accomplished by the horizontal shift in (d).
35. (a) Solve this graphically by finding the zero of the function
pt
P = 2t - 7 sin a b . The zero occurs at approximately
3
2.3. The function is positive to the right of the zero. So,
the shop began to make a profit in March.
[0, 13] by [–20, 50]
[0, 13] by [42, 88]
(e) There are several ways to find when the mean temperature will be 70°. Graphical solution: Graph the
line t=70 with the curve shown above, and find the
intersection of the two curves. The two intersections
are at t≠4.57 and t≠9.43.
(b) Solve this graphically by finding the maximum of the
pt
function P = 2 t - 7 sin a b . The maximum occurs
3
at approximately 10.76, so the shop enjoyed its greatest profit in November.
[0, 13] by [–20, 50]
Section 4.8
36. (a) Using the function W = 220 - 1.5t + 9.81 sin a
pt
b,
4
where t is measured in months after January 1 of the
first year and W is measured in pounds, we have t=0
at the beginning. This gives
p# 0
W = 220 - 1.5 10 2 + 9.81 sin a
b = 220 pounds.
4
At the end of two years, t=24, which gives
p # 24
W = 220 - 1.5 124 2 + 9.81 sin a
b = 184 pounds.
4
(b) Solve this graphically by finding the maximum of the
pt
function W = 220 - 1.5t + 9.81 sin a
b . The max4
imum occurs at t=1.75, where W L 227.
Solving Problems with Trigonometry
197
43. (a)
[0, 0.0062] by [–0.5, 1]
(b) The first is the best. This can be confirmed by graphing all three equations.
(c) About
1232
2464
L 392 oscillations/sec.
=
∏
2∏
44. (a) Newborn: about 6 hours. Four-year-old: about 24
hours. Adult: about 24 hours.
(b) The adult sleep cycle is perhaps most like a sinusoid,
though one might also pick the newborn cycle. At
least one can perhaps say that the four-year-old sleep
cycles is least like a sinusoid.
[0, 24] by [150, 250]
(c) Solve this graphically by finding the minimum of the
pt
function W = 220 - 1.5t + 9.81 sin a
b . The mini4
mum occurs at t L 22.2, where W L 177.
[0, 24] by [150, 250]
37. True. The frequency and the period are reciprocals:
f=1/T. So the higher the frequency, the shorter the
period.
38. False. One nautical mile equals about 1.15 statute miles,
and one knot is one nautical mile per hour. So, in the time
that the car travels 1 statute mile, the ship travels about
1.15 statute miles. Therefore the ship is traveling faster.
39. If the building height in feet is x, then tan 58°=x/50. So
x=50 tan 58° L 80. The answer is D.
45. The 7-gon can be split into 14 congruent right triangles
with a common vertex at the center. The legs of these triangles measure a and 2.5. The angle at the center is
2∏
∏
∏
= , so a=2.5 cot
L 5.2 cm
14
7
7
46. The 7-gon can be split into 14 congruent right triangles
with a common vertex at the center. The legs of these
triangles measure a and 2.5, while the hypotenuse has
∏
2∏
length r. The angle at the center is
= , so
14
7
∏
r=2.5 csc
L 5.8 cm
7
47. Choosing point E in the center of the rhombus, we have
¢AEB with right angle at E, and mjEAB=21°. Then
AE=18 cos 21° in., BE=18 sin 21° in., so that
AC=2AE≠33.6 in. and BD=2 BE≠12.9 in.
48. (a) BE=20 tan 50°≠23.8 ft.
(b) CD=BE+45 tan 20°≠40.2 ft.
(c) AE+ED=20 sec 50°+45 sec 20°≠79 ft, so the
total distance across the top of the roof is about 158 ft.
49. ¨=tan–1 0.06≠3.4
θ
40. By the Law of Cosines, the distance is
c= 2a2 + b2 - 2ab cos u
= 2402 + 202 + 2 1 40 2 120 2 cos 60 °≠53 naut. mi.
The answer is B.
41. Model the tide level as a sinusoidal function of time, t.
6 hr, 12 min=372 min is a half-period, and the amplitude
is half of 13-9=4. So use the model
f(t)=2 cos (∏t/372)-11 with t=0 at 8:15 PM. This
takes on a value of –10 at t=124. The answer is D.
42. The answer is A.
7 mi
(0.06)(7)
= 0.42 mi
50. Observe that there are two (congruent) right triangles
with hypotenuse 4100 mi (see figure below). The acute
4000
angle adjacent to the 4000 mi leg has measure cos–1
4100
–1 40
–1 40
=cos
, so ¨=2 cos
≠25.361°
41
41
≠0.4426 radians. The arc length is s=r¨
≠(4000 mi)(0.4426)≠1771 mi.
100 mi
4000 mi
θ
198
Chapter 4
Trigonometric Functions
51. (a)
10. 900° or 5∏ radians
[0, 0.0092] by [–1.6, 1.6]
(b) One pretty good match is
y=1.51971 sin[2467(t-0.0002)] (that is,
a=1.51971, b=2467, h=0.0002). Answers will
vary but should be close to these values. A good estimate of a can be found by noting the highest and lowest values of “Pressure” from the data. For the value
of b, note the time between maxima (approx.
0.0033728-0.0008256=0.0025472 sec); this is the
2p
period, so b L
L 2467. Finally, since
0.0025472
0.0008256 is the location of the first peak after t=0,
p
choose h so that 2467(0.0008256-h)≠ . This
2
gives h L 0.0002.
1
2467
L
L 393 Hz.
2p
0.0025472
It appears to be a G.
(c) Frequency: about
(d) Exercise 41 had b≠2464, so the frequency is again
about 392 Hz; it also appears to be a G.
■ Chapter 4 Review
1. On the positive y-axis (between quadrants I and II);
5p # 180°
=450°.
p
2
2. Quadrant II;
3p # 180°
=135°.
p
4
3. Quadrant III; –135° #
4. Quadrant IV; –45° #
5. Quadrant I; 78° #
28p
p
=
.
180°
45
p # 180°
=15°.
12 p
8. Quadrant II;
9. 270° or
p
p
=– .
180°
4
p
13p
=
.
180°
30
6. Quadrant II; 112° #
7. Quadrant I;
p
3p
=– .
180°
4
7p # 180°
=126°.
10 p
3p
radians
2
For #11–16, it may be useful to plot the given points and draw
the terminal side to determine the angle. Be sure to make
your sketch on a “square viewing window.”
11. ¨=tan–1 a
1
p
b =30°= radians
6
13
12. ¨=135°=
3p
radians
4
13. ¨=120°=
2p
radians
3
14. ¨=225°=
5p
radians
4
15. ¨=360°+tan–1 (–2)≠296.565≠5.176 radians
16. ¨=tan–1 2≠63.435°≠1.107 radians
17. sin 30°=
1
2
19. tan(–135°)=1
21. sin
5p
1
=
6
2
23. sec a -
p
b =2
3
18. cos 330°=
13
2
20. sec(–135°)=– 12
22. csc
2p
2
=
3
13
24. tan a -
2p
b = 13
3
25. csc 270°=–1
26. sec 180°=–1
27. cot(–90°)=0
28. tan 360°=0
p
= 30°; use a 30–60 right triangle with
6
side lengths 13, (–)1, and 2 (hypotenuse).
p
p
1
p
13
1
sin a - b =– , cos a - b = , tan a - b =–
;
6
2
6
2
6
13
p
p
p
2
csc a - b =–2, sec a - b =
; cot a - b =– 13.
6
6
6
13
p
30. Reference angle: = 45°; use a 45–45 right triangle with
4
side lengths (–)1, 1, and 12 (hypotenuse).
19p
19p
1
1
19p
sin
=
, cos
=–
, tan
=–1;
4
4
4
12
12
19p
19p
19p
csc
= 12, sec
=– 12; cot
=–1.
4
4
4
29. Reference angle:
31. Reference angle: 45°; use a 45–45 right triangle with side
lengths (–)1, (–)1, and 12 (hypotenuse).
1
1
,
sin(–135°)=–
, cos(–135°)=–
12
12
tan(–135°)=1; csc(–135°)=– 12,
sec(–135°)=– 12, cot(–135°)=1.
32. Reference angle: 60°; use a 30–60 right triangle with side
lengths 1, 13, and 2 (hypotenuse).
13
1
sin 420°=
, cos 420°= , tan 420°= 13;
2
2
1
2
csc 420°=
.
, sec 420°=2, cot 420°=
13
13
Chapter 4 Review
5
,
13
12
5
13
13
cos Å= , tan Å= , csc Å= , sec Å= ,
13
12
5
12
12
cot Å= .
5
33. The hypotenuse length is 13 cm, so sin Å=
For #34–35, since we are using a right triangle, we assume that ¨
is acute.
34. Draw a right triangle with legs 5 (adjacent) and
272 - 52= 124=2 16, and hypotenuse 7.
2 16
5
2 16
7
, cos ¨= , tan ¨=
; csc ¨=
,
sin ¨=
7
7
5
2 16
7
5
.
sec ¨= , cot ¨=
5
2 16
35. Draw a right triangle with legs 8 (adjacent) and 15, and
hypotenuse 282 + 152= 1289=17.
15
8
15
17
sin ¨= , cos ¨= , tan ¨= ; csc ¨= ,
17
17
8
15
17
8
sec ¨= , cot ¨= .
8
15
36. ¨≠64.623°
37. x≠4.075 radians
38. x≠0.220 or x≠2.922 radians
For #39–44, choose whichever of the following formulas is
appropriate:
a= 2c2 - b2=c sin Å=c cos ı=b tan Å=
b
tan ı
b= 2c2 - a2=c cos Å=c sin ı=a tan ı=
a
tan Å
c= 2a2 + b2=
44. c= 2a2 + b2 = 22.52 + 7.32 = 159.54≠7.716. For
2.5
the angles, we know tan Å= ; using a calculator, we
7.3
find Å≠18.90°, so that ı=90°-Å≠71.10°.
45. sin x 6 0 and cos x 6 0: Quadrant III
46. cos x 6 0 and
1
7 0: Quadrant II
sin x
47. sin x 7 0 and cos x 6 0: Quadrant II
48.
1
1
6 0 and
7 0: Quadrant II
cos x
sin x
2
1
, cos ¨=–
,
15
15
15
1
tan ¨=–2; csc ¨=
, sec ¨=– 15, cot ¨=– .
2
2
49. The distance OP=3 15, so sin ¨=
7
12
, cos ¨=
,
1193
1193
7
1193
1193
12
tan ¨= ; csc ¨=
, sec ¨=
, cot ¨= .
12
7
12
7
50. OP= 1193, so sin ¨=
3
5
, cos ¨=–
,
134
134
3
134
134
5
tan ¨= ; csc ¨=–
, sec ¨=–
, cot ¨= .
5
3
5
3
51. OP= 134, so sin ¨=–
9
4
, cos ¨=
,
197
197
9
197
197
4
tan ¨= ; csc ¨=
, sec ¨=
, cot ¨= .
4
9
4
9
52. OP= 197, so sin ¨=
53. Starting from y=sin x, translate left ∏ units.
b
a
a
b
=
=
=
cos ı
sin Å
sin ı
cos Å
If one angle is given, subtract from 90° to find the other angle.
If neither Å nor ı is given, find the value of one of the trigonometric functions, then use a calculator to approximate the
value of one angle, then subtract from 90° to find the other.
39. a=c sin Å=15 sin 35°≠8.604, b=c cos Å
=15 cos 35°≠12.287, ı=90°-Å=55°
[–2∏, 2∏] by [–1.2, 1.2]
54. Starting from y=cos x, vertically stretch by 2 then
translate up 3 units.
40. a= 2c2 - b2 = 2102 - 82 = 6. For the angles, we
8
4
know cos Å=
= ; using a calculator, we find
10
5
Å≠36.87°, so that ı=90°-Å≠53.13°.
41. b=a tan ı=7 tan 48°≠7.774, c=
=
a
cos ı
7
≠10.461, Å=90°-ı=42°
cos 48°
42. a=c sin Å=8 sin 28°≠3.756, b=c cos Å
≠8 cos 28°=7.064, ı=90°-Å=62°
43. a= 2c2 - b2 = 272 - 52 = 124 = 2 16≠4.90. For
5
the angles, we know cos Å= ; using a calculator, we find
7
Å≠44.42°, so that ı=90°-Å≠45.58°.
199
[–2∏, 2∏] by [–1, 6]
55. Starting from y=cos x, translate left
across x-axis, and translate up 4 units.
[–2∏, 2∏] by [–1, 6]
p
units, reflect
2
200
Chapter 4
Trigonometric Functions
56. Starting from y=sin x, translate right ∏ units, vertically
stretch by 3, reflect across x-axis, and translate down
2 units.
61. f(x)=2 sin 3x. Amplitude: 2; period:
2p
; phase shift: 0;
3
domain: (– q , q ); range: [–2, 2].
62. g(x)=3 cos 4x. Amplitude: 3; period:
p
; phase shift: 0;
2
domain: (– q , q ); range: [–3, 3].
63. f(x)=1.5 sin c 2 a x ∏; phase shift:
[–2∏, 2∏] by [–6, 2]
1
57. Starting from y=tan x, horizontally shrink by .
2
p
; domain: (– q , q ); range: [–1.5, 1.5].
8
64. g(x)=–2 sin c 3 a x phase shift:
p
b d . Amplitude: 1.5; period:
8
p
2p
b d . Amplitude: 2; period:
;
9
3
p
; domain: (– q , q ); range: [–2, 2].
9
65. y=4 cos c 2 a x -
1
b d . Amplitude: 4; period: ∏;
2
1
phase shift: ; domain: (– q , q ); range: [–4, 4].
2
66. g(x)=–2 cos c 3 a x +
[–0.5∏, 0.5∏] by [–5, 5]
1
58. Starting from y=cot x, horizontally shrink by , verti3
cally stretch by 2, and reflect across x-axis (in any order).
1
2p
b d . Amplitude: 2; period:
;
3
3
1
phase shift: – ; domain: (– q , q ); range: [–2, 2].
3
For #67–68, graph the function. Estimate a as the amplitude of
the graph (i.e., the height of the maximum). Notice that the
value of b is always the coefficient of x in the original functions. Finally, note that a sin[b(x-h)]=0 when x=h, so
estimate h using a zero of f(x) where f(x) changes from negative to positive.
67. a≠4.47, b=1, and h≠1.11, so f(x)
≠4.47 sin(x-1.11).
⎡–␲ , ␲ ⎤ by [–10, 10]
⎣ 3 3⎦
59. Starting from y=sec x, horizontally stretch by 2, vertically stretch by 2, and reflect across x-axis (in any order).
68. a≠3.61, b=2, and h≠–1.08, so f(x)
≠3.61 sin[2(x+1.08)].
69. L 49.996° L 0.873 radians
70. L 61.380° L 1.071 radians
[–4∏, 4∏] by [–8, 8]
60. Starting from y=csc px, horizontally shrink by
1
.
p
[–2, 2] by [–5, 5]
For #61–66, recall that for y=a sin[b(x-h)] or
y=a cos[b(x-h)], the amplitude is 0 a 0 , the period is
2p
,
0b0
and the phase shift is h. The domain is always (– q , q ), and
the range is [– 0 a 0 , 0 a 0 ].
71. 45°=
p
radians
4
72. 60°=
p
radians
3
1
73. Starting from y=sin–1x, horizontally shrink by .
3
1 1
p p
Domain: c - , d . Range: c - , d .
3 3
2 2
1
74. Starting from y=tan–1x, horizontally shrink by .
2
p p
Domain: (– q , q ). Range: a - , b .
2 2
75. Starting from y=sin–1x, translate right 1 unit, horizontally
1
2
shrink by , translate up 2 units. Domain: c 0, d .
3
3
p
p
Range: c 2 - , 2 + d .
2
2
76. Starting from y=cos–1x, translate left 1 unit, horizontally
1
shrink by , translate down 3 units. Domain: [–1, 0].
2
Range: [–3, ∏-3].
Chapter 4 Review
77. x=
5p
6
78. x=
79. x=
3p
4
80.
p
6
5p
3
5p
82.
6
3p
81.
2
sin x
83. As 0 x 0 S q , 2 S 0.
x
3
84. As x S q , e–x/12 sin(2x-3) S 0; as x S - q , the
5
function oscillates from positive to negative, and tends
to q in absolute value.
85. tan(tan–1 1)=tan
86. cos–1 a cos
p
= 1
4
96. tan 25°=
h
150 ft
150 ft
and tan 18°=
, so
x
d + x
x=150 cot 42° and d+x=150 cot 18°. Then
d=150 cot 18°-150 cot 42°≠295 ft.
97. tan 42°=
150 ft
42°
x
18°
d
PQ
, so PQ=4 tan 22°≠1.62 mi.
4
99. See figure below.
north tower
3
sin ¨
p p
, where ¨ is an angle in c - , d
87. tan(sin–1 )=
5
cos ¨
2 2
3
with sin ¨= . Then cos ¨= 21 - sin2 ¨= 10.64
5
=0.80 and tan ¨=0.75
88. cos–1 cos a -
h
, so h=51 tan 25°≠23.8 ft.
51 ft
25°
51 ft
98. tan 22°=
p
p
1
b =cos–1 =
3
2
3
23°
128°
p
p
b =7
7
south tower
p
+ np, n an integer.
89. Periodic; period ∏. Domain x Z
2
Range: [1, q ).
90. Not periodic. Domain: (– q , q ). Range: [–1, 1].
91. Not periodic. Domain: x Z
p
+ np, n an integer.
2
x
x + 855
100. tan 25°= and tan 33°=
, so x=d tan 25°
d
d
and x+855=d tan 33°. Then d tan 25°+855=
855
d tan 33°, so d=
≠4670 ft.
tan 33° - tan 25°
Range: [– q , q ).
25°
d
92. Periodic; period 2∏. Domain: (– q , q ). Range:
approximately [–5, 4.65].
33°
x
2p
4p
b =
93. s=r¨=(2) a
3
3
855 ft
94. Draw a right triangle with horizontal leg x (if x 7 0, draw
the horizontal leg right; if x 6 0, draw it left), vertically
leg 21 - x2, and hypotenuse 1. If x 7 0, let ¨ be the
acute angle adjacent to the horizontal leg; if x 6 0, let ¨
be the supplement of this angle. Then ¨=cos–1x, so
21 - x2
.
tan ¨ =tan(cos–1 x)=
x
101. tan 72°24'=
1
x
95. tan 78°=
1
h
, so h=62 tan 72°24'≠195.4 ft
62 ft
h
72°24′
x< 0
1 – x2
x> 0
θ
θ
x
h
, so h=100 tan 78°≠470 m.
100 m
62 ft
102. Let ¨ be the angle of elevation. Note that sin ¨=
so h=75 sin ¨.
(a) If ¨=22°, then h=75 sin 22°≠28 ft.
(b) If ¨=27°, then h=75 sin 27°≠34 ft.
75 ft
h
h
78°
100 m
201
θ
h
,
75 ft
202
Chapter 4
Trigonometric Functions
103. s=r¨=(44 in.) a 6° #
22p
p
b =
L 4.6 in.
180°
15
110
11
=
104. The blade sweeps out
of a circle; take this
360
36
fraction of (the area of a 20 in.-radius circle minus
the area of a 4 in.-radius circle):
352p
11
A= [∏(20)¤-∏(4)¤]=
≠368.6 in¤
36
3
105. Solve algebraically: Set T(x)=32 and solve for x.
2p
1 x - 114 2 d + 26 = 32
37.3 sin c
365
6
2p
1 x - 114 2 d =
sin c
365
37.3
6
2p
1x - 114 2 = sin - 1 a
b
365
37.3
2p
1 x - 114 2 L 0.162, 2.98
365
Note: sin ¨=sin(∏-¨)
x L 123, 287
Solve graphically: Graph T(x)=32 and
2p
T1 x2 = 37.3 sin c
1 x - 114 2 d + 25 on the same set
365
of axes, and then determine the intersections.
[0, 365] by [–50, 100]
[0, 365] by [–50, 100]
Using either method, we would expect the average temperature to be 32°F on day 123 (May 3) and day 287
(October 14).
106. Set h(x)=0 and solve for x.
x
0=35 cos a b + 17
35
x
17
= cos
35
35
x
17
= cos - 1 a - b L 2.078 rad
35
35
x=(35)(2.078)
x≠72.7 ft.
Chapter 4 Project
Solutions are based on the sample data shown in the table.
1.
[–0.1, 2.1] by [0, 1]
2. The peak value seems to occur between x=0.4 and
x=0.5, so let h=0.45. The difference of the two
extreme values is 0.931-0.495=0.436, so let
a L 0.436/2 L 0.22. The average of the two extreme
values is (0.931+0.495)/2=0.713, so let k=0.71.
The time interval from x=0.5 to x=1.3, which
equals 0.8, is right around a half-period, so let
b=∏/0.8 L 3.93. Then the equation is
y L 0.22 cos (3.93(x-0.45))+0.71.
3. The constant a represents half the distance the pendulum
bob swings as it moves from its highest point to its lowest
point. And k represents the distance from the detector to
the pendulum bob when it is in mid-swing.
4. Since the sine and cosine functions differ only by a phase
shift, only h would change.
5. The regression yields y L 0.22 sin (3.87 x-0.16)+0.71.
Most calculator/computer regression models are
expressed in the form y=a sin (bx+f)+k, where
–f/b=h in the equation y=a sin (b(x-h))+k.
Here, the regression equation can be rewritten as
y L 0.22 sin (3.87 (x-0.04))+0.71. The difference in
the two values of h for the cosine and sine models is 0.41,
which is right around a quarter-period, as it should be.
Section 5.1
Fundamental Identities
203
Chapter 5
Analytic Trigonometry
■ Section 5.1 Fundamental Identities
Exploration 1
1. cos ¨=1> sec ¨, sec ¨=1> cos ¨, and tan ¨=sin ¨> cos ¨
2. sin ¨=1> csc ¨ and tan ¨=1> cot ¨
3. csc ¨=1> sin ¨, cot ¨=1> tan ¨, and cot ¨=cos ¨> sin ¨
tan ¨ = - 115. And
cot ¨ = 1>tan ¨ = -1> 115 = - 115>15.
4. sin2 ¨ = 1 - cos2 ¨ = 1 - 10.8 2 2 = 0.36, so sin ¨ = ; 0.6.
But cos ¨>0, tan ¨<0 implies sin ¨ 6 0,
so sin ¨ = -0.6. Finally,
tan ¨ = sin ¨>cos ¨ = - 0.6>0.8 = -0.75.
5. cos(∏/2-¨)=sin ¨=0.45
6. cot ¨=tan(∏/2-¨)=–5.32
Quick Review 5.1
7. cos(–¨)=cos ¨=sin(∏/2-¨)
=–sin(¨-∏/2)=–0.73
For #1–4, use a calculator.
1. 1.1760 rad=67.380°
8. cot(–¨)=–cot ¨=–tan(∏/2-¨)
=tan(¨-∏/2)=7.89
2. 0.9273 rad=53.130°
3. 2.4981 rad=143.130°
9. tan x cos x=
4. –0.3948 rad=–22.620°
5. a2 - 2ab + b2 = 1 a - b2 2
6. 4u2 + 4u + 1 = 12u + 1 2 2
10. cot x tan x=
7. 2x2 - 3xy - 2y2 = 12x + y2 1 x - 2y 2
8. 2v - 5v - 3 = 12v + 1 2 1 v - 3 2
y - 2x
1 y
2 x
9. # - # =
x y
y x
xy
ay + bx
a y
b x
10. # + # =
x y
y x
xy
2
11.
12.
x + y
xy
= 1x + y2 # a
b = xy
1
1
x + y
+
x
y
2
2
y
x #x + y
# x - y = x2 + y2
x - y x + y
x + y x - y
x - y
1. sec2 ¨ = 1 + tan2 ¨ = 1 + 13>4 2 2 = 25>16, so
sec ¨ = ;5>4. Then cos ¨ = 1>sec ¨ = ; 4>5. But
sin ¨, tan ¨ 7 0 implies cos ¨ 7 0. So cos ¨ = 4>5. Finally,
3
4
3
sin ¨
=
cos ¨
4
3
3 4
3
sin ¨ = cos ¨ = a b = .
4
4 5
5
tan ¨ =
2. sec2 ¨ = 1 + tan2 ¨ = 1 +
But cos ¨ 7 0 implies sec ¨
tan ¨ = 3
sec ¨
= 3
csc ¨
1
csc ¨ = sec ¨ =
3
32 = 10, so sec ¨ = ; 110.
7 0, so sec ¨ = 110. Finally,
p
1 #
- yb =
cos y = 1
2
cos y
cos u #
sin u=cos u
sin u
1>cos2 x
1 + tan2 x
sec2 x
sin2 x
13.
= 2 =
= 2 = tan2 x
2
2
csc x
csc x
1>sin x
cos x
12. cot u sin u=
14.
1 - cos2 u
sin2 u
=
= sin u
sin u
sin u
15. cos x - cos3 x = cos x11 - cos2 x2 = cos x sin2 x
16.
sec2 u
sin2 u + tan2 u + cos2 u
1 + tan2 u
=
=
=sec u
sec u
sec u
sec u
2
1
=–1
sin1 –x2
1
# cos1 -x2 = 1
cos1–x2
p
cos a - x b
cos1 -x2
2
p
#
19. cot(–x) cot a - x b =
2
sin1 -x2
p
sin a - x b
2
cos 1 -x2 sin 1x2
#
=
=–1
sin 1 -x2 cos1x2
18. sec(–x) cos(–x)=
20. cot(–x) tan(–x)=
cos1 -x2 sin1 -x2
#
sin1 -x2 cos1 - x2
=1
21. sin2 1 - x2 + cos2 1 -x2 = 1
22. sec2 1 - x2 - tan2 x = sec2 x - tan2 x = 1
110
1
110 =
.
3
3
3. tan ¨ = sec ¨ - 1 = 4 - 1 = 15, so tan ¨ = ; 115.
But sec ¨ 7 0, sin ¨ 6 0 implies tan ¨ 6 0, so
2
cos x sin x
#
=1
sin x cos x
17. sin x csc(–x)=sin x #
Section 5.1 Exercises
2
11. sec y sin a
sin x
# cos x=sin x
cos x
23.
tan a
p
- x b csc x
2
cos x sin x
cot x
#
=
= cos x
=
csc x
sin x
1
csc2 x
204
24.
Chapter 5
Analytic Trigonometry
1 + tan x sin x cos x
sin x cos x + sin2 x
#
=
1 + cot x sin x cos x
sin x cos x + cos2 x
sin x1cos x + sin x2
=
=tan x
cos x1 sin x + cos x2
25. 1sec x + csc x2 - 1tan x + cot x2
= 1sec2 x - tan2 x2 + 1 csc2 x - cot2 x2 =1+1=2
2
26.
2
2
2
1
sec2 u - tan2 u
= = 1
1
cos2 v + sin2 v
27. (sin x)(tan x+cot x)=(sin x) a
=sin x a
cos x
sin x
+
b
cos x
sin x
1
sin2 x + cos2 x
b=
=sec x
1cos x2 1 sin x2
cos x
28. sin ¨-tan ¨ cos ¨+cos a
=sin ¨-
p
- ub
2
sin u
# cos u+sin ¨=sin ¨
cos u
29. (sin x)(cos x)(tan x)(sec x)(csc x)
1
1
sin x
sin x
ba
ba
b =
=(sin x)(cos x) a
cos x
cos x
sin x
cos x
=tan x
30.
1 sec y - tan y2 1 sec y + tan y2
=
sec y
sin y
sin y
1
1
ba
+
b
a
cos y
cos y
cos y
cos y
a
1
b
cos y
1 + sin y - sin y - sin2 y cos y
1 - sin2 y
#
=
=
1
cos y
cos2 y
cos2 y
= cos y
=
cos y
tan x
tan x
+
31.
2
csc x
sec2 x
sin x #
sin x
b 1sin2 x2 + a
b cos¤ x
=a
cos x
cos x
sin x
sin x
b 1sin2 x + cos2 x2 =
= tan x.
=a
cos x
cos x
a
1
# 1 b
cos2 x sin x
1
1
b + a 2 b
a
cos2 x
sin x
2
1
x sin2 x
sin x
# cos
= sin x
= 2
=
2
#
1
cos x sin x sin x + cos2 x
sec2 x csc x
32.
=
sec2 x + csc2 x
33.
sec2x
1
+ 2 =csc¤x+
2
sin x
tan x
1
cos2x¢
sin2x
≤
cos2x
1
=2csc¤x
sin2x
1
1
+
34.
1 - sin x
1 + sin x
1 - sin x
1 + sin x
+
=
11 - sin x2 11 + sin x2
11 - sin x2 1 1 + sin x2
2
2
=
= 2 sec2 x
=
1 - sin2 x
cos2 x
=csc¤x+
35.
36.
37.
38.
sin x
sin x
= 1 sin x2 1tan2 x2 - 1 sin x2 1 sec2 x2
cot2 x
cos2 x
= 1sin x2 1tan2 x - sec2 x2 = 1sin x2 1 - 12 = -sin x
1
sec x + 1 - sec x + 1
1
=
sec x - 1
sec x + 1
sec2 x - 1
2
= 2 = 2 cot2 x
tan x
sin x
sec x cos x - sin2 x
1 - sin2 x
sec x
=
=
sin x
cos x
sin x cos x
sin x cos x
cos x
cos2 x
=
= cot x
=
sin x cos x
sin x
sin2 x + 11 - cos x2 2
1 - cos x
sin x
+
=
1 - cos x
sin x
sin x1 1 - cos x2
2 11 - cos x2
sin2 x + cos x2 + 1 - 2 cos x
=
=
sin x11 - cos x2
sin x11 - cos x2
=2 csc x
39. cos2 x + 2 cos x + 1 = 1cos x + 12 2
40. 1 - 2 sin x + sin2 x = 11 - sin x2 2
41. 1 - 2 sin x + 1 1 - cos2 x2 = 1 - 2 sin x + sin2 x
= 11 - sin x2 2
42. sin x - cos2 x - 1 = sin x + sin2 x - 2
=(sin x-1)(sin x+2)
43. cos x - 2 sin2 x + 1 = cos x - 2 + 2 cos2 x + 1
=2 cos2 x + cos x - 1 = 12 cos x - 1 2 1cos x + 1 2
2
+ 1 = sin2 x + 2 sin x + 1
csc x
= 1sin x + 12 2
44. sin2 x +
45. 4 tan2 x -
4
+ sin x csc x
cot x
1
sin x
= 4 tan2 x - 4 tan x + 1 = 12 tan x - 12 2
=4 tan2 x - 4 tan x + sin x #
46. sec2 x - sec x + tan2 x = sec2 x - sec x + sec2 x - 1
=2 sec2 x - sec x - 1 = 12 sec x + 1 2 1sec x - 1 2
47.
1 1 - sin x2 11 + sin x2
1 - sin2 x
=
= 1 - sin x
1 + sin x
1 + sin x
1tan a - 12 1 tan a + 12
tan2 a - 1
=
= tan a - 1
1 + tan a
1 + tan a
11 - cos x2 11 + cos x2
sin2 x
1 - cos2 x
49.
=
=
1 + cos x
1 + cos x
1 + cos x
=1-cos x
1sec x - 1 2 1sec x + 1 2
sec2x - 1
tan2 x
50.
=
=
sec x + 1
sec x + 1
sec x + 1
=sec x-1
48.
51. 1cos x2 12 sin x - 12 = 0, so either cos x=0 or
1
p
p
sin x = . Then x =
+ np or x =
+ 2np or
2
2
6
5p
+ 2np, n an integer. On the interval:
x =
6
p p 5p 3p
x = e , ,
,
f
6 2 6 2
Section 5.1
52. 1tan x2 1 12 cos x - 1 2 = 0, so either tan x = 0 or
1
p
. Then x = np or x = ; + 2np, n an
cos x =
4
12
p
7p
integer. On the interval: x = e 0, , p,
f
4
4
53. 1tan x2 1sin2 x - 1 2 = 0, so either tan x = 0 or
p
sin2 x = 1. Then x = np or x =
+ np, n an interger.
2
p
However, tan x excludes x =
+ np, so we have only
2
x=n∏, n an integer. On the interval: x= 5 0, p6
54. 1sin x2 1 tan2 x - 12 = 0, so either sin x=0
or tan2 x = 1.
p
p
+ n , n an integer. Put another
Then x = np or x =
4
2
p
p 3p
way, all multiples of except for ; , ; , etc.
4
2
2
p 3p
5p 7p
On the interval: x = e 0, ,
, p,
,
f
4 4
4 4
p
55. tan x = ; 13, so x = ; + np, n an integer.
3
p 2p 4p 5p
On the interval: x = e ,
,
,
f
3 3 3 3
p
p
1
56. sin x = ;
, so x =
+ n , n an integer.
4
2
12
p 3p 5p 7p
On the interval: x = e ,
,
,
f
4 4 4 4
57. 12 cos x - 1 2 2 = 0, so cos x =
x= ;
1
; therefore
2
p
+ 2np, n an integer.
3
1
58. (2 sin x+1)(sin x+1)=0, so sin x= - or
2
p
5p
+ 2np or
sin x=–1. Then x= - +2n∏, x= 6
6
p
x = - + 2np, n an integer.
2
59. 1 sin u2 1 sin u - 2 2 = 0, so sin u = 0 or sin u = 2. Then
u = np, n an integer.
60. 3 sin t=2-2 sin¤ t, or 2 sin¤ t+3 sin t-2=0. This
1
factors to (2 sin t-1)(sin t+2)=0, so sin t= or
2
5p
p
+ 2np or t =
+ 2np,
sin t=–2. Then t =
6
6
n an integer.
61. cos(sin x)=1 if sin x=np. Only n=0 gives a value
between –1 and ±1, so sin x=0, or x=np, n an
integer.
62. This can be rewritten as (2 sin x-1)(sin x+2)=0, so
1
p
+ 2np or
sin x= or sin x = - 2. Then x =
2
6
5p
x =
+ 2np, n an integer. See also #60.
6
-1
63. cos 0.37 L 1.1918, so the solution set is
{ ; 1.1918+2n∏| n=0, ; 1, ; 2, . . . }.
Fundamental Identities
205
64. cos-1 0.75 L 0.7227, so the solution set is
{ ; 0.7227+2n∏| n=0, ; 1, ; 2, . . . }.
65. sin-1 0.30 L 0.3047 and ∏-0.3047≠2.8369, so the
solution set is {0.3047+2n∏ or 2.8369+2n∏| n=0,
; 1, ; 2, . . . }.
66. tan-1 5 L 1.3734, so the solution set is {1.3734+n∏|
n=0, ; 1, ; 2, . . . }.
67. 10.4 L 0.63246, and cos - 1 0.63246 L 0.8861, so the solution set is { ; 0.8861+n∏| n=0, ; 1, ; 2, . . . }.
68. 10.4 L 0.63246 and sin - 1 0.63246 L 0.6847, so the solution set is { ; 0.6847+n∏| n=0, ; 1, ; 2, . . . }.
69. 21 - cos2 u = @ sin u @
70. 2tan2 u + 1 = @ sec u @
71. 29 sec2 u - 9 = 3 @ tan [email protected]
72. 236 - 36 sin2 u = 6 @ cos u @
73. 281 tan2 u + 81 = [email protected] sec u @
74. 2100 sec2 u - 100 = 10 @ tan [email protected]
75. True. Since cosine is an even function, so is secant, and
thus sec (x-∏/2)=sec (∏/2-x), which equals csc x
by one of the cofunction identities.
76. False. The domain of validity does not include values of ¨
for which cos ¨=0 and tan ¨=sin ¨/cos ¨ is undefined,
namely all odd integer multiples of ∏/2.
77. tan x sec x=tan x/cos x=sin x/cos2 x Z sin x. The
answer is D.
78. sine, tangent, cosecant, and cotangent are odd, while
cosine and secant are even. The answer is A.
79. (sec ¨+1)(sec ¨-1)=sec2 ¨-1=tan2 ¨. The
answer is C.
80. By the quadratic formula, 3 cos2 x+cos x-2=0
implies
cos x =
-1 ; 11 - 4 132 1 -2 2
21 32
2
= -1 or
3
There are three solutions on the interval (0, 2∏). The
answer is D.
81. sin x, cos x = ; 21 - sin2 x , tan x = ;
csc x=
sin x
21 - sin2 x
,
1
1
, sec x= ;
sin x
21 - sin2 x
cot x = ;
21 - sin2 x
sin x
82. sin x = ; 21 - cos2 x , cos x, tan x = ;
csc x = ;
cot x = ;
1
21 - cos x
cos x
2
, sec x =
21 - cos2 x
,
cos x
1
,
cos x
21 - cos2 x
83. The two functions are parallel to each other, separated by
1 unit for every x. At any x, the distance between the two
graphs is sin2 x - 1 - cos2 x2 = sin2 x + cos2 x = 1.
206
Chapter 5
Analytic Trigonometry
90. Use the hint:
cos1 p - x2 =cos1p>2 - 1x - p>22 2
Cofunction identity
=sin1x - p>2 2
-sin
1p>2
x2
Since sin is odd
=
= -cos x
Cofunction identity
[–2∏, 2∏] by [–4, 4]
84. The two functions are parallel to each other, separated by
1 unit for every x. At any x, the distance between the two
graphs is sec2 x - tan2 x = 1.
[–2∏, 2∏] by [–4, 4]
85. (a)
91. Since A, B, and C are angles of a triangle, A+B=
∏-C. So: sin(A+B)=sin(∏-C)
=sin C
92. Using the identities from Exercises 69 and 70, we have:
sin1p - x2
tan(∏-x)=
cos1p - x2
sin x
=
-cos x
= - tan x
■ Section 5.2 Proving Trigonometric
Identities
Exploration 1
1. The graphs lead us to conclude that this is not an identity.
[–6, 70] by [220 000, 260 000]
(b) The equation is
y=13,111 sin(0.22997x+1.571)+238,855.
[–2∏, 2∏] by [–4, 4]
2. For example, cos(2 # 0)=1, whereas 2 cos(0)=2.
3. Yes.
4. The graphs lead us to conclude that this is an identity.
[–6, 70] by [220 000, 260 000]
(c) 12p2 >0.22998 L 27.32 days. This is the number of days
that it takes the Moon to make one complete orbit of
the Earth (known as the Moon’s sidereal period).
(d) 225,744 miles
(e) y = 13,111 cos1 - 0.22997x2 + 238,855, or
y = 13,111 cos10.22997x2 + 238,855.
86. Answers will vary.
87. Factor the left-hand side:
sin4 u - cos4 u = 1sin2 u - cos2 u 2 1 sin2 u + cos2 u 2
= 1sin2 u - cos2 u 2 # 1
=sin2 u - cos2 u
88. Any k satisfying k 2 or k -2.
89. Use the hint:
sin1p - x2 =sin1p>2 - 1 x - p>2 2 2
Cofunction identity
=cos1x - p>22
Since cos is even
=cos1p>2 - x2
=sin x
Cofunction identity
[–2∏, 2∏] by [–3, 3]
5. No. The graph window can not show the full graphs, so
they could differ outside the viewing window. Also, the
function values could be so close that the graphs appear
to coincide.
Quick Review 5.2
1. csc x+sec x=
1
1
sin x + cos x
+
=
sin x
cos x
sin x cos x
sin x
cos x
sin2 x + cos2 x
=
+
cos x
sin x
sin x cos x
1
=
sin x cos x
2. tan x+cot x=
1
1
cos2 x + sin2 x
+ sin x #
=
sin x
cos x
sin x cos x
1
=
sin x cos x
3. cos x #
Section 5.2
4. sin ¨ #
cos ¨
sin ¨
-cos ¨ #
=cos ¨-sin ¨
sin ¨
cos ¨
Proving Trigonometric Identities
12. (sin x)(cot x+cos x tan x)
cos x
sin x
=sin x #
+sin x cos x #
=cos x+sin2x
sin x
cos x
sin x
cos x
+
=sin2x+cos2x=1
1>sin x
1>cos x
1>cos Å
sin2Å
1
sin Å
6.
= 2 2
cos Å
cos Å>sin a
cos Å
cos2Å
2
1 - sin Å
=
= 1
cos2Å
5.
13. (1-tan x)2=1-2 tan x+tan2x
=(1+tan2x)-2 tan x=sec2x-2 tan x
14. (cos x-sin x)2=cos2x-2 sin x cos x+sin2x
=(cos2x+sin2x)-2 sin x cos x=1-2 sin x cos x
15. One possible proof:
11 - cos u 2 11 + cos u2
7. No. (Any negative x.)
cos2u
8. Yes.
9. No. (Any x for which sin x<0, e.g. x=–∏/2.)
10. No. (Any x for which tan x<0, e.g. x=–∏/4.)
11. Yes.
sin x
1
sin x + 1
+
=
cos x
cos x
cos x
cos x1sin x + 12
cos x1sin x + 12
cos x
=
=
=
1 - sin x
cos2x
1 - sin2x
Section 5.2 Exercises
x1x2 - x2
- 1x - 12
x
x - x - 1x2 - 12
-x + 1
1 - x
2
3. One possible proof:
x2 - 4
x2 - 9
x - 2
x + 3
1x + 32 1 x - 3 2
1x + 22 1x - 2 2
=
x - 2
x + 3
= x + 2 - 1x - 3 2
= 5
4. One possible proof:
1x - 12 1 x + 2 2 - 1x + 1 2 1x - 2 2
= x2 + x - 2 - 1x2 - x - 2 2
= x2 + x - 2 - x2 + x + 2
= 2x
6.
7.
8.
9.
10.
17.
2
2. One possible proof:
1
1 2
1 x
1
- = a b - a b
x
2
x 2
2 x
x
2
=
2x
2x
2 - x
=
2x
5.
1 - cos2 u
cos2 u
sin2 u
= 2
cos u
=tan2 u
=
16. tan x+sec x=
12. Yes.
1. One possible proof:
x3 - x2
- 1x - 1 2 1x + 1 2 =
x
=
=
=
207
1
sin2x + cos2x
=
=sin x. Yes.
csc x
csc x
tan x
sin x # cos x
=
=sin x. Yes.
sec x
cos x
1
cos x # cos x
cos2 x
=
. No.
cos x # cot x =
1
sin x
sin x
p
p
cos a x - b = cos a - x b = sin x. Yes.
2
2
sin3x
(sin3x)(1+cot2x)=(sin3x)(csc2x)= 2 =sin x. Yes.
sin x
No. Confirm graphically.
11. (cos x)(tan x+sin x cot x)
sin x
cos x
=cos x #
+cos x sin x #
=sin x+cos2x
cos x
sin x
18.
sin x
cos2x - 1
-sin2x
# sin x=–tan x sin x
=
=–
cos x
cos x
cos x
2
1
sec2 ¨ - 1
tan2 ¨
# a sin ¨ b = sin2¨
=
=
sin ¨
sin ¨
sin ¨
cos ¨
cos ¨
sin ¨
=
1 - sin2 ¨
19. Multiply out the expression on the left side.
11 + cos x2 + 11 - cos x2
1
1
+
20.
=
1 - cos x
1 + cos x
11 - cos x2 1 1 + cos x2
2
2
=
= 2 =2 csc2x
1 - cos2x
sin x
21. (cos t-sin t)2+(cos t+sin t)2
=cos2 t-2 cos t sin t+sin2 t+cos2 t
+2 cos t sin t+sin2 t= 2 cos2 t+2 sin2 t=2
22. sin2 Å-cos2 Å=(1-cos2 Å)-cos2 Å=1-2 cos2 Å
23.
24.
25.
sec2x
1 + tan2x
=
=sec2x
2
2
1
sin x + cos x
cos ı
sin b
cos2ı + sin2ı
1
+
+tan ı=
=
tan ı
sin ı
cos b
cos ı sin ı
1
= sec ı csc ı
=
cos ı sin ı
1 - sin2ı
cos ı
cos2ı
=
=
1 + sin ı
cos ı11 + sin ı2
cos ı1 1 + sin ı2
11 - sin ı2 11 + sin ı2
1 - sin ı
=
=
cos ı1 1 + sin ı2
cos ı
26. One possible proof:
1sec x + 12 1sec x - 1 2
sec x + 1
=
tan x
tan x 1sec x - 12
=
sec2x - 1
tan x1sec x - 12
sin x
tan2x
cos x # cos x
=
=
tan x1sec x - 12
1
cos x
- 1
cos x
sin x
=
1 - cos x
Chapter 5
208
27.
28.
Analytic Trigonometry
tan2x
sec2x - 1
1
=
=sec x-1=
-1
sec x + 1
sec x + 1
cos x
1 - cos x
=
cos x
cot v
cot v
1
=
1
+
+
1
cot v - 1 tan v
cot v tan v - tan v
#
=
=
1
cot v + 1 tan v
cot v tan v + tan v
tan v
cos v # sin v
(Note: cot v tan v=
=1)
tan v
sin v cos v
2
cos x
b -cos2x
sin x
cos2x11 - sin2x2
cos2x
= cos2x #
=
2
sin x
sin2x
=cos2x cot2x
29. cot2x-cos2x= a
sin ¨ 2
b -sin2 ¨
cos ¨
2
sin2 ¨ 1 1 - cos2 ¨2
2 # sin ¨
¨
=
=sin
cos2 ¨
cos2 ¨
=sin2 ¨ tan2 ¨
39.
1 - cos2 t + 1 - cos2 t
sin2 t + 1 - cos2 t
=
1 sin t2 1 1 - cos t2
1 sin t2 11 - cos t2
2 11 + cos t 2
211 - cos2 t2
=
=
1sin t 2 11 - cos t 2
sin t
=
40.
30. tan2 ¨-sin2 ¨= a
31. cos4 x-sin4 x=(cos2 x+sin2 x)(cos2 x-sin2 x)
=1(cos2 x-sin2 x)=cos2 x-sin2 x
32. tan4 t+tan2 t=tan2 t(tan2 t+1)=(sec2 t-1)(sec2 t)
=sec4 t-sec2 t
33. (x sin Å+y cos Å)2+(x cos Å-y sin Å)2
=(x2 sin2 Å+2xy sin Å cos Å+y2 cos2 Å)
+(x2 cos2 Å-2xy cos Å sin Å+y2 sin2 Å)
=x2 sin2 Å+y2 cos2 Å+x2 cos2 Å+y2 sin2 Å
=(x2+y2)(sin2 Å+cos2 Å)=x2+y2
34.
1 - cos ¨
1 - cos2 ¨
sin2 ¨
=
=
sin ¨
sin ¨1 1 + cos ¨ 2
sin ¨ 1 1 + cos ¨ 2
sin ¨
=
1 + cos ¨
tan x1sec x + 12
tan x1 sec x + 1 2
tan x
=
=
2
sec x - 1
sec x - 1
tan2x
sec x + 1
=
. See also #26.
tan x
sin2 t + 1 1 + cos t 2 2
sin t
1 + cos t
36.
+
=
1 + cos t
sin t
1 sin t2 11 + cos t 2
sin2 t + 1 + 2 cos t + cos2 t
2 + 2 cos t
=
=
1 sin t 2 11 + cos t2
1sin t 2 11 + cos t 2
2
=
=2 csc t
sin t
1sin x - cos x2 1 sin x + cos x2
sin x - cos x
37.
=
sin x + cos x
1 sin x + cos x2 2
38.
sin2 x - 11 - sin2 x2
sin2 x - cos2 x
=
2
1 + 2 sin x cos x
sin x + 2 sin x cos x + cos x
2 sin2 x - 1
=
1 + 2 sin x cos x
2
1 + cos x
1 + cos x # sec x
sec x + cos x sec x
=
=
1 - cos x
1 - cos x sec x
sec x - cos x sec x
1
sec x + 1
=
(Note: cos x sec x=cos x #
=1.)
sec x - 1
cos x
sin A cos B + cos A sin B
cos A cos B - sin A sin B
1
cos A cos B # sin A cos B + cos A sin B
=±
≤
1
cos A cos B - sin A sin B
cos A cos B
sin A
sin B
+
cos A
cos B
tan A + tan B
=
=
sin A sin B
1 - tan A tan B
1 cos A cos B
41. sin2 x cos3 x=sin2 x cos2 x cos x
=sin2 x(1-sin2 x)cos x=(sin2 x-sin4 x)cos x
42. sin5 x cos2 x=sin4 x cos2 x sin x
=(sin2 x)2 cos2 x sin x=(1-cos2 x)2 cos2 x sin x
=(1-2 cos2 x+cos4 x)cos2 x sin x
=(cos2x-2 cos4 x+cos6 x)sin x
43. cos5 x=cos4 x cos x=(cos2 x)2 cos x
=(1-sin2 x)2 cos x=(1-2 sin2x+sin4 x) cos x
44. sin3 x cos3 x=sin3 x cos2 x cos x
=sin3 x (1-sin2 x)cos x=(sin3 x-sin5 x)cos x
45.
35.
=
sin2 t + 11 + cos t 2 11 - cos t 2
sin t
1 + cos t
+
=
1 - cos t
sin t
1sin t 2 11 - cos t2
46.
tan x
cot x
+
1 - cot x
1 - tan x
cot x
tan x # sin x
# cos x
=
+
1 - cot x sin x
1 - tan x cos x
sin2x>cos x
cos2x>sin x sin x cos x
+
b
=a
sin x - cos x
cos x - sin x sin x cos x
3
3
sin x - cos x
=
sin x cos x1 sin x - cos x2
sin2 x + sin x cos x + cos2 x
=
sin x cos x
1 + sin x cos x
1
=
=
+1=csc x sec x +1.
sin x cos x
sin x cos x
3
3
This involves rewriting a -b as
(a-b)(a2+ab+b2), where a=sin x and b=cos x.
cos x
cos x
+
1 + sin x
1 - sin x
1cos x2 3 11 - sin x 2 + 11 + sin x2 4
2 cos x
=
=
11 + sin x 2 1 1 - sin x2
1 - sin2x
2 cos x
=
=2 sec x
cos2x
Section 5.2
47.
2 tan x
1
+
1 - tan2 x
2 cos2 x - 1
2 tan x
cos2 x
1
#
=
+ 2
2
2
1 - tan x cos x
cos x - sin2 x
2 sin x cos x
cos2 x + sin2 x
= 2
+ 2
cos x - sin2 x
cos x - sin2 x
2 sin x cos x + cos2 x + sin2 x
=
1cos x - sin x2 1 cos x + sin x2
=
1cos x + sin x2 2
1cos x - sin x2 1 cos x + sin x2
=
cos x + sin x
cos x - sin x
1 1 + cos x2 11 - 4 cos x2
1 - 3 cos x - 4 cos x
=
sin2 x
1 - cos2 x
11 + cos x2 11 - 4 cos x2
1 - 4 cos x
=
=
11 + cos x2 11 - cos x2
1 - cos x
2
48.
49. cos3x=(cos2 x)(cos x)=(1-sin2 x)(cos x)
50. sec4x=(sec2 x)(sec2 x)=(1+tan2 x)(sec2 x)
51. sin5x=(sin4 x)(sin x)=(sin2 x)2(sin x)
=(1-cos2 x)2(sin x)
=(1-2 cos2 x+cos4 x)(sin x)
52. (b) — divide through by cos x:
=
1 + sin x
cos x
1
sin x
+
=sec x+tan x.
cos x
cos x
53. (d) — multiply out: (1+sec x)(1-cos x)
=1-cos x+sec x-sec x cos x
1 #
1
=1 - cos x +
cos x
cos x
cos x
1
1 - cos2 x
sin2 x
=1-cos x+
-1=
=
cos x
cos x
cos x
sin x #
=
sin x=tan x sin x.
cos x
54. (a) — put over a common denominator:
2
2
1
1
sec 2 x+csc 2 x= a
b +a
b
cos x
sin x
2
sin2 x + cos2 x
1
1
# 1 b
=
= 2
=a
2
2
2
cos x sin x
cos x sin x
cos x sin x
=sec 2 x csc 2 x.
55. (c) — put over a common denominator:
1
1
1 - sin x + 1 + sin x
+
=
1 + sin x
1 - sin x
1 - sin2 x
2
= 2 =2 sec2 x.
cos x
1
tan x + cot x
sin x cos x
= 2
sin x + cos2 x
56. (e) — multiply and divide by sin x cos x:
sin x cos x
cos x
sin x
+
b 1sin x cos x2
a
cos x
sin x
sin x cos x
=
=sin x cos x.
1
=
57. (b) — multiply and divide by sec x+tan x:
1
# sec x + tan x = sec2 x + tan2x
sec x - tan x sec x + tan x
sec x - tan x
sec x + tan x
=
.
1
Proving Trigonometric Identities
209
58. False. There are numbers in the domain of both sides of
the equation for which equality does not hold, namely all
negative real numbers. For example, 2 1 -3 2 2 = 3, not –3.
59. True. If x is in the domain of both sides of the equation,
then x 0. The equation 1 1x2 2 = x holds for all x 0,
so it is an identity.
60. By the definition of identity, all three must be true. The
answer is E.
61. A proof is
sin x # 1 + cos x
sin x
=
1 - cos x
1 - cos x 1 + cos x
sin x 11 + cos x2
=
1 - cos2 x
sin x 11 + cos x2
=
sin2 x
1 + cos x
=
sin x
The answer is E.
62. One possible proof:
1
sin ¨
+
tan ¨ + sec ¨ =
cos ¨
cos ¨
sin ¨ + 1
=
cos ¨
sin ¨ + 1 # sin ¨ - 1
=
cos ¨
sin ¨ - 1
sin2 ¨ - 1
=
cos ¨ 1sin ¨ - 12
-cos2 ¨
=
cos ¨ 1sin ¨ - 12
- cos ¨
=
sin ¨ - 1
cos ¨
=
1 - sin ¨
The answer is C.
63. k must equal 1, so f(x) Z 0. The answer is B.
64. cos x; sin x cot x=sin x #
cos x
=cos x
sin x
65. sin x; cos x tan x=cos x #
sin x
=sin x
cos x
cos x
sin x
cos x
sin x
=
+
+
csc x
sec x
1>sin x
1>cos x
=sin2x+cos2x=1
cos x>sin2x
1>sin x
csc x
cot x csc x
67. 1;
=
sin x
sec x
sin x
1>cos x
1
cos2x
1 - cos2x
sin2x
= 2 - 2 =
= 2 =1
sin x
sin x
sin2x
sin x
66. 1;
68. cos x;
sin x
sin x
=
=cos x.
tan x
sin x>cos x
69. 1; (sec2x)(1-sin2x)= a
2
1
b (cos2x)=1
cos x
70. Since the sum of the logarithms is the logarithm of the
product, and since the product of the absolute values of
all six basic trig functions is 1, the logarithms sum to ln 1,
which is 0.
210
Chapter 5
Analytic Trigonometry
71. If A and B are complementary angles, then
sin2A+sin2B=sin2A+sin2(∏/2-A)
=sin2A+cos2A
=1
(b) One choice for h is 0.001 (shown). The function y3
is a combination of three sinusoidal functions
(1000 sin(x+0.001), 1000 sin x, and cos x), all with
period 2∏.
72. Check Exercises 11–51 for correct identities.
73. Multiply and divide by 1-sin t under the radical:
1 1 - sin t2 2
1 - sin t # 1 - sin t
=
C 1 + sin t 1 - sin t
C 1 - sin2 t
1 1 - sin t 2 2
ƒ 1 - sin t ƒ
=
=
since 2a2= ƒ a ƒ .
C
ƒ cos t ƒ
cos2 t
Now, since 1-sin t 0, we can dispense with the
absolute value in the numerator, but it must stay in the
denominator.
74. Multiply and divide by 1+cos t under the radical:
11 + cos t 2 2
1 + cos t # 1 + cos t
=
C 1 - cos t 1 + cos t
C 1 - cos2 t
11 + cos t 2 2
ƒ 1 + cos t ƒ
=
=
since 2a2= ƒ a ƒ .
C
ƒ sin t ƒ
sin2 t
Now, since 1+cos t 0, we can dispense with the
absolute value in the numerator, but it must stay in the
denominator.
75. sin6 x+cos6 x=(sin2 x)3+cos6 x
=(1-cos2 x)3+cos6 x
=(1-3 cos2 x+3 cos4 x-cos6 x)+cos6 x
=1-3 cos2 x(1-cos2 x)=1-3 cos2 x sin2 x.
76. Note that a3-b3=(a-b)(a2+ab+b2). Also
note that a2+ab+b2=a2+2ab+b2-ab
=(a+b)2-ab. Taking a=cos2 x and b=sin2 x,
we have cos6 x-sin6 x
=(cos2 x-sin2 x)(cos4 x+cos2 x sin2 x+sin4 x)
=(cos2 x-sin2 x)[(cos2 x+sin2 x)2-cos2 x sin2 x]
=(cos2 x-sin2 x)(1-cos2 x sin2 x).
77. One possible proof: ln|tan x|=ln
ƒ sin x ƒ
ƒ cos x ƒ
=ln|sin x|-ln|cos x|.
78. One possible proof:
ln|sec ¨+tan ¨|+ln|sec ¨-tan ¨|=ln|sec2 ¨-tan2 ¨|
=ln 1
=0
79. (a) They are not equal. Shown is the window
[–2∏, 2∏,] by [–2, 2]; graphing on nearly any viewing
window does not show any apparent difference —
but using TRACE, one finds that the y coordinates
are not identical. Likewise, a table of values will
show slight differences; for example, when x=1,
y1=0.53988 while y2=0.54030
[–2∏, 2∏] by [–2, 2]
[–2∏, 2∏] by [–0.001, 0.001]
1
1
80. (a) cosh2 x-sinh2 x= (ex+e–x)2- (ex-e–x)2
4
4
1 2x
= [e +2+e–2x-(e2x-2+e–2x)]
4
1
= (4)=1.
4
(b) 1-tanh2 x=1-
sinh2 x
cosh2 x - sinh2 x
=
2
cosh x
cosh2 x
1
, using the result from (a). This equals sech2 x.
cosh2 x
cosh2 x
cosh2 x - sinh2 x
-1=
(c) coth2x-1=
sinh2 x
sinh2 x
1
=
, using the result from (a). This equals csch2 x.
sinh2 x
81. In the decimal window, the x coordinates used to plot the
graph on the calculator are (e.g.) 0, 0.1, 0.2, 0.3, etc. — that
is, x=n/10, where n is an integer. Then 10 ∏x=∏n, and
the sine of integer multiples of ∏ is 0; therefore,
cos x+sin 10 ∏x=cos x+sin ∏n=cos x+0
1
=cos x. However, for other choices of x, such as x= ,
p
we have cos x+sin 10 ∏x=cos x+sin 10 Z cos x.
=
■ Section 5.3 Sum and Difference Identities
Exploration 1
1. sin 1u + v2 = -1, sin u + sin v = 1. No.
2. cos 1u + v2 = 1, cos u + cos v = 2. No.
3. tan 1p>3 + p>32 = - 13, tan p>3 + tan p>3 = 213.
(Many other answers are possible.)
Quick Review 5.3
1. 15° = 45° - 30°
2. 75° = 45° + 30°
3. 165° = 180° - 15° = 180° + 30° - 45° = 210° - 45°
4.
p
p
p
p
p
= 2# =
12
6
4
3
4
5.
5p
p
p
2p
p
= 4# =
12
6
4
3
4
6.
4p
3p
p
p
7p
=
+
=
+
12
12
12
3
4
7. No. 1 f1x2 + f1y2 = ln x + ln y = ln1xy2
= f1xy2 Z f1 x + y2 2
Section 5.3
8. No. 1 f1 x + y2 = ex + y = ex ey
= f1x2 f1y2 Z f1x + y2 2
9. Yes. 1f1x + y2 = 32 1x + y2 = 32x + 32y
= f1x2 + f1y2 2
Sum and Difference Identities
In #11–22, match the given expression with the sum and
difference identities.
11. sin142° - 17°2 = sin 25°
12. cos194° - 18°2 = cos 76°
10. No. 1 f1x + y2 = x + y + 10
13. sin a
p
7p
p
+ b = sin
5
2
10
Section 5.3 Exercises
14. sin a
p
p
4p
- b = sin
3
7
21
16. tan a
p
p
2p
- b = tan 5
3
15
17. cos a
p
p
- x b = cos a x - b
7
7
= f1x2 + y Z f1x2 + f1 y2 2
1. sin 15° = sin1 45° - 30°2
= sin 45° cos 30° - cos 45° sin 30°
12 13
12 # 1
16 - 12
#
=
=
2
2
2 2
4
tan 45° - tan 30°
2. tan 15° = tan145° - 30°2 =
1 + tan 45° tan 30°
13 - 132 2
3 - 13
=
=
= 2 - 13
=
9 - 3
1 + 13>3
3 + 13
1 - 13>3
3. sin 75° = sin1 45° + 30°2
= sin 45° cos 30° + cos 45° sin 30°
12 # 1
16 + 12
12 # 13
+
=
=
2
2
2 2
4
4. cos 75° = cos 145° + 30°2
= cos 45° cos 30° - sin 45° sin 30°
12 13
12 # 1
16 - 12
#
=
=
2
2
2 2
4
p
p
p
p
p
p
p
= cos a - b = cos cos + sin sin
12
3
4
3
4
3
4
1 12
13
12
12
+
16
#
= #
+
=
2 2
2
2
4
5. cos
7p
p
p
p
p
p
p
= sin a + b = sin cos + cos sin
12
3
4
3
4
3
4
13 12
1 12
16 + 12
#
=
+ #
=
2
2
2 2
4
tan1 2p>3 2 - tan1p>42
5p
2p
p
7. tan
= tan a
- b =
12
3
4
1 + tan1 2p>3 2 tan1p>4 2
1 13 + 1 2 2
- 13 - 1
13 + 1
=
=
=
= 2 + 13
3 - 1
1 - 13
13 - 1
6. sin
11p
2p
8. tan
= tan a
+
12
3
- 13 + 1
1 =
=
1 + 13
1 +
tan1 2p>3 2 + tan1p>42
p
b=
4
1 - tan12p>3 2 tan1 p>42
1 1 - 132 2
13
=
= 13 - 2
1 - 3
13
5p
p
7p
= cos a
- b
12
6
4
5p
5p
p
p
13 # 12
1 12
= cos
cos + sin
sin = + #
6
4
6
4
2
2
2 2
12 - 16
=
4
9. cos
10. sin a-
p
p
p
b = sin a - b
12
6
4
p
p
p
p
1 12
13 12
#
= sin cos - cos sin = #
6
4
6
4
2 2
2
2
12 - 16
=
4
211
15. tan 119° + 47°2 = tan 66°
18. cos a x +
p
b
7
19. sin13x - x2 = sin 2x
20. cos17y + 3y 2 = cos 10y
21. tan12y + 3x2
22. tan13a - 2b2
p
p
p
b = sin x cos - cos x sin
2
2
2
=sin x # 0 - cos x # 1 = -cos x
23. sin a x -
24. Using the difference identity for the tangent function, we
p
encounter tan , which is undefined. However, we can
2
sin1x - p>22
p
compute tan a x - b =
. From #23,
2
cos1x - p>22
p
sin a x - b = -cos x. Since the cosine function is even,
2
p
p
cos a x - b = cos a - x b = sin x (see Example 2,
2
2
-cos x
= -cot x.
or #25). Therefore this simplifies to
sin x
p
p
p
b = cos x cos + sin x sin
2
2
2
=cos x # 0 + sin x # 1 = sin x
25. cos a x -
26. The simplest way is to note that
p
p
p
- x - y =
- 1 x + y2 , so that
a - xb - y =
2
2
2
p
p
cos c a - x b - y d = cos c - 1x + y2 d . Now use
2
2
p
Example 2 to conclude that cos c - 1x + y2 d
2
=sin 1x + y2 .
p
p
p
27. sin a x + b = sin x cos + cos x sin
6
6
6
13
1
+ cos x #
=sin x #
2
2
p
p
p
b = cos x cos + sin x sin
4
4
4
12
12
12
+ sin x #
=
1cos x + sin x2
=cos x #
2
2
2
28. cos a x -
212
Chapter 5
Analytic Trigonometry
tan ¨ + tan1p>4 2
tan ¨ + 1
p
b =
=
4
1 - tan ¨ tan1p>4 2
1 - tan ¨ # 1
1 + tan ¨
=
1 - tan ¨
29. tan a ¨ +
p
p
p
b = cos ¨ cos - sin ¨ sin
2
2
2
=cos ¨ # 0 - sin u # 1 = -sin ¨
30. cos a ¨ +
31. Equations B and F.
32. Equations C and E.
33. Equations D and H.
34. Equations A and G.
35. Rewrite as sin 2x cos x-cos 2x sin x=0; the left side
equals sin(2x-x)=sin x, so x=np, n an integer.
36. Rewrite as cos 3x cos x-sin 3x sin x=0; the left side
p
equals cos(3x+x)=cos 4x, so 4x= + np; then
2
p
p
x =
+ n , n an integer.
8
4
p
p
p
- u b = sin cos u - cos sin u
2
2
2
=1 # cos u - 0 # sin u = cos u.
37. sin a
38. Using the difference identity for the tangent function, we
p
encounter tan , which is undefined. However, we can
2
sin1p>2 - u2
cos u
p
=
= cot u.
compute tan a - u b =
2
cos1 p>2 - u 2
sin u
Or, use #24, and the fact that the tangent function is odd.
cos1p>2 - u 2
p
sin u
- ub =
=
=tan u using
2
sin1p>2 - u 2
cos u
the first two cofunction identities.
39. cot a
1
1
p
- ub =
=
=csc u using the
2
cos1 p>2 - u2
sin u
first cofunction identity.
40. sec a
1
1
p
- ub =
=
=sec u using the
2
sin 1p>2 - u 2
cos u
second cofunction identity.
41. csc a
42. cos a x +
p
p
p
b =cos x cos a b - sin x sin a b
2
2
2
=cos x # 0 - sin x # 1
= -sin x
43. To write y = 3 sin x + 4 cos x in the form
y = a sin1bx + c2 , rewrite the formula using the formula
for the sine of a sum:
y = a1 1sin bx cos c2 + 1cos bx sin c2 2
= a sin bx cos c + a cos bx sin c
= 1a cos c2sin bx + 1a sin c2cos bx.
Then compare the coefficients: a cos c = 3, b=1,
a sin c = 4.
Solve for a as follows:
1a cos c2 2 + 1 a sin c2 2
a2 cos2 c + a2 sin2 c
a2 1 cos2 c + sin2 c2
a2
a
=
=
=
=
=
32 + 42
25
25
25
;5
If we choose a to be positive, then cos c = 3>5 and
sin c = 4>5. c = cos - 1 1 3>52 = sin - 1 14>5 2 . So the sinusoid
is y = 5 sin1x + cos - 1 13>5 2 2 L 5 sin 1 x + 0.92732 .
44. Follow the steps shown in Exercise 43 (using the formula
for the sine of a difference) to compare the coefficients in
y = 1a cos c2sin bx - 1a sin c2cos bx to the coefficients in
y = 5 sin x - 12 cos x: a cos c = 5, b=1, a sin c = 12.
Solve for a as follows:
1a cos c2 2 + 1a sin c2 2 = 52 + 122
a2 1cos2 c + sin2 c2 = 169
a = ; 13
If we choose a to be positive, then cos c = 5>13 and
sin c = 12>13. So the sinusoid is
y = 13 sin 1x - cos - 1 15>13 2 2 L 13 sin1x - 1.1762 .
45. Follow the steps shown in Exercise 43 to compare the
coefficients in y = 1a cos c2sin bx + 1a sin c2cos bx to
the coefficients in y = cos 3x + 2 sin 3x: a cos c = 2,
b=3, a sin c = 1.
Solve for a as follows:
1a cos c2 2 + 1a sin c2 2 = 12 + 22
a2 1cos2 c + sin2 c2 = 5
a = ; 15
If we choose a to be positive, then cos c = 2> 15 and
sin c = 1> 15. So the sinusoid is
y = 15 sin 13x - cos-1 12> 15 2 2 L 2.236 sin13x - 0.46362 .
46. Follow the steps shown in Exercise 43 to compare the
coefficients in y = 1a cos c2sin bx + 1a sin c2cos bx to
the coefficients in y = 3 cos 2x - 2 sin 2x
= -2 sin 2x + 3 cos 2x: a cos c = - 2,
b=2, a sin c = 3.
Solve for a as follows:
1a cos c2 2 + 1a sin c2 2 = 1 -2 2 2 + 32
a2 1cos2 c + sin2 c2 = 13
a = ; 113
If we choose a to be negative, then cos c = 2> 113 and
sin c = -3> 113. So the sinusoid is
y = - 113 sin 12x - cos -1 12> 1132 2
≠ -3.606 sin1 2x - 0.9828 2 .
47. sin(x-y)+sin(x+y)
=(sin x cos y-cos x sin y) + (sin x cos y+cos x sin y)
=2 sin x cos y
48. cos(x-y)+cos(x+y)
=(cos x cos y+sin x sin y) + (cos x cos y-sin x sin y)
=2 cos x cos y
49. cos 3x=cos[(x+x)+x]
=cos(x+x) cos x-sin(x+x) sin x
=(cos x cos x-sin x sin x) cos x
-(sin x cos x+cos x sin x) sin x
=cos‹ x-sin¤ x cos x-2 cos x sin¤ x
=cos‹ x-3 sin¤ x cos x
50. sin 3u=sin[(u+u)+u]=sin(u+u) cos u+
cos(u+u) sin u=(sin u cos u+cos u sin u) cos u+
(cos u cos u-sin u sin u) sin u=2 cos¤ u sin u+
cos¤ u sin u-sin‹ u=3 cos¤ u sin u-sin‹ u
51. cos 3x+cos x=cos(2x+x)+cos(2x-x); use #48
with x replaced with 2x and y replaced with x.
52. sin 4x+sin 2x=sin(3x+x)+sin(3x-x); use #47
with x replaced with 3x and y replaced with x.
Section 5.3
53. tan(x+y) tan(x-y)
tan x + tan y
tan x - tan y
=a
b#a
b
1 - tan x tan y
1 + tan x tan y
tan2 x - tan2 y
=
since both the numerator and
1 - tan2 x tan2 y
denominator are factored forms for differences of squares.
54. tan 5u tan 3u=tan(4u+u) tan(4u-u); use #53 with
x=4u and y=u.
sin 1x + y2
55.
sin 1x - y 2
sin x cos y + cos x sin y
=
sin x cos y - cos x sin y
sin x cos y + cos x sin y 1> 1 cos x cos y2
#
=
sin x cos y - cos x sin y 1> 1cos x cos y 2
1sin x cos y2> 1 cos x cos y2 + 1cos x sin y 2> 1cos x cos y2
=
1sin x cos y2 > 1cos x cos y2 - 1cos x sin y2 > 1cos x cos y 2
1sin x>cos x2 + 1sin y>cos y 2
=
1 sin x>cos x2 - 1sin y>cos y2
tan x + tan y
=
tan x - tan y
63. tan(u-v)=
64. The identity would involve tan a
3p
b , which does not exit.
2
3p
b
2
3p
cos a x b
2
3p
3p
sin x cos
- cos x sin
2
2
=
3p
3p
cos x cos
+ sin x sin
2
2
sin x # 0 - cos x # 1–12
=
cos x # 0 + sin x # 1–1 2
3p
tan a x b =
2
= sin 45 ° cos 30 ° - cos 45 ° sin 30 °
12 1
12 13
a
b a b
=
2
2
2 2
16 - 12
4
The answer is D.
=
tan u + tan v
. The answer is B.
1 - tan u tan v
cos1 u + v 2
sin u cos v + cos u sin v
=
cos u cos v - sin u sin v
cos u sin v
sin u cos v
+
cos u cos v
cos u cos v
=
cos u cos v
sin u sin v
cos u cos v
cos u cos v
sin v
sin u
+
cos u
cos v
=
sin u sin v
1 cos u cos v
tan u + tan v
=
1 - tan u tan v
sin a x +
65. The identity would involve tan a
60. Sin 15° = sin1 45 ° - 30 °2
62. tan(u+v)=
p
b , which does not exist.
2
p
b
2
p
cos a x + b
2
p
p
sin x cos + cos x sin
2
2
=
p
p
cos x cos - sin x sin
2
2
sin x # 0 + cos x # 1
=
cos x # 0 - sin x # 1
= -cot x
p
tan a x + b =
2
59. y=sin x cos 2x+cos x sin 2x=sin (x+2x)=sin 3x.
The answer is A.
sin 1u + v 2
sin1 u - v2
cos u sin v
sin u cos v
cos u cos v
cos u cos v
=
sin u sin v
cos u cos v
+
cos u cos v
cos u cos v
sin v
sin u
cos u
cos v
=
sin u sin v
1 +
cos u cos v
tan u - tan v
=
1 + tan u tan v
57. False. For example, cos 3∏+cos 4∏=0, but 3∏ and 4∏
are not supplementary. And even though
cos (3∏/2)+cos (3∏/2)=0, 3∏/2 is not supplementary
with itself.
61. For all u, v, tan1u + v 2 =
213
cos1 u - v2
sin u cos v - cos u sin v
=
cos u cos v + sin u sin v
56. True. If B=∏-A, then cos A+cos B
=cos A+cos (∏-A)
=cos A+cos ∏ cos A+sin ∏ sin A
=cos A+(–1) cos A+(0) sin A=0.
58. If cos A cos B=sin A sin B, then cos (A+B)=
cos A cos B-sin A sin B=0. The answer is A.
Sum and Difference Identities
sin a x -
= -cot x
66.
sin1x + h2 - sin x
h
=
=
sin x cos h + cos x sin h - sin x
h
sin x1 cos h - 12 + cos x sin h
=sin x a
h
cos h - 1
sin h
b + cos x
h
h
Chapter 5
214
67.
Analytic Trigonometry
cos1 x + h 2 - cos x
h
=
=
cos x cos h - sin x sin h - cos x
h
cos x1cos h - 12 - sin x sin h
h
cos h - 1
sin h
=cos x a
b - sin x
h
h
68. The coordinates of all 24 points must be
kp
kp
b , sin a
b b for k=0, 1, 2, », 23. We only
a cos a
12
12
need to find the coordinates of those points in Quadrant I,
because the remaining points are symmetric. We already
know the coordinates for the cases when k=0, 2, 3, 4, 6
since these correspond to the special angles.
p
p
p
p
p
k=1: cos a b = cos a - b = cos a b cos a b
12
3
4
3
4
p
p
1 12
13 # 12
b sin a b = #
+
3
4
2 2
2
2
12 + 16
=
4
p
p
p
p
p
sin a b = sin a - b = sin a b cos a b
12
3
4
3
4
p
p
13 # 12
12 # 1
- sin a b cos a b =
4
3
2
2
2 2
16 - 12
=
4
3p
p
5p
k=5: cos a
b = cos a
- b
12
4
3
3p
p
3p
p
=cos a
b cos a b + sin a
b sin a b
4
3
4
3
12 # 1
16 - 12
12 # 13
=+
=
2 2
2
2
4
3p
p
5p
b = sin a
- b
sin a
12
4
3
3p
p
p
3p
=sin a
b cos a b - sin a b cos a
b
4
3
3
4
12 # 1
12
12 + 16
13 #
=
ab =
2 2
2
2
4
Coordinates in the first quadrant are (1, 0),
+ sin a
a
13 1
12 12
12 + 16 16 - 12
,
b, a
, b, a
,
b,
4
4
2 2
2
2
1 13
16 - 12 12 + 16
a ,
b, a
,
b , 1 0, 1 2
2 2
4
4
69. sin1A + B2 =sin1 p - C 2
=sin p cos C - cos p sin C
=0 # cos C - 1 -1 2 sin C
=sin C
70. cos C=cos(∏-(A+B))
=cos ∏ cos(A+B)+sin ∏ sin(A+B)
=(–1)(cos A cos B-sin A sin B)
+0 # sin1A + B2
=sin A sin B-cos A cos B
sin A
sin B
sin C
+
+
cos A
cos B
cos C
sin A1 cos B cos C2 + sin B1cos A cos C 2
71. tan A+tan B+tan C=
=
+
=
=
=
=
cos A cos B cos C
sin C1cos A cos B2
cos A cos B cos C
cos C1sin A cos B + cos A sin B2 + sin C1cos A cos B2
cos A cos B cos C
cos C sin1 A + B2 + sin C1 cos1A + B2 + sin A sin B2
cos A cos B cos C
cos C sin1 p - C2 + sin C1cos1 p - C2 + sin A sin B2
cos A cos B cos C
cos C sin C + sin C1 -cos C2 + sin C sin A sin B
cos A cos B cos C
sin A sin B sin C
=
cos A cos B cos C
=tan A tan B tan C
72. cos A cos B cos C-sin A sin B cos C
-sin A cos B sin C-cos A sin B sin C
=cos A(cos B cos C-sin B sin C)
-sin A(sin B cos C+cos B sin C)
=cos A cos (B+C)-sin A sin(B+C)
=cos(A+B+C)
=cos ∏
=–1
73. This equation is easier to deal with after rewriting it as
cos 5x cos 4x+sin 5x sin 4x=0. The left side of this
equation is the expanded form of cos(5x-4x), which of
course equals cos x; the graph shown is simply y=cos x.
The equation cos x=0 is easily solved on the interval
p
3p
[–2∏, 2∏]: x = ; or x = ; . The original graph is so
2
2
crowded that one cannot see where crossings occur.
[–2∏, 2∏] by [–1.1, 1.1]
2pt
+ db
T
2pt
2pt
= a c cos a
b cos d - sin a
b sin d d
T
T
2pt
2pt
= 1a cos d 2 cos a
b + (–a sin d) sin a
b
T
T
74. x=a cos a
75. B = Bin + Bref
E0
E0
◊x
◊x
= cos a◊t b +
cos a◊t +
b
c
c
c
c
E0
◊x
◊x
=
a cos ◊t cos
+ sin ◊t sin
c
c
c
◊x
◊x
+cos ◊t cos
- sin ◊t sin
b
c
c
E0
E0
◊x
◊x
=
a 2 cos ◊t cos
b = 2
cos ◊t cos
c
c
c
c
Section 5.4
■ Section 5.4 Multiple-Angle Identities
2. sin
1 - cos 1p>42
p
=
8
2
1 - 1 12>2 2 2
#
=
2
2
2 - 12
=
4
4. sin
3. Starting with the result of #1: cos 2u=cos2 u-sin2 u
=(1-sin2 u)-sin2 u=1-2 sin2 u
1 - cos1 9p>4 2
9p
=
8
2
1 - 1 12>2 2 2
#
=
2
2
2 - 12
=
4
9p
2 - 12
- 22 - 12
= ;
=
.
8
C
4
2
p
We take the negative square root because is a third8
quadrant angle.
Quick Review 5.4
p
1. tan x=1 when x= +np, n an integer
4
p
2. tan x=–1 when x= - +np, n an integer
4
3. Either cos x=0 or sin x=1. The latter implies the
p
former, so x= + np, n an integer.
2
4. Either sin x=0 or cos x=–1. The latter implies the
former, so x=np, n an integer.
p
5. sin x=–cos x when x= - +np, n an integer
4
p
6. sin x=cos x when x= +np, n an integer
4
1
1
p
7. Either sin x= or cos x= - . Then x= + 2np or
2
2
6
2p
5p
+ 2np or x= ;
+ 2np, n an integer.
x=
6
3
or x= ;
12
3p
+ 2np
. Then x=
2
2
p
+ 2np , n an integer.
4
9. The trapezoid can be viewed as a rectangle and two trian1
1
gles; the area is then A=(2)(3)+ (1)(3)+ (2)(3)
2
2
=10.5 square units.
10. View the triangle as two right triangles with hypotenuse 3,
one leg 1, and the other leg — the height — equal to
232 - 12 = 28 = 222
tan u + tan u
2 tan u
=
1 - tan u tan u
1 - tan2 u
5. 2 sin x cos x-2 sin x=0, so 2 sin x(cos x-1)=0;
sin x=0 or cos x=1 when x=0 or x=p.
4. tan 2u=tan(u+u)=
p
2 - 12
22 - 12
= ;
=
.
8
B
4
2
8. Either sin x=–1 or cos x=
Section 5.4 Exercises
2. Starting with the result of #1: cos 2u=cos2 u-sin2 u
=cos2 u-(1-cos2 u)=2 cos2 u-1
p
We take the positive square root because is a first8
quadrant angle.
3. sin2
215
1. cos 2u=cos(u+u)=cos u cos u-sin u sin u
=cos2 u-sin2 u
Exploration 1
1. sin2
Multiple-Angle Identities
6. 2 sin x cos x-sin x=0
sin x(2 cos x-1)=0,
1
So sin x=0 or cos x=
2
5p
p
when x=0, ∏, , or
.
3
3
7. 2 sin2 x+sin x-1=0, so (2 sin x-1)(sin x+1)
1
p
=0; sin x= or sin x=–1 when x= ,
2
6
5p
3p
x=
or x= .
6
2
8. 2 cos2 x-cos x-1=0, so (2 cos x+1)(cos x-1)
1
=0; cos x= - or cos x=1 when x=0,
2
4p
2p
x=
or x=
3
3
9. 2 sin x cos x-
sin x
sin x
= 0, so
12 cos2 x - 12 = 0, or
cos x
cos x
sin x cos 2x
= 0. Then sin x=0 or cos 2x=0
cos x
p
3p
(but cos x Z 0), so x=0, x= , x= , x=p,
4
4
7p
5p
x=
or x= .
4
4
10. cos2 x-cos x-1=0, so cos x=
1 ; 15
. Only
2
1 - 15
1 - 15
b ≠2.2370
is in [–1, 1], so x=cos–1 a
2
2
or x=2p-cos–1 a
1 - 15
b ≠4.0461
2
For #11–14, any one of the last several expressions given is an
answer to the question. In some cases, other answers are possible, as well.
11. sin 2¨+cos ¨=2 sin ¨ cos ¨+cos ¨
=(cos ¨)(2 sin ¨+1)
12. sin 2¨+cos 2¨=2 sin ¨ cos ¨+cos2 ¨-sin2 ¨
=2 sin ¨ cos ¨+2 cos2 ¨-1
=2 sin ¨ cos ¨+1-2 sin2 ¨
13. sin 2¨+cos 3¨
=2 sin ¨ cos ¨+cos 2¨ cos ¨-sin 2¨ sin ¨
=2 sin ¨ cos ¨+(cos2¨-sin2 ¨) cos ¨-2 sin2 ¨ cos ¨
=2 sin ¨ cos ¨+cos3 ¨-3 sin2 ¨ cos ¨
=2 sin ¨ cos ¨+4 cos3 ¨-3 cos ¨
216
Chapter 5
Analytic Trigonometry
16. cos 6x=cos 2(3x)=2 cos2 3x-1
28. With u=2x, this becomes cos u+cos 2u=0, the same
5p
p
as #23. This means u= , u = p, u =
, etc. —
3
3
p
2p
p
5p
p
i.e., 2x= + n
. Then x = , x = , x =
,
3
3
6
2
6
3p
11p
7p
.
,x =
,x =
x =
6
2
6
2
2
=
17. 2 csc 2x=
sin 2x
2 sin x cos x
1
sin x
= csc2 x tan x
= 2 #
sin x cos x
29. Using results from #25, sin 2x-cos 3x
=(2 sin x cos x)-(cos x-4 sin2 x cos x)
=(cos x)(4 sin2 x+2 sin x-1)=0.
p
3p
cos x=0 when x= or x= , while the second
2
2
14. sin 3¨+cos 2¨
=sin 2¨ cos ¨+cos 2¨ sin ¨+cos2 ¨-sin2¨
=2 sin ¨ cos2 ¨+(cos2 ¨-sin2 ¨) sin ¨+cos2 ¨-sin2 ¨
=3 sin ¨ cos2 ¨-sin3 ¨+cos2 ¨-sin2 ¨
15. sin 4x=sin 2(2x)=2 sin 2x cos 2x
2 11 - tan2x2
2
1
=
=
- tan x
tan 2x
2 tan x
tan x
=cot x-tan x
18. 2 cot 2x=
19. sin 3x=sin 2x cos x+cos 2x sin x=2 sin x cos2 x
+(2 cos2 x-1) sin x=(sin x)(4 cos2 x-1)
20. sin 3x=sin 2x cos x+cos 2x sin x
=2 sin x cos2 x+(1-2 sin2 x) sin x
=(sin x)(2 cos2 x+1-2 sin2 x)
=(sin x)(3-4 sin2 x)
21. cos 4x=cos 2(2x)=1-2 sin2 2x
=1-2(2 sin x cos x)2=1-8 sin2 x cos2 x
22. sin 4x=sin 2(2x)=2 sin 2x cos 2x
=2(2 sin x cos x)(2 cos2 x-1)
=(4 sin x cos x)(2 cos2 x-1)
1
23. 2 cos2 x+cos x-1=0, so cos x=–1 or cos x= ,
2
5p
p
x= , x=p or x=
3
3
24. cos 2x+sin x=1-2 sin¤ x+sin x=0, so
11p
1
p
7p
sin x=1 or sin x= - , x= , x= , or x=
.
2
2
6
6
25. cos 3x=cos 2x cos x-sin 2x sin x
=(1-2 sin2x)cos x
-(2 sin x cos x)sin x
=cos x-2 sin2x cos x
-2 sin2x cos x
=cos x-4 sin2x cos x
Thus the left side can be written as 2(cos x)(1-2 sin2 x)
=2 cos x cos 2x. This equals 0 in [0, 2∏) when
p
p
3p
5p
3p
7p
x= , x = , x =
,x =
,x =
, or x =
.
4
2
4
4
2
4
-1 ; 15
. It turns
4
out — as can be observed by noting, e.g., that
-1 + 15
sin–1 a
b ≠ 0.31415926 — that this means
4
x=0.1p, x=0.9p, x=1.3p, or x=1.7p.
factor equals zero when sin x=
30. Using #14, the left side can be rewritten as
3 sin x cos2 x-sin3 x+cos2 x-sin2 x.
Replacing cos2 x with 1-sin2 x gives
–4 sin3 x-2 sin2 x+3 sin x+1
=(sin x+1)(–4 sin2 x+2 sin x+1).
3p
b , and
This equals 0 when sin x=–1 a x =
2
1 ; 15
when sin x=
. These values turn out to be
4
x=0.3p, x=0.7p, x=1.1∏, and x=1.9p,
as can be observed by noting, e.g., that
1 - 15
sin–1 a
b ≠–0.31415926.
4
1 - cos 30°
1
13
31. sin 15°=_
=_
a1 b
B
2
B2
2
1
=— 22 - 13 . Since sin 15° 7 0, take the positive
2
square root.
1 - 13>2
1 - cos 390°
32. tan 195°=
=
= 2 - 13. Note
sin 390°
1>2
that tan 195°=tan 15°.
1 + cos 150°
1
13
33. cos 75°=—
=—
a1 b
C
2
C2
2
1
=— 22 - 13. Since cos 75° 7 0, take the positive
2
square root.
1 - cos 15p>62
5p
1
13
=—
=—
a1 +
b
12
C
2
C2
2
1
5p
=— 22 + 13. Since sin
7 0, take the positive
2
12
square root.
34. sin
26. Using #19, this become 4 sin x cos2 x=0, so x=0,
p
3p
x= , x = p, or x =
.
2
2
27. sin 2x+sin 4x=sin 2x+2 sin 2x cos 2x
=(sin 2x)(1+2 cos 2x)=0. Then sin 2x=0 or
1
cos 2x= - ; the solutions in [0, 2p) are
2
p
p
x=0, x = , x = ,
3
2
2p
4p
3p
5p
.
x =
, x = p, x =
,x =
, or x =
3
3
2
3
35. tan
1 - cos17p>62
1 + 13>2
7p
=
=
= - 2 - 13.
12
sin17p>6 2
-1>2
1 + cos1 p>4 2
1
12
p
=—
=—
a1 +
b
8
C
2
C2
2
p
1
7 0, take the positive
=— 32 + 12. Since cos
2
8
square root.
36. cos
Section 5.4
1
37. (a) Starting from the right side: 11 - cos 2u2
2
1
1
= 31 - 11 - 2 sin2 u 2 4 = (2 sin2 u)=sin2 u.
2
2
1
(b) Starting from the right side: 11 + cos 2u2
2
1
1
= 31 + 12 cos2 u - 1 2 4 = 12cos2 u2 =cos2 u.
2
2
11 - cos 2u 2>2
sin2 u
1 - cos 2u
=
=
11 + cos 2u 2>2
1 + cos 2u
cos2 u
(b) The equation is false when tan u is a negative number.
It would be an identity if it were written as
1 - cos u
|tan u|=
.
B 1 + cos u
38. (a) tan2 u=
39. sin4 x=(sin2 x)2= c
1
1 1 - cos 2x2 d
2
1
= 11 - 2cos 2x + cos2 2x2
4
1
1
= c 1 - 2 cos 2x + 1 1 + cos 4x2 d
4
2
1
= 12 - 4 cos 2x + 1 + cos 4x2
8
1
= (3-4 cos 2x+cos 4x)
8
40. cos3 x=cos x cos2 x=cos x #
1 + cos x
, so 2 cos2 x+cos x-1=0.
2
1
Then cos x=–1 or cos x= . In the interval [0, 2p),
2
p
5p
x= , x=p, or x= . General solution:
3
3
p
x=— +2np or x=p+2np, n an integer.
3
45. The right side equals tan2(x/2); the only way that
tan(x/2) =tan2(x/2) is if either tan(x/2)=0 or
tan(x/2)=1. In [0, 2p), this happens when x=0 or
p
x= . The general solution is x=2np or
2
p
x= + 2np, n an integer.
2
1 + cos 2x
2
1
1
+ cos 2x
=
2
2
∏
1
1
= sin a - 2x b +
2
2
2
1
∏
1
= sin a -2 a x - b b +
2
4
2
The last expression is in the form for a sinusoid.
48. True. cos2 x =
1
= 1cos x2 11 + cos 2x2
2
41. sin3 2x=sin 2x sin2 2x=sin 2x #
1
11 - cos 4x2
2
49. f(2x)=sin 2x=2 sin x cos x=2 f(x)g(x). The
answer is D.
1
= 1sin 2x2 11 - cos 4x2
2
42. sin5 x=(sin x)(sin2 x)2=(sin x) c
1
11 - cos 2x2 d
2
2
1
= 1sin x2 11 - 2 cos 2x + cos2 2x2
4
1
1
= 1sin x2 c 1 - 2 cos 2x + 11 + cos 4x2 d
4
2
1
= 1sin x2 12 - 4 cos 2x + 1 + cos 4x2
8
1
= 1sin x2 13 - 4 cos 2x + cos 4x2 .
8
Alternatively, take sin5 x=sin x sin4 x and apply the
result of #39.
43. cos2 x=
1 - cos x
= 2 cos2 x - 1, so 4 cos2 x+cos x-3=0,
2
or (4 cos x-3)(cos x+1)=0. Then cos x=–1 or
3
3
cos x= . Let Å=cos–1 a b ≠0.7227. In the interval
4
4
[0, 2p), x=Å, x=∏, or x=2p-Å. General solution:
x=_Å+2np or x=p+2np, n an integer.
47. False. For example, f(x)=2 sin x has period 2∏ and
g(x)=cos x has period 2∏, but the product
f(x)g(x)=2 sin x cos x=sin 2x has period ∏.
1
11 + cos 2x2
2
1 - cos x
, so 2 cos2 x+cos x-1=0. Then
2
1
p
cos x=–1 or cos x= . In the interval [0, 2p), x= ,
2
3
5p
p
x=p, or x= . General solution: =— +2np or
3
3
x=p+2np, n an integer.
217
44. 1-cos2 x=
46.
2
Multiple-Angle Identities
50. sin 22.5° = sin a
=
45°
b
2
1 - cos 45°
C
2
1 - 12>2
=
C
=
C
2
2 - 12
4
22 - 12
2
The answer is E.
=
51. sin 2x=cos x
2 sin x cos x=cos x
2 sin x=1 or cos x=0
1
sin x= or cos x=0
2
∏
5∏
∏
3∏
x = or
x = or
6
6
2
2
The answer is E.
218
Chapter 5
Analytic Trigonometry
52. sin2x-cos2x=1-2 cos2x, which has the same period
as the function cos2 x, namely ∏. The answer is C.
53. (a) In the figure, the triangle with side lengths x/2 and R
is a right triangle, since R is given as the perpendicular
distance. Then the tangent of the angle ¨/2 is the ratio
x>2
¨
“opposite over adjacent”: tan =
Solving for x
2
R
gives the desired equation. The central angle ¨ is 2p/n
since one full revolution of 2p radians is divided
evenly into n sections.
u
(b) 5.87≠2R tan , where ¨=2p/11, so
2
p
R≠5.87/(2 tan )≠9.9957. R=10.
11
54. (a)
␪2
A
x
B
d 1 2
x
D
x
C
d1 2
Call the center of the rhombus E. Consider right ¢ABE,
with legs d2/2 and d1/2, and hypotenuse length x. jABE
has measure ¨/2, and using “sine
adj
opp
equals
” and “cosine equals
,” we have
hyp
hyp
d2>2
d1>2
d2
d1
¨
¨
cos =
=
=
and sin =
.
2
x
2x
2
x
2x
(b) Use the double angle formula for the sine function:
d1 d2
¨
¨
¨
# = d1d22
sin ¨ = sin 2 a b = 2 sin cos = 2
2
2
2
2x 2x
2x
55. (a)
1 ft
θ
θ
1 ft
1 ft
The volume is 10 ft times the area of the end. The end
is made up of two identical triangles, with area
1
(sin ¨) (cos ¨) each, and a rectangle with area
2
(1) (cos ¨). The total volume is then
10 # (sin ¨ cos ¨+cos ¨)=10 (cos ¨)(1+sin ¨).
p
p
Considering only - ¨ , the maximum value
2
2
occurs when ¨≠0.52 (in fact, it happens exactly at
p
¨= ). The maximum value is about 12.99 ft3.
6
56. (a)
p
, the maximum area occurs
2
p
when ¨= , or about 0.79. This gives
4
p
x=20 cos
= 10 12, or about 14.14, for a width of
4
about 28.28, and a height of y=1012 L 14.14
1
1
1
1
57. csc 2u=
=
= #
sin 2u
2 sin u cos u
2 sin u
1
= csc u sec u
2
#
1
cos u
1 - tan2 u
1
=
tan 2u
2 tan u
cot2 u - 1
cot2 u
1 - tan2 u
=a
b =
b a
2
2 tan u
2 cot u
cot u
58. cot 2u=
1
1
=
cos 2u
1 - 2 sin2 u
csc2 u
csc2 u
1
=a
b
a
b
=
1 - 2 sin2 u
csc2 u
csc2 u - 2
1
1
=
60. sec 2u=
cos 2u
2 cos2 u - 1
sec2 u
sec2 u
1
ba
b =
=a
2
2
2 cos u - 1
sec u
2 - sec2 u
1
1
=
61. sec 2u=
2
cos 2u
cos u - sin2 u
sec2 u csc2 u
1
b
a
b
=a
cos2 u - sin2 u
sec2 u csc2 u
sec2 u csc2 u
= 2
csc u - sec2 u
62. The second equation cannot work for any values of x for
which sin x 6 0, since the square root cannot be negative.
The first is correct since a double angle identity for the
cosine gives cos 2x=1-2 sin2 x; solving for sin x gives
1
sin2 x= 1 1 - cos 2x2 , so that
2
59. sec 2u=
x
E
(b) Considering 0 ¨ 1
1 1 - cos 2x2. The absolute value of both
A2
sides removes the“_.”
sin x=—
63. (a) The following is a scatter plot of the days past January 1
as x-coordinates (L1) and the time (in 24 hour mode)
as y-coordinates (L2) for the time of day that astronomical twilight began in northeastern Mali in 2005.
x 2 ⫹ y 2 ⫽ 400
(x, y)
[–30, 370] by [–60, 60]
y
2x
The height of the tunnel is y, and the width is 2x,
so the area is 2xy. The x- and y-coordinates of the
vertex are 20 cos ¨ and 20 sin ¨, so the area is
2(20 cos ¨)(20 sin ¨)=400(2 cos ¨ sin ¨)=400 sin 2¨.
Section 5.5
(b) The sine regression curve through the points defined by
L1 and L2 is y=41.656 sin(0.015x-0.825)-1.473.
This is a fairly good fit, but not really as good as one
might expect from data generated by a sinusoidal
physical model.
The Law of Sines
219
4. sin C2=sin (p-C1)=sin p cos C1-cos p sin C1
=sin C1.
5. If BC AB, then BC can only extend to the right of the
altitude, thus determining a unique triangle.
Quick Review 5.5
1. a=bc/d
2. b=ad/c
3. c=ad/b
4. d=bc/a
[–30, 370] by [–60, 60]
(c) Using the formula L2-Y1(L1) (where Y1 is the sine
regression curve), the residual list is:
{3.64, 7.56, 3.35, –5.94, –9.35, –3.90, 5.12, 9.43, 3.90,
–4.57, –9.72, –3.22}.
(d) The following is a scatter plot of the days past
January 1 as x-coordinates (L1) and the residuals (the
difference between the actual number of minutes (L2)
and the number of minutes predicted by the regression curve (Y1)) as y-coordinates (L3) for the time of
day that astronomical twilight began in northeastern
Mali in 2005.
The sine regression curve through the points defined
by L1 and L3 is
y=8.856 sin(0.0346x+ 0.576)-0.331. (Note:
Round L3 to 2 decimal places to obtain this answer.)
This is another fairly good fit, which indicates that the
residuals are not due to chance. There is a periodic
variation that is most probably due to physical causes.
[–30, 370] by [–15, 15]
(e) The first regression indicates that the data are
periodic and nearly sinusoidal. The second regression
indicates that the variation of the data around the
predicted values is also periodic and nearly sinusoidal.
Periodic variation around periodic models is a predictable consequence of bodies orbiting bodies, but
ancient astronomers had a difficult time reconciling
the data with their simpler models of the universe.
■ Section 5.5 The Law of Sines
Exploration 1
1. If BC AB, the segment will not reach from point B to
the dotted line. On the other hand, if BC 7 AB, then a
circle of radius BC will intersect the dotted line in a unique
point. (Note that the line only extends to the left of point A.)
2. A circle of radius BC will be tangent to the dotted line at
C if BC=h, thus determining a unique triangle. It will
miss the dotted line entirely if BC 6 h, thus determining
zero triangles.
3. The second point (C2) is the reflection of the first point
(C1) on the other side of the altitude.
5.
7 sin 48°
L 13.314°
sin 23°
6.
9 sin 121°
L 31.888°
sin 14°
7. x=sin–1 0.3≠17.458°
8. x=180°-sin–1 0.3≠162.542°
9. x=180°-sin–1(–0.7)≠224.427°
10. x=360°+sin–1(–0.7)≠315.573°
Section 5.5 Exercises
1. Given: b=3.7, B=45°, A=60° — an AAS case.
C=180°-(A+B)=75°;
b sin A
a
b
3.7 sin 60°
=
1 a=
=
L 4.5;
sin A
sin B
sin B
sin 45°
b
c
b sin C
3.7 sin 75°
=
1 c =
=
L 5.1
sin B
sin C
sin B
sin 45°
2. Given: c=17, B=15°, C=120° — an AAS case.
A=180°-(B+C)=45°;
c sin A
a
c
17 sin 45°
=
1 a =
=
L 13.9;
sin A
sin C
sin C
sin 120°
c sin B
b
c
17 sin 15°
=
1 b =
=
L 5.1
sin B
sin C
sin C
sin 120°
3. Given: A=100°, C=35°, a=22 — an AAS case.
B=180°-(A+C)=45°;
22 sin 45°
a sin B
=
L 15.8;
b =
sin A
sin 100°
22 sin 35°
a sin C
=
L 12.8
c =
sin A
sin 100°
4. Given: A=81°, B=40°, b=92 — an AAS case.
C=180°-(A+B)=59°;
92 sin 81°
b sin A
=
L 141.4;
a =
sin B
sin 40°
b sin C
92 sin 59°
c =
=
L 122.7
sin B
sin 40°
5. Given: A=40°, B=30°, b=10 — an AAS case.
C=180°-(A+B)=110°;
10 sin 40°
b sin A
=
L 12.9;
a =
sin B
sin 30°
b sin C
10 sin 110°
c =
=
L 18.8
sin B
sin 30°
6. Given: A=50°, B=62°, a=4 — an AAS case.
C=180°-(A+B)=68°;
4 sin 62°
a sin B
=
L 4.6;
b =
sin A
sin 50°
a sin C
4 sin 68°
c =
=
L 4.8
sin A
sin 50°
220
Chapter 5
Analytic Trigonometry
7. Given: A=33°, B=70°, b=7 — an AAS case.
C=180°-(A+B)=77°;
7 sin 33°
b sin A
=
L 4.1;
a =
sin B
sin 70°
b sin C
7 sin 77°
c =
=
L 7.3
sin B
sin 70°
8. Given: B=16°, C=103°, c=12 — an AAS case.
A=180°-(B+C)=61°;
12 sin 61°
c sin A
=
L 10.8;
a =
sin C
sin 103°
c sin B
12 sin 16°
b =
=
L 3.4
sin C
sin 103°
9. Given: A=32°, a=17, b=11 — an SSA case.
h=b sin A≠5.8; h<b<a, so there is one triangle.
b sin A
b = sin-1 10.342…2 L 20.1°
B = sin-1 a
a
C=180°-(A+B)≠127.9°;
a sin C
17 sin 127.9°
c =
=
L 25.3
sin A
sin 32°
10. Given: A=49°, a=32, b=28 — an SSA case.
h=b sin A≠21.1; h<b<a, so there is one triangle.
b sin A
b = sin-1 10.660…2 L 41.3°
B = sin-1 a
a
C=180°-(A+B)=89.7°;
a sin C
32 sin 89.7°
c =
=
L 42.4
sin A
sin 49°
11. Given: B=70°, b=14, c=9 — an SSA case.
h=c sin B≠8.5; h<c<b, so there is one triangle.
c sin B
b = sin-1 1 0.604…2 L 37.2°
C = sin-1 a
b
A=180°-(B+C)≠72.8°;
b sin A
14 sin 72.8°
a =
=
L 14.2
sin B
sin 70°
12. Given: C=103°, b=46, c=61 — an SSA case.
h=b sin C≠44.8; h<b<c, so there is one triangle.
b sin C
b = sin-1 10.734… 2 L 47.3°
B = sin-1 a
c
A=180°-(B+C)=29.7°;
c sin A
61 sin 29.7°
a =
=
L 31.0
sin C
sin 103°
13. Given: A=36°, a=2, b=7. h=b sin A≠4.1;
a<h, so no triangle is formed.
14. Given: B=82°, b=17, c=15. h=c sin B≠14.9;
h<c<b, so there is one triangle.
C1=180°-(A+B1)≠43.3°;
a sin C1
16 sin 43.3°
c1 =
=
L 12.2
sin A
sin 64°
Or (with B obtuse):
B2=180°-B1≠107.3°;
C2=180°-(A+B2)≠8.7°;
a sin C2
c2 =
L 2.7
sin A
20. Given: B=38°, b=21, c=25. h=c sin B≠15.4;
h<b<c, so there are two triangles.
c sin B
b = sin-1 1 0.732…2 L 47.1°
C1 = sin-1 a
b
A1=180°-(B+C1)≠94.9°;
b sin A1
21 sin 94.9°
a1 =
=
L 34.0
sin B
sin 38°
Or (with C obtuse):
C2=180°-C1≠132.9°;
A2=180°-(B+C2)≠9.1°;
b sin A2
a2 =
L 5.4
sin B
21. Given: C=68°, a=19, c=18. h=a sin C≠17.6;
h<c<a, so there are two triangles.
a sin C
b = sin-1 10.978…2 L 78.2°
A1 = sin-1 a
c
B1=180°-(A+C)≠33.8°;
c sin B1
18 sin 33.8°
b1 =
=
L 10.8
sin C
sin 68°
Or (with A obtuse):
A2=180°-A1≠101.8°;
B2=180°-(A2+C)≠10.2°;
c sin B2
b2 =
L 3.4
sin C
22. Given: B=57°, a=11, b=10. h=a sin B≠9.2;
h<b<a, so there are two triangles.
a sin B
b = sin-1 10.922… 2 L 67.3°
A1 = sin-1 a
b
C1=180°-(A1+B)≠55.7°;
b sin C1
10 sin 55.7°
c1 =
=
L 9.9
sin B
sin 57°
Or (with A obtuse):
A2=180°-A1≠112.7°;
C2=180°-(A2+B)≠10.3°;
b sin C2
L 2.1
c2 =
sin B
23. h=10 sin 42°≠6.691, so:
15. Given: C=36°, a=17, c=16. h=a sin C≠10.0;
h<c<a, so there are two triangles.
(a) 6.691 6 b 6 10.
16. Given: A=73°, a=24, b=28. h=b sin A≠26.8;
a<h, so no triangle is formed.
(c) b 6 6.691
17. Given: C=30°, a=18, c=9. h=a sin C=9;
h=c, so there is one triangle.
18. Given: B=88°, b=14, c=62. h=c sin B≠62.0;
b<h, so no triangle is formed.
19. Given: A=64°, a=16, b=17. h=b sin A≠15.3;
h<a<b, so there are two triangles.
b sin A
B1 = sin-1 a
b = sin-1 10.954… 2 L 72.7°
a
(b) b≠6.691 or b 10.
24. h=12 sin 53°≠9.584, so:
(a) 9.584 6 c 6 12.
(b) c≠9.584 or c 12.
(c) c 6 9.584
25. (a) No: this is an SAS case
(b) No: only two pieces of information given.
Section 5.5
26. (a) Yes: this is an AAS case.
B=180°-(A+C)=32°;
81° sin 32°
a sin B
b=
=
L 88.5;
sin A
sin 29°
a sin C
81 sin 119°
c=
=
L 146.1
sin A
sin 29°
28. Given: B=47°, a=8, b=21 — an SSA case.
h=a sin B≠5.9; h<a<b, so there is one triangle.
a sin B
b = sin-1 10.278… 2 L 16.2°
A = sin-1 a
b
C=180°-(A+B)=116.8°;
b sin C
21 sin 116.8°
c =
=
L 25.6
sin B
sin 47°
29. Given: A=136°, a=15, b=28 — an SSA case.
h=b sin A≠19.5; a<h, so no triangle is formed.
30. Given: C=115°, b=12, c=7 — an SSA case.
h=b sin C≠10.9; c<h, so no triangle is formed.
31. Given: B=42°, c=18, C=39° — an AAS case.
A=180°-(B+C)=99°;
18 sin 99°
c sin A
=
L 28.3;
a =
sin C
sin 39°
18 sin 42°
c sin B
=
L 19.1
b =
sin C
sin 39°
32. Given: A=19°, b=22, B=47° — an AAS case.
C=180°-(A+B)=114°;
22 sin 19°
b sin A
=
L 9.8;
a =
sin B
sin 47°
b sin C
22 sin 114°
c =
=
L 27.5
sin B
sin 47°
33. Given: C=75°, b=49, c=48. — an SSA case.
h=b sin C≠47.3; h<c<b, so there are two
triangles.
b sin C
b = sin-1 1 0.986…2 L 80.4°
B1 = sin-1 a
c
A1=180°-(B+C)≠24.6°;
c sin A1
48 sin 24.6°
a1 =
=
L 20.7
sin C
sin 75°
Or (with B obtuse):
B2=180°-B1≠99.6°;
A2=180°-(B2+C)≠5.4°;
c sin A2
L 4.7
a2 =
sin C
34. Given: A=54°, a=13, b=15. — an SSA case.
h=b sin A≠12.1; h<a<b, so there are two triangles.
b sin A
b = sin-1 10.933… 2 L 69.0°
B1 = sin-1 a
a
C1=180°-(A+B1)≠57.0°;
a sin C1
13 sin 57.0°
c1 =
=
L 13.5
sin A
sin 54°
Or (with B obtuse):
B2=180°-B1≠111.0°;
C2=180°-(A+B2)≠15.0°;
a sin C2
L 4.2
c2 =
sin A
221
35. Cannot be solved by law of sines (an SAS case).
36. Cannot be solved by law of sines (an SAS case).
37. Given: c=AB=56, A=72°, B=53° — an ASA case,
so C=180°-(A+B)=55°
(a) AC=b=
(b) No: this is an SAS case.
27. Given: A=61°, a=8, b=21 — an SSA case.
h=b sin A=18.4; a<h, so no triangle is formed.
The Law of Sines
c sin B
56 sin 53°
=
L 54.6 ft.
sin C
in 55°
(b) h=b sin A(= a sin B)≠51.9 ft.
38. Given: c=25, A=90°-38°=52°,
B=90°-53°=37° — an ASA case, so
C=180°-(A+B)=91° and
25 sin 52°
c sin A
=
L 19.7 mi.
a=
sin C
sin 91°
c sin B
25 sin 37°
=
L 15.0 mi,
b=
sin C
sin 91°
and finally h=b sin A=a sin B≠11.9 mi.
39. Given: c=16, C=90°-62=28°,
B=90°+15°=105° — an AAS case.
A=180°-(B+C)=47°, so
c sin A
16 sin 47°
a=
=
L 24.9 ft.
sin C
sin 28°
40. Given: c=2.32, A=28°, B=37° — an ASA case.
C=180°-(A+B)=115°;
2.32 sin 28°
c sin A
a=
=
L 1.2 mi.
sin C
sin 115°
c sin B
2.32 sin 37°
b=
=
L 1.5 mi.
sin C
sin 115°
Therefore, the altitude is h=b sin A≠(1.5) sin 28°
≠0.7 mi — or a sin B≠(1.2) sin 37° mi≠0.7 mi.
41.
4 ft
x
18˚
10˚
The length of the brace is the leg of the larger triangle.
x
sin 28° = , so x=1.9 ft.
4
42.
78.75˚
22.5˚
B
15.5 ft
C
The center of the wheel (A) and two adjacent chairs
360°
(B and C) form a triangle with a=15.5, A=
16
=22.5°, and B=C=78.75°. This is an ASA case, so
a sin B
15.5 sin 78.75°
the radius is b=c=
=
L 39.7 ft.
sin A
sin ____
22.5°
Alternatively, let D be the midpoint of BC, and consider
right ^ABD, with mjBAD=11.25° and BD=7.75 ft;
then r is the hypotenuse of this triangle, so
7.75
r =
L 39.7 ft.
sin 11.25°
222
Chapter 5
Analytic Trigonometry
43. Consider the triangle with vertices at the top of the flagpole (A) and the two observers (B and C). Then a=600,
B=19°, and C=21° (an ASA case), so
A=180°-(B+C)=140°;
a sin B
600 sin 19°
b=
=
L 303.9;
sin A
sin 140°
600 sin 21°
a sin C
c=
=
L 334.5
sin A
sin 140°
and finally h=b sin C=c sin B≠108.9 ft.
44. Consider the triangle with vertices at the top of the tree
(A) and the two observers (B and C). Then a=400,
B=15°, and C=20° (an ASA case), so
A=180°-(B+C)=145°;
a sin B
400 sin 15°
b=
=
L 180.5;
sin A
sin 145°
a sin C
400 sin 20°
c=
=
L 238.5;
sin A
sin 145°
and finally h=b sin C=c sin B≠61.7 ft.
45. Given: c=20, B=52°, C=33° — an AAS case.
A=180°-(B+C)=95°, so
20 sin 95°
c sin A
a=
=
L 36.6 mi, and
sin C
sin 33°
20 sin 52°
c sin B
b=
=
L 28.9 mi.
sin C
sin 33°
46. We use the mean (average) measurements for A, B, and
AB, which are 79.7°, 83.9°, and 25.9 feet, respectively. This
gives 16.4° for angle C. By the Law of Sines,
25.9 sin 83.9 °
AC =
L 91.2 feet.
sin 16.4 °
sin B
sin A
,
=
a
b
sin A
a
which is equivalent to
= (since sin A, sin B Z 0).
sin B
b
47. True. By the law of sines,
48. False. By the law of sines, the third side of the triangle
10 sin 100 °
measures
, which is about 15.32 inches. That
sin 40 °
makes the perimeter about 10+10+15.32=35.32,
which is less than 36 inches.
49. The third angle is 32°. By the Law of Sines,
sin 32°
sin 53°
, which can be solved for x.
=
12.0
x
The answer is C.
50. With SSA, the known side opposite the known angle sometimes has two different possible positions. The answer is D.
length — say, a. Suppose that a=ka for some constant k. Then for this new triangle, we have
sin A
sin B
sin C
sin A
sin A
. Since
=
=
=
=
a
b
c
a
ka
sin B
1 # sin A
1 sin B
, we can see that
,
= #
k
a
b¿
k
b
so that b=kb and similarly, c=kc. So for any
choice of a positive constant k, we can create a
triangle with angles A, B, and C.
(b) Possible answers: a=1, b= 13, c=2 (or any set of
three numbers proportional to these).
(c) Any set of three identical numbers.
54. In each proof, assume that sides a, b, and c are opposite
angles A, B, and C, and that c is the hypotenuse.
(a)
(b)
(c)
sin 90°
sin A
=
a
c
1
sin A
=
a
c
opp
a
sin A =
=
c
hyp
sin B
sin 90°
=
b
c
cos1p>2 - B2
1
=
b
c
adj
b
cos A= =
c
hyp
sin B
sin A
=
a
b
a
sin A
=
sin B
b
a
sin A
=
cos A
b
opp
a
tan A =
=
b
adj
55. (a) h=AB sin A
(b) BC 6 AB sin A
(c) BC AB or BC=AB sin A
(d) AB sin A 6 BC 6 AB
C
51. The longest side is opposite the largest angle, while the
shortest side is opposite the smallest angle. By the Law of
sin 50°
sin 70 °
Sines,
, which can be solved for x.
=
9.0
x
The answer is A.
52. Because BC>AB, only one triangle is possible. The
answer is B.
53. (a) Given any triangle with side lengths a, b, and c, the
sin A
sin B
sin C
=
=
law of sines says that
.
a
b
c
But we can also find another triangle(using ASA)
with two angles the same as the first (in which case
the third angle is also the same) and a different side
____
56. Drawing the line suggested in the hint, and extending BC
to meet that line at, say, D, gives right ^ADC and right
^ADB.
5
8
22°
B
A
D
Then AD=8 sin 22°≠3.0 and DC=8 cos 22°≠7.4,
so DB=DC-5 and
c=AB= 2AD2 + DB2 L 3.9. Finally,
DB
b L 29.1° and
A=(90°-22°)-sin–1 a
AB
B=180°-A-C≠128.9°.
Section 5.6
57. Given: c=4.1, B=25°, C=36.5°-25°=11.5°. An
AAS case: A=180°-(B+C)=143.5°, so
4.1 sin 25°
c sin B
AC=b=
=
L 8.7 mi, and
sin C
sin 11.5°
c sin A
4.1 sin 143.5°
BC=a=
=
L 12.2 mi.
sin C
sin 11.5°
The height is h=a sin 25°=b sin 36.5°≠5.2 mi.
The Law of Cosines
223
Section 5.6 Exercises
1. Given: B=131°, c=8, a=13 — an SAS case.
b = 2a2 + c2 - 2ac cos B L 1369.460 L 19.2;
a2 + b2 - c2
b L cos-1 1 0.9492 L 18.3°;
C = cos-1 a
2ab
A = 180° - 1B + C 2 L 30.7°.
2. Given: C=42°, b=12, a=14 — an SAS case.
■ Section 5.6 The Law of Cosines
Exploration 1
1. The semiperimeters are 154 and 150.
A= 1154 1154 - 115 2 1154 - 812 1 154 - 112 2
+ 1150 1 150 - 1122 1150 - 102 2 1150 - 862
=8475.742818 paces2
2. 41,022.59524 square feet
3. 0.0014714831 square miles
4. 0.94175 acres
5. The estimate of “a little over an acre” seems questionable,
but the roughness of their measurement system does not
provide firm evidence that it is incorrect. If Jim and
Barbara wish to make an issue of it with the owner, they
would be well-advised to get some more reliable data.
6. Yes. In fact, any polygonal region can be subdivided into
triangles.
Quick Review 5.6
3
1. A=cos–1 a b ≠53.130°
5
3. A=cos–1(–0.68)≠132.844°
1.92
b ≠50.208°
3
5. (a) cos A =
x2 + y2 - 81
81 - x2 - y2
=
.
-2xy
2xy
x2 + y2 - 81
b
(b) A=cos a
2xy
–1
6. (a) cos A =
3. Given: a=27, b=19, c=24 — an SSS case.
b2 + c2 - a2
b L cos-1 1 0.2282 L 76.8°;
A = cos-1 a
2bc
a2 + c2 - b2
B = cos-1 a
b L cos-1 1 0.7282 L 43.2°;
2ac
C = 180° - 1A + B2 L 60°.
4. Given: a=28, b=35, c=17 — an SSS case.
b2 + c2 - a2
b L cos-1 1 0.6132 L 52.2°;
A = cos-1 a
2bc
a2 + c2 - b2
b L cos-1 1 –0.1592 L 99.2°;
B = cos-1 a
2ac
C = 180° - 1A + B2 L 28.6°.
5. Given: A=55°, b=12, c=7 — an SAS case.
a = 2b2 + c2 - 2bc cos A L 196.639 L 9.8;
a2 + c2 - b2
b L cos-1 10.011 2 L 89.3°;
B = cos-1 a
2ac
C = 180° - 1A + B2 L 35.7°.
6. Given: B=35°, a=43, c=19 — an SAS case.
2. C=cos–1(–0.23)≠103.297°
4. C=cos–1 a
c = 2a2 + b2 - 2ab cos C L 190.303 L 9.5;
b2 + c2 - a2
A = cos-1 a
b L cos-1 1 0.1672 L 80.3°;
2bc
B = 180° - 1A + C 2 L 57.7°.
y2 - x2 - 25
x2 - y2 + 25
=
-10
10
x2 - y2 + 25
b
(b) A=cos–1 a
10
7. One answer: (x-1)(x-2)=x2-3x+2.
Generally: (x-a)(x-b)=x2-(a+b)x+ab
for any two positive numbers a and b.
8. One answer: (x-1)(x+1)=x2-1.
Generally, (x-a)(x+b)=x2-(a-b)x-ab
for any two positive numbers a and b.
2
9. One answer: (x-i)(x+i)=x +1
10. One answer: (x-1)2=x2-2x+1.
Generally: (x-a)2=x2-2ax+a2
for any positive number a.
b = 2a2 + c2 - 2ac cos B L 1871.505 L 29.5;
a2 + b2 - c2
b L cos-1 1 0.9292 L 21.7°;
C = cos-1 a
2ab
A = 180° - 1B + C 2 L 123.3°.
7. Given: a=12, b=21, C=95° — an SAS case.
c = 2a2 + b2 - 2ab cos C L 1628.926 L 25.1;
b2 + c2 - a2
b L cos-1 1 0.8792 L 28.5°;
A = cos-1 a
2bc
B = 180° - 1A + C 2 L 56.5°.
8. Given: b=22, c=31, A=82° — an SAS case.
a = 2b2 + c2 - 2bc cos A L 11255.167 L 35.4;
a2 + c2 - b2
b L cos-1 10.788 2 L 37.9°;
B = cos-1 a
2ac
C = 180° - 1A + B2 L 60.1°.
9. No triangles possible (a+c=b)
10. No triangles possible (a+b<c)
11. Given: a=3.2, b=7.6, c=6.4 — an SSS case.
b2 + c2 - a2
b L cos-1 1 0.9092 L 24.6°;
A = cos-1 a
2bc
a2 + c2 - b2
b L cos-1 1 –0.160 2 L 99.2°;
B = cos-1 a
2ac
C = 180° - 1A + B2 L 56.2°.
224
Chapter 5
Analytic Trigonometry
12. No triangles possible (a+b<c)
Exercises 13–16 are SSA cases, and can be solved with either
the Law of Sines or the Law of Cosines. The law of cosines
solution is shown.
13. Given: A=42°, a=7, b=10 — an SSA case. Solve
the quadratic equation 72=102+c2-2(10)c cos 42°,
or c2-(14.862…)c+51=0; there are two positive
a2 + c2 - b2
solutions: L 9.487 or 5.376. Since cos B=
:
2ac
–1
c1≠9.487, B1≠cos (0.294)≠72.9°, and
C1=180°-(A+B1)≠65.1°,
or
c2≠5.376, B2≠cos–1(–0.294)≠107.1°, and
C2=180°-(A+B2)≠30.9°.
14. Given: A=57°, a=11, b=10 — an SSA case. Solve
the quadratic equation 112=102+c2-2(10)C cos 57°,
or c2-(10.893)c-21=0; there is one positive
a2 + c2 - b2
solution c=12.564. Since cos B=
,
2ac
B≠cos–1(0.647)≠49.7° and C=180°-(A+B)
≠73.3°.
15. Given: A=63°, a=8.6, b=11.1 — an SSA case. Solve
the quadratic equation
8.62=11.12+c2-2(11.1)c cos 63°, or
c2-(10.079)c+49.25=0; there are no real solutions,
so there is no triangle.
16. Given: A=71°, a=9.3, b=8.5 — an SSA case. Solve
the quadratic equation
9.32=8.52+c2-2(8.5)c cos 71°, or
c2-(5.535)c-14.24=0; there is one positive
a2 + c2 - b2
solution: c≠7.447. Since cos B=
,
2ac
–1
B≠cos (0.503)≠59.8° and C=180°-(A+B)
≠49.2°.
For #21–28, a triangle can be formed if a+b<c, a+c<b,
and b+c<a.
21. s=
17
; Area = 166.9375 L 8.18
2
22. s=
21
; Area = 1303.1875 L 17.41
2
23. No triangle is formed (a+b=c).
24. s=27; Area = 112,960 = 36 110 L 113.84
25. a=36.4; Area = 146,720.3464 L 216.15
26. No triangle is formed (a+b<c)
27. s=42.1; Area = 198,629.1856 L 314.05
28. s=23.8; Area 110,269.224 L 101.34
29. Let a=4, b=5, and c=6. The largest angle is
opposite the largest side, so we call it C. Since
a2 + b2 - c2
1
cos C =
, C = cos-1 a b L 82.819°
2ab
8
≠1.445 radians.
30. The shorter diagonal splits the parallelogram into two
(congruent) triangles with a=26, B=39°, and c=18.
The diagonal has length b = 2a2 + c2 - 2ac cos B
≠ 1272.591 L 16.5 ft.
31. Following the method of Example 3, divide the hexagon into
6 triangles. Each has two 12-inch sides that form a 60° angle.
1
6 * 112 2 1122 sin 60° = 21613 L 374.1 square inches
2
32. Following the method of Example 3, divide the nonagon
into 9 triangles. Each has two 10-inch sides that form a
40° angle.
1
9 * 110 2 1102 sin 40° L 289.3 square inches
2
33.
17. Given: A=47°, b=32, c=19 — an SAS case.
a = 2b2 + c2 - 2bc cos A L 1555.689 L 23.573,
so Area L 149431.307 L 222.33 ft2 (using Heron’s
1
formula). Or, use A= bc sin A.
2
30°
a
s
18. Given: A=52°, b=14, c=21 — an SAS case.
a = 2b2 + c2 - 2bc cos A L 1274.991 L 16.583,
so Area L 113418.345 L 115.84 m2 (using Heron’s
1
formula). Or, use A= bc sin A.
2
19. Given: B=101°, a=10, c=22 — an SAS case.
b = 2a2 + c2 - 2ac cos B L 1667.955 L 25.845,
so Area L 111659.462 L 107.98 cm (using Heron’s
1
formula). Or, use A= ac sin B.
2
s
In the figure, a=12 and so s=12 sec 30° = 813.
The area of the hexagon is
1
6 * 18 132 18132sin 60° = 28813
2
L 498.8 square inches.
34.
2
20. Given: C=112°, a=1.8, b=5.1 — an SAS case.
c = 2a2 + b2 - 2ab cos C L 136.128 L 6.011,
so Area L 118.111 L 4.26 in.2 (using Heron’s
1
formula). Or, use A= ab sin C.
2
20°
s
s
a
In the figure, a=10 and so s=10 sec 20°. The area of
the nonagon is
1
9 * 110 sec 20°2 110 sec 20°2 sin 40°
2
L 327.6 square inches.
Section 5.6
35. Given: C=54°, BC=a=160, AC=b=110 — an
SAS case. AB = c = 2a2 + b2 - 2ab cos C
≠ 117,009.959 L 130.42 ft.
36. (a) The home-to-second segment is the hypotenuse of
a right triangle, so the distance from the pitcher’s
rubber to second base is 9012 - 60.5 L 66.8 ft.
This is a bit more than
c = 260.52 + 902 - 2 160.5 2 190 2 cos 45°
≠ 14059.857 L 63.7 ft.
(b) B = cos-1 a
2
2
2
a + c - b
b L cos-1 1 -0.0492
2ac
≠92.8°.
37. (a) c = 2402 + 602 - 2 140 2 160 2 cos 45°
(b) The home-to-second segment is the hypotenuse of a
right triangle, so the distance from the pitcher’s
rubber to second base is 6012 - 40 L 44.9 ft.
a2 + c2 - b2
b L cos-1 1 -0.0572
2ac
≠93.3 °.
38. Given: a=175, b=860, and C=78°. An SAS case, so
AB=c= 2a2 + b2 - 2ab cos C L 1707,643.581
≠841.2 ft.
39. (a) Using right ¢ACE, mjCAE = tan-1 a
225
44. ¢ABC is a right triangle (C=90°), with BC=a
= 222 + 22 = 2 12 and AC=b=1, so AB=c
1
= 2a2 + b2 = 3 and B=m j ABC=sin–1 a b
3
≠19.5°.
45. True. By the Law of Cosines, b2+c2-2bc cos A=a2,
which is a positive number. Since b2+c2-2bc cos A>0,
it follows that b2+c2>2bc cos A.
46. True. The diagonal opposite angle ¨ splits the
parallelogram into two congruent triangles, each with area
1
ab sin ¨.
2
a
≠ 11805.887 L 42.5 ft.
(c) B = cos-1 a
The Law of Cosines
6
b
18
1
=tan a b L 18.435°.
3
-1
(b) Using A L 18.435°, we have an SAS case, so
DF = 292 + 122 - 2 1 92 1 122 cos A L 120.084
≠4.5 ft.
(c) EF = 2182 + 122 - 2 118 2 112 2 cos A L 158.168
≠7.6 ft.
40. After two hours, the planes have traveled 700 and 760
miles, and the angle between them is 22.5°, so the
distance is 27002 + 7602 - 2 1 700 2 1760 2 cos 22.5°
≠ 184,592.177 L 290.8 mi.
41. AB = 2732 + 652 - 2 173 2 165 2 cos 8°
≠ 1156.356 L 12.5 yd.
42. m jHAB = 135 ° , so
HB = 2202 + 202 - 21 202 1 20 2 cos 135°
≠ 11365.685
_____ L 37.0 ft.
Note that AB is the hypotenuse of an equilateral right
_____
20
= 10 12 , and HC is the
triangle with leg length
12
hypotenuse of an equilateral right triangle with leg length
20 + 10 12 , so HC = 22 120 + 10 122 2 L 48.3 ft.
Finally, using right ¢HAD with leg lengths
HA = 20 ft and AD = HC L 48.3 ft , we have
HD = 2HA2 + AD2 L 52.3 ft.
43. AB = c = 222 + 32 = 113, AC = b = 212 + 32
= 110 , and BC = a = 212 + 22 = 15, so
b2 + c2 - a2
mjCAB = A = cos-1 a
b
2bc
9
b ≠37.9°.
= cos-1 a
1130
θ
b
47. Following the method of Example 3, divide the
dodecagon into 12 triangles. Each has two 12-inch sides
that form a 30° angle.
1
12 * 1 122 1122 sin 30° = 432
2
The answer is B.
48. The semiperimeter is s=(7+8+9)/2=12. Then by
Heron’s Formula, A = 112 112 - 72 112 - 82 112 - 9 2
= 12 15. The answer is B.
49. After 30 minutes, the first boat has traveled 12 miles and the
second has traveled 16 miles. By the Law of Cosines, the
two boats are 2122 + 162 - 21122 116 2 cos 110 ° L 23.05
miles apart. The answer is C.
50. By the Law of Cosines, 122=17 2+252-2(17)(25)
172 + 252 - 122
≤ ≠25.06°.
cos ¨, so ¨=cos–1 ¢
2 117 2 1252
The answer is E.
51. Consider that a n-sided regular polygon inscribed within a
circle can divide into n equilateral triangles, each with
r2
360°
equal area of sin
. (The two equal sides of the
2
n
equilateral triangle are of length r, the radius of the
circle.) Then, the area of the polygon is exactly
nr2
360°
.
sin
2
n
b2 + c2 - 1b2 + c2 - 2bc cos A2
b2 + c2 - a2
52. (a)
=
2abc
2abc
Law of Cosines
2bc cos A
=
2abc
cos A
=
a
(b) The identity in (a) has two other equivalent forms:
cos B
a2 + c2 - b2
=
b
2abc
cos C
a2 + b2 - c2
=
c
2abc
226
Chapter 5
Analytic Trigonometry
We use them all in the proof:
cos A
cos B
cos C
+
+
a
b
c
a2 + c2 - b2
a2 + b2 - c2
b2 + c2 - a2
+
+
=
2abc
2abc
2abc
b2 + c2 - a2 + a2 + c2 - b2 + a2 + b2 - c2
=
2abc
a2 + b2 + c2
=
2abc
30.2 - 15.1
= 15.1 knots;
1 hr
37.2 - 12.4
Ship B:
= 12.4 knots
2 hrs
53. (a) Ship A:
(b) cos A =
=
b2 + c2 - a2
2bc
115.1 2 2 + 112.42 2 - 18.7 2 2
A = 35.18°
6. cos2 2x-cos2 x=(1-sin2 2x)-(1-sin2 x)
=sin2 x-sin2 2x
7. tan2 x-sin2 x=sin2 x a
sin2 x
=sin2 x tan2 x
cos2 x
8. 2 sin ¨ cos3 ¨+2 sin3 ¨ cos ¨
=(2 sin ¨ cos ¨)(cos2 ¨+sin2 ¨)
=(2 sin ¨ cos ¨)(1)=sin 2¨.
=sin2 x
54. Use the area formula and the Law of Sines:
1
A¢ = ab sin C
2
1
a sin B
a sin B
r
= aa
b sin C qLaw of Sines 1 b=
sin A
2
sin A
2
a sin B sin C
=
2 sin a
55. Let P be the center of the circle. Then,
52 + 52 - 72
= 0.02 , so P≠88.9°. The area
cos P =
2152 15 2
88.9 °
of the segment is pr2 #
L 25p # 10.247 2 L 19.39 in2.
360 °
1
The area of the triangle, however, is 1 52 152 sin 188.92
2
≠12.50 in2, so the area of the shaded region is approx.
6.9 in2.
1
- cos x
sin x
2
2
1 - cos x
sin x
=
=
= sin x
sin x
sin x
10.
=
1 + tan u
1 + cot u
+
1 - tan u
1 - cot u
11 + tan u 2 1 1 - cot u 2 + 1 1 + cot u2 11 - tan u2
11 - tan u2 11 - cot u2
11 + tan u - cot u - 12 + 1 1 + cot u - tan u - 1 2
1 1 - tan u 2 11 - cot u2
0
= 0
=
11 - tan u 2 11 - cot u 2
=
12. sin 3¨=sin(2¨+¨)=sin 2¨ cos ¨+cos 2¨ sin ¨
=2 sin ¨ cos2 ¨+(cos2 ¨-sin2 ¨) sin ¨
=3 sin ¨ cos2 ¨-sin3 ¨
2
t
1
1
= c;
11 + cos t 2 d = 11 + cos t2
2
B2
2
sec t
1 + sec t
1 + cos t
=a
b a
b =
2
sec t
2 sec t
tan3 g - cot3 g
13. cos2
14.
tan2 g + csc2 g
=
=
1. 2 sin 100° cos 100°=sin 200°
15.
3. 1; the expression simplifies to (cos 2¨)2+(2 sin ¨ cos ¨)2
=(cos 2¨)2+(sin 2¨)2=1.
4. cos2 2x; the expression can be rewritten
1-(2 sin x cos x)2=1-(sin 2x)2=cos2 2x.
5. cos 3x=cos(2x+x)=cos 2x cos x-sin 2x sin x
=(cos2 x-sin2 x) cos x-(2 sin x cos x) sin x
=cos3 x-3 sin2 x cos x
=cos3 x-3(1-cos2 x)cos x
=cos3 x-3 cos x+3 cos3 x
=4 cos3 x-3 cos x
cos x
sin x
11. Recall that tan ¨ cot ¨=1.
■ Chapter 5 Review
2 tan 40°
= tan 80°
1 - tan2 40°
#
tan u + sin u
1 + cos u
1 + cos u 2
=
= a;
b
2 tan u
2
A
2
2
u
= a cos b
2
=
2.
#
9. csc x-cos x cot x=
2 1 15.12 1 12.4 2
(c) c2=a2+b2-2ab cos C
=(49.6)2+(60.4)2-2(49.6)(60.4) cos (35.18°)
≠1211.04, so the boats are 34.8 nautical miles
apart at noon.
1 - cos2 x
b
cos2 x
16.
1tan g - cot g2 1tan2 g + tan g cot g + cot2 g2
tan2 g + csc2 g
1tan g - cot g2 1tan2 g + 1 + cot2 g2
tan2 g + csc2 g
1tan g - cot g2 1tan2 g + csc2 g2
tan2 g + csc2 g
= tan g - cot g
cos f
sin f
+
1 - tan f
1 - cot f
sin f
cos f
cos f
sin f
= a
ba
b + a
ba
b
1 - tan f
cos f
1 - cot f
sin f
cos2 f
cos2 f - sin2 f
sin2 f
=
+
=
cos f - sin f
sin f - cos f
cos f - sin f
= cos f + sin f
cos1 -z2
cos 1 - z2
sec 1 -z2 + tan 1 -z2
31 + sin1 -z2 4>cos1 -z2
cos2 1 -z2
cos2 z
1 - sin2 z
=
=
=
= 1 + sin z
1 + sin1 -z2
1 - sin z
1 - sin z
=
Chapter 5
17.
11 - cos y2 2
1 - cos y
=
B 1 + cos y
B 1 1 + cos y 2 11 - cos y2
1 1 - cos y 2 2
11 - cos y 2 2
=
=
2
B 1 - cos y
B
sin2 y
0 1 - cos y 0
1 - cos y
— since 1-cos y 0,
=
0 sin y 0
0 sin y 0
we can drop that absolute value.
=
18.
11 - sin g2 1 1 + sin g2
1 - sin g
=
B 1 + sin g
B
11 + sin g2 2
1 - sin2 g
cos2 g
=
=
2
B 11 + sin g2
B 1 1 + sin g2 2
0 cos g 0
0 cos g 0
=
— since 1+sin ˝ 0,
=
0 1 + sin g 0
1 + sin g
we can drop that absolute value.
tan u + tan13p>4 2
3p
b =
4
1 - tan u tan13p>4 2
tan u + 1 -1 2
tan u - 1
=
=
1 - tan u1 -1 2
1 + tan u
19. tan a u +
20.
1
1
1
sin 4˝= sin 2(2˝)= (2 sin 2˝ cos 2˝)
4
4
4
1
= (2 sin ˝ cos ˝)(cos2 ˝-sin2 ˝)
2
=sin ˝ cos3 ˝-cos ˝ sin3 ˝
21. tan
1 - cos b
cos b
1
1
b =
=
= csc b - cot b
2
sin b
sin b
sin b
22. Let ¨=arctan t, so that tan ¨=t. Then
2t
2 tan u
=
tan 2¨=
. Note also that since
1 - tan2 u
1 - t2
p
p
p
p
–1<t<1, - 6 u 6 , and therefore - 6 2u 6 .
4
4
2
2
That means that 2¨ is in the range of the arctan function,
2t
and so 2¨=arctan
, or equivalently
1 - t2
1
2t
¨= arctan
— and of course, ¨=arctan t.
2
1 - t2
23. Yes: sec x-sin x tan x=
=
1
sin2 x
cos x
cos x
25. Many answers are possible, for example,
sin 3x+cos 3x
=(3 sin x-4 sin3 x)+(4 cos3 x-3 cos x)
=3(sin x-cos x)-4(sin3 x-cos3 x)
=(sin x-cos x)[3-4(sin2 x+sin x cos x+cos2 x)]
=(sin x-cos x) (3-4-4 sin x cos x)
=(cos x-sin x)(1+4 sin x cos x). Check
other answers with a grapher.
227
26. Many answers are possible, for example,
sin 2x+cos 3x=2 sin x cos x+4 cos3 x-3 cos x
=cos x(2 sin x+4 cos2 x-3)
=cos x(2 sin x+1-4 sin2 x). Check other answers
with a grapher.
27. Many answers are possible, for example,
cos2 2x-sin 2x=1-sin2 2x-sin 2x
=1-4 sin2 x cos2 x-2 sin x cos x. Check other
answers with a grapher.
28. Many answers are possible, for example (using Review
Exercise #12), sin 3x-3 sin 2x
=3 cos2 x sin-sin3 x-6 sin x cos x
=sin x(3 cos2 x-sin2 x-6 cos x)
=sin x(4 cos2 x-1-6 cos x). Check other answers
with a grapher.
In #29–33, n represents any integer.
p
5p
29. sin 2x=0.5 when 2x= +2n∏ or 2x= +2n∏,
6
6
5p
p
so x= +n∏ or x= +n∏.
12
12
30. cos x=
p
13
when x= ; +2n∏
2
6
p
31. tan x=–1 when x= - +n∏
4
32. If sin–1 x=
12
12
, then x=sin
.
2
2
33. If tan–1 x=1, then x=tan 1.
34.
2 cos 2x=1
1
cos 2x=
2
1
2
3
cos2x=
4
2 cos2x-1=
cos x=—
So x=
13
2
p
5p
+ 2np or x=
+ 2np for n any integer.
6
6
35. x≠1.12 or x≠5.16
cos2 x
1 - sin2 x
=
= cos x .
cos x
cos x
24. Yes: (sin2 Å-cos2 a)(tan2 Å+1)
=(sin2 Å-cos2 Å)(sec2 Å)
sin2 a
sin2 a - cos2 a
=
=
- 1 = tan2 a - 1.
2
cos a
cos2 a
Review
[0, 2␲ ] by [–4, 4]
36. x≠0.14 or x≠3.79
[0, 2␲ ] by [–3, 2]
228
Chapter 5
Analytic Trigonometry
46. 2 sin x cos x=2 cos x is equivalent to
(cos x)(sin x-1)=0, so the solutions in (0, 2∏] are
p
3p
and x =
. The solution set for the inequality
x =
2
2
p 3p
p
3p
is
; that is, a ,
6 x 6
b.
2
2
2 2
37. x≠1.15
1
p
5p
has solutions x =
and x =
in the
2
3
3
interval [0, 2∏). The solution set for the inequality is
p 5p
p
5p
; that is, a ,
6 x 6
b.
3
3
3 3
47. cos x =
[0, 2␲ ] by [–3, 2]
38. x≠1.85 or x≠3.59
48. tan x=sin x is equivalent to (sin x)(cos x-1)=0,
p p
so the only solution in a - , b is x=0. The solution
2 2
p
p
set for the inequality is - 6 x 6 0 ; that is, a - , 0 b .
2
2
49. y=5 sin (3x+cos–1(3/5))≠5 sin (3x+0.9273)
[0, 2␲ ] by [–3, 2]
1
p
5p
39. cos x= , so x= or x= .
2
3
3
50. y=13 sin (2x-cos–1(5/13))≠13 sin (2x-1.176)
51. Given: A=79°, B=33°, a=7 — an AAS case.
C = 180° - 1A + B2 = 68°
a sin B
7 sin 33°
b =
=
L 3.9;
sin A
sin 79°
7 sin 68°
a sin C
=
L 6.6.
c =
sin A
sin 79°
40. sin 3x=(sin x)(4 cos2 x-1). This equation
becomes (sin x)(4 cos2 x-1)=sin x, or
2(sin x)(2 cos2 x-1)=0, so sin x=0 or
12
p
3p
cos x=—
; x=0, x= , x= , x=∏,
2
4
4
7p
5p
x= , or x= .
4
4
41. The left side factors to (sin x-3)(sin x+1)=0;
3p
only sin x=–1 is possible, so x= .
2
42. 2 cos2 t-1=cos t, or 2 cos2 t-cos t-1=0,
or (2 cos t+1)(cos t-1)=0. Then cos t=–
or cos t=1: t=0, t=
1
2
4p
2p
or t= .
3
3
p
+ 2np. No choice of
2
n gives a value in [–1, 1], so there are no solutions.
43. sin(cos x)=1 only if cos x=
44. cos(2x)+5 cos x-2=2 cos2 x-1+5 cos x-2
=2 cos2 x+5 cos x-3=0. (2 cos x-1)(cos x+3)
1
=0, so cos(x)= and cos(x)=–3. The latter is
2
p 5p
f
extraneous so x = e ,
3 3
For #45–48, use graphs to suggest the intervals. To find the
endpoints of the intervals, treat the inequalities as equations
and solve.
p
1
5p
7p
,x =
,
has solutions x = , x =
2
6
6
6
11p
and x =
in interval [0, 2∏). The solution
6
p
set for the inequality is, 0 x 6
6
5p
7p
11p
6 x 6
6 x 6 2p;
or
or
6
6
6
p
5p 7p
11p
b ´ a
, 2p b .
that is, c 0, b ´ a ,
6
6 6
6
45. cos 2x =
52. Given: a=5, b=8, B=110° — an SSA case. Using the
Law of Sines: h=a sin B=4.7; h<a<b, so there is
one triangle.
A = sin-1 a
a sin B
b L sin-1 1 0.5872 L 36.0°;
b
C = 180° - 1A + B2 L 34.0°;
b sin C
8 sin 34.0°
c =
L
L 4.8.
sin B
sin 110°
Using Law of Cosines: Solve the quadratic
equation 82=52+c2-2(5)c cos 110°, or
c2+(3.420)c-39=0; there is one positive solution:
b2 + c2 - a2
c≠4.8. Since cos A=
:
2bc
–1
A≠cos (0.809)≠36.0° and C=180°-(A+B)
≠34.0°.
53. Given: a=8, b=3, B=30° — an SSA case. Using the Law
of Sines: h=a sin B=4; b<h, so no triangle is formed.
Using the Law of Cosines: Solve the quadratic equation
32=82+c2-2(8)c cos 30°, or c2- 18 132 c + 55 = 0;
there are no real solutions.
54. Given: a=14.7, A=29.3°, C=33° — an AAS case.
B=180°-(A+C)=117.7°, and
a sin B
14.7 sin 117.7°
b =
=
L 26.6;
sin A
sin 29.3°
14.7 sin 33°
a sin C
=
L 16.4.
c =
sin A
sin 29.3°
55. Given: A=34°, B=74°, c=5 — an ASA case.
C=180°-(A+B)=72°;
c sin A
5 sin 34°
a =
=
L 2.9;
sin C
sin 72°
c sin B
5 sin 74°
b =
=
L 5.1.
sin C
sin 72°
Chapter 5
56. Given: c=41, A=22.9°, C=55.1° — an AAS case.
B=180°-(A+C)=102°;
c sin A
41 sin 22.9°
a =
=
L 19.5;
sin C
sin 55.1°
c sin B
41 sin 102°
b =
=
L 48.9.
sin C
sin 55.1°
57. Given: a=5, b=7, c=6 — an SSS case.
b2 + c2 - a2
b L cos-1 1 0.714 2 L 44.4°;
A = cos-1 a
2bc
a2 + c2 - b2
b L cos-1 10.2 2 L 78.5°;
B = cos-1 a
2ac
C = 180° - 1 A + B2 L 57.1°.
58. Given: A=85°, a=6, b=4 — an SSA case. Using the
Law of Sines: h=b sin A≠4.0; h<b<a, so there is
one triangle.
b sin A
b L sin-1 10.664 2 L 41.6°;
B = sin-1 a
a
C = 180° - 1 A + B2 L 53.4°;
6 sin 53.4°
a sin C
=
L 4.8.
c =
sin A
sin 85°
Using the Law of Cosines: Solve the quadratic equation
62=42+c2-2(4)c cos 85°, or c2-(0.697)c-20=0;
there is one positive solution:
a2 + c2 - b2
c≠4.8. Since cos B=
:
2ac
B≠cos–1(0.747)≠41.6° and C=180°-(A+B)
≠53.4°.
1
59. s= (3+5+6)=7;
2
Area = 1s1 s - a 2 1 s - b2 1 s - c2
= 17 1 7 - 32 1 7 - 5 2 17 - 6 2
= 156 L 7.5
60. c≠7.672 so area≠ 1528.141 L 23.0 (using Heron’s
1
formula). Or, use A= ab sin C.
2
Review
229
65. Let a=8, b=9, and c=10. The largest angle is
opposite the largest side, so we call it C.
a2 + b2 - c2
5
Since cos C=
, C=cos–1 a b
2ab
16
≠71.790°, 1.253 rad.
66. The shorter diagonal splits the parallelogram into
two (congruent) triangles with a=15, B=40°,
and c=24. The shorter diagonal has length
b= 2a2 + b2 - 2ac cos B L 1249.448 L 15.794 ft.
Since adjacent angles are supplementary, the other angle
is 140°. The longer diagonal splits the parallelogram
into (two) congruent triangles with a=15, B=140°,
and c=24, so the longer diagonal length is
b= 2a2 + c2 - 2ac cos B L 11352.552 L 36.777 ft.
67. (a) The point (x, y) has coordinates (cos ¨, sin ¨), so the
bottom is b1=2 units wide, the top is
b2=2x=2 cos ¨ units wide, and the height is
h=y=sin ¨ units. Either use the formula for the
1
area of a trapezoid, A= (b1+b2)h, or notice that
2
the trapezoid can be split into two triangles and a
rectangle. Either way:
A(¨)=sin ¨+sin ¨ cos ¨=sin ¨(1+cos ¨)
1
=sin ¨+ sin 2¨.
2
p
(b) The maximizing angle is ¨= =60°; the maximum
3
3
area is 13 L 1.30 square units.
4
68. (a) Substituting the values of a and b:
S1u2 = 6.825 + 0.63375 1 -cot u + 13 # csc u2
0.633751 13 - cos u2
= 6.825 +
sin u
61. h=12 sin 28°≠5.6, so:
(a) ≠5.6<b<12.
(b) b≠5.6 or b 12.
(c) b<5.6.
62. (a) C=180°-(A+B)=45°, so
c sin B
80 sin 65°
=
L 102.5 ft.
AC=b=
sin C
sin 45°
(b) The distance across the canyon is b sin A≠96.4 ft.
63. Given: c=1.75, A=33°, B=37° — an ASA case, so
C=180°-(A+B)=110°;
c sin A
1.75 sin 33°
a =
=
L 1.0;
sin C
sin 110°
c sin B
1.75 sin 37°
b =
=
L 1.1,
sin C
sin 110°
and finally, the height is h=b sin A=a sin b≠0.6 mi.
64. Given: C=70°, a=225, b=900 — an SAS case, so
AB=c= 2a2 + b2 - 2ab cos C
≠ 1722,106.841 L 849.77 ft.
[0, 9.4] by [–6.2, 6.2]
(b) Considering only angles between 0 and ∏, the
minimum occurs when ¨≠0.96≠54.74°.
(c) The minimum value of S is approximately
S(0.96)≠7.72 in.2
69. (a) Split the quadrilateral in half to leave two
(identical) right triangles, with one leg 4000 mi,
hypotenuse 4000+h mi, and one acute angle ¨/2.
u
4000
Then cos =
; solve for h to leave
2
4000 + h
4000
u
h =
- 4000 = 4000 sec - 4000 miles.
cos 1u>22
2
(b) cos
u
4000
20
=
, so u = 2 cos-1 a b L 0.62 L 35.51°.
2
4200
21
230
Chapter 5
Analytic Trigonometry
70. Rewrite the left side of the equation as follows:
sin x-sin 2x+sin 3x
=sin x-2 sin x cos x+sin(2x+x)
=sin x-2 sin x cos x+sin 2x cos x+cos 2x sin x
=sin x-2 sin x cos x+2 sin x cos2 x+
(cos2 x-sin2 x) sin x
=sin x-2 sin x cos x+2 sin x cos2 x+
sin x cos2 x-sin3 x
=sin x-2 sin x cos x+3 sin x cos2 x-sin3 x
=sin x (1-2 cos x+3 cos2 x-sin2 x)
=sin x ((1-sin2 x)-2 cos x+3 cos2 x)
=sin x (cos2 x-2 cos x+3 cos2 x)
=sin x (4 cos2 x-2 cos x)
=2 sin x cos x (2 cos x-1)
=sin 2x (2 cos x-1)
p
So sin 2x = 0 or 2 cosx - 1 = 0. So x = n or
2
p
x = ;
+ 2np, n an integer.
3
71. The hexagon is made up of 6 equilateral triangles;
using Heron’s formula (or some other method), we
find that each triangle has area 2241 24 - 16 2 3
= 112,288 = 64 13 . The hexagon’s area is therefore,
384 13 cm2 , and the radius of the circle is 16 cm, so the
area of the circle is 256∏ cm2, and the area outside the
hexagon is 256p - 384 13 L 139.140 cm2 .
72. The pentagon is made up of 5 triangles with base length
6
12 cm and height
L 8.258 cm, so its area is about
tan 36°
2
247.749 cm . The radius of the circle is the height of those
triangles, so the desired area is about 247.749-∏(8.258)2
≠33.51 cm2.
73. The volume of a cylinder with radius r and height h
is V=∏r2h, so the wheel of cheese has volume
∏(92)(5)=405∏ cm3; a 15° wedge would have
405p
15
1
fraction
of that volume, or
=
L 53.01 cm3.
360
24
24
1
74. (a) (cos(u-v)-cos(u+v))
2
1
= (cos u cos v+sin u sin v
2
-(cos u cos v-sin u sin v))
1
= (2 sin u sin v)
2
=sin u sin v
1
(b) (cos(u-v)+cos(u+v))
2
1
= (cos u cos v+sin u sin v+cos u cos v
2
-sin u sin v)
1
= (2 cos u cos v)
2
=cos u cos v
1
(c) (sin(u+v)+sin(u-v))
2
1
= (sin u cos v+cos u sin v+sin u cos v
2
-cos u sin v)
1
= (2 sin u cos v)
2
=sin u cos v
75. (a) By the product-to-sum formula in 74 (c),
u + v
u - v
2 sin
cos
2
2
1
u
+
v
+ u - v
= 2 # a sin
2
2
u + v - 1 u - v2
+sin
b
2
=sin u+sin v
(b) By the product-to-sum formula in 74 (c),
u - v
u + v
2 sin
cos
2
2
1
u
v
+ u + v
= 2 # a sin
2
2
u - v - 1 u + v2
+sin
b
2
=sin u+sin(–v)
=sin u-sin v
(c) By the product-to-sum formula in 74 (b),
u + v
u - v
2 cos
cos
2
2
u + v - 1u - v2
1
= 2 # a cos
2
2
u + v + u - v
+cos
b
2
=cos v+cos u
=cos u+cos v
(d) By the product-to-sum formula in 74 (a),
u + v
u - v
–2 sin
sin
2
2
u + v - 1u - v2
1
= -2 # a cos
2
2
u + v + u - v
-cos
b
2
=–(cos v-cos u)
=cos u-cos v
76. Pat faked the data. The Law of Cosines can be solved to
2
. Only Carmen’s values
B 1 - cos u
are consistent with the formula.
show that x = 12
77. (a) Any inscribed angle that intercepts an arc of 180° is
a right angle.
(b) Two inscribed angles that intercept the same arc are
congruent.
opp
a
= .
(c) In right ¢A¿BC, sin A¿ =
hyp
d
(d) Because jA¿ and jA are congruent,
a>d
sin A
sin A¿
1
=
=
= .
a
a
a
d
(e) Of course. They both equal
sin A
by the Law of Sines.
a
Chapter 5
Chapter 5 Project
1.
[–2, 34] by [–0.1, 1.1]
2. We set the amplitude as half the difference between the
maximum value, 1.00, and the minimum value, 0.00, so
a=0.5. And we set the average value as the average of
the maximum and minimum values, so k=0.5. Since
cos (b(x-h)) takes on its maximum value at h, we set
h=29. Experimenting with the graph suggests that
b should be about 2∏/30.5. So the equation is
2∏
1x - 292 b + 0.5 .
y L 0.5 cos a
30.5
Review
5. Using the identities from Questions 3 and 4 together,
2∏
y L 0.5 cos a
1x - 292 b + 0.5
30.5
2∏
= 0.5 cos a
1 29 - x2 b + 0.5
30.5
∏
2∏
= 0.5 sin a 1 29 - x2 b + 0.5
2
30.5
2∏
= 0.5 sin a
1x - 21.3752 b + 0.5,
30.5
which is equivalent to
y L 0.5 sin 10.21x - 4.4 2 + 0.5.
Sinusoidal regression yields
y L 0.5 sin 10.21x + 1.87 2 + 0.49.
231
232
Chapter 6
Applications of Trigonometry
Chapter 6
Applications of Trigonometry
■ Section 6.1 Vectors in the Plane
9
7. ¨=tan–1 a b L 60.95°
5
Exploration 1
7
8. ¨=360°+tan–1 a - b L 305.54°
5
1. Use the HMT rule, which states that if an arrow has initial
point 1 x1, y1 2 and terminal point 1 x2, y2 2 , it represents the
vector ˚x2-x1, y2-y1¬. If the initial point is (2, 3) and
the terminal point is (7, 5), the vector is ˚7-2, 5-3¬=
˚5, 2¬.
2. Use the HMT rule, which states that if an arrow has initial
point 1 x1, y1 2 and terminal point 1 x2, y2 2 , it represents the
vector ˚x2-x1, y2-y1¬. If the initial point is (3, 5) and
the terminal point is 1 x2, y2 2 , the vector is ˚x2-3, y2-5¬.
Using the given vector ˚–3, 6¬, we have x2-3=–3 and
y2-5=6.
x2-3=–3 1 x2=0; y2-5=6 1 y2=11.
The terminal point is (0, 11).
3. Use the HMT rule, which states that if an arrow has initial
point 1 x1, y1 2 and terminal point 1 x2, y2 2 , it represents the
vector ˚x2-x1, y2-y1¬. If the initial point P is (4, –3)
and the terminal point Q is 1 x2, y2 2 , the vector PQ is
¡
¡
˚x2-4, y2-(–3)¬. Using the given vector PQ˚2, –4¬,
we have x2-4=2 and y2+3=–4.
x2-4=2 1 x2=6; y2+3=–4 1 y2=–7.
The point Q is (6, –7).
4. Use the HMT rule, which states that if an arrow has initial
point 1 x1, y1 2 and terminal point 1 x2, y2 2 , it represents the
vector ˚x2-x1, y2-y1¬. If the initial point P is 1x1, y1 2
¡
and the terminal point Q is (4, –3), the vector PQ is
¡
˚4-x1, –3-y1¬. Using the given vector PQ˚2, –4¬,
we have 4-x1=2 and –3-y1=–4.
10. After 3 hours, the ship has traveled (3)(42 sin 40°) naut mi
east and (3)(42 cos 40°) naut mi north. Five hours later, it
is (3)(42 sin 40°)+(5)(42 sin 125°)≠253.013 naut mi
east and (3)(42 cos 40°)+(5)(42 cos 125°)
≠–23.929 naut mi north (about 23.93 naut mi south) of
Port Norfolk.
253.013
Bearing: 180°+tan–1 a b L 95.40°
23.929
Distance: 21 253.0132 2 + 1 - 23.929 2 2 L 254.14 naut mi.
Section 6.1 Exercises
For #1–4, recall that two vectors are equivalent if they have
the same magnitude and direction. If R has coordinates (a, b)
¡
and S has coordinates (c, d), then the magnitude of RS is
@ RS @ = 2 1c - a2 2 + 1 d - b2 2 = RS, the distance from
¡
¡
R to S. The direction of RS is determined by the coordinates
(c-a, d-b).
1. If R=(–4, 7) and S=(–1, 5), then, using the HMT rule,
¡
RS =˚–1-(–4), 5-7¬=˚3, –2¬.
If P=(0, 0) and Q=(3, –2), then, using the HMT rule,
¡
PQ=˚3-0, –2 –0¬=˚3, –2¬.
Both vectors represent ˚3, –2¬ by the HMT rule.
2. If R=(7, –3) and S=(4, –5), then, using the HMT rule,
4-x1=2 1 x1=2; –3-y1=–4 1 y1=1.
The point P is (2, 1).
¡
RS =˚4-7, –5-(–3)¬=˚–3, –2¬.
If P=(0, 0) and Q=(–3, –2), then, using the HMT rule,
¡
Quick Review 6.1
1. x=9 cos 30°=
5
9. ¨=180°+tan–1 a b L 248.20°
2
PQ=˚–3 –0, –2 –0¬=˚–3, –2¬.
Both vectors represent ˚–3, –2¬ by the HMT rule.
9 13
, y=9 sin 30°=4.5
2
3. If R=(2, 1) and S=(0, –1), then, using the HMT rule,
¡
15 13
2. x=15 cos 120°=–7.5, y=15 sin 120°=
2
RS =˚0-2, –1 –1¬=˚–2, –2¬.
If P=(1, 4) and Q=(–1, 2), then, using the HMT rule,
3. x=7 cos 220°≠–5.36, y=7 sin 220°≠–4.5
PQ=˚–1-1, 2-4¬=˚–2, –2¬.
Both vectors represent ˚–2, –2¬ by the HMT rule.
4. x=6 cos(–50°)≠3.86, y=6 sin(–50°)≠–4.60
¡
4. If R=(–2, –1) and S=(2, 4), then, using the HMT rule,
For #5–6, use a calculator.
¡
RS =˚2-(–2), 4-(–1)¬=˚4, 5¬.
If P=(–3, –1) and Q=(1, 4), then, using the HMT rule,
5. ¨≠33.85°
6. ¨≠104.96°
¡
For #7–10, the angle determined by P(x, y) involves tan 1y>x2.
Since this will always be between –180° and+180°, you may
need to add 180° or 360° to put the angle in the correct
quadrant.
–1
PQ=˚1-(–3), 4-(–1)¬=˚4, 5¬.
Both vectors represent ˚4, 5¬ by the HMT rule.
¡
5. PQ=˚3-(–2), 4-2¬=˚5, 2¬,
@ PQ @ = 252 + 22 = 229
¡
Section 6.1
¡
6. RS =˚2-(–2), –8-5¬=˚4, –13¬,
27. (a)
@ RS @ = 24 + 1 -132 = 2185
¡
2
2
4
¡
28. (a)
¡
8. PS =˚2-(–2), –8-2¬=˚4, –10¬,
@ PS @ = 242 + 1 -102 2 = 2116=2229
¡
(b)
¡
9. 2 QS =2˚2-3, –8-4¬=˚–2, –24¬,
@ 2 QS @ = 2 1 -2 2 2 + 1 -24 2 2 = 2580 = 2 2145
5
4
5
4
5
i + abj = i j
141
141
141
141
(b) -
@ QR @ = 2 1 -5 2 2 + 12 = 226
233
˚ - 141, - 141¬
¡
7. QR=˚–2-3, 5-4¬=˚–5, 1¬,
Vectors in the Plane
4
3
˚5 , - 5¬
3
4
3
4
i + a - b j = i - j.
5
5
5
5
¡
10. 1 222 PR= 22 ˚–2-(–2), 5-2¬=˚0 , 3 22¬,
¡
@ 22 QR @ = 302 + 13 122 2 = 322
¡
¡
¡
11. 3 QR + PS =3˚–5, 1¬+˚4, –10¬=˚–11, –7¬,
@ 3 QR + PS @ = 21 -11 2 2 + 1 -7 2 2 = 2170
¡
¡
¡
¡
12. PS - 3 PQ=˚4, –10¬-3 ˚5, 2¬=˚–11, –16¬,
@ PS - 3 PQ @ = 21 -11 2 2 + 1 -16 2 2 = 2377
¡
¡
13. ˚–1, 3¬+˚2, 4¬=˚1, 7¬
14. ˚–1, 3¬-˚2, 4¬=˚–3, –1¬
15. ˚–1, 3¬-˚2, –5¬=˚–3, 8¬
17. 2 ˚–1, 3¬+3 ˚2, –5¬=˚4, –9¬
18. 2 ˚–1, 3¬-4 ˚2, 4¬=˚–10, –10¬
20. –˚–1, 3¬-˚2, 4¬=˚–1, –7¬
22.
23.
˚
¬
¬
For #25–28, the unit vector in the direction of v=˚a, b¬ is
1
v =
0v0
a
˚
,
b
¬
2a + b 2a + b2
a
b
i +
j.
=
2
2
2
2a + b
2a + b2
25. (a)
(b)
26. (a)
2
2
2
2
1
˚ 15, 15¬
1
2
i +
j
15
15
3
2
˚ - 113, 113¬
(b) -
3
2
i +
j
113
113
34. |u|= 21 -1 2 2 + 22 = 25, Å=cos–1 a
-1
b ≠116.57°
15
3
35. |u|= 232 + 1 -4 2 2 = 5, Å=360°-cos–1 a b
5
≠306.87°
39. v= 0 v 0
w
5
5
,
≠0.71i+0.71j
24.
=
2
2
2
0w0
25 + 5 25 + 52
˚
3
33. |u|= 232 + 42 = 5, Å=cos–1 a b ≠53.13°
5
For #39–40, first find the unit vector in the direction of u.
Then multiply by the magnitude of v, |v|.
¬
-2
w
-1
,
=
2
2
0w0
21 -1 2 + 1 -2 2 21 -1 2 2 + 1 - 2 2 2
≠–0.45i-0.89j
˚
32. v=˚33 cos 136°, 33 sin 136°¬≠˚–23.74, 22.92¬
38. Since (2 cos 60°)i+(2 sin 60°)j=(|u| cos Å)i
+(|u| sin Å)j, |u|=2 and Å=60°.
¬
v
-1
1
,
=
0v0
212 + 1 -1 2 2 212 + 1 -1 2 2
≠0.71i-0.71j
˚
31. v=˚47 cos 108°, 47 sin 108°¬≠˚–14.52, 44.70¬
37. Since (7 cos 135°)i+(7 sin 135°)j=(|u| cos Å)i
+(|u| sin Å)j, |u|=7 and Å=135°.
19. –2 ˚–1, 3¬-3 ˚2, 4¬=˚–4, –18¬
u
4
-2
,
=
2
2
0u0
2 1 -2 2 + 4 21 -2 2 2 + 42
≠–0.45i+0.89j
30. v=˚14 cos 55°, 14 sin 55°¬≠˚8.03, 11.47¬
36. |u|= 21 -3 2 2 + 1 -52 2 = 234,
-3
Å=360°-cos–1 a
b ≠239.04°
134
16. 3 ˚2, 4¬=˚6, 12¬
21.
29. v=˚18 cos 25°, 18 sin 25°¬≠˚16.31, 7.61¬
˚
=2
#
u
-3
3
= 2
,
0u0
232 + 1 -3 2 2 232 + 1 -3 2 2
˚
¬
˚
¬
12 - 12
,
=˚ 12, - 12¬
2
2
40. v= 0 v 0
u
-5
7
= 5
,
0u0
21 -52 2 + 72 2 1 -5 2 2 + 72
≠5˚–0.58, 0.81¬=˚–2.91, 4.07¬
#
¬
41. A bearing of 335° corresponds to a direction angle of 115°.
v=530 ˚cos 115°, sin 115°¬≠˚–223.99, 480.34¬
42. A bearing of 170° corresponds to a direction angle of –80°.
v=460 ˚cos (–80°), sin (–80°)¬≠˚79.88, –453.01¬
43. (a) A bearing of 340° corresponds to a direction angle of
110°. v=325 ˚cos 110°, sin 110°¬≠˚–111.16, 305.40¬
(b) The wind bearing of 320° corresponds to a direction
angle of 130°. The wind vector is
w=40 ˚cos 130°, sin 130°¬ ≠˚–25.71, 30.64¬.
Actual velocity vector: v+w ≠˚–136.87, 336.04¬.
Actual speed: ||v+w||≠ 2136.872 + 336.042
≠362.84 mph.
336.04
b
Actual direction: ¨=180°+tan–1 a
- 136.87
≠112.16°, so the bearing is about 337.84°.
234
Chapter 6
Applications of Trigonometry
44. (a) A bearing of 170° corresponds to a direction angle
of –80°:
v=460 ˚cos (–80°), sin (–80°)¬≠˚79.88, –453.01¬.
(b) The wind bearing of 200° corresponds to a direction
angle of –110°. The wind vector is
w=80 ˚cos (–110°), sin (–110°)¬ ≠˚–27.36, –75.18¬.
Actual velocity vector: v+w ≠˚52.52, –528.19¬.
Actual speed: ||v+w||≠ 252.522 + 528.192
≠530.79 mph.
52.52
Actual direction: ¨=180°+tan–1 ab
528.19
≠–84.32°, so the bearing is about 174.32°.
45. (a) v=10 ˚cos 70°, sin 70°¬≠˚3.42, 9.40¬
(b) The horizontal component is the (constant) horizontal
speed of the basketball as it travels toward the basket.
The vertical component is the vertical velocity of the
basketball, affected by both the initial speed and the
downward pull of gravity.
46. (a) v=2.5 ˚cos 15°, sin 15°¬≠˚2.41, 0.65¬
(b) The horizontal component is the force moving the box
forward. The vertical component is the force moving
the box upward against the pull of gravity.
47. We need to choose w=˚a, b¬=k ˚cos 33°, sin 33°¬, so
that k cos(33°-15°)=k cos 18°=2.5. (Redefine
“horizontal” to mean the parallel to the inclined plane;
then the towing vector makes an angle of 18° with the
2.5
“horizontal.”) Then k=
≠2.63 lb, so that
cos 18°
w≠˚2.20, 1.43¬.
48. Juana’s force can be represented by
23 ˚cos 18°, sin 18°¬≠˚21.87, 7.11¬, while Diego’s force
is 27 ˚cos(–15)°, sin (–15°)¬≠˚26.08, –6.99¬. Their
total force is therefore ˚47.95, 0.12¬, so Corporal must be
pulling with an equal force in the opposite direction:
˚–47.95, –0.12¬. The magnitude of Corporal’s force is
about 47.95 lb.
49. F=˚50 cos 45°, 50 sin 45°¬+˚75 cos (–30°),
75 sin (–30°)¬≠˚100.31, –2.14¬, so |F|≠100.33 lb
and ¨≠–1.22°.
50. F=100˚cos 50°, sin 50°¬+50˚cos 160°, sin 160°¬
+80˚cos(–20), sin(–20°)¬≠˚92.47, 66.34¬, so
|F|≠113.81 lb and ¨≠35.66°.
53. Let w be the speed of the ship. The ship’s velocity
(in still water) is ˚w cos 270°, w sin 270°¬=˚0, –w¬.
Let z be the speed of the current. Then, the current
velocity is ˚z cos 135°, z sin 135°¬≠˚–0.71z, 0.71z¬.
The position of the ship after two hours is
˚20 cos 240°, 20 sin 240°¬≠˚–10, –17.32¬. Putting all
this together we have:
2˚0, –w¬+2˚–0.71z, 0.71z¬ =˚–10, –17.32¬,
˚–1.42z, –2w+1.42z¬=˚–10, –17.32¬, so z≠7.07
and w≠13.66. The speed of the ship is about 13.66 mph,
and the speed of the current is about 7.07 mph.
54. Let u=˚u1, u2¬, v=˚v1, v2¬, and w=˚w1, w2¬.
(a) u+v=˚u1, u2¬+˚v1, v2¬=˚u1+v1, u2+v2¬
=˚v1+u1, v2+u2¬=˚v1, v2¬+˚u1, u2¬=v+u
(b) (u+v)+w=˚u1+v1, u2+v2¬+˚w1, w2¬
=˚u1+v1+w1, u2+v2+w2¬
=˚u1, u2¬+˚v1+w1, v2+w2¬
=u+(v+w)
(c) u+0=˚u1, u2¬+˚0, 0¬=˚u1+0, u2+0¬
=˚u1, u2¬=u
(d) u+(–u)=˚u1, u2¬+˚–u1, –u2¬
=˚u1+(–u1), u2+(–u2)¬=˚0, 0¬=0
(e) a(u+v)=a˚u1+v1, u2+v2¬
=˚a(u1+v1), a(u2+v2)¬
=˚au1+av1, au2+av2¬=˚au1, au2¬+˚av1, av2¬
=a ˚u1, u2¬+a ˚v1, v2¬=au+av
(f) (a+b)u=˚(a+b)u1, (a+b)u2¬
=˚au1+bu1, au2+bu2¬
=˚au1, au2¬+˚bu1, bu2¬=a ˚u1, u2¬+b ˚u1, u2¬
=au+bu
(g) (ab)u=˚(ab)u1, (ab)u2¬=˚a(bu1), a(bu2)¬
=a ˚bu1, bu2¬=a(bu)
(h) a0=a ˚0, 0¬=˚a0, a0¬=˚0, 0¬=0
0u=0 ˚u1, u2¬=˚0u1, 0u2¬=˚0, 0¬=0
(i) (1)u=˚(1)u1, (1)u2¬=˚u1, u2¬=u
(–1)u=˚(–1)u1, (–1)u2¬=˚–u1, –u2¬=–u
(j) |au|=|˚au1, au2¬|= 2 1au1 2 2 + 1au2 2 2
= 2a2u21 + a2u22 = 2a2 1u21 + u22 2
= 0 a 0 2u21 + u22 = |a| |u|
55. True. Vectors u and –u have the same length but opposite
directions. Thus, the length of –u is also 1.
51. Ship heading: ˚12 cos 90°, 12 sin 90°¬=˚0, 12¬
Current heading: ˚4 cos 225°, 4 sin 225°¬≠˚–2.83, –2.83¬
The ship’s actual velocity vector is ˚–2.83, 9.17¬, so its
speed is≠ 21 -2.83 2 2 + 9.172≠9.6 mph and the
-2.83
direction angle is cos–1 a
b ≠107.14°, so the
9.6
bearing is about 342.86°.
56. False. 1/u is not a vector at all.
52. Let v=˚0, 20¬ be the velocity of the boat and w=˚8, 0¬
be the velocity vector of the current. If the boat travels
t minutes to reach the opposite shore, then its position, in
vector form, must be
59. The x-component is 3 cos 30°, and the y-component is
3 sin 30°. The answer is A.
˚0, 20t¬+˚8t, 0¬=˚8t, 20t¬=˚8t, 1¬.
1
So 20t=1 1 t=
1 8t=0.4 mi. The boat meets
20
the shore 0.4 mi downstream.
57. 0 82, -19 0 = 222 + 1 -12 2 = 15
The answer is D.
58. u - v = 8 -2, 3 9 - 8 4, -19
= 8 -2 - 4, 3 - 1 -12 9
= 8 -6, 49
The answer is E.
60. 0 v 0 = 11 - 12 2 + 32 = 110, so the unit vector is
8 -1, 3 9> 110. The answer is C.
Section 6.2
OC = t OA + 1 1 - t 2 OB. We also know
¡
¡
(b1, b2). Then OA is the vector ˚a1, a2¬ and OB is
the vector ˚b1, b2¬.
¡
¡
¡
¡
¡
¡
¡
¡
64.
¡
0 BC 0
¡
=OC+y
¡
# CA+y CB
0 CA 0
¡
¡
#
CA
since x=y
@ [email protected]
@ [email protected]
¡
A
P
BC
0 BC 0
¡
¡
+ CB b
¡
=OC+y (BC+CB)
¡
=OC
¡
¡
62. (a) By Exercise 61, OM1=x OA+y OB, where
x+y=1. Since M1 is the midpoint, ƒ BM1 ƒ = 0 M1A 0 .
We know from Exercise 61, however that
¡
@ BM1 @
x
1
= 1 = . So x=y= . As a result,
¡
y
2
@ M1A @
¡
¡
¡
¡
1
1
OM1= OA +
OB. The proof for OM2
2
2
¡
¡
¡
■ Section 6.2 Dot Product of Vectors
Exploration 1
1. u=˚–2-x, 0-y¬=˚–2-x, –y¬
v=˚2-x, 0-y¬=˚2-x, –y¬
2. u # v=(–2-x)(2-x)+(–y)(–y)=–4+x2+y2
and OM3 are similar.
=–4+4=0
¡
1 ¡
1¡
(b) 2 OM1 + OC = 2 a OA + OB b + OC
2
2
¡
¡
¡
¡
Therefore, ¨=90°.
3. Answers will vary.
¡
=OA+OB+OC. Use the same method for
the other proofs.
¡
¡
¡
¡
(c) Part (b) implies that 2 OM1 + OC = 2 OM2 + OA
¡
¡
= 2 OM3 + OB. Each of the three vectors lies
along a different median (that is, if nonzero, the
three vectors have three different directions).
Hence they can only be equal if all are equal to 0.
¡
¡
¡
¡
¡
0 OM1 0
0 OM2 0
¡
=
0 BC 0
¡
¡
0 OM3 0
¡
¡
=
¡
2. |u|= 2 1 -3 2 2 + 1 -42 2 = 5
5. AB=˚1-(–2), 13-0¬=˚3, 13¬
=
1
.
2
0 OC 0
0 OA 0
0 OB 0
63. Use the result of Exercise 61. First we show that if C is on
the line segment AB, then there is a real number t so
¡
1. |u|= 222 + 1 -3 2 2 = 113
4. |u|=2 2cos2 75° + sin2 75° = 2
¡
¡
Quick Review 6.2
3. |u|= 2cos2 35° + sin2 35° = 1
Thus 2OM1 = -OC, 2OM2 = -OA, and
2OM3 = - OB, so
x
The line segment OM is a median of ^ABO since M is a
midpoint of AB. The line segment AQ is a median of
^ABO since diagonals of a parallelogram bisect each other.
By the result of Exercise 62, since P is the intersection
AP
2
CR
2
point of 2 medians,
= . Similarly,
= . This
PQ
1
RQ
1
implies that AP=PR=RC, so the diagonal has been
trisected.
¡
¡
10
0
¡
=OC+y a 0 BC 0 #
¡
N
Q
C
@ CA @
same direction.
¡
M
R
is a unit vector, and BC points in the
¡
¡
B
¡
¡
¡
¡
0 CA 0
¡
¡
¡
+ CB b
CA
¡
y
¡
=OC+y a 0 BC 0
¡
¡
¡
¡
10
¡
=OC+x CA+y CB (since x+y=1)
¡
¡
¡
¡
=(x+y) OC+x CA+y CB
¡
¡
¡
Hence BC and AC have the same or opposite directions, so
C must lie on the line L through the two points A and B.
=x OC+x CA+y OC+y CB
¡
¡
¡
t 1OA - OB 2 = BC and 1t - 1 2 1OA - OB 2 = AC.
¡
(b) x OA+y OB=x(OC+CA)+y(OC+CB)
(from part (a))
¡
¡
¡
¡
¡
¡
t OA + 1 1 - t 2OB = OA + AC. Therefore,
=OA-OB
¡
¡
t OA + 1 1 - t 2OB = OB + BC and
¡
¡
¡
¡
So, BA=˚a1-b1, a2-b2¬
=˚a1, a2¬-˚b1, b2¬
¡
¡
OC = OB + BC and OC = OA + AC. So we have
¡
¡
235
Suppose there is a real number t so that
61. (a) Let A be the point (a1, a2) and B be the point
¡
Dot Product of Vectors
6. AB=˚1-2, 13-0¬=˚–1, 13¬
¡
7. AB=˚1-2, - 13 - 0¬=˚–1, - 13¬
¡
8. AB=˚1-(–2), - 13-0¬=˚3, - 13¬
9. u=|u| #
¡
t
. (Convince yourself that
that ¡ =
1 - t
0 CA 0
¡
¡
¡
BC
t=
works.) Then OC=t OA + 1 1 - t 2 OB.
BC + CA
A similar argument can be used in the cases where B is
on the line segment AC or A is on the line segment BC.
v
@ [email protected]
=
2 # 2, 3
2
˚
10. u=|u| #
v
@ [email protected]
4, 6
=
113
22 + 3
4
6
,
=
113 113
2
=
¬
3 # -4, 3
21 -4 2 + 3
12 9
= - ,
5 5
˚
2
2
¬
=
-12, 9
5
236
Chapter 6
Applications of Trigonometry
24. u # v=˚–4, –1¬ # ˚1, –4¬ = -4112 + 1 -1 2 1 -4 2
= -4 + 4 = 0
Since u # v=0, u and v are orthogonal.
Section 6.2 Exercises
1. 60+12=72
2. –40+26=–14
For #25–28, first find projvu. Then use the fact that u # v=0
when u and v are orthogonal.
3. –12-35=–47
4. 10-56=–46
25. projvu= a
5. 12+18=30
6. –16-28=–44
-8, 3 # -6, -2
b -6, -2
36 + 4
21
42
b -6, -2 =
-6, -2
40
20
21
3, 1
=10
21
17
3, 1 +
-1, 3
u= 10
10
=a
7. –14+0=–14
8. 0+33=33
9. |u|= 1u # u = 125 + 144 = 13
10. |u|= 1u # u = 164 + 225 = 17
11. |u|= 1u # u = 116 = 4
12. |u|= 1u # u = 19 = 3
13. u # v=4-15=–11, |u|= 116
+ 9 = 5,
-11
b L 115.6°
|v|= 11 + 25 = 126, ¨=cos a
5 126
3, -7 # -2, -6
b -2, -6
4 + 36
9
36
= a b -2, -6 = - 1, 3
40
5
9
8
u= - 1, 3 + 3, -1
5
5
26. projvu= a
-1
14. u # v= -6 + 6 = 0, u = 90 °
15. u # v=–6+15=9,
|u|= 14 + 9 = 113,
|u|= 19 + 25 = 134,
9
b L 64.65 °
¨=cos-1 a
113 # 134
16. u # v=–30-2=–32, |u|= 125 + 4 = 129,
0 v 0 = 136 + 1 = 137,
- 32
b L 167.66 °
¨=cos-1 a
129 # 137
17. u # v=–6-6 13, 0 u 0 = 19 + 9 = 118,
|v|= 14 + 12 = 116 = 4,
-6 - 6 13
b =165°
¨=cos-1 a
118 # 4
18. u # v=0, ¨=90°
p
3p
and v has direction angle
4
2
p
(which is equivalent to - ), so the angle between the
2
p
p
3p
- a- b =
vectors is
or 135°.
4
2
4
8, 5 # - 9, -2
b -9, -2
81 + 4
82
-82
b -9, -2 =
9, 2
=a
85
85
82
29
u= 9, 2 +
-2, 9
85
85
27. projvu= a
-2, 8 # 9, -3
b 9, -3
81 + 9
7
-42
b 9, -3 = -3, 1
=a
90
5
7
1
u= -3, 1 + 11, 33
5
5
28. projvu= a
29.
y
10
A(1, 10)
19. u has direction angle
p
5p
and v has direction angle
,
3
6
5p
p
p
=
so the angle between the vectors is
or 90°.
6
3
2
B(–4, 5)
C(3, 1)
x
5
20. u has direction angle
21. u # v=–24+20=–4, 0 u 0 = 164 + 25 = 189,
0 v 0 = 19 + 16 = 5,
-4
¨ = cos-1 a
b L 94.86°
5 189
22. u # v=3-72=–69, 0 u 0 = 19 + 64 = 173,
0 v 0 = 11 + 81 = 182,
-69
¨ = cos-1 a
b L 153.10°
173 # 182
3
3
23. u # v=˚2, 3¬ # h , -1 i = 2 a b + 3 1 -1 2
2
2
= 3 - 3 = 0
Since u # v=0, u and v are orthogonal.
¡
¡
CA # CB=˚1-3, 10-1¬ # ˚–4-3, 5-1¬
=˚–2, 9¬ # ˚–7, 4¬=14+36=50,
0 CA 0 = 14 + 81 = 185, 0 CB 0 = 149 + 16 = 165,
40
b L 47.73°
C=cos–1 a
185 # 165
¡ ¡
BC # BA=˚7, –4¬ # ˚1-(–4), 10-5¬
¡
¡
=˚7, –4¬ # ˚5, 5¬=35-20=15,
0 BC 0 = 165, 0 BA 0 = 150, B=cos–1 a
¡
¡
15
b
165 # 150
≠74.74°
A=180°-B-C≠180°-74.74°-47.73°=57.53°
Section 6.2
30.
¡
y
AP=
2
A(–4, 1)
Dot Product of Vectors
237
3, -4
3 4
u
, - or
=
=
5
5
ƒuƒ
19 + 16
˚
¬
-3, 4
v
3 4
= - , .
=
5 5
ƒvƒ
19 + 16
So, P is (4.6, –0.8) or (3.4, 0.8).
x
5
B(5, –1)
¡
˚
AP =
¬
40. (a) A is (–5, 0) and B is (0, 2).
(b) The line is parallel to
C(1, –6)
¡
AB=˚0-(–5), 2-0¬=˚5, 2¬, so the direction
¡
¡
¡
CA # CB=˚–4-1, 1-(–6)¬ # ˚5-1, –1-(–6)¬
=˚–5, 7¬ # ˚4, 5¬=–20+35=15,
0 CA 0 = 125 + 49 = 174, 0 CB 0 = 116 + 25 = 141,
¡
¡
C=cos–1 a
15
b L 74.20°
174 # 141
¡ ¡
BA # BC=˚–4-5, 1-(–1)¬ # ˚1-5, –6-(–1)¬
=˚–9, 2¬ # ˚–4, –5¬=36-10=26,
0 BA 0 = 181 + 4 = 185, 0 BC 0 = 141,
¡
¡
B=cos–1 a
26
b ≠63.87°
185 # 141
A=180°-B-C≠180°-63.87°-74.20°=41.93°
of AP is u=˚–2, 5¬ or v=˚2, –5¬.
-2, 5
2
5
or
= ,
14 + 25
129 129
¡
2, -5
2
5
.
AP =
=
,=
129
129
129
@ [email protected]
2
5
,
b ≠(–5.37, 0.93) or
So P is a -5 129 129
2
5
a -5 +
,b ≠(–4.63, –0.93).
129
129
41. (a) A is (7, 0) and B is (0, –3).
¡
AP=
u
@ [email protected]
v
˚
=
¬
˚
¬
(b) The line is parallel to
¡
AB=˚0-7, –3-0¬=˚–7, –3¬, so the direction
¡
For #31–32, use the relationship u # v=|u| |v| cos ¨.
of AP is u=˚3, –7¬ or v=˚–3, 7¬.
31. u # v = 3 # 8 cos 150° L -20.78
AP=
32. u # v = 12 # 40 cos a
¡
p
b = 240
3
˚
33. Parallel: –2 -
10
3
10
=
,, 3 =˚5, 3¬
4
2
2
¬ ˚
¬
40
Z 0 and
3
4
3
3 10 4
Zu
v =
, = 2,
5
5 3 3
5
3
7
˚
¡
of AP is u=˚3, 6 or v=˚–3, –6.
¡
¬
36. Orthogonal: u # v = -60 + 60 = 0
AP=
1
1
38. Parallel: - v = - -4, 14 = 2, -7 = u
2
2
¡
For #39–42 part (b), first find the direction(s) of AP and then
find the unit vectors. Then find P by adding the coordinates of
A to the components of a unit vector.
39. (a) A is (4, 0) and B is (0, –3).
(b) The line is parallel to
¡
AB=˚0-4, –3-0¬=˚–4, –3¬, so the direction
u
@ [email protected]
=
3, 6
19 + 36
=
3, 6
315
˚ 15, 15 or
=
1
2
-3, - 6
-3, -6
v
=
=
19 + 36
3 15
@ [email protected]
1
1
2
2
. So P is a 6 +
= ,,
b
15
15
15 15
2
1
,b ≠(5.55, –0.89).
≠(6.45, 0.89) or a 6 15
15
3
43. 2v1+3v2=10, v12+v22=17. Since v1=5- v2,
2
2
3
9
a 5 - v2 b +v22=17, 25-15v2+ v22 +v22=17,
2
4
13 2
v2 -15v2+8=0, 13v22-60v2+32=0,
4
8
(v2-4)(13v2-8)=0, so v2=4 or v2= .
13
53 8
≠˚4.07, 0.62¬.
,
Therefore, v≠˚–1, 4¬ or v=
13 13
¡
AP =
˚
37. Orthogonal: u # v = -60 + 60 = 0
¡
˚ 158, - 158 or
¡
35. Neither: u # v = -120 Z 0 and
-75
-15
-15
Zu
˚–4, 5¬= 15,
v =
4
4
4
of AP is u=˚3, –4¬ or v=˚–3, 4¬.
=
AB=˚0-6, 3-0=˚–6, 3, so the direction
¬ ˚ ¬
˚
3, -7
(b) The line is parallel to
34. Neither: u # v =
˚
@ [email protected]
v
=
19 + 49
-3, 7
3
7
.
AP =
= ,
=
158
158 158
@ [email protected]
7
3
,b ≠(–7.39, 0.92) or
So P is a 7 +
158
158
3
7
a7 ,
b ≠(6.61, 0.92).
158 158
42. (a) A is (6, 0) and B is (0, 3).
¡
For #33–38, vectors are orthogonal if u # v = 0 and are
parallel if u=kv for some constant k.
u
˚
¬
238
Chapter 6
Applications of Trigonometry
5
11
44. –2v1+5v2=–11, v12+v22=10. Since v1= v2+ ,
2
2
2
1
a 15v2 + 112 b +v22=10,
2
25v2 2
110v2
121
+
+
+v22=10,
4
4
4
110
81
29 2
v +
v +
= 0, 29v22+110v2+81=0,
4 2
4 2
4
81
(v2+1)(29v2+81)=0, so v2=–1 or v2=– .
29
43
81
Therefore, v=˚3, –1¬ or v= - , 29
29
≠˚–1.48, –2.79¬.
˚
¬
1
13
45. v=(cos 60°)i+(sin 60°)j= i+
j
2
2
1
13
jb b v
F1=projvF=(F # v)v= a - 160 j # a i +
2
2
= - 80 13 a
1
13
i +
j b = -40 13 i - 120 j.
2
2
The magnitude of the force is
ƒ F1 ƒ = 21 - 40132 2 + 1 -120 2 2 = 119,200
≠138.56 pounds.
46. In this case, F=–125 j and v remains the same as in
Example 6.
12
b v=62.5 (i+j).
F1=projvF=(F # v)v=–125 a
2
The magnitude of the force is |F1|=62.5 12≠88.39
pounds.
47. (a) v=(cos 12°)i+(sin 12°)j
F=–2000j
F1=projvF=(F # v)v
=(˚0, –2000¬ # ˚cos 12°, sin 12°¬) ˚cos 12°, sin 12°¬
=(–2000 sin 12°)˚cos 12°, sin 12°¬.
Since ˚cos 12°, sin 12°¬ is a unit vector,the
magnitude of the force being extended is
|F1|=2000 sin 12°≠415.82 pounds.
(b) We are looking for the gravitational force exerted
perpendicular to the street. A unit vector perpendicular to the street is w=˚cos (–78°), sin (–78°)¬,
so F2=projwF=(F # w)w
=(–2000 sin (–78°)) ˚cos (–78°), sin (–78°)¬
Since ˚cos (–78°), sin (–78°)¬ is a unit vector, the
magnitude of the force perpendicular to the street
is –2000 sin (–78°)≠1956.30 pounds.
48. We want to determine “how much” of the 60 pound
force is projected along the inclined plane.
F=60 ˚cos 43°, sin 43°¬≠˚43.88, 40.92¬ and
v=˚cos 18°, sin 18°¬ ≠˚0.95, 0.31¬
1 43.88, 40.92 # 0.95, 0.312 0.95, 0.31
projv F=
1 112 2
54.38 0.95, 0.31
≠
≠˚51.72, 16.80¬. The magnitude of
1
this force is 0 F1 0 = 2151.72 2 2 + 1 16.80 2 2
≠54.38 pounds. Of note, it is also possible to evaluate
this problem considering the x-axis parallel to the inclined
plane and the y-axis perpendicular to the plane. In this
case F=60 ˚cos 25°, sin 25°¬≠˚54.38, 25.36¬. Since we
only want the force in the x-direction, we immediately
find our answer of about 54.38 pounds.
49. Since the car weighs 2600 pounds, the force needed to lift
the car is ˚0, 2600¬.
¡
W=F # AB = ˚0, 2600¬ # ˚0, 5.5¬=14,300 foot-pounds
50. Since the potatoes weigh 100 pounds, the force needed to
lift the potatoes is ˚0, 100¬.
¡
W=F # AB = ˚0, 100¬ # ˚0, 3¬=300 foot-pounds
1 ,2
12
˚1, 2¬
=
0 1, 2 0
15
¡
12
48
W=F # AB =
˚1, 2¬ # ˚4, 0¬=
15
15
≠21.47 foot-pounds
51. F=12 #
4, 5
24
24
˚4, 5¬≠
˚4, 5¬
=
0 4, 5 0
141
242 + 52
¡
24
120
W=F # AB =
˚5, 0¬=
141
141
¡
24
480
W=F # AB =
˚4, 5¬ # ˚5, 0¬=
141
141
≠74.96 foot-pounds
52. F=24 #
2, 2
30
˚2, 2¬=15 12 ˚1, 1¬
=
2
0 2, 2 0
22 + 22
1
Since we want to move 3 feet along the line y= x, we
2
solve for x and y by using the Pythagorean theorem:
1 2
5
x2+y2=32, x2+ a x b =9, x2=9,
2
4
6
3
x=
, y=
15
15
¡
6
3
AB=
,
15 15
¡
6
3
,
W=F # AB = ˚15 12, 15 12¬ #
15 15
2
=135
=27 110≠85.38 foot-pounds
B5
53. F=30 #
˚
¬
˚
¬
2, 3
50
50
=
˚2, 3¬=
˚2, 3¬
0 2, 3 0
113
222 + 32
Since we want to move the object 5 feet along the line
y=x, we solve for x and y by using the Pythagorean
theorem: x2+y2=5¤, x2+x2=25, 2x2=25,
x=2.5 12, y=2.5 12.
54. F=50 #
¡
AB=˚2.5 12, 2.5 12¬.
¡
50
2
W=F # AB =
˚2, 3¬ # ˚2.5 12, 2.5 12¬=625
B 13
113
≠245.15 foot-pounds
55. W=F # AB = 0 F 0 0 AB 0 cos ¨=200 113 cos 30°
13
= 100139 L 624.5 foot-pounds
=200 113 #
2
¡
¡
¡
56. AB=˚4, 3¬-˚–1, 1¬=˚5, 2¬
W=F # AB = 0 F 0 0 AB 0 cos ¨=75 129 cos 60°
1
75 129
L 201.9 foot-pounds
=75 129 # =
2
2
¡
¡
Section 6.2
57. (a) Let u=˚u1, u2¬, v=˚v1, v2¬, and w=˚w1, w2¬.
0 # u=˚0, 0¬ # ˚u1, u2¬=0 # u1+0 # u2=0
(b) u # (v+w)=˚u1, u2¬ # (˚v1, v2¬+˚w1, w2¬)
=˚u1, u2¬ # ˚v1+w1, v2+w2¬
=u1(v1+w1)+u2(v2+w2)
=u1v1+u2v2+u1w1+u2w2
=˚u1, u2¬ # ˚v1, v2¬+˚u1, u2¬ # ˚w1, w2¬
=u # v+u # w
(c) (u+v) # w=(˚u1, u2¬+˚v1, v2¬) # ˚w1, w2¬
=˚u1+v1, u2+v2¬ # ˚w1, w2¬
=(u1+v1) w1+(u2+v2)w2
=u1w1+u2w2+v1w1+v2w2
=˚u1, u2¬ # ˚w1, w2¬+˚v1, v2¬ # ˚w1, w2¬
=u # w+v # w
(d) (cu) # v=(c˚u1, u2¬) # ˚v1, v2¬=˚cu1, cu2¬ # ˚v1, v2¬
=cu1v1+cu2v2=˚u1, u2¬ # ˚cv1, cv2¬=u # (cv)
=c(u1v1+u2v2)=c(˚u1, u2¬ # ˚v1, v2¬)=c(u # v)
58. (a) When we evaluate the projection of u onto v we are
actually trying to determine “how much” of u is
“going” in the direction of v. Using Figure 6.19,
imagine that v runs along our x-axis, with the y-axis
perpendicular to it. Written in component form,
u=(|u| cos ¨, |u| sin ¨) and we see that the projection
of u onto v is exactly |u| cos ¨ times v’s unit vector
v
v
. Thus, projvu=|u| cos ¨ #
0v0
0v0
=0u0 a
1u # v 2 v
u#v
v
u#v
b
=
= a 2bv
0u0 0v0 0v0
0v0 0v0
0v0
¡
(b) Recall Figure 6.19 and let w1 be PR=projvu
¡
and w2=RQ=u-projvu. Then,
(u-projvu) # (projvu)=w2 # w1. Since w1 and w2 are
perpendicular, w1 # w2=0.
59.
v
Dot Product of Vectors
239
63. u # v=0, so the vectors are perpendicular.
The answer is D.
64. u # v=˚4, –5¬ # ˚–2, –3¬
=4(–2)+(–5)(–3)
=–8+15
=7
The answer is C.
u#v
bv
0v02
3 + 0
=a
b ˚2, 0¬
22
3
= a b ˚2, 0¬
4
3
=k h , 0 i
2
The answer is A.
65. projvu= a
1
˚–1, 1¬. The
12
force is represented by 5 times the unit vector. The answer
is B.
66. The unit vector in the direction of u is
67. (a) 2 # 0+5 # 2=10 and 2 # 5+5 # 0=10
¡
(b) AP=˚3-0, 7-2¬=˚3, 5¬
¡
AB=˚5-0, 0-2¬=˚5, –2¬
¡
3, 5 # 5, -27
¡
AP= a 2
b ˚5, –2¬
w1=projAB
5 + 1 -22 2
15 - 10
5
=a
b ˚5, –2¬= ˚5, –2¬
29
29
¡
¡
¡
w2=AP-projAB
AP
5
=˚3, 5¬- ˚5, –2¬
29
25
10
1
= 3- , 5+
= ˚62, 155¬
29
29
29
˚
¬
¡
u-v
(c) w2 is a vector from a point on AB to point P. Since w2
¡
u
is perpendicular to AB, |w2| is the shortest distance
¡
u+v
As the diagram indicates, the long diagonal of the parallelogram can be expressed as the vector u+v while the
short diagonal can be expressed as the vector u-v. The
sum of the squares of the diagonals is
|u+v|2+|u-v|2=(u+v) # (u+v)
+(u-v) # (u-v)=u # u+u # v+v # u+v # v
+u # u-u # v-u # v+v # v=2|u|2+2|v|2
+2u # v-2u # v=2|u|2+2|v|2, which is the sum of the
squares of the sides.
60. Let u=˚u1, u2¬.
(u # i)i+(u # j)j
=(˚u1, u2¬ # ˚1, 0¬)i+(˚u1, u2¬ # ˚0, 1¬)j
=(u1)i+(u2)j
=u1i+u2 j
=˚u1, u2¬
=u
61. False. If either u or v is the zero vector, then u # v=0 and
so u and v are orthogonal, but they do not count as
perpendicular.
62. True. u # u= 0 u 0 2 = 1 1 2 2 = 1.
from AB to P.
62 2
155 2
27,869
31 129
b =
=
|w2|= a b + a
B 29
29
B 292
29
(d) Consider Figure 6.19. To find the distance from a
point P to a line L, we must first find u1=projvu. In
this case,
¡
x0, y0 - 2 # 5, -2
¡
AP= a
b ˚5, –2¬
projAB
1 252 + 1 -22 2 2 2
5x0 - 2y0 + 4
b ˚5, –2¬
=a
29
25x0 - 10y0 + 20 -10x0 + 4y0 - 8
,
and
=
29
29
˚
¡
¬
¡
¡
AP-projAB
AP=˚x0, y0-2¬
25x0 - 10y0 + 20 -10x0 + 4y0 - 8
,
29
29
1
= (˚29x0, 29 (y0-2)¬
29
-˚25x0-10y0+20, –10x0+4y0-8¬)
1
= ˚4x0+10y0-20, 10x0+25y0-50¬
29
˚
¬
240
Chapter 6
Applications of Trigonometry
So, the distance is the magnitude of this vector.
1
d= 214x0 + 10y0 - 20 2 2 + 1 10x0 + 25y0 - 50 2 2
29
22 2 1 2x0 + 5y0 - 10 2 2 + 52 12x0 + 5y0 - 10 2 2
=
29
229 1 2x0 + 5y0 - 10 2
=
Exploration 1
1.
2
29
0 12x0 + 5y0 - 102 0
=
129
¡
˚
¡
AP= x0, y0a
˚
=
c -c
˚a, b ¬ and
(e) In the general case, AB=
=
■ Section 6.3 Parametric Equations and
Motion
[–10, 5] by [–5, 5]
2. 0.5(17)+1.5=10, so the point (17, 10) is on the graph,
t=–8
c
¡
, so projAB
AP
b
¡
¬
1 bcy0 - c2 2
x0c
c
c
bh ,- i
a
a b
b2
3. 0.5(–23)+1.5=–10, so the point (–23, –10) is on the
graph, t=12.
a
1
- .
2
2
Alternatively, b=2-t, so t=2-b.
2
2
2 c 2 + 2 -c 2
a
b
-abx0 + a2y0 -
2
b x0 - aby0 + ac
a2 + b2
4. x=a=1-2t, 2t=1-a, t =
,
a2c
b
a2 + b2
¡
¡
c
¡
AP-projAB
AP= x0, y0b
˚
¬
Exploration 2
¬
1. It looks like the line in Figure 6.32.
a2c
-abx0 + a2y0 2
b x0 - aby0 + ac
b
,
a2 + b2
a2 + b2
a2x0 + aby0 - ac abx0 + b2 y0 - bc
=
,
a2 + b2
a2 + b2
˚
˚
5. Choose Tmin and Tmax so that
Tmin -2 and Tmax 5.5.
¬
¬
The magnitude of this vector, @ AP-projAB AP @ , is
ax0 + by0 - c
the distance from point P to L:
.
2a2 + b2
¡
2. The graph is a vertical line segment that extends from
(400, 0) to (400, 20).
3. For 19° and 20°, the ball does not clear the fence, as
shown below.
19°:
¡
¡
68. (a) Yes, if v=˚0, 0¬ or t=n∏, n=any integer
(b) Yes, if u=˚0, 0¬ or t=
np
, n=odd integer
2
[0, 450] by [0, 80]
20°:
(c) Generally, no, because sin t Z cos t for most t.
p
Exceptions, however, would occur when t= + np,
4
n=any integer, or if u=˚0, 0¬ and/or v=˚0, 0¬.
69. One possible answer:
If au+bv=cu+dv
au-cu+bv-dv=0
(a-c)u+(b-d)v=0
Since u and v are not parallel, the only way for this equality to hold true for all vectors u and v is if (a-c)=0
and (b-d)=0, which indicates that a=c and b=d.
[0, 450] by [0, 80]
For 21° and 22°, the ball clears the fence, as shown below.
21°
[0, 450] by [0, 80]
22°:
[0, 450] by [0, 80]
Section 6.3
Quick Review 6.3
Parametric Equations and Motion
(b)
241
y
¡
1. (a) OA =˚–3, –2¬
5
¡
(3, 4)
¡
a4,
(b) OB =˚4, 6¬
(c) AB =˚4-(–3), 6-(–2)¬=˚7, 8¬
¡
5
b
2
2. (a) OA =˚–1, 3¬
¡
x
8
a0, – 1 b
2
¡
(b) OB =˚4, –3¬
(1, –2)
(c) AB =˚4-(–1), –3-3¬=˚5, –6¬
6 - 1 -2 2
8
=
4 - 1 -3 2
7
8
8
y+2= 1x + 3 2 or y - 6 = 1x - 4 2
7
7
3. m =
6. (a) t
-3 - 3
6
= 4 - 1 -1 2
5
6
6
y-3= - 1x + 1 2 or y + 3 = - 1x - 42
5
5
x
4. m =
y
p
2
0
∞
1
∞
0
0
∞
–1
1
(b)
5. Graph y = ; 18x.
∏
0
∞
3p
2
0
–1
2∏
∞
1
0
y
2
(0, 1)
(–1, 0)
(1, 0)
2
x
(0, –1)
[–3, 7] by [–7, 7]
6. Graph y = ; 1-5x.
7.
[–10, 10] by [–10, 10]
[–7, 2] by [–7, 7]
2
2
8.
7. x +y =4
2
2
8. (x+2) +(y-5) =9
9.
600 rotations 1 min
#
# 2p rad = 20p rad>sec
1 min
60 sec 1 rotation
10.
70
700 rotations # 1 min # 2p rad
=
p rad>sec
1 min
60 sec 1 rotation
3
Section 6.3 Exercises
[–10, 10] by [–10, 10]
9.
1. (b) [–5, 5] by [–5, 5]
2. (d) [–5, 5] by [–5, 5]
3. (a) [–5, 5] by [–5, 5]
4. (c) [–10, 10] by [–12, 10]
5. (a) t
x †
y p
–2
0 †
1
2
p
–1
1 †
0
2
†
–2 p undef. p
1
3 †
4 p
2
4
[–10, 10] by [–10, 10]
10.
5
2
[–10, 10] by [–10, 10]
11. x=1+y, so y=x-1: line through (0, –1) and (1, 0)
Chapter 6
242
Applications of Trigonometry
1
17
12. t=y-5, so x=2-3(y-5); y = - x +
: line
3
3
17
through a 0,
b and 1 17, 0 2
3
33. In Quadrant I, we need x>0 and y>0, so 2-|t|>0
and t-0.5>0. Then –2<t<2 and t>0.5, so
0.5<t<2. This is not changed by the additional
requirement that - 3 t 3.
3
1
3
1
13. t= x + , so y = 9 - 4 a x + b ;
2
2
2
2
y=–2x+3, 3 x 7: line segment with endpoints
(3, –3) and (7, –11)
34. In Quadrant II, we need x<0 and y>0, so 2-|t|<0
and t-0.5>0. Then (t<–2 or t>2) and t>0.5, so
t>2. With the additional requirement that
-3 t 3, this becomes 2 6 t 3.
14. t=y-2, so x=5-3(y-2);
11
1
y= - x +
, -4 x 8: line segment with
3
3
endpoints (8, 1) and (–4, 5)
35. In Quadrant III, we need x<0 and y<0, so 2-|t|<0
and t-0.5<0. Then (t<–2 or t>2) and t<0.5, so
t<–2. With the additional requirement that
-3 t 3, this becomes -3 t 6 -2.
15. x=(y-1)2: parabola that opens to right with vertex at
(0, 1)
16. y=x2-3: parabola that opens upward with vertex at
(0, –3)
17. y=x3-2x+3: cubic polynomial
18. x=2y2-1; parabola that opens to right with vertex at
(0, –1)
19. x=4-y2; parabola that opens to left with vertex at
(4, 0)
20. t=2x, so y=16x3-3: cubic, -1 x 1
21. t=x+3, so
2
y =
, on domain : 3 -8, -3 2 ´ 1 -3, 2 4
x + 3
4
,x 4
22. t=x-2, so y =
x - 2
23. x2+y2=25, circle of radius 5 centered at (0, 0)
24. x2+y2=16, circle of radius 4 centered at (0, 0)
25. x2+y2=4, three-fourths of a circle of radius 2 centered
at (0, 0) (not in Quadrant II)
26. x2+y2=9, semicircle of radius 3, y 0 only
¡
¡
¡
27. OA=˚–2, 5¬, OB = ˚4, 2¬, OP=˚x, y¬
OP - OA = t 1OB - OA 2
˚x+2, y-5¬=t˚6, –3¬
x+2=6t 1 x=6t-2
y-5=3t 1 y=–3t+5
¡
¡
¡
¡
¡
¡
¡
28. OA = ˚–3, –3¬, OB=˚5, 1¬, OP=˚x, y¬
OP - OA = t 1OB - OA 2
˚x+3, y+3¬=t˚8, 4¬
x+3=8t 1 x=8t-3
y+3=4t 1 y=4t-3
¡
¡
¡
¡
For #29–32, many answers are possible; one or two of the
simplest are given.
29. Two possibilities are x=t+3,
7
y = 4 - t, 0 t 3, or x = 3t + 3, y = 4 - 7t,
3
0 t 1.
6
30. Two possibilities are x=5-t, y=2- t,
7
0 t 7, or x=5-7t, y=2-6t, 0 t 1.
31. One possibility is x=5+3 cos t, y=2+3 sin t,
0 t 2p.
32. One possibility is x=–2+2 cos t, y=–4+2 sin t,
0 t 2p.
36. In Quadrant IV, we need x>0 and y<0, so 2-|t|>0
and t-0.5<0. Then –2<t<2 and t<0.5, so
–2<t<0.5. This is not changed by the additional
requirement that –3 t 3.
37. (a) One good window is [–20, 300] by [–1, 10]. If your
grapher allows, use “Simultaneous” rather than
“Sequential” plotting. Note that 100 yd is 300 ft. To
show the whole race, use 0 t 13 (upper limit
may vary), since Ben finishes in 12.916 sec. Note that
it is the process of graphing (during which one
observes Ben passing Jerry and crossing “the finish
line” first), not the final product (which is two horizontal lines) which is needed; for that reason, no
graph is shown here.
(b) After 3 seconds, Jerry is at 20(3)=60 ft and Ben is
at 24(3)-10=62 ft. Ben is ahead by 2 ft.
38. (a) If your grapher allows, use “Simultaneous” rather
than “Sequential” plotting. To see the whole race, use
0 t 5.1 (upper limit may vary), since the faster
runner reaches the flag after 5.1 sec. Note that it is the
process of graphing, not the final product (which
shows a horizontal line) which is needed; for that
reason, no graph is shown here.
(b) The faster runner (who is coming from the left in the
simulation) arrives at t=5.1 sec. At this instant, the
slower runner is 4.1 ft away from the flag; the slower
runner doesn’t reach the flag until t = 5.5 sec. This
can be observed from the simulation, or by solving
algebraically x1=50 and x2=50.
39. (a) y=–16t2+v0t+s0=–16t2+0t+1000
=–16t2+1000
(b) Graph and trace: x=1 and y=–16t2+1000 with
0 t 6, on the window [0, 2] by [0, 1200]. Use
something like 0.2 or less for Tstep. This graph will
appear as a vertical line from (1, 424) to (1, 1000); it
is not shown here because the simulation is accomplished by the tracing, not by the picture.
(c) When t=4, y=–16(4)2+1000=744 ft; the food
containers are 744 ft above the ground after 4 sec.
40. (a) y=–16t2+v0t+s0=–16t2+80t+5
(b) Graph and trace: x=6 and y=–16t2+80t+5
with 0 t 5.1 (upper limit may vary) on
[0, 7] by [0, 120]. This graph will appear as a vertical
line from about (6, 0) to about (6, 105). Tracing shows
how the ball begins at a height of 5 ft, rises to over
100 ft, then falls back to the ground.
Section 6.3
(c) Graph x=t and y=–16t2+80t+5 with
0 t 5.1 (upper limit may vary).
[0, 7] by [0, 120]
(d) When t=4, y=–16(4)2+80(4)+5=69 ft. The
ball is 69 ft above the ground after 4 sec.
(e) From the graph in (b), when t=2.5 sec, the ball is at
its maximum height of 105 ft.
41. Possible answers:
(a) 0 6 t 6
p
1t in radians 2
2
(b) 0 6 t 6 p
(c)
p
3p
6 t 6
2
2
42. (a) Both pairs of equations can be changed to
x2+y2=9 — a circle centered at the origin with
radius 3. Also, when one chooses a point on this
circle and swaps the x and y coordinates, one obtains
another point on the same circle.
(b) The first begins at the right side (when t=0) and
traces the circle counterclockwise. The second begins
at the top (when t=0) and traces the circle clockwise.
43. (a) x=400 when t L 2.80 — about 2.80 sec.
(b) When t L 2.80 sec, y L 7.18 ft.
(c) Reaching up, the outfielder’s glove should be at or
near the height of the ball as it approaches the wall. If
hit at an angle of 20°, the ball would strike the wall
about 19.74 ft up (after 2.84 sec) — the outfielder
could not catch this.
44. (a) No: x=(120 cos 30°)t; this equals 350 when
t L 3.37. At this time, the ball is at a height of
y=–16t2+(120 sin 30°)t + 4 L 24.59 ft.
(b) The ball hits the wall about 24.59 ft up when t L 3.37
(see part a) — not catchable.
Parametric Equations and Motion
48. Yes: x=(25 cos 55°)t and y=–16t2+(25 sin 55°)t+4.
The dart lands when y=0, which happens when
t L 1.45 sec. At this point, the dart is about 20.82 ft from
Sue, inside the target.
49. The parametric equations for this motion are
x=(v+160 cos 20°) t and
y=–16t2+(160 sin 20°)t +4, where v is the velocity
of the wind (in ft/sec) — it should be positive if the wind
is in the direction of the hit, and negative if the wind is
against the ball.
To solve this algebraically, eliminate the parameter t as
follows:
2
x
x
. So y = -16 a
b
v + 160 cos 20°
v + 160 cos 20°
x
+ 160 sin 20° a
b + 4.
v + 160 cos 20°
Substitute x=400 and y=30:
t =
2
400
b
v + 160 cos 20°
400
b + 4.
+ 160 sin 20° a
v + 160 cos 20°
400
Let u =
, so the equation becomes
v + 160 cos 20°
-16u2 + 54.72u - 26 = 0. Using the quadratic formula,
30 = -16 a
we find that u =
- 54.72 ; 254.72 2 - 4 1 -16 2 1 -26 2
L
-32
400
0.57, 2.85. Solving 0.57 =
and
v + 160 cos 20°
400
, v≠551, v≠–10. A wind speed
2.85 =
v + 160 cos 20°
of 551 ft/sec (375.7 mph) is unrealistic, so we eliminate that
solution. So the wind will be blowing against the ball in order
for the ball to hit within a few inches of the top of the wall.
To verify this graphically, graph the equation
2
400
b
v + 160 cos 20°
400
+ 160 sin 20° a
b + 4, and find
v + 160 cos 20°
the zero.
30 = -16 a
45. (a) Yes: x=(5+120 cos 30°)t; this equals 350 when
t L 3.21. At this time, the ball is at a height of
y=–16t2+(120 sin 30°)t+4 L 31.59 ft.
(b) The ball clears the wall with about 1.59 ft to spare
(when t L 3.212 .
46. For Linda’s ball, x1=(45 cos 44°)t and
y1=–16t2+(45 sin 44°)t+5. For Chris’s ball,
x2=78-(41 cos 39°)t and y2=–16t2
+(41 sin 39°)t+5. Find (graphically)
the minimum of d(t)= 2 1 x1 - x2 2 2 + 1y1 - y2 2 2.
It occurs when t L 1.21 sec; the minimum distance is
about 6.60 ft.
47. No: x=(30 cos 70°)t and y=–16t2+(30 sin 70°)t+3.
The dart lands when y=0, which happens when
t L 1.86 sec. At this point, the dart is about 19.11 ft from
Tony, just over 10 in. short of the target.
243
[–15, 5] by [–3, 10]
50. Assuming the course is level, the ball hits the ground
when y=–16t2+(180 sin ¨)t equals 0, which
180 sin u
happens when t =
=11.25 sin ¨ sec. At that
16
time, the ball has traveled x=(180 cos ¨)t
=2025(cos ¨)(sin ¨) feet. The answers are therefore
approximately:
(a) 506.25 ft.
(b) 650.82 ft.
(c) 775.62 ft.
(d) 876.85 ft.
244
Chapter 6
Applications of Trigonometry
p
p
51. x = 35 cos a t b and y = 50 + 35 sin a t b
6
6
1
2
1
2
21
3
x + , y = 3 + 3a x + b= x +
5
5
5
5
5
5
21
15
3
=
= 3 and
Since 3= 1 - 2 2 +
5
5
5
3
21
30
6 = 132 +
=
= 6, both (–2, 3) and (3, 6) are
5
5
5
on the line.
52. t =
62. The parametrization describes a circle of radius 2, centered at the origin and t represents the angle traveled
counterclockwise from (1, 0). The answer is A.
63. Set -16t2 + 80t + 7 equal to 91 and solve either graphically or using the quadratic formula. The answer is D.
64. The equations are both linear, so the answer is either (a),
(b), or (c). Since t has a minimum value and no maximum
value, the answer is C.
65. (a)
53. (a) When t=∏(or 3∏, or 5∏, etc.), y=2. This
corresponds to the highest points on the graph.
(b) The x-intercepts occur where y=0, which happens
when t=0, 2∏, 4∏, etc. The x coordinates at those
times are (respectively) 0, 2∏, 4∏, etc., so these are
2∏ units apart.
54. (a)
[–6, 6] by [–4, 4]
(b) x2+y2=(a cos t)2+(a sin t)2
=a2 cos2 t+a2 sin2 t
=a2
The radius of the circles are a={1, 2, 3, 4},
centered at (0, 0).
(c)
[–4.7, 4.7] by [–3.1, 3.1]
(b) All 2’s should be changed to 3’s.
55. The particle begins at –10, moves right to ±2.25
(at t=1.5), then changes direction and ends at –4.
56. The particle begins at –5, moves right to ±4 (at time
t=2), then changes direction and returns to –5.
57. The particle begins at –5, moves right to about + 0.07
(at time t≠0.15), changes direction and moves left to
about –20.81 (at time t≠4.5), then changes direction
and ends at + 7.
58. The particle begins at –10, moves right to about + 0.88
(at t≠0.46), changes direction and moves left to about
–6.06 (at time t≠2.9), then changes direction and ends
at + 20.
59. True. Eliminate t from the first set:
t = x1 + 1
y1 = 31x1 + 1 2 + 1
y1 = 3x1 + 4
Eliminate t from the second set:
3
t = x2 + 2
2
3
y2 = 2 a x2 + 2 b
2
y2 = 3x2 + 4
Both sets correspond to the rectangular equation
y=3x+4.
60. True. x=0 and y=1 when t=1, and x=2 and y=5
when t=3. Eliminating t,
t=x+1.
y=2(x+1)-1.
y=2x+1, 0 x 2, 1 y 5.
1
61. x = 1 -12 2 - 4 = -3, y = - 1 +
= -2
-1
The answer is A.
[–6, 6] by [–4, 4]
(d) x-h=a cos t and y-k=a sin t, so
(x-h)2+(y-k)2
=(a cos t)2+(a sin t)2
=a2 cos2 t+a2 sin2 t
=a2
The graph is the circle of radius a centered at (h, k).
(e) If (x+1)2+(y-4)2=9, then a=3, h=–1,
and k=4.
As a result, x=3 cos t-1 and y=3 sin t+4.
66. (a)
[–2, 8] by [–4, 6]
(b)
[–2, 8] by [–4, 6]
b
1
b
1
x - , y = ca x - b + d
a
a
a
a
b - ad
c
, a Z 0.
= x a
a
(c) t =
Section 6.4
c
, if a Z 0;
a
- b + ad
y-intercept: a 0,
b , if a Z 0;
a
b - ad
x-intercept: a
, 0 b , if c Z 0.
c
(d) Slope:
(e) The line will be horizontal if c=0. The line will
be vertical if a=0.
67. (a) Jane is traveling in a circle of radius 20 feet and
center (0, 20), which yields x1=20 cos (nt) and
y1=20+20 sin (nt). Since the Ferris wheel is
making one revolution (2∏) every 12 seconds,
2p
p
2∏=12n, so n =
= .
12
6
Thus,
p
p
x1=20 cos a t b and y1=20+20 sin a t b , in
6
6
radian mode.
(b) Since the ball was released at 75 ft in the positive
x-direction and gravity acts in the negative
y-direction at 16 ft/s2, we have x2=at+75 and
y2=–16t2+bt, where a is the initial speed of the
ball in the x-direction and b is the initial speed of the
ball in the y-direction. The initial velocity vector of the
ball is 60 ˚cos 120°, sin 120°¬ =˚–30, 30 13¬, so
a=–30 and b=30 13. As a result x2=–30t+75
and y2=–16t2+30 13t are the parametric equations for the ball.
Polar Coordinates
245
68. Assuming that the bottom of the ferris wheel and the ball’s
initial position are at the same height, the position
p
of Matthew is x1=71 cos a t b and y1
10
p
=71 + 71 sin a t b . The ball’s position is
10
x2=90+(88 cos 100)t and y2=–16t2+(88 sin 100)t.
Find (graphically) the minimum of
d1t 2 = 2 1 x1 - x2 2 2 + 1y1 - y2 2 2. It occurs when
t L 2.19 sec; the minimum distance is about 3.47 ft.
p
69. Chang’s position: x1=20 cos a t b and y1=20
6
p
p
+20 sin a t b . Kuan’s position: x2=15+15 cos a t b
6
4
p
and y2=15+15 sin a t b .
4
Find (graphically) the minimum of
2
2
d(t)= 2 1x1 - x2 2 + 1y1 - y2 2 . It occurs when
t≠21.50 sec; the minimum distance is about 4.11 ft.
p
70. Chang’s position: x1=20 cos a t b and y1=20
6
p
p
+20 sin a t b . Kuan’s position: x2=15+15 sin a t b
6
4
p
and y2=15-15 cos a t b . Find (graphically) the mini4
mum of d(t)= 2 1x1 - x2 2 2 + 1y1 - y2 2 2. It occurs
when t≠12.32 sec; the minimum distance is about
10.48 ft.
(c)
71. (a) x(0)=0c+(1-0)a=a and
y(0)=0d+(1-0)b=b
(b) x(1)=1c+(1-1)a=c and
y(1)=1d+(1-1)b=d
[–50, 100] by [–50, 50]
Our graph shows that Jane and the ball will be close
to each other but not at the exact same point at
t=2.1 seconds.
(d)
d(t)= 21 x1 - x2 2 2 + 1y1 - y2 2 2
2
p
=C a 20 cos a 6 t b + 30t - 75 b
2
p
+ a 20 + 20 sin a t b + 16t2 - 1 30132 t b
6
(e)
72. x(0.5)=0.5c+(1-0.5)a=0.5(a+c)=(a+c)/2,
while y(0.5)=(b+d)/2 — the correct coordinates for
the midpoint.
73. Since the relationship between x and y is linear and one
unit of time (t=1) separates the two points,
1
t= L 0.33 will divide the segment into three equal
3
1
pieces. Similarly, t= = 0.25 will divide the segment
4
into four equal pieces.
■ Section 6.4 Polar Coordinates
Exploration 1
2.
[0, 5] by [–5, 25]
The minimum distance occurs at t=2.2, when
d(t)=1.64 feet.
a 2,
p
b = 11, 132
3
p
a -1, b =(0, –1)
2
(2, ∏)=(–2, 0)
3p
a -5,
b =(0, 5)
2
(3, 2∏)=(3, 0)
Chapter 6
246
3.
Applications of Trigonometry
1 -1, - 132 = a - 2,
p
b
3
(b)
y
5
(0, 2)= a 2,
p
b
2
(3, 0)=(3, 0)
(–1, 0)=(1, ∏)
3p
(0, –4)= a 4,
b
2
a4,
a2,
5
b
6
1. (a) Quadrant II
7.
(b) Quadrant III
4π
3
2. (a) Quadrant I
O
3
(b) Quadrant III
3. Possible answers: 7∏> 4, –9∏> 4
a3,
4. Possible answers: 7∏> 3, –5∏> 3
8.
5. Possible answers: 520 °, -200 °
a2,
4π
b
3
5π
b
6
6. Possible answers: 240 °, -480 °
9.
8. x2+(y+4)2=9
1 O
10. a2=92+62-2(9)(6) cos 40 °
a≠5.85
10.
Section 6.4 Exercises
a–1,
2π
b
5
a–3,
17π
b
10
3
3 313
b
1. a - ,
2 2
17π
10
2. 12 12, 2 122
3. 1 -1, - 13 2
11.
r
∞
p
4
3 12
2
∞
O
(2, 30°)
30°
2
12 12
,
b
2
2
5. (a) ¨
O
2π
5
2
9. a =12 +10 -2(12)(10)cos 60 °
a≠11.14
4. a -
5π
6
2
7. (x-3)2+y2=4
2
O
p
2
∞
3
(b)
5p
6
∏
∞
1.5
0
∞
4p
3
-3 13
2
12.
∞
210°
2∏
3
O
(3, 210°)
0
13.
120°
y
5
a3, b
2
5
3
a ,
b
2 6
O 2
3 3
,
a–
2
3 2
,
a
2
(0, )
(0, 2 )
4
b
3
(–2, 120°)
b
4
5
14.
x
135°
O
3
(–3, 135°)
6. (a) ¨
r
p
4
p
2
5p
6
2 12
2
4
a–
4 3 4
, 3 b
3
a2 2 ,
b
4
5
Quick Review 6.4
2
b
2
3 3
15. a , 13 b
4 4
∏
4p
3
2∏
5
5
16. a 12, 12 b
4
4
undefined
- 413
2
undefined
17. (–2.70, 1.30)
18. (1.62, 1.18)
19. (2, 0)
20. (0, 1)
21. (0, –2)
x
Section 6.4
22. (–3, 0)
p
p
+ 2np b and a -2, + 1 2n + 1 2 p b ,
6
6
n an integer
23. a 2,
p
p
+ 2np b and a - 1, - + 1 2n + 1 2p b ,
4
4
n an integer
24. a 1, -
25. 11.5, -20 ° + 360n°2 and 1 - 1.5, 160 ° + 360n °2,
n an integer
26. 1 -2.5, 50 ° + 360n°2 and 12.5, 230 ° + 360n°2 ,
n an integer
27. (a) a 12,
p
5p
b or a - 12,
b
4
4
a - 12,
247
38. r2+4r cos ¨=0, or x2+y2+4x=0. Completing the
square gives (x+2)2+y2=4 — a circle centered at
(–2, 0) with radius 2
39. r2-r sin ¨=0, or x2+y2-y=0. Completing the
1 2
1
square gives x2 + a y - b = — a circle centered
2
4
1
1
at a 0, b with radius .
2
2
40. r2-3r cos ¨=0, or x2+y2-3x=0. Completing the
9
3 2
square gives a x - b + y2 = — a circle centered at
2
4
3
3
a , 0 b with radius .
2
2
41. r2-2r sin ¨+4r cos ¨=0, or x2+y2-2y+4x=0.
Completing the square gives (x+2)2+(y-1)2=5
— a circle centered at (–2, 1) with radius 15.
p
3p
b
(b) a 12, b or a - 12, 4
4
(c) The answers from (a), and also a 12,
Polar Coordinates
9p
b or
4
13p
b
4
28. (a) 1 110, tan-1 32 L 1 110, 1.252 or
1 - 110, tan-1 3 + p2 L 1 - 110, 4.39 2
42. r2-4r cos ¨+4r sin ¨=0, or x2+y2-4x+4y=0.
Completing the square gives (x-2)2+(y+2)2=8
— a circle centered at (2, –2) with radius 2 12.
43. r=2/cos ¨=2 sec ¨ — a vertical line
(b) 1 110, tan-1 32 L 1 110, 1.252 or
1 - 110, tan-1 3 - p2 L 1 - 110, -1.892
(c) The answers from (a), and also
1 110, tan-1 3 + 2p 2 L 1 110, 7.532 or
1 - 110, tan-1 3 + 3p2 L 1 - 110, 10.67 2
29. (a) 1 129, tan-1 1 -2.52 + p2 L 1 129, 1.95 2 or
1 - 129, tan-1 1 -2.52 + 2p 2 L 1 - 129, 5.092
[–5, 5] by [–5, 5]
44. r=5/cos ¨=5 sec ¨
(b) 1 - 129, tan-1 1 -2.52 2 L 1 - 129, -1.19 2 or
1 129, tan-1 1 -2.52 + p2 L 1 129, 1.95 2
(c) The answers from (a), plus
1 129, tan-1 1 -2.52 + 3p2 L 1 129, 8.23 2 or
1 - 129, tan-1 1 -2.52 + 4p 2 L 1 - 129, 11.382
30. (a) 1 - 15, tan-1 2 2 L 1 - 15, 1.112 or
1 15, tan-1 2 + p2 L 1 15, 4.25 2
[0, 10] by [–5, 5]
45. r=
(b) 1 - 15, tan-1 22 L 1 - 15, 1.11 2 or
1 15, tan-1 2 - p2 L 1 15, -2.03 2
5
2 cos u - 3 sin u
(c) The answers from (a), plus
1 - 15, tan-1 2 + 2p 2 L 1 - 15, 7.39 2 or
1 15, tan-1 2 + 3p2 L 1 15, 10.53 2
31. (b)
[–5, 5] by [–5, 5]
32. (d)
33. (c)
34. (a)
46. r=
2
3 cos u + 4 sin u
35. x=3 — a vertical line
36. y=–2 — a horizontal line
37. r2+3r sin ¨=0, or x2+y2+3y=0. Completing the
3 2
9
square gives x2 + a y + b = — a circle centered at
2
4
3
3
a 0, - b with radius .
2
2
[–5, 5] by [–5, 5]
Chapter 6
248
Applications of Trigonometry
47. r2-6r cos ¨=0, so r=6 cos ¨
56. True. For (r1, ¨) and (r2, ¨+∏) to represent the same
point, (r2, ¨+∏) has to be the reflection across the orgin
of (r1, ¨+∏), and this is accomplished by setting
r2=–r1.
57. For point (r, ¨), changing the sign on r and adding 3∏ to ¨
constitutes a twofold reflection across the orgin.
The answer is C.
[–3, 9] by [–4, 4]
48. r2-2r sin ¨=0, so r=2 sin ¨
58. The rectangular coordinates are
(–2 cos(–∏/3), –2 sin(–∏/3))=(–1, 13).
The answer is C.
59. For point (r, ¨), changing the sign on r and subtracting
180° from ¨ constitutes a twofold reflection across the
origin. The answer is A.
60. (–2, 2) lies in Quadrant III, whereas 1 -2 12, 135°2 lies in
Quadrant IV. The answer is E.
[–3, 3] by [–1, 3]
49. r2+6r cos ¨+6r sin ¨=0, so r=–6 cos ¨-6 sin ¨
[–12, 6] by [–9, 3]
2
50. r -2r cos ¨+8r sin ¨=0, so r=2 cos ¨-8 sin ¨
61. (a) If ¨1-¨2 is an odd integer multiple of ∏, then
the distance is @ r1 + r2 @ . If ¨1-¨2 is an even integer
multiple of ∏, then the distance is @ r1 - r2 @ .
(b) Consider the triangle formed by O1, P1, and P2
(ensuring that the angle at the origin is less than
180°), then by the Law of Cosines,
2
2
2
P1P2 =OP1 +OP2 -2 # OP1 # OP2 cos u,
where ¨ is the angle between OP1 and OP2. In polar
coordinates, this formula translates very nicely into
d2 = r12 + r22 - 2r1r2 cos 1u2 - u1 2 (or cos (¨1-¨2)
since cos (¨2-¨1)=cos (¨1-¨2)), so
d = 3r12 + r22 - 2r1 r2 cos 1 u2 - u1 2.
(c) Yes. If ¨1-¨2 is an odd integer multiple of ∏, then
cos (¨1-¨2)=–1 1 d = 3r12 + r22 + 2r1r2
= @ r1 + r2 @ . If ¨1-¨2 is an even integer multiple
of ∏, then cos1 u1 - u2 2 =1 1
[–8, 10] by [–10, 2]
51. d= 242 + 22 - 2 # 4 # 2 cos 1 12 ° - 72 °2
= 120 - 16 cos 60 ° = 112 = 2 13 mi
52. d= 232 + 52 - 2 # 3 # 5 cos 1 170 ° - 150 ° 2
d= 134 - 30 cos 20 °≠2.41 mi
53. Using the Pythagorean theorem, the center-to-vertex
a p
a
. The four vertices are then a
, b,
distance is
12
12 4
a 3p
a 5p
a 7p
a
,
b, a
,
b , and a
,
b . Other polar
12 4
12 4
12 4
coordinates for these points are possible, of course.
54. The vertex on the x-axis has polar coordinates (a, 0).
All other vertices must also be a units from the origin;
2p
4p
6p
b , a a,
b , a a,
b,
their coordinates are a a,
5
5
5
8p
b . Other polar coordinates for these points
and a a,
5
are possible, of course.
55. False. Point (r, ¨) is the same as point (r, ¨+2n∏) for
any integer n. So each point has an infinite number of
distinct polar coordinates.
d = 3r12 + r22 - 2r1r2 = @ r1 - r2 @ .
62. (a) The right half of a circle centered at (0, 2) of radius 2.
(b) Three quarters of the same circle, starting at (0, 0)
and moving counterclockwise.
(c) The full circle (plus another half circle found through
the TRACE function).
(d) 4 counterclockwise rotations of the same circle.
63. d = 222 + 52 - 2 12 2 152 cos 120° L 6.24
64. d = 242 + 62 - 2 14 2 162 cos 45° L 4.25
65. d = 2 1 - 32 2 + 1 -52 2 - 21 -3 2 1 -5 2 cos 135 °≠7.43
66. d = 262 + 82 - 2 16 2 182 cos 30°≠4.11
67. Since x=r cos ¨ and y=r sin ¨, the parametric
equation would be x=f(¨) cos (¨) and y=f(¨) sin (¨).
68. x=2 cos2 ¨
y=2(cos ¨)(sin ¨)
69. x=5(cos ¨)(sin ¨)
y=5 sin2 ¨
70. x=2(cos ¨)(sec ¨)=2
y=2(sin ¨)(sec ¨)=2 tan ¨
71. x=4(cos ¨)(csc ¨)=4 cot ¨
y=4(sin ¨)(csc ¨) =4
Section 6.5
■ Section 6.5 Graphs of Polar Equations
Graphs of Polar Equations
249
2. (a)
Exploration 1
Answers will vary.
¨
0
∏/6
∏/3
∏/2
r
0
2
0
–2
(b)
Exploration 2
2∏/3 5∏/6
0
2
∏
0
y
5
1. If r2=4 cos(2¨), then r does not exist when
cos(2¨)<0. Since cos(2¨)<0 whenever ¨ is in the
p
3p
interval ¢ + np,
+ np≤ , n is any integer, the
4
4
domain of r does not include these intervals.
2. -r1cos 12¨ 2 draws the same graph, but in the opposite
direction.
3. (r)2-4 cos(–2¨)=r2-4 cos(2¨)
(since cos (¨)=cos (–¨))
5
a0, 0,
a2,
5∏
b
6
6
∏ 2∏
, ,∏ b
3 3
a2, ∏ b
6
5
x
a–2, ∏ b
2
3. k=∏
4. (–r)2-4 cos(–2¨)=r2-4 cos(2¨)
5. (–r)2-4 cos(2¨)=r2-4 cos(2¨)
Quick Review 6.5
For #1–4, use your grapher’s TRACE function to solve.
p 3p
1. Minimum: –3 at x= b ,
r ; Maximum: 3 at
2 2
x= E 0, p, 2p F
[–5, 5] by [–4, 3]
4. k=2∏
2. Minimum: –1 at x=∏; Maximum: 5 at x= E0, 2pF
3. Minimum: 0 at x= b
x= E 0, p, 2p F
p 3p 5p 7p
,
,
,
r ; Maximum: 2 at
4 4 4 4
p
3p
4. Minimum: 0 at x= ; Maximum: 6 at x=
2
2
5. (a) No
(b) No
(c) Yes
6. (a) No
(b) Yes
(c) No
[–5, 5] by [–3, 3]
5. k=2∏
7. sin(∏-¨)=sin ¨
8. cos(∏-¨)=–cos ¨
9. cos 2(∏+¨)=cos(2∏+2¨)=cos 2¨
=cos2¨-sin2¨
10. sin 2(∏+¨)=sin(2∏+2¨)=sin 2¨
=2 sin ¨ cos ¨
[–5, 5] by [–3, 3]
Section 6.5 Exercises
6. k=∏
1. (a)
¨
0
∏/4
∏/2
3∏/4
∏
r
3
0
–3
0
3
(b)
a0,
5
6
∏ 3 ∏ 5∏ 7∏
,
, ,
b 5
4 4
4
4
(3, ∏)
a–3, ∏ b
2
5∏/4 3∏/2 7∏/4
0
–3
0
y
a–3,
3∏
b
2
(3, 0)
x
5
[–5, 5] by [–3, 3]
7. r1 is not shown (this is a 12-petal rose). r2 is not shown
(this is a 6-petal rose), r3 is graph (b).
8. 6 cos 2¨ sin 2¨=3(2 cos u sin u) where u=2¨; this
equals 3 sin 2u=3 sin 4¨. r=3 sin 4¨ is the equation for
the eight-petal rose shown in graph (a).
250
Chapter 6
Applications of Trigonometry
p
9. Graph (b) is r=2-2 cos ¨: Taking ¨=0 and ¨= ,
2
we get r=2 and r=4 from the first equation, and r=0 and
r=2 from the second. No graph matches the first of
these (r, ¨) pairs, but (b) matches the latter (and any
others one might choose).
10. Graph (c) is r=2+3 cos ¨: Taking ¨=0, we get
r=–1 from the other equation, which matches nothing.
Any (r, ¨) pair from the first equation matches (c), however.
p
11. Graph (a) is r=2-2 sin ¨ — where ¨= ,
2
p
2+2 cos ¨=2, but a 2, b is clearly not
2
p
p
on graph (a); meanwhile 2-2 sin =0, and a 0, b
2
2
(the origin) is part of graph (a).
p
12. Graph (d) is r=2-1.5 sin ¨ — where ¨= ,
2
p
2+1.5 cos ¨=2, but a 2, b is clearly not
2
p
on graph (d); meanwhile 2-1.5 sin =0.5, and
2
p
a 0.5, b is part of graph (d).
2
25. Domain: All reals
Range: [–3, 3]
Symmetric about the x-axis, y-axis, and origin
Continuous
Bounded
Maximum r-value: 3
No asymptotes
[–6, 6] by [–4, 4]
26. Domain: All reals
Range: [–2, 2]
Symmetric about the x-axis, y-axis, and origin
Continuous
Bounded
Maximum r-value: 2
No asymptotes
13. Symmetric about the y-axis: replacing (r, ¨)
with (r, ∏-¨) gives the same equation, since
sin(∏-¨)=sin ¨.
14. Symmetric about the x-axis: replacing (r, ¨) with
(r, –¨) gives the same equation, since cos(–¨)=cos ¨.
15. Symmetric about the x-axis: replacing (r, ¨) with (r, –¨)
gives the same equation, since cos(–¨)=cos ¨.
16. Symmetric about the y-axis: replacing (r, ¨) with
(r, ∏-¨) gives the same equation, since
sin(∏-¨)=sin ¨.
17. All three symmetries. Polar axis: replacing (r, ¨) with
(r, –¨) gives the same equation, since cos(–2¨)=cos 2¨.
y-axis: replacing (r, ¨) with (r, ∏-¨) gives the same
equation, since cos [2(∏-¨)]=cos (2∏-2¨)
=cos(–2¨)=cos 2¨. Pole: replacing (r, ¨) with
(r, ¨+∏) gives the same equation, since cos[2(¨+∏)]
=cos(2¨+2∏)=cos 2¨.
18. Symmetric about the y-axis: replacing (r, ¨) with (–r, –¨)
gives the same equation, since sin(–3¨)=–sin 3¨.
19. Symmetric about the y-axis: replacing (r, ¨) with
(r, ∏-¨) gives the same equation, since
sin(∏-¨)=sin ¨.
20. Symmetric about the x-axis: replacing (r, ¨) with (r, –¨)
gives the same equation, since cos(–¨)=cos ¨.
21. Maximum r is 5 — when ¨=2n∏ for any integer n.
22. Maximum r is 5 (actually, –5) — when ¨=
[–4.5, 4.5] by [–3, 3]
27. Domain: ¨=∏/3+n∏, n is an integer
Range: (–q, q)
Symmetric about the origin
Continuous
Unbounded
Maximum r-value: none
No asymptotes
[–4.7, 4.7] by [–3.1, 3.1]
28. Domain: ¨=–∏/4+n∏, n is an integer
Range: (–q, q)
Symmetric about the origin
Continuous
Unbounded
Maximum r-value: none
No asymptotes
3p
+2n∏
2
for any integer n.
23. Maximum r is 3 (along with –3) — when ¨=2n∏/3 for
any integer n.
24. Maximum r is 4 (along with –4) — when ¨=n∏/4 for
any odd integer n.
[–4.7, 4.7] by [–3.1, 3.1]
Section 6.5
29. Domain: All reals
Range: [–2, 2]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 2
No asymptotes
[–3, 3] by [–2, 2]
30. Domain: All reals
Range: [–3, 3]
Symmetric about the x-axis, y-axis, and origin
Continuous
Bounded
Maximum r-value: 3
No asymptotes
[–4.7, 4.7] by [–3.1, 3.1]
31. Domain: All reals
Range: [1, 9]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 9
No asymptotes
[–9, 9] by [–2.5, 9.5]
32. Domain: All reals
Range: [1, 11]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 11
No asymptotes
[–16, 8] by [–8, 8]
Graphs of Polar Equations
33. Domain: All reals
Range: [0, 8]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 8
No asymptotes
[–6, 12] by [–6, 6]
34. Domain: All reals
Range: [0, 10]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 10
No asymptotes
[–9, 9] by [–10.5, 1.5]
35. Domain: All reals
Range: [3, 7]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 7
No asymptotes
[–7, 11] by [–6, 6]
36. Domain: All reals
Range: [2, 4]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 4
No asymptotes
[–6, 6] by [–6, 3]
251
252
Chapter 6
Applications of Trigonometry
37. Domain: All reals
Range: [–3, 7]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 7
No asymptotes
Continuous
No symmetry
Unbounded
Maximum r-value: none
No asymptotes
Graph for ¨ 0:
[–45, 45] by [–30, 30]
[–4, 8] by [–4, 4]
38. Domain: All reals
Range: [–1, 7]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 7
No asymptotes
[–7.5, 7.5] by [–8, 2]
39. Domain: All reals
Range: [0, 2]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 2
No asymptotes
42. Domain: All reals
Range: [0, q)
Continuous
No symmetry
Unbounded
Maximum r-value: none
No asymptotes
Graph for ¨ 0:
[–6, 6] by [–4, 4]
43. Domain: B 0,
p
3p
R ª B , 2pR
2
2
Range: [0, 1]
Symmetric about the origin
Continuous on each interval in domain
Bounded
Maximum r-value: 1
No asymptotes
[–3, 1.5] by [–1.5, 1.5]
40. Domain: All reals
Range: [1, 3]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 3
No asymptotes
[–1.5, 1.5] by [–1, 1]
44. Domain: B 0,
p
3p 5p
7p
RªB ,
R ª B , 2pR
4
4 4
4
Range: [0, 3]
Symmetric about the x-axis, y-axis, and origin
Continuous on each interval in domain
Bounded
Maximum r-value: 3
No asymptotes
[–3.75, 3.75] by [–1.5, 3.5]
41. Domain: All reals
Range: [0, q)
[–3.3, 3.3] by [–2.2, 2.2]
Section 6.5
For #45–48, recall that the petal length is the maximum
r-value over the interval that creates the petal.
45. r=–2 when ¨= b
¨= b
3p 7p
,
r and r=6 when
4 4
p 5p
,
r . There are four petals with lengths {6, 2, 6, 2}.
4 4
Graphs of Polar Equations
253
58. Symmetry about y-axis: r-3 sin(4¨)=0 1
–r-3 sin(4(–¨))=–r+3 sin(4¨) (since sin (¨) is
odd, i.e., sin (–¨)=–sin(¨))=r-3 sin(4¨)=0.
Symmetry about the origin: r-3 sin (4¨)=0
1 r-3 sin (4¨+4∏)=r-3 sin (4¨)=0.
59.
46. r=–2 when ¨={0, ∏} and r=8 when
p 3p
¨= b ,
r . There are four petals with lengths {2, 8, 2, 8}.
2 2
47. r=–3 when ¨= b 0,
2p 4p 6p 8p
,
,
,
r and
5 5 5 5
[0, 2] by [0, 6]
p 3p
7p 9p
r=5 when ¨= b ,
, p,
,
r.
5 5
5 5
There are ten petals with lengths {3, 5, 3, 5, 3, 5, 3, 5, 3, 5}.
48. r=7 when ¨= b
p p 9p 13p 17p
, ,
,
,
r and
10 2 10 10 10
y=3-3 sin x has minimum and maximum values of 0
and 6 on [0, 2∏]. So the range of the polar function
r=3-3 sin ¨ is also [0, 6].
60.
3p 7p 11p 3p 19p
,
,
,
,
r.
10 10 10 2 10
There are ten petals with lengths {7, 1, 7, 1, 7, 1, 7, 1, 7, 1}.
r=–1 when ¨= b
49. r1 and r2 produce identical graphs — r1 begins at (1, 0)
and r2 begins at (–1, 0).
50. r1 and r3 produce identical graphs — r1 begins at (3, 0)
and r2 begins at (1, 0).
51. r2 and r3 produce identical graphs — r1 begins at (3, 0)
and r3 begins at (–3, 0).
52. r1 and r2 produce identical graphs — r1 begins at (2, 0)
and r2 begins at (–2, 0).
53. (a) A 4-petal rose curve with 2 short petals of length
1 and 2 long petals of length 3.
(b) Symmetric about the origin.
(c) Maximum r-value: 3.
54. (a) A 4-petal rose curve with petals of about length 1, 3.3,
and 4 units.
(b) Symmetric about the y-axis.
(c) Maximum r-value: 4.
[0, 2] by [–1, 5]
y=2+3 cos x has minimum and maximum values of
–1 and 5 on [0, 2∏]. So the range of the polar function
r=2+3 cos ¨ is also [–1, 5].
In general, this works because any polar graph can also be
plotted using rectangular coordinates. Here, we have y
representing r and x representing ¨ on a rectangular coordinate graph. Since y is exactly equal to r, the range of y
and range of r will be exactly the same.
61. False. The spiral r=¨ is unbounded, since a point on the
curve can be found at any arbitrarily large distance from
the origin by setting ¨ numerically equal to that distance.
62. True. If point (r, ¨) satisfies the equation r=2+cos ¨, then
point (r, –¨) does also, since 2+cos(–¨)=2+cos ¨=r.
63. With r=a cos n¨, if n is even there are 2n petals.
The answer is D.
(b) Symmetric about the x-axis.
64. The four petals lie along the x- and y-axis, because cos 2¨
takes on its extreme values at multiples of ∏/2.
The answer is D.
(c) Maximum r-value: 4.
65. When cos ¨ =–1, r=5. The answer is B.
55. (a) A 6-petal rose curve with 3 short petals of length
2 and 3 long petals of length 4.
56. (a) A 6-petal rose curve with 3 short petals of length 2
and 3 long petals of length 4.
(b) Symmetric about the y-axis.
(c) Maximum r-value: 4.
57. Answers will vary but generally students should find that
a controls the length of the rose petals and n controls
both the number of rose petals and symmetry. If n is odd,
n rose petals are formed, with the cosine curve symmetric
about the polar x-axis and sine curve symmetric about the
y-axis. If n is even, 2n rose petals are formed, with both
the cosine and sine functions having symmetry about the
polar x-axis, y-axis, and origin.
66. With r=a sin n¨, if n is odd there are n petals.
The answer is B.
67. (a) Symmetry about the polar x-axis: r-a cos (n¨)=0
1 r-a cos (–n¨)=r-a cos (n¨) (since cos (¨)
is even, i.e., cos (¨)=cos (–¨) for all ¨)=0.
(b) No symmetry about y-axis: r-a cos (n¨)=0 1
–r-a cos (–n¨)=–r-a cos (n¨) (since cos (¨)
is even) Z r-a cos (n¨) unless r=0. As a result,
the equation is not symmetric about the y-axis.
(c) No symmetry about origin: r-a cos (n¨)=0 1
–r-a cos (n¨) Z –r-a cos (n¨) unless r=0.
As a result, the equation is not symmetric about
the origin.
(d) Since |cos (n¨)| 1 for all ¨, the maximum r-value
is |a|.
254
Chapter 6
Applications of Trigonometry
(e) Domain: All reals
Range: [–|a|, |a|]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 2
No asymptotes
68. (a) Symmetry about the y-axis: r-a sin (n¨)=0 1
–r-a sin (–n¨)=–r+a sin (n¨) (since sin (¨) is
odd, sin (–¨)=–sin (¨))=–1(r-a sin (n¨))
=(–1)(0)=0.
(b) Not symmetric about polar x-axis: r-a sin (n¨)=0
1 r-a sin (–n¨)=r+a sin (n¨). The two functions are equal only when –sin (n¨)=sin (n¨)=0,
or ¨={0, ∏}, so r-a sin (n¨) is not symmetric
about the polar x-axis.
(c) Not symmetric about origin: r-a sin (n¨) 1
–r-a sin (n¨)=–(r+a sin (n¨)). The two functions are equal only when r=0, so r-a sin (n¨))
is not symmetric about the origin.
(a)
[–3, 3] by [–3, 3]
(b)
[–3, 3] by [–3, 3]
(c)
(d) Since |sin (n¨)| 1 for all ¨, the maximum r-value
is |a|.
(e) Domain: All reals
Range: [–|a|, |a|]
Symmetric about y-axis
Continuous
Bounded
Maximum r-value: |a|
No asymptotes
[–3, 3] by [–3, 3]
71. Starting with the graph of r1, if we rotate counterclockwise (centered at the origin) by ∏/4 radians (45°), we get
the graph of r2; rotating r1 counterclockwise by ∏/3 radians (60°) gives the graph of r3.
(a)
69. (a) For r1: 0 ¨ 4∏ (or any interval that is 4∏ units
long). For r2: same answer.
(b) r1: 10 (overlapping) petals. r2: 14 (overlapping)
petals.
[–5, 5] by [–5, 5]
(b)
[–4, 4] by [–4, 4]
[–5, 5] by [–5, 5]
(c)
[–4, 4] by [–4, 4]
70. Starting with the graph of r1, if we rotate clockwise (centered at the origin) by ∏/12 radians (15°), we get the
graph of r2; rotating r1 clockwise by ∏/4 radians (45°)
gives the graph of r3.
[–5, 5] by [–5, 5]
Section 6.6
72. Starting with the graph of r1, if we rotate clockwise (centered at the origin) by ∏/4 radians (45°), we get the graph
of r2; rotating r1 clockwise by ∏/3 radians (60°) gives the
graph of r3.
(a)
[–7.05, 7.05] by [–4.65, 4.65]
(b)
x =
DeMoivre’s Theorem and nth Roots
255
- 1 - 62 ; 2 1 -6 2 2 - 4 15 2 15 2
21 52
6 ; 136 - 100
6 ; 1-64
6 ; 8i
=
=
10
10
10
6
8i
6 + 8i
=
+
= 0.6 + 0.8i and
x =
10
10
10
6 - 8i
6
8i
x =
=
+
= 0.6 - 0.8i
10
10
10
The roots are 0.6+0.8i and 0.6-0.8i.
=
3. (1+i)5=(1+i) # [(1+i)2]2=(1+i) # (2i)2
=–4(1+i)=–4-4i
4. (1-i)4=[(1-i)2]2=(–2i)2=–4=–4+0i
For #5–8, use the given information to find a point P on the
terminal side of the angle, which in turn determines the quadrant of the terminal side.
5. P(– 13, 1), in quadrant II: ¨=
[–7.05, 7.05] by [–4.65, 4.65]
6. P(1, –1), in quadrant IV: ¨=
(c)
5p
6
7p
4
7. P(–1, – 13,), in quadrant III: ¨=
8. P(–1, –1), in quadrant III: ¨=
[–7.05, 7.05] by [–4.65, 4.65]
73. The second graph is the result of rotating the first graph
clockwise (centered at the origin) through an angle of Å.
The third graph results from rotating the first graph counterclockwise through the same angle. One possible explanation: the radius r achieved, for example, when ¨=0 in
the first equation is achieved instead when ¨=–Å for the
second equation, and when ¨=Å for the third equation.
4p
3
5p
4
9. x3=1 when x=1
10. x4=1 when x= ; 1
Section 6.6 Exercises
1.
y
–2 + 2i
1 + 2i
i
x
■ Section 6.6 DeMoivre’s Theorem and
nth Roots
Quick Review 6.6
1. Using the quadratic equation to find the roots of
x2 + 13 = 4x, we have x2 - 4x + 13 = 0 with a=1,
b=–4, and c=13.
- 1 -4 2 ; 21 -4 2 2 - 4 11 2 113 2
x =
2112
4 ; 116 - 52
4 ; 1- 36
4 ; 6i
=
=
2
2
2
4
6i
4 + 6i
=
+
= 2 + 3i and
x =
2
2
2
4 - 6i
4
6i
x =
=
= 2 - 3i
2
2
2
The roots are 2+3i and 2-3i.
3–i
2.
y
1+i
x
3
–2 – i
=
2. Using the quadratic equation to find the roots of
5 1x2 + 12 = 6x, we have 5x2 - 6x + 5 = 0 with a=5,
b=–6, and c=5.
2 – 3i
For #3–12, a+bi=r(cos ¨+i sin ¨), where r=|a+bi|
a
= 2a2 + b2 and ¨ is chosen so that cos ¨=
2
2a + b2
b
and sin ¨=
.
2
2a + b2
256
Chapter 6
Applications of Trigonometry
p
3. r=|3i|=3; cos ¨=0 and sin ¨=1, so ¨= :
2
p
p
3i=3 a cos + i sin b
2
2
3p
4. r=|–2i|=2; cos ¨=0 and sin ¨=–1, so ¨= :
2
3p
3p
+ i sin
b
–2i=2 a cos
2
2
12
12
and sin ¨=
,
2
2
p
p
p
so ¨= : 2+2i=2 12 a cos + i sin b
4
4
4
19. (2 # 7)[cos(25°+130°)+i sin(25°+130°)]
=14 (cos 155°+i sin 155°)
20. ( 12 # 0.5)[cos(188°-19°)+i sin(118°-19°)]
=
21. (5 # 3) c cos a
=15 a cos
5. r=|2+2i|=2 12; cos ¨=
13
1
and sin ¨= ,
2
2
p
p
p
so ¨= : 13+i=2 a cos + i sin b
6
6
6
6. r=| 13+i|=2; cos ¨=
24.
12
12
and sin ¨=–
,
2
2
7p
7p
7p
so ¨= : 3-3i=3 12 a cos
+ i sin
b
4
4
4
25.
8. r=|3-3i|=3 12; cos ¨=
3
and
113
2
sin ¨=
, so ¨≠0.588: 3+2i
113
≠ 113(cos 0.59+i sin 0.59)
4
10. r=|4-7i|= 165; cos ¨=
and
165
7
sin ¨=–
, so ¨≠5.232: 4-7i
165
≠ 165 (cos 5.23+i sin 5.23)
9. r=|3+2i|= 113; cos ¨=
p
p
p
11. r=3; 30°= ; 3 a cos + i sin b
6
6
6
12. r=3; 225°=
13. 3 a
5p
5p
5p
+ i sin
b
; 4 a cos
4
4
4
3
3
13
1
- i b = 13- i
2
2
2
2
14. 8 a -
13
1
- i b =–4 13-4i
2
2
15. 5 a
1
13
5
5
i b = - 13 i
2
2
2
2
16. 5 a
12
12
5
5
i b = 12+ 12 i
2
2
2
2
17. 12 a -
13
1
16
12
- ib=
i
2
2
2
2
18. ≠2.56+0.68i
p
5p
p
5p
+
b + i sin a +
bd
4
3
4
3
23p
23p
+ i sin
b
12
12
1
p
3p
p
3p
+ b + i sin a
+ bd
22. a 13 # b c cos a
3
4
6
4
6
13
11p
11p
a cos
+ i sin
b
=
3
12
12
2
23. [cos(30°-60°)+i sin(30°-60°)]
3
2
2
= [cos(–30°)+i sin(–30°)]= (cos 30°-i sin 30°)
3
3
1
13
and sin ¨=
,
2
2
2p
2p
2p
so ¨= : –2+2i 13=4 a cos
+ i sin
b
3
3
3
7. r=|–2+2i 13|=4; cos ¨= -
12
(cos 99°+i sin 99°)
2
5
[cos(220°-115°)+i sin(220°-115°)]
2
5
= (cos 105°+ i sin 105°)
2
6
[cos(5∏-2∏)+i sin(5∏-2∏)]
3
=2(cos 3∏+i sin 3∏)=2(cos ∏ +i sin ∏)
26. 1 ccos a
p
p
p
p
p
p
- b +i sin a - b d=cos + i sin
2
4
2
4
4
4
27. (a) 3-2i≠ 113 [cos(5.695)+i sin(5.695)]
and 1+i= 12 a cos
p
p
+ i sin b , so
4
4
113 ccos(5.695)+i sin(5.695)d ? 12 ccos
p
p
+ i sin d
4
4
p
p
5 126 ccosa5.695+ b + i sin a5.695 + b d = 5 + i
4
4
1133cos1 5.695 2 + i sin1 5.695 2 4
123cos 1p>42 + i sin1p>4 2 4
p
p
L 16.5 c cos a 5.695 - b + i sin a 5.695 - b d
4
4
5
1
= - i
2
2
(b) (3-2i)(1+i)=3+3i-2i-2i2=5+i
3 - 2i # 1 - i
1 - 5i
1
5
3 - 2i
=
=
=
- i
1 + i
1 + i 1 - i
2
2
2
Section 6.6
7p
7p
+ i sin
b
4
4
p
p
and 13 + i = 2 a cos + i sin b , so
6
6
7p #
p
7p
p
12 a cos
+ i sin
b 2 a cos
+ i sin b
4
4
6
6
23p
23p
+ sin
b L 2.73 - 0.73i
= 2 12 a cos
12
12
28. (a) 1-i= 12 a cos
12 3cos17p>42 + i sin17p>4 2 4
2 3cos 1p>62 + i sin1p>62 4
19p
1
19p
+ i sin
b L 0.18 - 0.68i
=
a cos
12
12
12
(b) (1-i)( 13+i)= 13+i- 13i-i2
(1+ 13)+(1- 13)i≠2.73-0.73i
11 - i2 1 13 - i2
1 - i
1 - i # 13 - i
=
=
4
13 + i
13 + i 13 - i
=
1
B 13 - 1 - 1 13 + 1 2 iR ≠0.18-0.68i
4
29. (a) 3+i≠ 110[cos(0.321)+i sin(0.321)]
and 5-3i≠ 134[cos(–0.540)+i sin(–0.540)], so
110[cos(0.321)+i sin(0.321)] ? 134[cos(–0.540)
+i sin(–0.540)]
=2 185[cos(–0.219)+i sin(–0.219)]=18-4i
1103 cos 10.321 2 + i sin1 0.321 2 4
1343cos1 - 0.540 2 + i sin1 -0.540 2 4
5
L
3cos 10.862 2 + i sin1 0.862 2 4
A 17
≠ 0.35+0.41i
(b) (3+i)(5-3i)=15-9i+5i-3i2=18-4i
13 + i2 15 + 3i2
3 + i
3 + i # 5 + 3i
=
=
5 - 3i
5 - 3i 5 + 3i
34
1
1 6 + 7i2 ≠0.35+0.41i
17
30. (a) 2-3i≠ 113[cos(–0.982)+i sin(–0.982)],
p
p
and 1- 13i=2 ccos a - b +i sin a - b d , so
3
3
p
113[cos(–0.983)+i sin(–0.983)] ? 2ccos a - b
3
p
p
+i sin a - b d =2 113 ccos a–0.983- b
3
3
p
+i sin a–0.983- bd
3
≠–3.20-6.46i
1133cos1 - 0.983 2 + i sin1 -0.983 2 4
2 3cos1 - p>3 2 + i sin1 -p>3 2 4
113
3 cos1 0.064 2 + i sin10.064 2 4
L
2
≠1.80+0.12i
(b) (2-3i)(1- 13i)=2-2 13i-3i+3 13i2
=(2-3 13)-(2 13+3)i≠–3.196-6.464i
2 - 3i
2 - 3i # 1 + 13i
=
1 - 13i
1 - 13i 1 + 13i
12 - 3i2 11 + 13i2
=
4
1
= c 2 + 3 13 + 1213 - 3 2i d ≠1.80+0.12i
4
DeMoivre’s Theorem and nth Roots
257
p 3
3p
3p
p
+ i sin b = cos
+ i sin
4
4
4
4
12
12
=+ i
2
2
31. a cos
32. c 3 a cos
15p
3p
3p 5
15p
+ i sin
b d =243 a cos
+ i sin
b
2
2
2
2
=243i
3p
3p 3
9p
9p
+ i sin
b d =8 a cos
+ i sin
b
4
4
4
4
= 4 12 + 4 12i
5p 4
5p
34. c 6 a cos
+ i sin
bd
6
6
20p
20p
=1296 a cos
+ i sin
b
6
6
= -648 - 648 13i
33. c 2 a cos
p
p 5
+ i sin b d
4
4
5p
5p
5
= 1 12 2 a cos
+ i sin
b
4
4
5p
5p
=4 12 a cos
+ i sin
b =–4-4i
4
4
35. (1+i)5= c 12 a cos
36. (3+4i)20
20
4
4
= e 5 c cos tan-1 a b + i sin tan-1 a b d f
3
3
4
20
-1 4
=5 e cos c 20 tan a b d + i sin c 20 tan-1 a b d f
3
3
=520[cos(5.979)+i sin(5.979)]≠520(0.95-0.30i)
5p
5p 3
+ i sin
bd
3
3
=8(cos 5∏+i sin 5∏)=8(cos ∏+i sin ∏)=–8
37. 11 - 13i2 3= c 2 a cos
38. a
1
13 3
p
p 3
+ i
b = a cos + i sin b
2
2
3
3
= cos p + i sin p = -1
For #39–44, the cube roots of r(cos ¨+i sin ¨) are
u + 2kp
u + 2kp
3
= 1r
a cos
+ i sin
b , k = 0, 1, 2.
3
3
3
39. 12 a cos
2kp + 2p
2kp + 2p
+ i sin
b
3
3
2p 1 k + 1 2
2p1k + 1 2
3
+ i sin
b,
= 12 a cos
3
3
k=0, 1, 2:
2p
1
2p
13
3
3
12
a cos
+ i sin
b = 12 a - +
ib
3
3
2
2
1 + 13 i
=
,
3
14
4p
4p
3
12
a cos
+ i sin
b
3
3
1
13
-1 - 13 i
3
= 12
,
a- ib =
3
2
2
14
3
3
12 1cos 2p + i sin 2p2 = 12
3
40. 12
a cos
2kp + p>4
= 12 a cos
3
k=0, 1, 2:
+ i sin
3
p18k + 1 2
12
b
3
p 1 8k + 1 2
2kp + p>4
+ i sin
12
b,
Chapter 6
258
Applications of Trigonometry
p
3p
3p
p
+ i sin , cos
+ i sin
, –1,
5
5
5
5
7p
9p
9p
7p
+ i sin
, cos
+ i sin
cos
5
5
5
5
p
p
+ i sin b ,
12
12
3p
3p
3
12
a cos
+ i sin
b
4
4
1
1
-1 + i
3
= 12
a+
ib =
,
6
12
12
12
17p
17p
3
12
a cos
+ i sin
b
12
12
3
12
a cos
3
41. 13
a cos
2kp + 4p>3
= 13 a cos
3
+ i sin
3
2p1 3k + 2 2
9
b
3
2p1 3k + 2 2
cos
5
46. 132 a cos
=2 a cos
2kp + 4p>3
+ i sin
9
b,
k=0, 1, 2:
3
13
a cos
4p
4p
10p
10p
3
+ i sin
b , 13
acos
+ i sin
b,
9
9
9
9
16p
16p
3
13
a cos
+ i sin
b
9
9
3
42. 127
a cos
=3 a cos
2kp + 11p>6
3
p 112k + 11 2
+ i sin
+ i sin
2kp + 11p>6
3
p 112k + 11 2
b
b,
18
18
k=0, 1, 2:
11p
11p
23p
23p
3 a cos
+ i sin
b , 3 a cos
+ i sin
b,
18
18
18
18
35p
35p
3 a cos
+ i sin
b
18
18
43. 3-4i≠5 (cos 5.355+i sin 5.355)
2kp + 5.355
2kp + 5.355
3
15
a cos
+ i sin
b
3
3
k=0, 1, 2:
3
≠ 15
1cos 1.79 + i sin 1.792 ,
3
≠ 15
1cos 3.88 + i sin 3.882 ,
3
≠ 15 1cos 5.97 + i sin 5.972
44. –2+2i=2 12 a cos
3p
3p
+ i sin
b . Note that
4
4
3
22
12 = 12.
2kp + 3p>4
2kp + 3p>4
12 a cos
+ i sin
b
3
3
p18k + 3 2
p18k + 3 2
+ i sin
b ,
= 12 a cos
12
12
k=0, 1, 2:
p
p
12
i12
12 a cos + i sin b = 12 a
+
b = 1 + i,
4
4
2
2
11p
11p
12 a cos
+ i sin
b,
12
12
19p
19p
+ i sin
b
12 a cos
12
12
For #45–50, the fifth roots of r (cos ¨+i sin ¨) are
5
a cos
1r
u + 2kp
u + 2kp
+ i sin
b , k=0, 1, 2, 3, 4.
5
5
2kp + p
2kp + p
+ i sin
5
5
p1 2k + 1 2
p12k + 1 2
+ i sin
, k = 0, 1, 2, 3, 4:
= cos
5
5
45. cos
2kp + p>2
5
p14k + 1 2
+ i sin
2kp + p>2
5
p14k + 12
b
+ i sin
b,
10
10
k=0, 1, 2, 3, 4:
p
p
p
p
+ i sin b , 2 a cos + i sin b =2i,
2 a cos
10
10
2
2
9p
9p
13p
13p
2 a cos
+ i sin
b , 2 a cos
+ i sin
b,
10
10
10
10
17p
17p
+ i sin
b
2 a cos
10
10
5
47. 12 a cos
2kp + p>6
5
= 12
a cos
+ i sin
5
p112k + 12
30
b
5
p112k + 12
2kp + p>6
+ sin
30
b,
k=0, 1, 2, 3, 4:
p
p
13p
13p
5
5
12
a cos
+ i sin b , 12
a cos
+ i sin
b,
30
30
30
30
5p
5p
37p
37p
5
5
12
a cos
+ i sin
b , 12
a cos
+ i sin
b,
6
6
30
30
49p
49p
5
12
a cos
+ i sin
b
30
30
5
48. 12 a cos
2kp + p>4
= 12 a cos
5
+ i sin
5
p18k + 12
20
b
5
p18k + 1 2
2kp + p>4
+ i sin
20
b,
k=0, 1, 2, 3, 4:
p
p
9p
9p
5
5
12
a cos
+ i sin b , 12
a cos
+ i sin
b,
20
20
20
20
17p
5p
17p
5p
5
5
12
a cos
+ i sin
b , 12 a cos
+ i sin
b =
20
20
4
4
1>5
1>5
1
1
-2 - 2 i
5
12
aib=
=
12
12
2 1>2
-1 - i
-1 - i
=
,
10
2 3>10
18
33p
33p
5
+ i sin
b
12
a cos
20
20
5
49. 12 a cos
2kp + p>2
+ i sin
b
5
5
p14k + 12
p14k + 1 2
5
+ i sin
b,
= 12 a cos
10
10
k=0, 1, 2, 3, 4:
p
p
p
p
5
5
5
12
a cos
+ i sin b , 12
a cos + i sin b = 12
i,
10
10
2
2
9p
13p
9p
13p
5
5
+ i sin
b , 12
+ i sin
b,
12
a cos
a cos
10
10
10
10
17p
17p
5
+ i sin
b
12
a cos
10
10
2kp + p>2
Section 6.6
5
50. 12
a cos
2kp + p>3
= 12 a cos
5
+ i sin
5
p1 6k + 1 2
15
2kp + p>3
+ i sin
15
b,
k=0, 1, 2, 3, 4:
p
p
7p
7p
5
5
12
a cos
+ i sin b , 12
a cos
+ i sin
b,
15
15
15
15
13p
19p
13p
19p
5
5
+ i sin
b , 12
+ i sin
b,
12
a cos
a cos
15
15
15
15
5p
5p
1
13i
5
12
a cos
+ i sin
b =2 1>5 a b =
3
3
2
2
1 - 13i
1 - 13i
=
5
2 4>5
116
For #51–56, the nth roots of r(cos ¨+i sin ¨)are
n
a cos
1r
259
3p
3p
+ i sin
b , so the roots are
4
4
2kp + 3p>4
2kp + 3p>4
4
2212
a cos
+ i sin
b
4
4
p18k + 32
p18k + 3 2
8
= 18
a cos
+ i sin
b,
16
16
k=0, 1, 2, 3:
3p
3p
11p
11p
8
8
18
a cos
+ i sin
b , 18
a cos
+ i sin
b,
16
16
16
16
27p
19p
19p
27p
8
8
18
a cos
+ i sin
b , 18 a cos
+ i sin
b
16
16
16
16
54. -2 + 2i = 212 a cos
b
5
p16k + 1 2
DeMoivre’s Theorem and nth Roots
u + 2kp
u + 2kp
+ i sin
b , k=0, 1, 2, . . . , n-1.
n
n
51. 1 + i = 12 a cos
p
p
+ i sin b , so the roots are
4
4
2kp + p>4
2kp + p>4
4
212 a cos
+ i sin
b
4
4
p1 8k + 1 2
p18k + 1 2
8
= 12
a cos
+ i sin
b,
16
16
k=0, 1, 2, 3:
p
p
9p
9p
8
8
12
a cos
+ i sin b , 12
a cos
+ i sin
b,
16
16
16
16
25p
17p
17p
25p
8
8
a cos
12
a cos
+ i sin
b , 12
+ i sin
b
16
16
16
16
52. 1-i= 12 a cos
7p
7p
+ i sin
b , so the roots are
4
4
2kp + 7p>4
2kp + 7p>4
6
212
a cos
+ i sin
b
6
6
p1 8k + 7 2
p 18k + 7 2
12
= 12 a cos
+ i sin
b,
24
24
k=0, 1, 2, 3, 4, 5:
7p
5p
7p
5p
12
12
12 a cos
+ i sin
b , 12 a cos
+ i sin
b,
24
24
8
8
23p
31p
23p
31p
12
12
+ i sin
b , 12 a cos
+ i sin
b,
12 a cos
24
24
24
24
47p
13p
13p
47p
12
12
12 a cos
+ i sin
b , 12 a cos
+ i sin
b
8
8
24
24
p
p
+ i sin b , so the roots are
4
4
2kp
+
p>4
2kp + p>4
3
22
12 a cos
+ i sin
b
3
3
p18k + 1 2
p1 8k + 1 2
= 12 a cos
+ i sin
b , k=0, 1, 2:
12
12
p
p
+ i sin b , –1+i,
12 a cos
12
12
17p
17p
12 a cos
+ i sin
b
12
12
55. -2i = 2 a cos
3p
3p
+ i sin
b , so the roots are
2
2
2kp + 3p>2
2kp + 3p>2
6
12
a cos
+ i sin
b
6
6
p14k + 32
p14k + 3 2
6
= 12
a cos
+ i sin
b,
12
12
k=0, 1, 2, 3, 4, 5:
1 + i 6
7p
7p
, 12 a cos
+ i sin
b,
6
12
12
14
11p
11p
5p
5p
6
6
12
a cos
+ i sin
b , 12
a cos
+ i sin
b,
12
12
4
4
19p
19p
23p
23p
6
6
12
a cos
+ i sin
b , 12
a cos
+ i sin
b
12
12
12
12
56. 32=32(cos 0+i sin 0), so the roots are
2kp + 0
2kp + 0
5
132
+ i sin
b
a cos
5
5
2kp
2kp
+ i sin
b , k=0, 1, 2, 3, 4:
= 2 a cos
5
5
2p
2p
+ i sin
b,
21 cos 0 + i sin 02 = 2, 2 a cos
5
5
4p
4p
6p
6p
2 a cos
+ i sin
b , 2 a cos
+ i sin
b,
5
5
5
5
8p
8p
2 a cos
+ i sin
b
5
5
For #57–60, the nth roots of unity are
2kp
2kp
cos
+ i sin
, k=0, 1, 2, . . . , n-1.
n
n
57. 1, -
13
1
;
i
2
2
y
53. 2 + 2i = 2 12 a cos
0.5
0.5
x
260
Chapter 6
Applications of Trigonometry
58. —1, —i
60. 1, –1
y
y
0.5
0.5
0.5
59. _1,
x
1
13
1
13
;
i, - ;
i
2
2
2
2
y
x
3
p
p
+ i sin b R
3
3
=8(cos ∏+i sin ∏)=–8; the cube roots are
–2 and 1_ 13i.
61. z = 11 + 13i2 3 = B2 a cos
4
5p
5p
+ i sin
bR
4
4
=64(cos 5∏+i sin 5∏)=–64; the fourth roots are
2_2i and –2_2i.
62. z=(–2-2i)4= B 212 a cos
0.5
0.5
63.
0.5
x
r1 1cos u1 + i sin u1 2
r1 cos u1 + i sin u1 cos u2 - i sin u2
z1
#
=
= #
z2
r2 1cos u2 + i sin u2 2
r2 cos u2 + i sin u2 cos u2 - i sin u2
r1 cos u1 cos u2 + sin u1 sin u2 + i1sin u1 cos u2 - cos u1 sin u2 2
= #
r2
1cos u2 2 2 + 1sin u2 2 2
r1
= # 3 cos u1 cos u2 + sin u1 sin u2 + i1sin u1 cos u2 - cos u1 sin u2 2 4.
r2
Now use the angle difference formulas:
cos(¨1-¨2)=cos ¨1 cos ¨2+sin ¨1 sin ¨2 and sin(¨1-¨2)=sin ¨1 cos ¨2-cos ¨1 sin ¨2.
r1
z1
So
= 3cos1u1 - u2 2 + i sin1 u1 - u2 2 4 .
z2
r2
n
64. The n nth roots are given by 1r
a cos
2kp + u
2kp + u
+ i sin
b , k=0, 1, 2, . . . n-1:
n
n
u
u
u + 2p
u + 2p
n
n
1r
a cos + i sin b , 1r
a cos
+ i sin
b,
n
n
n
n
u + 2 1n - 1 2p
u + 21 n - 12p
u + 4p
u + 4p
n
n
1r
a cos
+ i sin
b , . . . , 1r
+ i sin
b.
a cos
n
n
n
n
2p
2p
The angles between successive values in this list differ by
radians, while the first and last roots differ by 2∏- ,
n
n
2p
n
which also makes the angle between them
. Also, the modulus of each root is 1r, placing it at that distance from
n
n
the origin, on the circle with radius 1r
.
65. False. If z=r(cos ¨+i sin ¨), then it is also true that z=r[cos (¨+2n∏)+i sin (¨+2n∏)] for any integer n.
∏
9∏
∏
9∏
For example, here are two trigonometric forms for 1 + i : 12 a cos + i sin b , 12 a cos
+ i sin
b.
4
4
4
4
66. True. i3 = i2 # i = -i, so i is a cube root of -i.
67. 2 a cos
2∏
1
13
2∏
+ i sin
b = 2a - + i
b = -1 + 13i
3
3
2
2
The answer is B.
Section 6.6
DeMoivre’s Theorem and nth Roots
68. Any complex number has n distinct nth roots, so 1+i has five 5th roots. The answer is E.
69. 12 a cos
∏
∏
7∏
7∏
+ i sin b # 12 a cos
+ i sin
b
4
4
4
4
∏
7∏
∏
7∏
= 1 12 # 122 c cos a +
b + i sin a +
bd
4
4
4
4
= 21 cos 2∏ + i sin 2∏2
=2
The answer is A.
70. 1 1i2 4 = 3 1 1i2 2 4 2 = i2 = - 1 Z 1; The answer is E.
b
71. (a) a+bi=r(cos ¨+i sin ¨), where r = 2a2 + b2 and ¨=tan–1 a b . Then a+(–bi)=rœ(cos ¨œ+i sin ¨œ), where
a
b
-b
b
b . Since rœ= 2a2 + b2 = r and ¨œ=tan–1 a
b =–tan–1 a b =–¨, we
rœ= 2a2 + 1 -b 2 2 and ¨=tan–1 a
a
a
a
have a-bi=r(cos (–¨)+i sin (–¨))
(b) z # z=r[cos ¨+i sin ¨] # r[cos (–¨)+i sin (–¨)]
=r2[cos ¨ cos (–¨)+i (sin (–¨))(cos ¨)+i (sin ¨)(cos (–¨))-(sin ¨)(sin (–¨))]
Since sin ¨ is an odd function (i.e., sin(–¨)=–sin(¨)) and cos ¨ is an even function (i.e., cos (–¨)=cos ¨), we have
z # z=r2[cos2 ¨-i(sin ¨)(cos ¨)+i(sin ¨)(cos ¨)+sin2 ¨]
=r2[cos2 ¨+sin2 ¨]
=r2
(c)
r3 cos u + i sin u4
z
=
=cos[¨-(–¨)]+i sin[¨-(–¨)]
z
r3 cos 1 -u 2 + i sin1 -u 2 4
=cos (2¨)+i sin (2¨)
(d) –z=–(a+bi)=–a+(–bi)=r(cos ¨+i sin ¨), where r = 2 1 - a2 2 + 1 -b 2 2 and
-b
b
¨=tan–1 a
b = tan-1 a b
-a
a
Recall, however, that (–a, –b) is in the quadrant directly opposite the quadrant that holds (a, b) (i.e., if (a, b) is in
Quadrant I, (–a, –b) is in Quadrant III, and if (a, b) is in Quadrant II, (–a, –b) is in Quadrant IV). Thus,
–z= 2a2 + b2 (cos(¨+∏)+i sin (¨+∏))=r(cos(¨+∏)+i sin(u+∏)).
72. (a) @ [email protected] = 21r cos u 2 2 + 1r sin u 2 2
= 2r2 cos2 u + r2 sin2 u
= @ [email protected] 2cos2 u + sin2 u .
= @ [email protected]
(b) @ z1 # z2 @ =
=
=
=
@ 3 r1 cos u1 + 1r1 sin u1 2i4 # 3 r2 cos u2 + 1r2 sin u2 2 i4 @
@ r1r2 cos u1 cos u2 + 1r1r2 cos u1 sin u2 2i + 1r1r2 sin u1 cos u2 2 i + 1 r1r2 sin u1 u2 2i2 @
@ r1r2 3cos ¨1 cos ¨2 - sin ¨1 sin ¨2 + 1cos ¨1 sin ¨2 + sin ¨1 cos ¨2 2i4 @
@ r1r2 3cos 1 u1 + u2 2 + 1sin 1u1 + u2 2 2i4 @
= 21 r1r2 2 2 3cos2 1u1 + u2 2 + sin2 1 u1 + u2 2 4
= 2 1 r1r2 2 2
= @ r1 @ # @ r2 @
= @ z1 @ # @ z1 @
73. (a)
(b)
by part (a)
(c)
261
262
Chapter 6
Applications of Trigonometry
74. (a)
y
z1z2
␪1 ⫹ ␪2
z2
(b) Yes. 6∏/8, 10∏/8, 14∏/8
(c) For fifth and seventh roots of unity, all of the roots
except the complex number 1 generate the
corresponding roots of unity. For the sixth roots
of unity, only 2∏/6 and 10∏/6 generate the sixth
roots of unity.
(d) 2∏k/n generates the nth roots of unity if and only if k
and n have no common factors other than 1.
75. Using tan–1 a
1
b L 0.62, we have
12
12 + i L 13 (cos 0.62+i sin 0.62), so graph
x(t)= 1 132 t cos(0.62t) and
y(t)= 1 132 t sin(0.62t).
Use Tmin=0, Tmax=4, Tstep=1.
Shown is [–7, 2] by [–0, 6].
z1
␪2
[–2.4, 2.4] by [–1.6, 1.6]
0
␪1
1
x
78. Construct an angle with vertex at z2, and one ray from z2
to 0, congruent to the angle formed by 0, 1, and z1, with
vertex 1. Also, be sure that this new angle is oriented in
the appropriate direction: e.g., if z1 is located “counterclockwise” from 1 then the points on this new ray should
also be located counterclockwise from z2. Now similarly
construct an angle with vertex at 0, and one ray from 0 to
z2, congruent to the angle formed by 1, 0, and z1, with
vertex 0. the intersection of the two newly constructed
rays is z1z2.
79. The solutions are the cube roots of 1:
2pk
2pk
cos a
b + i sin a
b , k=0, 1, 2 or
3
3
1
13
1
13
1, - +
i, - i
2
2
2
2
80. The solutions are the fourth roots of 1:
pk
pk
cos a
b + i sin a
b , k=0, 1, 2, 3 or
2
2
–1, 1, –i, i
[–7, 2] by [0, 6]
3p
3p
76. –1+i= 12 a cos
+ i sin b , so graph
4
4
x(t)= 1 122 t cos(0.75∏t) and y(t)= 1 122 t sin(0.75∏t)
Use Tmin=0, Tmax=4, Tstep=1.
Shown is [–5, 3] by [–3, 3].
[–5, 3] by [–3, 3]
77. Suppose that z1=r1(cos ¨1+i sin ¨1) and
z2=r2(cos ¨2+i sin ¨2). Each triangle’s angle at the origin has the same measure: For the smaller triangle, this
angle has measure ¨1; for the larger triangle, the side from
the origin out to z2 makes an angle of ¨2 with the x-axis,
while the side from the origin to z1z2 makes an angle of
¨1+¨2, so that the angle between is ¨1 as well.
The corresponding side lengths for the sides adjacent to
these angles have the same ratio: two longest side have
lengths |z1|=r1 (for the smaller) and |z1z2|=r1r2, for a
ratio of r2. For the other two sides, the lengths are 1 and
r2, again giving a ratio of r2. Finally, the Law of Sines can
be used to show that the remaining side have the same
ratio.
81. The solutions are the cube roots of –1:
p + 2pk
p + 2pk
cos a
b + i sin a
b , k=0, 1, 2 or
3
3
1
13 1
13
-1, +
i, i
2
2 2
2
82. The solutions are the fourth roots of –1:
p + 2pk
p + 2pk
b + i sin a
b , k=0, 1, 2, 3 or
cos a
4
4
12
12
12 12
12
12
12
12
+
i,
i, +
i, i
2
2
2
2
2
2
2
2
83. The solutions are the fifth roots of –1:
p + 2pk
p + 2pk
b + i sin a
b , k=0, 1, 2, 3, 4, or
cos a
5
5
–1, ≠0.81+0.59i, 0.81-0.59i, –0.31+0.95i,
–0.31-0.95i
84. The solutions are the fifth roots of 1:
2pk
2pk
cos a
b + i sin a
b , k=0, 1, 2, 3, 4 or
5
5
1, ≠0.31+0.95i, 0.31-0.95i, –0.81+0.59i,
–0.81-0.59i
■ Chapter 6 Review
1. u-v=˚2-4, –1-2¬=˚–2, –3¬
2. 2u-3w=˚4-3, –2+9¬=˚1, 7¬
3. |u+v|= 21 2 + 4 2 2 + 1 -1 + 2 2 2 = 137
4. |w-2u|= 211 - 4 2 2 + 1 -3 + 2 2 2 = 110
Chapter 6
5. u # v=8-2=6
¡
7. 3AB =3˚3-2, 1-(–1)¬=˚3, 6¬;
@ 3AB @ = 232 + 62 = 145 = 315
¡
¡
8. AB +CD=˚3-2, 1-(–1)¬+˚1-(–4), –5-2¬
=˚6, -5¬; @ AB +CD @ = 262 + 52 = 161
¡
¡
¡
¡
9. AC +BD =˚–4-2, 2-(–1)¬+˚1-3, –5-1¬
=˚–8, –3¬; @ AC +BD @ = 282 + 32 = 173
¡
¡
¡
¡
10. CD -AB =˚1-(–4), –5-2¬-˚3-2, 1-(–1)¬
=˚4, –9¬; @ CD +AB @ = 242 + 92 = 197
¡
¡
11. (a)
AB
@ AB @
(b) - 3 #
¡
12. (a)
=
¡
-2, 1
=
15
2
2
1
,
=X Y
15 15
¡
-2 1
6
3
,
,Y=X
Y
¡ =–3 X
15
15
15
15
@ BA @
AB
@ AB @
(b) - 3 #
2 1 -22 + 1
2
AB
¡
¡
-2, 1
2, 0
=
=
22 + 0
2
2
2, 0
2
=˚1, 0¬
¡
AB
@ BA @
¡
=–3 ˚1, 0¬=˚–3, 0¬
b
For #13–14, the direction angle ¨ of ˚a, b¬ has tan ¨= ;
a
b
start with tan–1 a b , and add (or subtract) 180° if
a
the angle is not in the correct quadrant. The angle between two
vectors is the absolute value of the difference between their
angles; if this difference is greater than 180°, subtract it from 360°.
3
5
13. (a) ¨u=tan–1 a b ≠0.64, ¨v=tan–1 a b ≠1.19
4
2
(b) ¨v-¨u≠0.55.
14. (a) ¨u=∏+tan–1(–2)=cos–1 a -
1
b ≠2.03,
15
2
¨v=tan–1 a b ≠0.59
3
(b) ¨u-¨v≠1.45
15. (–2.5 cos 25°, –2.5 sin 25°)≠(–2.27, –1.06)
16. (–3.1 cos 135°, –3.1 sin 135°)=(1.55 12, –1.55 12)
17. (2 cos(–∏/4), 2 sin(–∏/4))=( 12, – 12)
18. (3.6 cos(3∏/4), 3.6 sin(3∏/4))=(–1.8 12, 1.8 12)
19. a1, –
2p
2p
+(2n+1)∏b and a–1, –
+2n∏b, n
3
3
an integer
20. a2,
5p
5p
+(2n+1)∏b and a–2,
+2n∏b, n an
6
6
integer
3
21. (a) a– 113, ∏+tan–1 a - b b≠(– 113, 2.16) or
2
3
–1
a 113, 2∏+tan a - b b≠( 113, 5.30)
2
3
(b) a 113, tan a - b b≠( 113, –0.98) or
2
3
a– 113, ∏+tan–1 a - b b≠(– 113, 2.16)
2
–1
263
(c) The answers from (a), and also
3
a– 113, 3∏+tan–1 a - b b≠(– 113, 8.44) or
2
3
a113, 4∏+tan–1 a - b b≠( 113, 11.58)
2
6. u # w=2+3=5
¡
Review
22. (a) (–10, 0) or (10, ∏) or (–10, 2∏)
(b) (10, –∏) or (–10, 0) or (10, ∏)
(c) The answers from (a), and also (10, 3∏) or (–10, 4∏)
23. (a) (5, 0) or (–5, ∏) or (5, 2∏)
(b) (–5, –∏) or (5, 0) or (–5, ∏)
(c) The answers from (a), and also (–5, 3∏) or (5, 4∏)
24. (a) a -2,
3p
p
b or a 2,
b
2
2
(b) a 2, -
p
p
b or a -2, b
2
2
(c) The answers from (a), and also a -2,
or a 2,
7p
b
2
5p
b
2
1
3
1
3
25. t= - x + , so y=4+3 a - x + b
5
5
5
5
29
3
:
=- x +
5
5
29
3
b with slope m = Line through a 0,
5
5
26. t=x-4, so y=–8-5(x-4)=–5x+12,
1 x 9: segment from (1, 7) to (9, –33).
27. t=y+1, so x=2(y+1)2+3: Parabola that opens to
right with vertex at (3, –1).
28. x2+y2=(3 cos t)2+(3 sin t)2=
9 cos2 t+9 sin2 t=9, so x2+y2=9: Circle of radius 3
centered at (0, 0).
ln 1 x + 1 2
1
29. x+1=e2t, t =
, so y= e 2ln1x + 12 = eln1x + 1
2
= 1x + 1 : square root function starting at (–1, 0)
3
3
30. t = 1x, so y = ln 1 1x2 = ln x1>3 =
1
ln x : the
3
logarithmic function, with asymptote at x=0
4 - 1 -22
6
3
= = , so ¢ x=2 when ¢ y=3.
3 - 1 -12
4
2
One possibility for the parametrization of the line is:
x=2t+3, y=3t+4.
31. m=
1 - 3
-2
, so ¢ x=7 when ¢ y=–2.
5 - 1 -22
7
One possibility for the parametrization of the segment is:
x=7t+5, y=–2t+1, –1 t 0. Another possibility is x=7t-2, y=–2t+3, 0 t 1.
32. m=
33. a=–3, b=4, |z1|= 232 + 42=5
34. z1=5 e cos c cos-1 a -
3
3
b d + i sin c cos-1 a - b d f
5
5
≠5[cos (2.21)+i sin (2.21)]
35. 6 (cos 30°+i sin 30°)=6 a
13
1
+ i b = 3 13 + 3i
2
2
36. 3 (cos 150°+i sin 150°)=3 a = -1.513 + 1.5i
1
13
+ ib
2
2
264
Chapter 6
Applications of Trigonometry
4p
4p
13
1
+ i sin
b =2.5 a - ib
3
3
2
2
=–1.25-1.25 13 i
37. 2.5 a cos
6
p
p
p
p
+ i sin b d = 76 a cos + i sin b
24
24
4
4
p
p
= 117, 649 a cos + i sin b
4
4
48. (a) c 7 a cos
38. 4 (cos 2.5+i sin 2.5)≠–3.20+2.39i
39. 3-3i=3 12 a cos
7p
7p
+ i sin
b . Other
4
4
7p
representations would use angles
+2n∏, n an integer.
4
40. –1+i12 = 13 e cos c cos-1 a -
1
bd
13
+ i sin c cos-1 a -
1
b d f ≠ 13[cos (2.19)
13
+i sin (2.19)]. Other representations would use
angles 2.19+2n∏, n an integer.
41. 3-5i= 134 e cos c tan-1 a + i sin c tan-1 a -
5
bd
3
5
bdf
3
L 134 3 cos 1 -1.03 2 +i sin(–1.03)]
≠ 134[cos (5.25+i sin (5.25)]
Other representations would use angles≠5.25+2n∏,
n an integer.
42. –2-2i=2 12 a cos
5p
5p
+ i sin
b . Other
4
4
5p
representations would use angles
+2n∏, n an integer.
4
43. z1 z2 =(3)(4) [cos (30°+60°)+i sin (30°+60°)]
=12 (cos 90°+i sin 90°)
3
z1/z2= [cos (30°-60°)+i sin (30°-60°)]
4
3
= [cos (–30°)+i sin (–30°)]
4
3
= (cos 330°+i sin 330°)
4
44. z1 z2 =(5)(–2) [cos (20°+45°)+i sin (20°+45°)]
=–10 (cos 65°+i sin 65°)
=10 (cos 245°+i sin 245°)
5
z1/z2=
[cos (20°-45°)+i sin (20°-45°)]
-2
=–2.5[cos (–25°)+i sin (–25°)]
=2.5 (cos 155°+i sin 155°)
5p
p 5
5p
p
+ i sin b d = 35 a cos
+ i sin
b
4
4
4
4
5p
5p
= 243 a cos
+ i sin
b
4
4
45. (a) c 3 a cos
(b)
117, 649 12
117, 64912
+
i
2
2
For #49–52, the nth roots of r (cos ¨+i sin ¨) are
2kp + u
2kp + u
n
1r
a cos
+ i sin
b , k=0, 1, 2, . . . ,
n
n
n-1.
49. 3+3i=312 a cos
p
p
+ i sin b , so the roots are
4
4
2kp + p>4
2kp + p>4
4
2312 a cos
+ i sin
b
4
4
p18k + 1 2
p1 8k + 12
8
= 118
+ i sin
b,
a cos
16
16
k=0, 1, 2, 3:
p
p
8
118
acos
+ i sin b ,
16
16
9p
9p
8
118
acos
+ i sin
b,
16
16
17p
17p
8
118
acos
+ i sin
b,
16
16
25p
25p
8
118
acos
+ i sin
b
16
16
y
1
1
50. 8=8 (cos 0+i sin 0), so the roots are
2kp + 0
2kp + 0
3
18
a cos
+ i sin
b
3
3
2kp
2kp
+ i sin
b , k = 0, 1, 2 :
= 2 a cos
3
3
2 1cos 0 + i sin 02 = 2 ,
2p
2p
+ i sin
b,
2 a cos
3
3
4p
4p
2 a cos
+ i sin
b
3
3
y
243 12
243 12
i
(b) 2
2
8
p
2p
p
2p
+ i sin b d = 28 a cos
+ i sin
b
12
12
3
3
2p
2p
= 256 a cos
+ i sin
b
3
3
46. (a) c 2 a cos
(b) –128+128 13 i
5p
5p 3
+ i sin
b d = 53 1 cos 5p + i sin 5p2
3
3
= 125 1cos p + i sin p 2
47. (a) c 5 a cos
(b) -125 + 0i = -125
x
1
1
x
Chapter 6
51. 1=cos 0+i sin 0, so the roots are
2kp
2kp + 0
2kp + 0
2kp
cos
+ i sin
+ i sin
=cos
,
5
5
5
5
k=0, 1, 2, 3, 4:
2p
2p
,
cos 0 + i sin 0 = 1 , cos
+ i sin
5
5
4p
4p
6p
6p
, cos
cos
+ i sin
+ i sin
5
5
5
5
8p
8p
cos
+ i sin
5
5
y
63. r2+3r cos ¨+2r sin ¨=x2+y2+3x+2y=0.
13
3 2
Completing the square: a x + b + 1y + 12 2 =
2
4
3
113
— a circle of radius
with center a - , -1 b
2
2
3
3
= 1 = 0 1 x - 3 = 0, x = 3 — a verr cos u
x
tical line through (3, 0)
64. 1 -
0.5
-4
= - 4 csc u
sin u
x
52. –1=1 (cos ∏+i sin ∏), so the roots are
2kp + p
2kp + p
+ i sin
cos
6
6
p 12k + 1 2
p 12k + 1 2
=cos
,
+ i sin
6
6
k=0, 1, 2, 3, 4, 5:
p
p
p
p
cos + i sin , cos + i sin =i,
6
6
2
2
5p
5p
7p
7p
, cos
,
cos
+ i sin
+ i sin
6
6
6
6
3p
3p
11p
11p
=–i, cos
cos
+ i sin
+ i sin
2
2
6
6
y
[–10, 10] by [–10, 10]
66. r=
5
= 5 sec u
cos u
[–10, 10] by [–10, 10]
67. (r cos ¨-3)2+(r sin ¨+1)2=10, so
r2(cos2 ¨+sin2 ¨)+r(–6 cos ¨+2 sin ¨)+10
=10, or r=6 cos ¨-2 sin ¨
0.5
0.5
x
[–3, 9] by [–5, 3]
53. Graph (b)
68. 2r cos ¨-3r sin ¨=4, r =
54. Not shown
55. Graph (a)
56. Not shown
57. Not shown
58. Graph (d)
59. Graph (c)
60. Not shown
61. x2+y2=4 — a circle with center (0, 0) and radius 2
265
62. r2+2r sin ¨=x2+y2+2y=0. Completing the
square: x2+(y2+1)2=1 — a circle of radius 1 with
center (0, –1)
65. r=
0.5
Review
[–4.7, 4.7] by [–3.1, 3.1]
4
2 cos u - 3 sin u
266
Chapter 6
Applications of Trigonometry
69.
73. (a) r=a sec ¨ 1
r
=a 1 r cos ¨=a 1 x=a.
sec u
(b) r=b csc ¨ 1
r
=b 1 r sin ¨=b 1 y=b.
csc u
(c) y=mx+b 1 r sin ¨=mr cos ¨+b 1
b
r(sin ¨-m cos ¨)= b 1 r=
.
sin u - m cos u
[–7.5, 7.5] by [–8, 2]
The domain of r is any value of ¨ for which
sin ¨ Z m cos ¨ 1 tan ¨ Z m 1 ¨ Z arctan (m).
Domain: All reals
Range: [–3, 7]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 7
No asymptotes
(d)
70.
[–9, 2] by [–6, 6]
74. (a) v=540 ˚sin 80°, cos 80°¬≠˚531.80, 93.77¬
(b) The wind vector is w=55 ˚sin 100°, cos 100°¬
≠˚54.16, –9.55¬. Actual velocity vector:
v+w≠˚585.96, 84.22¬. Actual speed: ßv+wß
[–12, 6] by [–6, 6]
Domain: All reals
Range: [0, 8]
Symmetric about the x-axis
Continuous
Bounded
Maximum r-value: 8
No asymptotes
≠ 2585.962 + 84.222≠591.98 mph. Actual
585.96
b ≠82.82°
bearing: tan–1 a
84.22
75. (a) v=480 ˚sin 285°, cos 285°¬≠˚–463.64, 124.23¬
(b) The wind vector is w=30 ˚sin 265°, cos 265°¬
≠˚–29.89, –2.61¬. Actual velocity vector:
v+w≠˚–493.53, 121.62¬. Actual speed: ßv+wß
71.
≠ 2493.532 + 121.622≠508.29 mph. Actual
-493.53
b ≠283.84°
bearing: 360°+tan–1 a
121.62
76. F=˚120 cos 20°, 120 sin 20°¬+˚300 cos(–5°),
300 sin(–5°)¬≠˚411.62, 14.90¬, so
ßFß≠ 2411.622 + 14.902 ≠411.89 lb and
14.90
b ≠2.07°
¨=tan–1 a
411.62
[–3, 3] by [–2.5, 1.5]
Domain: All reals
Range: [–2, 2]
Symmetric about the y-axis
Continuous
Bounded
Maximum r-value: 2
No asymptotes
72.
77.
16˚ –3000
Domain: c 0,
p
3p
d ´ c p,
d
2
2
Range: [0, 12]
Symmetric about the origin
Bounded
Maximum r-value: 12
No asymptotes
F2
F1
F2 Force perpendicular to
the street
(a) F1=–3000 sin 16°≠–826.91, so the force required
to keep the car from rolling down the hill is approximately 826.91 pounds.
(b) F2=–3000 cos 16°≠2883.79, so the force perpendicular to the street is approximately 2883.79 pounds.
78. F=36 #
[–2.35, 2.35] by [–1.55, 1.55]
F1 Force to keep car from
going downhill
16˚
3, 5
23 + 5
2
2
=
108, 180
134
Since AB =˚10, 0¬, F # AB =(10) a
¡
=
¡
1080
≠185.22 foot-pounds.
134
108
b +0
134
Chapter 6
79. (a) h=–16t2+v0 t+s0=–16t2+245t+200
(b) Graph and trace: x=17 and
y=–16t2+245t+200 with 0 t 16.1 (upper
limit may vary) on [0, 18] by [0, 1200]. This graph will
appear as a vertical line from about (17, 0) to about
(17, 1138). Tracing shows how the arrow begins at a
height of 200 ft, rises to over 1000 ft, then falls back to
the ground.
-16 a
(d) When t=4, h=924 ft.
800 2
800
b + v0 sin 30° a
b + 2.5 = 15
13v0
13v0
-161640,0002
400
+
= 12.5
3v20
13
(e) When t≠7.66, the
arrow is at its peak:
about 1138 ft.
-161640,0002
[0, 18] by [0, 1200]
80. x=35 cos a
p
p
t b , y=50+35 sin a
t b , assuming the
10
10
wheel turns counterclockwise.
2p
2p
t b , y=50-40 cos a
t b , assuming
15
15
the wheel turns counterclockwise.
82. x=–40 sin a
p
p
t b , y=50+40 cos a t b , assuming
9
9
the wheel turns counterclockwise.
400v20
= 12.5v20
3
13
-161 640,0002
400
= v20 a 12.5 b
3
13
-161640,0002
= v20
400
3 a 12.5 b
13
; 125.004 L v0
+
81. x=40 sin a
267
88. x = 1v0 cos 30 °2t and y = -16t2 + 1v0 sin 30° 2t + 2.5.
v0 must be (at least) just over 125 ft/sec. This can be
found graphically (by trial-and-error), or algebraically: the
ball is 400 ft from the plate (i.e., x=400) when
800> 13
400
Substitute this value of t in
t =
=
v0 cos 30 °
v0
the parametric equation for y. Then solve to see what value
of v0 makes y equal to 15 ft.
(c) Graph x=t and y=–16t2+245t+200 with
0 t 16.1 (upper
limit may vary).
(f) The arrow hits the
ground (h=0) after about
16.09 sec.
Review
The negative root doesn’t apply to this problem, so the
initial velocity needed is just over 125 ft/sec.
89. Kathy’s position: x1 = 60 cos a
y1 = 60 + 60 sin a
∏
t b and
6
∏
tb
6
Ball’s position:
x2 = -80 + 1 100 cos 70°2t and
y2 = -16t2 + 1100 sin 70°2t
Find (graphically) the minimum of
d1 t 2 = 2 1x1 - x2 2 2 + 1y1 - y2 2 2 . It occurs when
t L 2.64 sec; the minimum distance is about 17.65 ft.
83. (a)
[–7.5, 7.5] by [–5, 5]
(b) All 4’s should be changed to 5’s.
84. x=(66 cos 5°)t and y=–16t2+(66 sin 5°)t+4.
y=0 when t≠0.71 sec (and also when t≠–0.352,
but that is not appropriate in this problem). When
t≠0.71 sec, x≠46.75 ft.
2
85. x=(66 cos 12°)t and y=–16t +(66 sin 12°)t+3.5.
y=0 when t≠1.06 sec (and also when t≠–0.206,
but that is not appropriate in this problem). When
t≠1.06 sec, x≠68.65 ft.
86. x=(70 cos 45°)t and y=–16t2+(70 sin 45°)t. The
ball traveled 40 yd (120 ft) horizontally after about
2.42 sec, at which point it is about 25.96 ft above the
ground, so it clears the crossbar.
87. If we assume that the initial height is 0 ft, then
x=(85 cos 56°)t and y=–16t2+(85 sin 56°)t.
[If the assumed initial height is something other than
0 ft, add that amount to y.]
(a) Find graphically: the maximum y value is about
77.59 ft (after about 2.20 seconds).
(b) y=0 when t L 4.404 sec
90. x = 1 20 cos 50°2t and y = -16t2 + 120 sin 50°2t + 5.
y=0 when t=1.215 sec (and also when t=–0.257,
but that is not appropriate in this problem). When
t=1.215 sec, x=15.62 ft. The dart falls several feet
short of the target.
Chapter 6 Project
Answers are based on the sample data shown in the table.
1.
[–0.1, 2.1] by [0, 1.1]
2. Sinusoidal regression produces
y≠0.28 sin(3.46x+1.20)+0.75 or, with a phase shift
of 2∏, y L 0.28 sin13.46x - 5.092 + 0.75
L 0.28 sin1 3.461 x - 1.472 2 + 0.75.
268
Chapter 6
Applications of Trigonometry
5.
3.
[–0.1, 2.1] by [–1.1, 1.1]
The curve y L 0.9688 cos 13.46 1x - 1.472 2 closely fits the
data.
4. The distance and velocity both vary sinusoidally, with
the same period but a phase shift of 90° — like the x- and
y-coordinates of a point moving around a circle. A scatter
plot of distance versus time should have the shape of a
circle (or ellipse).
[0, 1.1] by [–1.1, 1.1]
[0, 1.1] by [–1.1, 1.1]
Section 7.1
Solving Systems of Two Equations
269
Chapter 7
Systems and Matrices
■ Section 7.1 Solving Systems of Two
Equations
Exploration 1
6.
x3+x2-6x=0
x(x2+x-6)=0
x(x+3)(x-2)=0
x=0, x=–3, x=2
4
4
7. m = - , y - 2 = - 1x + 1 2
5
5
4
4
y= - x - + 2
5
5
-4x + 6
y=
5
1.
[0, 10] by [–5, 5]
8. m =
2.
5
5
, y - 2 = 1x + 1 2
4
4
5
5
y = x + + 2
4
4
5x + 13
y =
4
9. –2(2x+3y)=–2(5)
–4x-6y=–10
[0, 10] by [–5, 5]
[0, 10] by [–5, 5]
10.
3. The function ln x is only defined for x 7 0, so all
solutions must be positive. As x approaches infinity,
x2-4x+2 is going to infinity much more quickly than
ln x is going to infinity, hence will always be larger than
ln x for x-values greater than 4.
[–4, 4] by [–15, 12]
Quick Review 7.1
[–4, 4] by [–15, 12]
1. 3y=5-2x
5
2
y = - x
3
3
2. x(y+1)=4
4
y + 1 = ,x Z 0
x
4
y =
- 1
x
3. (3x+2)(x-1)=0
3x+2=0
or x-1=0
3x=–2
x=1
2
x=–
3
4. x =
-5 ; 252 - 4 1 2 2 1 -10 2
4
-5; 2105
=
4
-5 + 2105 -5 - 2105
x =
,
4
4
5.
x3-4x=0
x(x2-4)=0
x(x-2)(x+2)=0
x=0, x=2, x=–2
[–4, 4] by [–15, 12]
Section 7.1 Exercises
1. (a) No: 5102 - 2 14 2 Z 8.
(b) Yes: 5(2)-2(1)=8 and 2(2)-3(1)=1.
(c) No: 21 -2 2 - 31 -9 2 Z 1.
2. (a) Yes: -3 = 2 2 - 612 2 + 5 and -3 = 2 122 - 7.
(b) No: -5 Z 12 - 611 2 + 5.
(c) Yes: 5 = 62 - 616 2 + 5 and 5 = 2 16 2 - 7.
In #3–12, there may be more than one good way to choose
the variable for which the substitution will be made. One
approach is given. In most cases, the solution is only shown up
to the point where the value of the first variable is found.
3. (x, y)=(9, –2): Since y=–2, we have x-4=5,
so x=9.
4. (x, y)=(3, –17): Since x=3, we have 3-y=20,
so y=–17.
270
Chapter 7
5. (x, y)= a
Systems and Matrices
50 10
, - b : y=20-3x,
7
7
so x-2 (20-3x)=10, or 7x=50, so x=
1
113 - 7y2 to get
4
1
x =
152 < 728712.
65
Substitute into x =
50
.
7
6. (x, y)= a -
23 23
,
b : y=–x,
5 5
23
so 2x+3x=–23, or x= - .
5
1
7. (x, y)= a - , 2 b : x = 13y - 72 >2,
2
so 2(3y-7)+5y=8, or 11y=22, so y=2.
8. (x, y)=(–3, 2): x=(5y-16)> 2, so
1.5(5y-16)+2y=–5, or 9.5y=19, so y=2.
9. No solution: x=3y+6, so –2(3y+6)+6y=4,
or –12=4 — not true.
10. There are infinitely many solutions, any pair (x, 3x+2):
From the first equation, y=3x+2, so –9x+
3(3x+2)=6, or 6=6 — always true.
11. (x, y)= 1 ; 3, 9 2; The second equation gives y=9,
so, x2=9, or x = ; 3.
12. (x, y)=(0, –3) or (x, y)=(4, 1): Since x=y+3,
we have y+3-y2=3y, or y2+2y-3=0.
Therefore y=–3 or y=1.
3 27
1 2
b or (x, y)= a , b :
13. (x, y)= a - ,
2 2
3 3
3
1
6x2+7x-3=0, so x = - or x = . Substitute these
2
3
values into y=6x2.
5
14. (x, y)=(–4, 28) or (x, y)= a , 15 b :
2
In the following, E1 and E2 refer to the first and second
equations, respectively.
19. (x, y)=(8, –2): E1+E2 leaves 2x=16,
so x=8.
20. (x, y)=(3, 4): 2E1+E2 leaves 5x=15,
so x=3.
21. (x, y)=(4, 2): 2E1+E2 leaves 11x=44,
so x=44.
22. (x, y)=(–2, 3): 4E1+5E2 leaves 31x=–62,
so x=–2.
23. No solution: 3E1+2E2 leaves 0=–72, which is false.
1
24. There are infinitely many solutions, any pair a x, x - 2 b :
2
E1+2E2 leaves 0=0, which is always true. As long as
(x, y) satisfies one equation, it will also satisfy the other.
2
5
25. There are infinitely many solutions, any pair ¢x, x - ≤ :
3
3
3E1+E2 leaves 0=0, which is always true. As long as
(x, y) satisfies one equation, it will also satisfy the other.
26. No solution: 2E1+E2 leaves 0=11, which is false.
27. (x, y)=(0, 1) or (x, y)=(3, –2)
28. (x, y)=(1.5, 1)
29. No solution
30. (x, y)=(0, –4) or (x, y)= 1 ; 27, 3 2 L 1 ;2.65, 32
31. One solution
5
.
2
2
Substitute these values into y=2x +x.
2x2+3x-20=0, so x=–4 or x =
15. (x, y)=(0, 0) or (x, y)=(3, 18): 3x2=x3, so x=0 or
x=3. Substitute these values into y=2x2.
16. (x, y)=(0, 0) or (x, y)=(–2, –4): x3+2x2=0, so
x=0 or x=–2. Substitute these values into y=–x2.
[–5, 5] by [–5, 5]
-1 + 3 189 3 + 189
,
b and
10
10
- 1 - 3 189 3 - 189
,
b : x - 3y = - 1, so x = 3y + 1.
a
10
10
Substitute x = 3y + 1 into x2 + y2 = 9:
1 3y - 1 2 2 + y2 = 9 1 10y2 - 6y - 8 = 0. Using the
3 ; 189
quadratic formula, we find that y =
.
10
32. No solution
52 + 7 2871 91 - 42871
,
b
65
65
≠(3.98, –0.42) or
52 - 72871 91 + 42871
(x, y)= a
,
b
65
65
1
1 13 - 7y 2 2 + y2 = 16, so
≠(–2.38, 3.22):
16
1
65y2-182y-87=0. Then y= 1 91 ; 4 28712.
65
33. Infinitely many solutions
17. 1 x, y2 = a
18. (x, y)= a
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
Section 7.1
34. One solution
[–4.7, 4.7] by [–3.1, 3.1]
35. 1 x, y2 L 1 0.69, - 0.372
[–5, 5] by [–5, 5]
36. 1 x, y2 L 1 1.13, 1.27 2
37. 1 x, y2 L 1 - 2.32, -3.16 2 or (0.47, –1.77) or (1.85, –1.08)
Solving Systems of Two Equations
271
40. 1x, y2 L (–1.2, –1.6) or (2, 0)
[–4.7, 4.7] by [–3.1, 3.1]
41. 1x, y2 L 1 2.05, 2.192 or (–2.05, 2.19)
[–4, 4] by [–4, 4]
42. 1x, y2 L 1 2.05, -2.19 2 or (–2.05, –2.19)
[–4, 4] by [–4, 4]
43. (x, p)=(3.75, 143.75): 200-15x=50+25x, so
40x=150.
44. (x, p)=(130, 5.9): 15-0.07x=2+0.03x, so
0.10x=13.
[–5, 5] by [–5, 5]
38. 1 x, y2 L 1 - 0.70, -2.40 2 or (5.70, 10.40)
45. In this problem, the graphs are representative of the
expenditures (in billions of dollars) for benefits and
administrative costs from federal hospital and medical
insurance trust funds for several years, where x is the
number of years past 1980.
(a) The following is a scatter plot of the data with the
quadratic regression equation
y=–0.0938x2+15.0510x-28.2375 superimposed
on it.
[–9, 11] by [–10, 14]
39. (x, y)=(–1.2, 1.6) or (2, 0)
[0, 30] by [–100, 500]
[–3, 3] by [–3, 3]
(b) The following is a scatter plot of the data with the
353.6473
logistic regression equation y =
11 + 8.6873e -0.1427x 2
superimposed on it.
[0, 30] by [–100, 500]
272
Chapter 7
Systems and Matrices
(c) Quadratic regression model
Graphical solution: Graph the line y=300 with the quadratic regression curve
y=–0.0938x2+15.0510x-28.2375 and find the intersection of the two curves. The two intersect at x≠26.03.
The expenditures will be 300 billion dollars sometime in
the year 2006.
[0, 30] by [–100, 500]
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
Note: The quadratic and the line intersect in two points,
but the second point is an unrealistic answer. This would
be sometime in the year 2114.
[0, 200] by [–100, 800]
x =
Algebraic solution: Solve
300=–0.0938x2+15.0510x-28.2375 for x.
Use the quadratic formula to solve the equation
–0.0938x2+15.0510x-328.2375=0.
a=–0.0938
b=15.0510
c=–328.2375
- 115.0510 2 ; 2 1 15.0510 2 2 - 4 1 -0.0938 2 1 - 328.23752
x =
=
2 1 -0.0938 2
- 115.05102 ; 1226.5326 - 123.1547
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
353.6473
Algebraic solution: Solve 300 =
for x.
11 + 8.6873e -0.1427x 2
30011 + 8.86873e -0.1427x 2 = 353.6473
353.6473
- 1
8.8687e -0.1427x =
300
-0.1427x
= 1.1788 - 1 = 0.1788
8.8687e
0.1788
- 0.1427x
=
= 0.0202
e
8.8687
-0.1427x = ln 0.0202
ln 0.0202
x =
L 27.34
-0.1427
The expenditures will be 300 billion dollars sometime in
the year 2007.
(d) The long-range implication of using the quadratic
regression equation is that the expenditures will eventually fall to zero.
(The graph of the function is a parabola with vertex at
about (80, 576) and it opens downward. So, eventually
the curve will cross the x-axis and the expenditures
will be 0. This will happen when x≠158.)
(e) The long-range implication of using the logistic
regression equation is that the expenditures will
eventually level off at about 354 billion dollars.
(We notice that as x gets larger, e -0.1427x approaches 0.
Therefore, the denominator of the function approaches
1 and the function itself approaches 353.65, which is
about 354.)
46. In this problem, the graphs are representative of the total
personal income (in billions of dollars) for residents of
the states of (a) Iowa and (b) Nevada for several years,
where x is the number of years past 1990.
(a) The following is a scatter plot of the Iowa data with
the linear regression equation y=2.8763x+48.4957
superimposed on it.
-0.1876
- 115.0510 2 ; 10.1675
- 0.1876
x≠26.03 and x≠134.43
We select x=26.03, which indicates that the expenditures will be 300 billion dollars sometime in the year 2006.
Logistic regression model
Graphical solution: Graph the line y=300 with the
353.6473
logistic regression curve y =
and
1 1 + 8.6873e -0.1427x 2
find the intersection of the two curves. The two intersect
at x≠27.21.
The expenditures will be 300 billion dollars sometime in
the year 2007.
[–5, 20] by [–10, 100]
(b) The following is a scatter plot of the Nevada data with
the linear regression equation y=3.5148x+25.0027
superimposed on it.
[–5, 20] by [–10, 100]
[0, 50] by [–100, 500]
(c) Graphical solution: Graph the two linear equations
y=2.8763x+48.4957 and y=3.5148x+25.0027
on the same axes and find the point of intersection.
The two curves intersect at x≠36.8.
Section 7.1
Solving Systems of Two Equations
273
The personal incomes of the two states will be the same
sometime in the year 2026.
[–5, 50] by [0, 8000]
[–5, 70] by [–10, 250]
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
Algebraic solution:
Solve 2.8763x+48.4957=3.5148x+25.0027 for x.
2.8763x+48.4957=3.5148x+25.0027
0.6385x=23.4930
23.4930
x =
L 36.8
0.6385
The personal incomes of the two states will be the same
sometime in the year 2026.
47. In this problem, the graphs are representative of the population (in thousands) of the states of Arizona and
Massachusetts for several years, where x is the number of
years past 1980.
(a) The following is a scatter plot of the Arizona data
with the linear regression equation
y=127.6351x+2587.0010 superimposed on it.
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
Algebraic solution:
Solve 127.6351x+2587.0010=31.3732x+5715.9742 for x.
127.6351x+2587.0010=31.3732x+5715.9742
96.2619x=3128.9732
3128.9732
L 32.5
x =
96.2619
The population of the two states will be the same sometime in the year 2012.
48. (a) None: the line never crosses the circle.
One: the line touches the circle at only one point—
a tangent line.
Two: the line intersects the circle at two points.
(b) None: the parabola never crosses the circle.
One, two, three, or four: the parabola touches the
circle in one, two, three or four points.
49. 200=2(x+y) and 500=xy. Then y=100-x,
so 500=x(100-x), and therefore x=50 ; 20 25,
and y=50 < 20 25. Both answers correspond to a rectangle with approximate dimensions 5.28 m*94.72 m.
50. 220=2(x+y) and 3000=xy. Then y=110-x,
so 3000=x(110-x), and therefore x=50 or 60.
That means y=60 or 50; the rectangle has dimensions
50 yd *60 yd.
[–5, 30] by [0, 8000]
(b) The following is a scatter plot of the Massachusetts
data with the linear regression equation
y=31.3732x+5715.9742 superimposed on it.
[–5, 30] by [0, 8000]
(c) Graphical solution: Graph the two linear equations
y=127.6351x+2587.0010 and
y=31.3732x+5715.9742 on the same axes and find
the point of intersection. The two curves intersect at
x≠32.5.
The population of the two states will be the same
sometime in the year 2012.
51. If r is Hank’s rowing speed (in miles per hour) and c is
24
the speed of the current, 1r - c2 = 1 and
60
13
5
1r + c2 = 1. Therefore r = c + (from the first
60
2
5
13
equation); substituting gives
a 2c + b = 1, so
60
2
60
5
55
55
2c =
- =
, and c =
L 1.06 mph. Finally,
13
2
26
52
5
185
r = c + =
L 3.56 mph.
2
52
52. If x is airplane’s speed (in miles per hour) and y is the
wind speed, 4.4(x-y)=2500 and 3.75(x+y)=2500.
Therefore x=y+568.18; substituting gives
3.75(2y+568.18)=2500, so 2y=98.48, and
y=49.24 mph. Finally, x=y+568.18=617.42 mph.
53. m + / = 1.74 and / = m + 0.16, so 2m+0.16=1.74.
Then m=$0.79 (79 cents) and / = $0.95 (95 cents).
274
Chapter 7
Systems and Matrices
54. p+c=5 and 1.70p+4.55c=2.80 # 5. Then
1.70(5-c)+4.55c=14, so 2.85c=5.5. That means
175
110
c =
L 1.93 lb of cashews and p=
L 3.07 lb of
57
57
peanuts.
65. (a)
55. 4=–a+b and 6=2a+b, so b=a+4 and
2
14
.
6=3a+4. Then a = and b =
3
3
56. 2a-b=8 and –4a-6b=8, so b=2a-8 and
8=–4a-6(2a-8)=–16a+48. Then
40
5
a =
= and b=–3.
16
2
57. (a) Let C(x)=the amount charged by each rental
company, and let x=the number of miles driven
by Pedro.
Company A: C(x)=40+0.10x
Company B: C(x)=25+0.15x
Solving these two equations for x,
40+0.10x=25+0.15x
15=0.05x
300=x
Pedro can drive 300 miles to be charged the same
amount by the two companies.
(b) One possible answer: If Pedro is making only a short
trip, Company B is better because the flat fee is less.
However, if Pedro drives the rental van over 300 miles,
Company A’s plan is more economical for his needs.
58. (a) Let S(x)=Stephanie’s salary, and let x=total sales
from household appliances sold weekly.
Plan A: S(x)=300+0.05x
Plan B: S(x)=600+0.01x
Solving these equations, we find:
300+0.05x=600+0.01x
0.04x=300
x=7500
Stephanie’s sales must be exactly $7500 for the plans
to provide the same salary.
(b)
[–3, 3] by [–4, 4]
1 x, y2 L 1 -1.29, 2.29 2 or 11.91, -0.912
(c)
66. (a)
60. False. The system would have no solutions, because any
solution of the original system would have to be a solution of 7=0, which has no solutions.
61. Using (x, y)=(3, –2),
2(3)-3(–2)=12
3+2(–2)=–1
The answer is C.
62. A parabola and a circle can intersect in at most 4 places.
The answer is E.
63. Two parabolas can intersect in 0, 1, 2, 3, or 4 places, or
infinitely many places if the parabolas completely
coincide. The answer is D.
64. When the solution process leads to an identity (an equation that is true for all (x, y), the original system has
infinitely many solutions. The answer is E.
1 -1.29 2 2
1 2.292 2
L 0.9987 L 1 and
4
9
(–1.29)+(2.29)=1, so the first solution checks.
1 1.912 2
1 -0.91 2 2
+
L 1.004 L 1 and
4
9
11.912 + 1 -0.912 = 1, so the second solution checks.
+
y2
x2
= 1
4
9
–9x2+4y2=–36
4y2=9x2-36
9x2 - 36
y2 =
4
3
3
2
y = 2x - 4, y = - 2x2 - 4
2
2
(b)
(b) One possible answer: If Stephanie expects that her
sales will generally be above $7500 each week, then
Plan A provides a better salary. If she believes that
sales will not reach $7500/week, however, Plan B will
maximize her salary.
59. False. A system of two linear equations in two variables
has either 0, 1, or infinitely many solutions.
y2
x2
+ =1
4
9
9x2 + 4y2=36
4y2=36 - 9x2
36 - 9x2
y2=
4
3
3
y= 24 - x2, y = - 24 - x2
2
2
[–5, 5] by [–5, 5]
1 x, y2 L 12.68, 2.682 or 1 -2.68, -2.68 2
(c)
1 2.682 2
-
12.682 2
=
1 - 2.682 2
-
1 -2.68 2 2
4
9
4
9
≠0.9976 L 1, so both solutions check.
67. Subtract the second equation from the first, leaving
10
2
10
–3y=–10, or y =
. Then x2 = 4 = , so
3
3
3
2
.
x = ;
B3
68. Add the two equations to get 2x2 = 2, so x2 = 1, and
therefore x = ;1. Then y=0.
69. The vertex of the parabola R=(100-4x)x
=4x(25-x) has first coordinate x=12.5 units.
70. The local maximum of R=x(80-x2)=80x-x3
has first coordinate x L 5.16 units.
Section 7.2
■ Section 7.2 Matrix Algebra
2. (a) (–3, –2)
Exploration 1
3. (a) (–2, 3)
Matrix Algebra
275
(b) (–x, y)
1. a11
a12
a21
a22
=
=
=
=
311 2
311 2
312 2
312 2
-
2
So, A = c
5
112
122
112
122
=
=
=
=
Set i=j=1.
Set i=1, j=2.
Set i=2, j=1.
Set i=j=2.
2
1
5
4
(b) (y, x)
4. (a) (2, –3)
(b) (–y, –x)
1
d . Similar computations show that
4
-1
B = c
2
2
d.
5
2 1
-2
d + c
5 4
-5
The order of [0] is 2*2.
-1
0
d = c
-4
0
1
-1
d - 2c
4
2
2
d
5
6
=c
15
3
-2
d - c
12
4
4
8
d = c
10
11
9. cos 1 a + b2 = cos a cos b - sin a sin b
10. cos 1 a - b2 = cos a cos b + sin a sin b
0
d = 304
0
5. 3*1; not square
6. 1*1; square
7. a13=3
a 13
a
a23 † =a11 1 -1 2 2 ` 22
a 32
a 33
a23
`
a33
+a12 1 -1 2 3 `
a21 a23
a21 a22
` +a13 1 -1 2 4 `
`
a31 a33
a31 a32
=a11 1a22 a33 - a23 a32 2-a12 1 a21 a33 - a23 a31 2
+a13 1a21 a 32 - a22 a31 2
=a11 a22 a33 - a11 a23 a32 - a12 a21a33 + a12 a23 a31
+a13 a21 a32 - a13 a22 a31
The two expressions are exactly equal.
3. Recall that Aij is (–1)i+jMij where Mij is the determinant of the matrix obtained by deleting the row and column containing aij. Let A=k*k square matrix with
zeros in the ith row. Then: det(A)=
a11
a21
6 o
ith row S 0
o
ak1
a12
a22
o
0
o
ak2
…
…
…
…
= 0 # A i1 + 0 # A i2 + …
=0
Quick Review 7.2
(b) (x, –y)
1. 2*3; not square
4. 1*3; not square
-1
d
2
1. det (A)= - a12 a21 a33 + a13 a21 a32 + a11 a22 a33
-a13 a22 a31 - a11 a23 a32 + a12 a23 a31
Each element contains an element from each row and
each column due to a definition of a determinant.
Regardless of the row or column “picked” to apply the
definition, all other elements of the matrix are eventually
factored into the multiplication.
1. (a) (3, 2)
Section 7.2 Exercises
3. 3*2; not square
Exploration 2
a12
a22
a32
7. sin 1a + b2 = sin a cos b + cos a sin b
2. 2*2; square
2
3. 3A-2B=3 c
5
a11
2. † a21
a31
6. (r cos ¨, r sin ¨)
8. sin 1a - b2 = sin a cos b - cos a sin b
2. The additive inverse of A is –A and
-2 - 1
-A = c
d.
-5 - 4
A + 1 -A2 = c
5. (3 cos ¨, 3 sin ¨)
a1k
a2k
o6
0
o
akk
+ 0 # A ik=0+0+…+0
8. a24=–1
9. a32=4
10. a33=–1
11. (a) c
3
-3
(b) c
1
1
(c) c
6
-3
0
d
1
6
d
9
9
d
15
2
-1
1
3
d -3 c
5
-2
-3
d=
-4
3
6
d-c
10
-6
-9
1
d=c
-12
4
15
d
22
(d) 2A-3B=2 c
c
4
-2
1
12. (a) £ 3
6
(b) £
-3
5
-2
-3
(c) £ 12
6
1
1
-3
2
1§
0
2
-3 §
2
-1
1
3
0
3
0
6
-3 §
3
-1
(d) 2A-3B=2 £ 4
2
=£
-2
8
4
-8
= £ 11
-8
0
2
0
-3
2
9
0
1
0
2
2
- 1 § -3 £ - 1
1
4
4
6
-2 § - £ - 3
2
12
4
-8 §
5
3
0
-9
0
6§
-3
1
0
-3
0
2§
-1
276
Chapter 7
1
13. (a) £ -2
-1
Systems and Matrices
1
0§
0
-3
1§
4
-1
(b) £ 1 §
-4
15. (a) £
(b) £
-7
2
5
1
-2 §
2
(c) £
-9
0
6
3
-3 §
3
(c) £
(d) 2A-3B=2 £
3
3
1
0
1
4
-1 § -3 £ - 2
1
-3
2
12
- 2 § - £ -6
2
-9
-6
=£ 0
4
14. (a) c
-3
0
2
4
1
0
1§
-1
0
-18
3§=£
6
-3
13
(d) 2A-3B=2 £
2
-5 §
5
1
d
0
7
(b) c
-5
-5
0
2
3
1
d
4
15
(c) c
-3
-6
0
9
6
3
d
6
-2
-1
-4
-3
1 § -3 £ 0 § = £ 2 § - £ 0 §
0
4
0
12
-1
=£
2§
-12
0
(c) 3 -3
34
-2
-4
2
34
-6
0
94
(d) 2A-3B=2 3 -1 -2 0
= 3 -2 -4 0 64 - 33
=3 -5 -10 6 64
5
-1
-2
0
3c
-2
4
3
0
1
-1
0
d
-2
-4
0
6
4
2
d4
10
-2
16. (a) 30
(b) 3 -2
(d) 2A-3B=2 c
=c
-6
3§
0
34 -3 31 2
6 -6 04
1
d –
2
3
2
-6 9
3
0
d
12 0 - 3 -6
16 -13 3
2
d
=c
- 14
0 7 10
c
17. (a) AB = c
(b) BA = c
18. (a) AB = c
(b) BA = c
19. (a) AB = c
1 22 1 12 + 1 32 1 - 22
1 -12 112 + 152 1 -22
12 2 1 -3 2 + 13 2 1 -4 2
-4
d = c
1 -1 2 1 -3 2 + 15 2 1 -4 2
- 11
11 2 15 2 + 1 - 42 1 - 2 2
1 2 2 15 2 + 16 2 1 - 22
11 2 1 1 2 + 1 -42 1 -32
13
d = c
-2
12 2 11 2 + 16 2 1 -3 2
1 1 2 1 2 2 + 1 -3 2 1 -1 2
1 - 22 1 22 + 1 -4 2 1 -1 2
15 2 11 2 + 11 2 12 2
1 - 2 2 1 12 + 1 -32 1 2 2
7
15 2 1 - 4 2 + 112 1 62
d = c
-8
1 -2 2 1 -4 2 + 1 -3 2 162
1 22 1 1 2 + 1 02 1 -3 2 + 11 2 10 2
11 2 11 2 + 14 2 1 -3 2 + 1 -3 2 10 2
1 12 1 2 2 + 1 2 2 11 2
(b) BA = £ 1 - 3 2 1 2 2 + 1 1 2 11 2
1 0 2 12 2 + 1 - 22 1 12
20. (a) AB = B
5
11 2 13 2 + 1 - 32 1 52
d = c
1 -2 2 13 2 + 1 - 42 15 2
0
-18
d
-17
- 12
d
- 26
13
d
-16
-14
d
-10
12 2 122 + 10 2 11 2 + 112 1 -2 2
2
d = c
1 1 2 12 2 + 14 2 11 2 + 1 -32 1 -2 2
-11
11 2 1 0 2 + 12 2 14 2
1 - 3 2 1 0 2 + 11 2 142
1 0 2 10 2 + 1 - 2 2 14 2
1 1 2 112 + 12 2 1 -3 2
4
1 -3 2 11 2 + 11 2 1 -3 2 § = £ -5
10 2 11 2 + 1 -2 2 1 -32
-2
2
d
12
8
4
-8
-5
-6 §
6
112 1 5 2 + 10 2 10 2 + 1 - 2 2 1 -1 2 + 13 2 1 42 11 2 1 -1 2 + 10 2 12 2 + 1 -2 2 13 2 + 13 2 122
19
R = c
12 2 152 + 11 2 10 2 + 1 4 2 1 -1 2 + 1 -1 2 14 2 12 2 1 -1 2 + 112 1 2 2 + 14 2 13 2 + 1 -12 12 2
2
1 52 1 12 + 1 -1 2 122
10 2 11 2 + 12 2 122
(b) BA = ≥
1 -12 11 2 + 132 122
142 11 2 + 122 12 2
3
4
= ≥
5
8
-1
2
3
2
-14
8
14
0
1 5 2 10 2 + 1 -1 2 112 152 1 -22 + 1 -12 1 42
10 2 1 -2 2 + 1 22 14 2
10 2 10 2 + 12 2 112
1 -1 2 10 2 + 13 2 11 2 1 -1 2 1 -22 + 132 1 4 2
142 1 -2 2 + 1 22 14 2
14 2 10 2 + 12 2 112
16
-2
¥
-6
10
15 2 132 + 1 -1 2 1 -12
10 2 132 + 12 2 1 -1 2
T
1 -12 13 2 + 13 2 1 -12
14 2 132 + 1 22 1 -1 2
-1
d
10
-2
04
Section 7.2
1 -1 2 12 2 + 1 0 2 1 -1 2 + 12 2 14 2
21. (a) AB = £ 1 4 2 12 2 + 1 1 2 1 -1 2 + 1 -1 2 14 2
1 2 2 12 2 + 1 02 1 -1 2 + 11 2 14 2
6
= £3
8
(b) BA = £
= £
-7
7
-1
1 -1 2 112 + 10 2 10 2 + 1 22 1 -32 1 -1 2 1 02 + 10 2 1 22 + 1 22 1 -1 2
1 4 2 112 + 11 2 10 2 + 1 - 12 1 -3 2 14 2 102 + 11 2 12 2 + 1 -1 2 1 -1 2 S
1 2 2 11 2 + 10 2 10 2 + 112 1 -3 2
12 2 10 2 + 10 2 1 22 + 1 12 1 -1 2
-2
3§
-1
12 2 1 -1 2 + 1 12 14 2 + 10 2 12 2
1 -1 2 1 -12 + 1 02 14 2 + 1 2 2 12 2
1 4 2 1 -1 2 + 1 -32 14 2 + 1 - 1 2 1 2 2
2
5
- 18
1
0
-3
3
0§
10
1 -2 2 14 2 + 1 32 1 02 + 10 2 1 -1 2
22. (a) AB = £ 1 1 2 14 2 + 1 - 22 1 0 2 + 14 2 1 -1 2
1 3 2 14 2 + 1 22 1 02 + 11 2 1 -1 2
= £
(b) BA = £
-8
0
11
8
7
4
1 22 10 2 + 1 12 11 2 + 10 2 1 0 2
122 12 2 + 112 1 -1 2 + 10 2 1 12
1 -1 2 10 2 + 10 2 11 2 + 12 2 1 02
1 -1 2 12 2 + 10 2 1 -1 2 + 1 22 1 1 2 S
142 10 2 + 1 - 3 2 112 + 1 -1 2 10 2 14 2 122 + 1 -32 1 -1 2 + 1 -1 2 1 12
1 -2 2 1 -1 2 + 13 2 12 2 + 102 1 3 2 1 -2 2 1 22 + 1 32 1 3 2 + 102 1 -1 2
1 1 2 1 -1 2 + 1 - 2 2 122 + 1 42 1 3 2 11 2 12 2 + 1 -2 2 13 2 + 14 2 1 -1 2 S
1 32 1 - 12 + 12 2 12 2 + 11 2 13 2
13 2 1 22 + 1 22 1 32 + 112 1 -1 2
5
-8 §
11
1 42 1 - 22 + 1 -1 2 11 2 + 1 2 2 13 2
10 2 1 -2 2 + 1 2 2 11 2 + 13 2 1 3 2
1 -1 2 1 -2 2 + 132 11 2 + 1 - 1 2 13 2
-3
= £ 11
2
18
2
- 11
1 42 132 + 1 -1 2 1 -2 2 + 12 2 12 2
1 0 2 132 + 12 2 1 -2 2 + 132 12 2
1 -1 2 13 2 + 1 32 1 - 22 + 1 -12 12 2
24. (a) AB = £
1 - 52 1 2 2
1 42 1 2 2
1 22 1 2 2
1 -2 2 1 -1 2
1 3 2 1 -1 2
1 -4 2 1 -1 2
142 10 2 + 1 -1 2 142 + 12 2 1 12
10 2 10 2 + 1 22 14 2 + 13 2 11 2 S
1 -12 10 2 + 13 2 14 2 + 1 -12 1 1 2
-2
11 §
11
23. (a) AB = 3 1 2 2 1 - 52 + 1 - 12 14 2 + 13 2 12 2 4 = 3 -8 4
(b) BA = £
Matrix Algebra
1 -5 2 1 - 1 2
142 1 -12
12 2 1 - 1 2
1 -5 2 13 2
-10
14 2 13 2 § = £
8
12 2 13 2
4
1 -2 2 1 2 2
13 2 12 2
1 -4 2 1 2 2
5
-4
-2
1 -2 2 14 2
2
13 2 14 2 § = £ -3
1 -4 2 14 2
4
-15
12 §
6
-4
6
-8
(b) BA = 3 1 -1 2 1 -22 + 1 2 2 13 2 + 1 4 2 1 - 4 2 4 = 3 - 84
-8
12 §
-16
25. (a) AB is not possible.
(b) BA=[(–3)(–1)+(5)(3)
(–3)(2)+(5)(4)]=[18 14]
1 - 12 15 2 + 1 32 12 2
1 0 2 15 2 + 1 12 122
26. (a) AB = ≥
1 12 152 + 1 02 122
1 -3 2 15 2 + 1 - 12 12 2
1 -1 2 1 -6 2 + 13 2 13 2
1
1 0 2 1 -62 + 11 2 13 2
2
¥ = ≥
1 1 2 1 -6 2 + 10 2 13 2
5
1 - 3 2 1 -6 2 + 1 -1 2 13 2
-17
(b) BA is not possible.
1 0 2 112 + 1 0 2 12 2 + 1 1 2 1 -1 2
27. (a) AB = £ 1 02 112 + 1 1 2 12 2 + 1 0 2 1 - 1 2
1 12 112 + 1 0 2 12 2 + 1 0 2 1 -1 2
(b) BA = £
1 0 2 12 2 + 1 02 10 2 + 11 2 132
1 0 2 122 + 11 2 102 + 10 2 13 2
1 1 2 1 2 2 + 1 02 10 2 + 10 2 13 2
1 1 2 10 2 + 1 22 10 2 + 11 2 1 1 2
1 2 2 10 2 + 1 02 10 2 + 11 2 11 2
1 - 1 2 10 2 + 1 32 1 02 + 14 2 11 2
0
0
28. (a) AB = ≥
-1
0
0
0
(b) BA = ≥
0
0
+
+
+
+
+
+
+
+
0
2
0
0
+
+
+
3
0
0
0
+
+
+
+
0
0
0
0
+
+
+
+
3
0
1
2
+
+
+
+
0
0
0
0
0
0
0
4
0
0
0
0
1 1 2 102 + 12 2 11 2 + 11 2 1 0 2
1 2 2 10 2 + 102 1 12 + 11 2 1 02
1 - 1 2 10 2 + 13 2 112 + 14 2 1 02
0
0
2
0
+
+
+
+
0
1
0
0
+
+
+
+
2
0
0
0
+
+
+
+
+
+
+
+
2
1
2
0
+
+
+
+
0
0
0
0
+
+
+
+
0
0
0
0
0
0
0
0
0
0
3
0
+
+
+
+
-1
2
-3
4
0
0
0
0
+
+
+
+
+
+
+
+
0
0
0
0
1
0
0
0
+
+
+
+
+
+
+
+
0
0
0
0
0
0
0
2
+
+
+
+
0
0
0
0
15
3
¥
-6
15
102 11 2 + 10 2 11 2 + 11 2 142
-1
102 11 2 + 11 2 11 2 + 10 2 142 S = £ 2
11 2 11 2 + 102 11 2 + 10 2 14 2
1
0
0
-4
0
0
0
0
0
112 1 1 2 + 12 2 10 2 + 11 2 102
1
12 2 1 12 + 1 02 1 0 2 + 11 2 10 2 S = £ 1
1 -12 1 1 2 + 13 2 10 2 + 14 2 102
4
+
+
+
+
+
+
+
0
1
0
0
0
0
0
0
+
+
+
+
+
+
+
+
3
0
0
0
0
0
0
0
+
+
+
+
-
-3
0
2
0
¥ = ≥
-1
0
4
1
3
4
0
1
¥ = ≥
1
3
2
1
2
1
2
0
2
1
2
0
1
0
3
2
-1
2
-3
4
3
0
2
3
-1
¥
-4
-1
-4
-1
¥
3
-1
4
1§
1
2
0
3
1
2§
-1
277
Chapter 7
278
Systems and Matrices
In #29–32, use the fact that two matrices are equal only if all entries are equal.
29. a=5, b=2
30. a=3, b=–1
31. a=–2, b=0
32. a=1, b=6
33. AB = c
BA = c
122 10.8 2 + 1 12 1 -0.6 2
13 2 10.8 2 + 1 42 1 -0.6 2
10.82 122 + 1 - 0.2 2 1 32
1 - 0.2 2 1 22 + 1 0.42 1 3 2
12 2 1 -0.2 2 + 11 2 10.4 2
1
d = c
13 2 1 - 0.2 2 + 1 4 2 1 0.42
0
1
10.8 2 11 2 + 1 -0.22 1 42
d = c
10.6 2 11 2 + 1 -0.4 2 14 2
0
1 -2 2 102 + 1 12 1 0.252 + 3 10.25 2
34. AB = £ 1 1 2 10 2 + 1 22 10.252 + 1 -2 2 1 0.25 2
1 0 2 10 2 + 1 1 2 10.25 2 + 1 -1 2 10.25 2
1
= £0
0
0
1
0
1 02 1 3 2 + 11 2 1 -2 2 + 1 - 2 2 1 -1 2
1 0.252 132 + 10.5 2 1 -2 2 + 1 - 0.25 2 1 -1 2 S
1 0.252 132 + 10.5 2 1 -2 2 + 1 - 1.25 2 1 -1 2
1 0 0
= £ 0 1 0 § , so A and B are inverses.
0 0 1
2
2
3 -1
2
1
d =
c
2
1 22 1 2 2 - 12 2 1 3 2 -2
2
1
= - c
2 -2
1.5
d
-1
41. Use row 2 or column 2 since they have the greatest number of zeros. Using column 2:
2
3 -1
1
36. No inverse: The determinant is (6)(5)-(10)(3)=0.
37. No inverse: The determinant (found with a calculator) is 0.
-1
2 3 -1
4§
38. Using a calculator: £ -1 0
0 1
1
1
= £ -0.25
0.25
1
-0.5
0.5
1 -2 2 1 -2 2 + 11 2 1 -0.25 2 + 13 2 1 -1.252
11 2 1 -22 + 12 2 1 -0.252 + 1 -2 2 1 -1.252 S
10 2 1 -22 + 11 2 1 -0.25 2 + 1 -1 2 1 -1.252
1 0 2 11 2 + 1 12 1 22 + 1 -22 11 2
10.25 2 11 2 + 10.5 2 122 + 1 -0.25 2 112
10.25 2 112 + 10.52 122 + 1 -1.25 2 11 2
-3
d
2
-3
-1
d = c
2
1
0
d , so A and B are inverses.
1
1 -2 2 11 2 + 1 1 2 10.52 + 13 2 1 0.52
1 1 2 11 2 + 12 2 10.5 2 + 1 -2 2 1 0.5 2
1 0 2 11 2 + 1 1 2 10.5 2 + 1 -1 2 10.5 2
0
0§
1
10 2 1 - 22 + 1 12 1 1 2 + 1 -2 2 10 2
BA = £ 1 0.252 1 -2 2 + 1 0.5 2 11 2 + 1 -0.25 2 10 2
1 0.252 1 - 22 + 1 0.5 2 11 2 + 1 -1.25 2 10 2
35. c
0
d,
1
-3
1.75 § ;
-0.75
to confirm, carry out the multiplication.
1 -1
1 -1
-1
1 -1
1
¥;
39. A = ≥
1 -1
1 -1
-1
1 -1
1
No inverse, det(A)=0 (found using a calculator)
0 1 2
40. B = £ 1 0 1 §
2 1 0
-0.25
0.5
0.25
0.5 -1.0
0.5 §
B-1 = £
0.25
0.5 -0.25
(found using a calculator, use multiplication to confirm)
1
0
3
1
-1
2 † = 1 12 1 -1 2 3 2
1
-1
+ 102 1 -1 2 4 `
2
`
-1
2
1
2
` + 132 1 -1 2 5 `
1 -1
-1
= 1 - 12 11 - 2 2 + 0 + 1 -32 1 4 + 12
= 1 + 0 - 15
= -14
1
`
2
42. Use row 1 or 4 or column 2 or 3 since they have the
greatest number of zeros. Using column 3:
1
40
1
1
0
1
-1
0
2
2
0
0
0
0
1 3
34
= 122 1 -1 2 4 3 1 -1 2 3
2
1
0 3
3
1
0 0
+ 12 2 1 -1 2 5 3 1 -1 2 † + 0 + 0
1
0 3
1 3
1 3
` + 11 -12 4 `
`d
= 2 # c 0 + 11 -12 3 `
0 3
-1 2
-1 2
` + 0 + 0d
- 2 c 1 1 -1 2 2 `
0 3
= 2 1 1 -1 2 13 - 02 + 11 2 1 2 + 3 2 2 - 2 1 11 2 1 -3 - 02 2
=2(–3+5)-2(–3)
=4+6
=10
Section 7.2
43. 3X=B-A
B - A
1 4
1
3
1
1
X =
= ac d - c db = c
d=c 1d
3
3 2
3
3 -1
-3
44. 2X=B-A
X=
B - A
1 1
= ac
2
2 1
1 2
= c
2 1
1
=c 1
2
4
-1
d - c
-1
0
5
(d) RC= C 7
6
2
db
3
22
20
27
14
10
8
Matrix Algebra
$1,600
17 $ 900
21 S E $ 500 U
13 $ 100
$1,000
7
9
5
$52,500
= C $56,100 S
$51,400
2
d
-4
(e) BRC=(BR)C
1
d
-2
45. (a) The entries aij and aji are the same because each gives
the distance between the same two cities.
= 3163
650
266
$1,600
$ 900
4314 E $ 500 U
$ 100
$1,000
175
(b) The entries aii are all 0 because the distance between
a city and itself is 0.
1.1 # 120
46. B = £ 1.1 # 150
1.1 # 80
132
1.1 # 70
1.1 # 110 § = £ 165
88
1.1 # 160
77
121 §
176
B=1.1A
47. (a) BTA= 3$0.80
0.80 11002
£ + 0.85 11202
1 12002
= 3382
$0.85
227.504
100
$1.004 £ 120
200
60
70 §
120
0.80 160 2
+ 0.85170 2 §
+ 1 1120 2
(b) b1j in matrix BTA represents the income Happy Valley
Farms makes at grocery store j, selling all three types
of eggs.
16
48. (a) SP= £ 12
4
10
0
12
$15,550
= £ $8,070
$8,740
8
10
0
$180
12
$275
14 § ≥
$355
8
$590
$269.99
$399.99
¥
$499.99
$799.99
$21,919.54
$11,439.74 §
$12,279.76
(b) The wholesale and retail values of all the inventory at
store i are represented by ai1 and ai2, respectively, in
the matrix SP.
49. (a) Total revenue=sum of (price charged)(number sold)
=ABT or BAT
(b) Profit=Total revenue-Total Cost
=ABT-CBT
=(A-C)BT
50. (a) B= 36
7
279
14 4
5 22 14 7 17
(b) BR= 36 7 144 £ 7 20 10 9 21 §
6 27
8 5 13
= 3163 650 266 175 4314
$1,600
$ 900
(c) C= E $ 500 U
$ 100
$1,000
=[$1,427,300]
This is the building contractor’s total cost of building
all 27 houses.
51. (a) 3x¿
y¿ 4 = 3x
y4 B
cos a
sin a
-sin a
R,
cos a
x, y=1, a = 30°
= 31
=c
(b) 3x
13
2
14 D
1
2
13 + 1
2
y4 = 3x¿
1
2
T
13
2
-
13 - 1
d L 31.37
2
y¿ 4 B
cos a
-sin a
x, y=1, a = 30°
13
2
= 31 1 4 D
1
2
=c
13 - 1
2
0.37 4
sin a
R,
cos a
1
2
T
13
2
13 + 1
d L 30.37
2
1.374
52. Answers will vary. One possible answer is given.
(a) A+B= 3aij + bij 4 = 3bij + aij 4 = B + A
(b) (A+B)+C=[aij+bij]+C=[aij+bij+cij]
=[aij+(bij+cij)]=A+[bij+cij]
=A+(B+C)
(c) A(B+C)=A[bij+cij]=[ a aik(bkj+ckj)]
k
(following the rules of matrix multiplication)
=[ a (aikbkj+aikckj)]
k
=[ a aikbkj+ a aikckj]
k
k
=[ a aikbkj]+[ a aikckj]=AB+AC
k
k
Chapter 7
280
Systems and Matrices
(d) (A-B)C=[aij-bij]C=[ a (aik-bik)cki]
k
55. A # A-1 = B
=[ a (aikcki+bikcki)]
= a
k
=[ a aikcki- a bikcki]
k
k
k
1
a
bB
ad - bc
c
(since
=[ a aikcki]-[ a bikcki]
= a
k
=AC-BC
1
d
b
Ra
bB
-c
d
ad - bc
a
c
d
b
RB
d -c
-b
R
a
-b
R
a
1
is a scalar)
ad - bc
1
ad - bc
bB
ad - bc
cd - cd
-ab + ba
1
R = a
b
- bc + ad
ad - bc
53. Answers will vary. One possible answer is provided for each.
(a) c(A+B)=c 3 aij + bij 4 = 3caij + cbij 4 = cA + cB
(b) (c+d)A=(c+d)[aij]=c[aij]+d[aij]
=cA+dA
(c) c(dA)=c[daij]=[cdaij]=cd[aij]=cdA
(d) 1 # A = 1 # 3 aij 4 = 3aij 4 = A
c
= ≥
54. One possible answer: If the definition of determinant is
followed, the evaluation of the determinant of any n*n
square matrix 1 n 7 2 2 eventually involves the evaluation
of a number of 2*2 sub-determinants. The determinant
of the 2*2 matrix serves as the building block for all
other determinants.
a11
a21
56. AIn= ≥
o
an1
a12
a22
o
an2
…
…
…
a1n
a2n
¥
o
ann
1
0
≥
o
0
a11 + 0 # a12 + … + 0 # a1n
a21 + 0 # a22 + … + 0 # a2n
=≥
o
#
an1 + 0 an2 + … + 0 # ann
0
1
o
0
…
…
∞
…
ad - bc
0
0
d
ad - bc
ad - bc
ad - bc
0
0
ad - bc
ad - bc
= c
1
0
¥
0
d = I2
1
0
0
¥
o
1
0 # a11 + a12 + 0 # a13 + … + 0 # a1n
0 # a21 + a22 + 0 # a23 + … + 0 # a2n
o
#
#
0 an1 + an2 + 0 an3 + … + 0 # ann
…
…
0 # a11 + 0 # a12 + … + a1n
0 # a21 + 0 # a22 + … + a2n
…
0 # an1 + 0 # an2 + … + ann
T
a11 a12 … a1n
a21 a22 … a2n
¥ = A
=≥
o
o
o
an1 an2 … ann
Use a similar process to show that InA=A.
57. If (x, y) is reflected across the y-axis, then
(x, y) 1 (–x, y).
3x¿
y¿ 4 = 3x
y4 c
-1
0
0
d
1
58. If (x, y) is reflected across the line y=x, then
(x, y) 1 (y, x).
3x¿
y¿ 4 = 3x
y4 c
0
1
y¿ 4 = 3x
y4 c
-1
d
0
0
-1
60. If (x, y) is vertically stretched (or shrunk) by a factor of
a, then (x, y) 1 1x, ay2 .
3x¿
y¿ 4 = 3x
1
y4 c 0
0
ad
3 x¿
y¿ 4 = 3 x
y4 c
0
d
1
c
0
62. False. A square matrix A has an inverse if and only if
det A Z 0.
63. False. The determinant can be negative. For example, the
1
d
0
59. If (x, y) is reflected across the line y=–x, then
(x, y) 1 (–y, –x).
3x¿
61. If (x, y) is horizontally stretched (or shrunk) by a factor
of c, then (x, y) 1 (cx, y).
determinant of A = c
1
2
0
d is 1(–1)-2(0)=–1.
-1
64. 2(–1)-(–3)(4)=10. The answer is C.
65. The matrix AB has the same number of rows as A and
the same number of columns as B. The answer is B.
1
4
2 7 -1
c
d =
21 42 - 1 17 2 -1
1 4
. The answer is E.
66. c
-7
4
d = c
2
-1
67. The value in row 1, column 3 is 3. The answer is D.
-7
d
2
Section 7.2
68. (a) Recall that Aij is (–1)i+jMij where Mij is the determinant of the matrix obtained by deleting the row and
column containing aij. Let A=3*3 square matrix.
Then:
a11 a12 a13
det(A)= † a21 a22 a23 †
a31 a32 a33
=a11A 11 + a12A 12 + a13A 13
=a11 0 A 11 0 - a12 0 A 12 0 + a13 0 A 13 0
Now let B be the matrix A with rows 1 and 2
interchanged. Then:
a21 a22 a23
det(B)= † a11 a12 a13 †
a31 a32 a33
= a11B21 + a12B22 + a13B23
= -a11 0 A 11 0 + a12 0 A 12 0 - a13 0 A 13 0
= 1 - 1 2 1a11 0 A 11 0 - a12 0 A 12 0 + a13 0 A 13 0 2
= -det1 A2
To generalize, we would say that by the definition of
a determinant, the determinant of any k*k square
matrix is ultimately dependent upon a series of 3*3
determinants. (In the 4*4 case, for example, we
would have the expansion — using the first row — of
a11A 11 + a12A 12 + a13A 13 + a14A 14. If a row of matrix
A is interchanged with another, the elements of all of
matrix A’s 3*3 matrices will be affected, resulting in
a sign change to the determinant.
(b) Let A be a k*k square matrix with two rows exactly
the same, and B be the matrix A with those exact
same rows interchanged. From #4, we know that
det(A)=–det(B). However, since A=B elementwise (i.e., aij=bij for 1 i, j k), we also know
that det(A)=det(B). These two properties can hold
true only when det(A)=det(B)=0.
(c) Let A=3*3 square matrix. Then:
det(A)=a 11 a 22 a33 - a11 a23 a32 - a12 a21 a33
+ a12 a23 a31 + a13 a 21 a 32 - a13 a22 a31
Now, let B be the 3*3 square matrix A, with the
following exception: the first row of B is replaced with k
times the second row of A plus the first row of A. Then:
a11 + ka21 a12 + ka22 a13 + ka23
det(B)= †
a21
a22
a23
†
a31
a32
a33
a
a
a23 2
a23 2
= 1a11 + ka21 2 2 22
- 1 a12 + ka22 2 2 21
a32 a33
a31 a33
a
a22 2
+ 1a13 + ka23 2 2 21
a31 a32
a23
a23
a22 a23
a
a
=a11 `
` + ka21 ` 22
` - a12 ` 21
`
a32 a33
a32 a33
a31 a33
a
a
a
a23
a22
a22
- ka22 ` 21
` + a13 ` 21
` + ka23 ` 21
`
a31 a33
a31 a32
a31 a32
=a11 1a22 a 33 - a23 a32 2 - a12 1 a21 a33 - a23 a31 2
+ a13 1a21 a32 - a22 a31 2 + ka21 1 a22 a33 - a23 a32 2
- ka22 1 a21 a33 - a23 a31 2 + ka23 1 a21 a32 - a22 a 31 2
= a11 a22 a33 - a11 a23 a32 - a12 a21 a33 + a12 a23 a 31
+ a13 a21 a32 - a13 a 22 a 31 + ka21 a22 a33 - ka21 a23 a32
Matrix Algebra
281
- ka21 a22 a33 + ka22 a23a31 + ka21 a22 a32 - ka22 a23 a31
= a11 a22 a33 - a11 a23 a32 - a12 a21 a33 + a12 a23 a31
+ a13 a21 a32 - a13 a22 a31 + 0
= det1A2.
This result holds in general.
69. (a) Let A =[aij] be an n*n matrix and let B be the
same as matrix A, except that the ith row of B is the
ith row of A multiplied by the scalar c. Then:
a11
a21
6 o
cai1
o
an1
det(B)=
ith row →
=cai1 1 -1 2
a12
a22
…
…
a1n
a2n
cai2
…
cain
an2
…
ann
i+1
6
@ A i1 @ + cai2 1 -12 i + [email protected] A i2 @ + …
+cain 1 -1 2 @ A in @
in
=c1ai1 1 -1 2 i + 1 @ A i1 @ + ai2 1 -1 2 i + [email protected] A i2 @ + …
+ain 1 -12 i + n @ A in @ 2
=c det(A) (by definition of determinant)
(b) Use the 2*2 case as an example
det(A)= 2
a11
a21
02
=a11a22-0=a11a22
a22
which is the product of the diagonal elements.
Now consider the general case where A is an n*n
matrix. Then:
a11
a21
det(A)= 4
o
an1
0
a22
o
an2
0
0
a 22
a32
=a11 (–1)2 4
o
an2
…
…
0
0
∞
o
…
a nn
0
0
0
o
a33
o
an3
a33
a43
=a11 (a22)(–1)2 4
o
an3
=a11 a22 … an-2
n-2
an4
0
a44
o
an4
…
…
0
04
o
…
ann
…
…
0
0
∞
o
…
ann
a
1 -12 2 2 n - 1
an
n-1
n-1
02
ann
=a11 a22 … an–2 n–2 an–1 n–1 ann, which is
exactly the product of the diagonal elements
(by induction).
1 x
y
x
y1 2
1 y1 2
70. (a) 3 1 x1 y1 3 =1(–1)2 2 1
+x(–1)3 2
x2 y2
1 y2
1 x2 y2
+y(–1)4 2
1
1
x1
2
x2
=(x1y2-y1x2)-x(y2-y1)+y(x2-x1)
Since (y2-y1) is not a power of x and (x2-x1) is
not a power of y, the equation is linear.
282
Chapter 7
Systems and Matrices
(b) If (x, y)=(x1, y1), then det(A)
=x1y2-x2y1-x1y2+x1y1+x2y1-x1y1
=0, so (x1, y1) lies on the line.
If (x, y)=(x2, y2), then, det(A)
=x1y2-x2y1-x2y2+x2y1+x2y2-x1y2=0,
so (x2, y2) lies on the line.
1 x3 y3
(c) 3 1 x1 y1 3 =0
1 x2 y2
1
(d) 3 1
1
x3
x1
x2
cos a
sin a
–sin a cos a
dc
cos a –sin a
sin a
d
cos a
cos2 a + sin2 a
sin a cos a - cos a sin a
cos Å sin Å - sin Å cos Å
d
sin2 Å + cos2 Å
= c
1
0
0
d = I2
1
(b) From the diagram, we know that:
x=r cos ¨
y=r sin ¨
x=r cos (¨-Å)
y=r sin (¨-Å)
y¿
x¿
or cos (¨-Å)=
sin (¨-Å)=
r
r
From algebra, we know that:
x=r cos (¨+Å-Å)=r cos (Å+(¨-Å)) and
y=r sin (¨+Å-Å)=r sin (Å+(¨-Å))
Using the trigonometric properties and substitution,
we have:
x=r(cos Å cos (¨-Å)-sin Å sin (¨-Å))
=r cos Å cos (¨-Å)-r sin Å sin (¨-Å)
y¿
x¿
=(r cos Å) a b -(r sin Å) a b
r
r
=x cos Å-y sin Å
y=r (sin Å cos (¨-Å)+cos Å sin (¨-Å))
=r sin Å cos (¨-Å)+r cos Å sin (¨-Å)
y¿
x¿
=(r sin Å) a b +(r cos Å) a b
r
r
=x sin Å+y cos Å.
(c) 3 x
y4 = 3x¿
which is 3x¿
y¿ 4 B
cos a
-sin a
sin a
R
cos a
y¿ 4 A-1, the inverse of A.
72. (a) det1 xI2 - A2 = det c
x - a11
-a21
-a12
d
x - a22
=(x-a11)(x-a22)-(a12)(a21)
=x2-a22x-a11x+a11a22-a12a21
=x2+(–a22-a11)x+(a11a22-a12a21)
f(x) is a polynomial of degree 2.
(b) They are equal.
(c) The coefficient of x is the opposite of the sum of the
elements of the main diagonal in A.
(d) f(A)=det(AI-A)=det(A-A)
=det([0])=0.
x - a11
-a21
-a31
-a12
x - a22
-a32
=(x-a11)(–1)2 2
y3
y1 3 Z 0
y2
71. (a) A # A–1 = c
= c
73. det(xI3-A)= †
-a13
-a23 †
x - a33
x - a22
-a32
-a23 2
x - a33
+(–a12)(–1)3 †
-a21
-a31
-a23
†
x - a33
+(–a13)(–1)4 †
- a21
-a31
x - a22
†
-a32
=(x-a11)((x-a22)(x-a33)-a23a32)
+a12((–a21)(x-a33)-a23a31)
-a13(a21a32+(a31)(x-a22))
=(x-a11)(x2-a33x-a22x+a22a33-a23a32)
+a12(–a21x+a21a33-a23a31)
-a13(a21a32+a31x-a22a31)
=x3-a33x2-a22x2+a22a33x-a23a32x-a11x2
+a11a33x+a11a22x-a11a22a33+a11a23a32-a12a21x
+a12a21a33-a12a23a31-a13a21a32-a13a31x
+a13a22a31
=x3+(–a33-a22-a11)x2+(a22a33-a23a32+a11a33
+a11a22-a12a21-a13a31)x+(–a11a22a33+a11a23a32
+a12a21a33-a12a23a31-a13a21a32+a13a22a31)
(b) The constant term equals –det(A).
(c) The coefficient of x2 is the opposite of the sum of the
elements of the main diagonal in A.
(d) f(A)=det (AI-A)=det (A-A)
=det([0])=0
■ Section 7.3 Multivariate Linear Systems and
Row Operations
Exploration 1
1. x+y+z must equal the total number of liters in the
mixture, namely 60 L.
2. 0.15x+0.35y+0.55z must equal total amount of acid
in the mixture; since the mixture must be 40% acid and
have 60 L of solution, the total amount of acid must be
0.40(60)=24 L.
3. The number of liters of 35% solution, y, must equal twice
the number of liters of 55% solution, z. Hence y=2z.
1
4. £ 0.15
0
1
1
x
60
0.35 0.55 § £ y § = £ 24 §
1
-2
z
0
1
A = £ 0.15
0
1
1
x
60
0.35 0.55 § , X = £ y § , B = £ 24 §
1
-2
z
0
Section 7.3
Multivariate Linear Systems and Row Operations
x - y + z = 0
-2y + z = -3
2z = 2
3.75
5. X = A B = £ 37.5 §
18.75
-1
x - y + z = 0
1
3
y - z =
2
2
6. 3.75 L of 15% acid, 37.5 L of 35% acid, and 18.75 L of 55%
acid are required to make 60 L of a 40% acid solution.
Quick Review 7.3
z = 1
1. (40)(0.32)=12.8 liters
4. (80)(1-0.70)=24 milliliters
4.
5. (–1, 6)
6. (0, –1)
7. y=w-z+1
0
1
2
1
2 -1
3 § = £ 0.5
0.5
-2
-0.75
d
0.25
- 0.5
0
0
0.25
0.5 §
0
1. x - 3y + z = 0
(1)
(2)
2y + 3z = 1
z = -2 (3)
Use z=–2 in equation (2).
2y + 31 - 22 = 1
2y = 7
7
y =
2
Use z=–2, y=7/2 in equation (1).
7
x - 3 a b + 1 -2 2 = 0
2
25
x =
2
So the solution is (25/2, 7/2, –2).
So the solution is (–3, –3, 2).
3.
x - y + z = 0
2x - 3z = -1
-x - y + 2z = - 1
x - y + z = 0
-2y + z = - 3
-x - y + 2z = -1
x - y + z = 0
-2y + z = - 3
-2y + 3z = -1
= 6
= -3
= 8
74
=
3
= -3
= 8
x + 3y - z = -3
1
8
y + z =
3
3
74
z =
13
x-y+z=0
–x-y+2z=–1
–7y+2z=6
7
(3y+z=8)
3
1
13y + z = 82
3
3 13
74
a z =
b
13 3
3
8
10
1 74
a b = ;y =
3 13
3
13
74
5
10
= -3; x =
x + 3a b 13
13
13
The solution is (5/13, 10/13, 74/13).
5.
x + y + z = -3
4x - y = - 5
-3x + 2y + z = 4
x + y + z = -3
4x - y = - 5
-4x + y = 7
x + y + z = -3
4x - y = - 5
0 = 2
The system has no solution.
6.
–1(x+y+z=–3)
–3x+2y+z=4
4x-y=–5
–4x+y=7
x + y - 3z = -1
2x - 3y + z = 4
3x - 7y + 5z = 4
x + y - 3z = -1
-5y + 7z = 6
3x - 7y + 5z = 4
x + y - 3z = -1
-5y + 7z = 6
-10y + 14z = 7
2x-3z=–1
2(–x-y+2z=–1)
2x-y=0
–2(x+3y-z=–3)
y +
(1)
(2)
(3)
From equation (3), z=2. Use this in equation (2).
y + 312 2 = 3
y = -3
Use z=2, y=–3 in equation (1).
3x - 1 -3 2 + 21 22 = -2
3x = -9
x = -3
2x - y = 0
x + 3y - z = -3
3y + z = 8
-7y + 2z
x + 3y - z
3y + z
13
z
3
x + 3y - z
3y + z
Section 7.3 Exercises
2. 3x - y + 2z = -2
y + 3z = 3
2z = 4
1
- 1 -2y + z = -32
2
1
12z = 2 2
2
3
1
11 2 = ; y = 2
2
2
x - 2 + 1 = 0; x = 1
The solution is (1, 2, 1).
3. (50)(1-0.24)=38 liters
0
10. £ - 2
0
–1(–2y+z=–3)
–2y+3z=–1
y -
2. (60)(0.14)=8.4 milliliters
8. x=2z-w+3
1
3 -1
-0.5
d = c
9. c
- 2 -2
0.5
283
x + y - 3z = -1
-5y + 7z = 6
0 = -5
The system has no solution.
–2(x+y-3z=–1)
2x-3y+z=4
–3(x+y-3z=–1)
3x-7y+5z=4
–2(–5y+7z=6)
–10y+14z=7
284
Chapter 7
7.
x + y - z
y + w
x - y
x + z + w
=
=
=
=
4
-4
1
1
2y - z
y + w
x - y
x + z + w
=
=
=
=
3
-4
1
1
2y - z
y + w
x - y
y + z + w
=
=
=
=
3
-4
1
0
- z - 2w
y + w
x - y
y + z + w
=
=
=
=
11
-4
1
0
- z - 2w
y + w
x - y
z
=
=
=
=
11
-4
1
4
x - y
y + w
1
w + z
2
z
Systems and Matrices
x+y-z=4
–1(x-y=1)
1
x - y + z - w
2
w
x - z
y + w
–1(x-y=1)
x+z+w=1
2y-z=3
–2(y+w=–4)
1
- 1 -z - 2w = 11 2
2
11
15
1
14 2 = - ; w = 2
2
2
7
15
y + a - b = -4; y =
2
2
7
9
x - = 1; x =
2
2
9 7
15
So the solution is a , , 4, - b .
2 2
2
w +
8.
3
z - w
2
w
x - z
y + w
-y +
x - z
2
2
z - y - w
3
3
y + w
w
–1(y+w=–4)
y+z+w=0
= 1
= -4
11
= 2
= 4
1
x - y + z - w
2
- x + y + z + 2w
x - z
y + w
1
x - y + z - w
2
y + 2w
x - z
y + w
= 1
= -1
= 2
= 0
–x+y+z+2w=–3
x-z=2
= 1
y+2w=–1
–1(y+w=0)
= -1
= 2
= 0
1
x-y+z-w=1
2
1
– (x-z=2)
2
= 0
= -1
= 2
= 0
= 2
= 0
3
2
a -y + z - w = 0 b
3
2
= 0
= -1
y + 1 -12 = 0; y = 1
2
2
z - 11 2 - 1 -12 = 0; z = 0
3
3
x - 0 = 2; x = 2
So the solution is (2, 1, 0, –1).
2
9. £ 1
0
-6
2
-8
4
-3 §
4
10. £
1
1
-3
-3
2
1
= 1
11. £
0
1
-3
-10
2
1
= -3
= 2
= 0
12. £
2
3
-3
-6
-4
1
2
-3 §
-2
10
-3 §
-2
4
1§
-2
13. R12
14. (2)R2+R1
15. (–3)R2+R3
16. (1/4)R3
For #17–20, answers will vary depending on the exact sequence of row operations used. One possible sequence of row operations
(not necessarily the shortest) is given. The answers shown are not necessarily the ones that might be produced by a grapher or
other technology. In some cases, they are not the ones given in the text answers.
17. £
1 3
2 1
-3 0
1
18. £ - 3
-2
2
-6
-4
-1
1
1 - 2 2 R1 + R2
4 § ———————" £ 0
13 2 R1 + R3
1
0
-3
1
1 3 2R1 + R2
10 § ———————" £ 0
1 22R1 + R3
7
0
3
-5
9
-1
1
19>52R2 + R3
6 § ———————" £ 0
1 -1>52R2
-2
0
3
1
0
1
-1
1 5>44 2R3
-1.2 § —————" £ 0
0
8.8
2
0
0
-3
1
1 -12R2 + R3
1 § ———————" £ 0
1
0
2
0
0
-3
1§
0
3
1
0
-1
-1.2 §
1
Section 7.3
1
19. £ - 2
3
2
6
12
1 2 3
£0 1 0
0 0 1
20. c
3
2
6
5
-4
1
12 2R1 + R2
2 § ———————" £ 0
1 -3 2R1 + R3
12
0
3
-6
6
-4
- 0.6 §
- 9.2
11>3 2R1
-6
1 2 3
d —————" c
-3
2 5 5
9
5
2
10
6
Multivariate Linear Systems and Row Operations
-4
1
1 1>10 2R2
-6 § —————" £ 0
1 -1>3 2R3
24
0
3
0
-3
1 -2 2R1 + R2
-2
1
d ———————" c
-3
0
2
1
3
-1
2
1
-2
-4
122R2 + R3
- 0.6 § ———————"
-8
3
0
1
-2
d
1
In #21–24, reduced row echelon format is essentially unique, though the sequence of steps may vary from those shown.
1
21. £ 3
2
0
2
1
1
3
22. ≥
-2
3
1
1
1 - 3 2R1 + R2
7 § ———————" £ 0
1 -2 2R1 + R3
4
0
2
4
3
-2
-5
4
-5
2
6
-3
6
1
3
-2
4
0
2
1
2 1
1
1 1>2 2R2
-2 4 § —————" £ 0
-1 2
0
1
1
+
R
1
32R
-1
0
1
2
¥ ———————" ≥
5
0
1 22 R1 + R3
-3
3
1
1 -1 2R2 + R4
0
———————" ≥
0
1 -2 2R3 + R1
0
0
1
0
0
0
0
1
0
1
0
0
1
1
-3
2
-5
3
-7
13 2 R1 + R2
1
1
d ———————" c
-4
0
24. £
3
2
-3
-6
-4
6
3
2
-3
-3
1
1 1>3 2R1
- 2 § —————" £ 2
3
-3
2
25. £ - 1
3
-3
1
0
1
-4
-1
1
-3 §
2
3
1
-4
0
1
1
2
27. £ 1
0
-5
0
2
1
-2
-3
28. c
3
-1
-2
5
-1
-2
2
0
1
6
1
0
0
4
1
-21
1 -12R4 + R1
0
-4
¥ ———————" ≥
0
7
0
-2
23. c
26. c
-2
1
0
-5
2
1
-2
-4
6
0
1
1
2
-1
-1
1
1
1 -1 2R2 + R3
2 § ———————" £ 0
0
2
1
1 0 2 1
+
R
1
-3
2R
-4
0 1 0 0
1
4
¥ ———————" ≥
7
0 0 1 0
12 2R2 + R1
-3
0 1 0 1
0
1
0
0
0
0
1
0
0
0
0
1
- 19
-4
¥
7
-2
1 -2 2R2 + R1
1 0
1
d ———————" c
0 1
-1
3
2
1
2
-3
1
-1
1 -2 2R1 + R2
-2 § ———————" £ 0
13 2R1 + R3
0
3
-1
2
3
d
-1
-2 1
0 0
0 0
-1
0§
0
1
d
4
-1
1
-1
-3
4§
5
5
d
7
In #29–32, the variable names (x, y, etc.) are arbitrary.
3x+2y=–1
–4x+5y=2
30. x-z+2w=–3
2x+y-w=4
–x+y+2z=0
29.
2x+z=3
–x+y=2
2y-3z=–1
31.
32. 2x+y-2z=4
–3x+2z=–1
1 -2
1
8
1
1 -2 2 R1 + R2
2
1 - 3 - 9 § ———————" £ 0
13 2R1 + R3
-3
1
3
5
0
x-2y+z=8
y-z=–5
z=4
y-4=–5; y=–1
x-2(–1)+4=8; x=2
So the solution is (2, –1, 4).
33. £
-2
5
-5
1
-5
6
8
1
11 2R2 + R3
- 25 § ———————" £ 0
11>5 2R2
29
0
-2
1
0
1
-1
1
8
-5 §
4
0
1
0
2
-1
0
-7
-4
¥
7
-6
1
2§
0
285
Chapter 7
286
Systems and Matrices
3 7 -11 44
0 1
1 -3 2 R2 + R1
34. £ 1 2
-3 12 § ———————" £ 1 2
1 -4 2 R2 + R3
4 9 -13 53
0 1
x+2y-3z=12
y-2z=8
z=–3
y-2(–3)=8; y=2
x+2(2)-3(–3)=12; x=–1
So the solution is (–1, 2, –3).
£
35. (x, y, z)=(–2, 3, 1):
1
3
-2
2
1
0
-2 1
-3 2
-8 5
-2
1
1 -22 R1 + R2
2 § ———————" £ 0
1 -4 2R1 + R3
-5
0
-2 1
1 0
0 1
1
4
2
3
10
4
2
1
1 -3 2R1 + R2
5 § ———————" £ 0
1 -1 2R1 + R3
3
0
-3
-2
1
12
8§
-3
1
-1 3
1 -2 2R2 + R1
0 3 § ———————" £ 0
11 2R3 + R1
0
1 1
3
1
1 -32R1 + R2
12 § ———————" £ 0
12 2R1 + R3
-5
0
1
£3
1
37. No solution:
2
1
0
-1
-3
3
2
7
-4
1
£2
4
36. (x, y, z)=(7, 6, 3):
1
8
1 - 12R1 + R3
12 § ———————" £ 0
R12
0
5
-2
-3
-1
1
1
1
3
1
1
0
1
0
1
-2
12 2R2 + R1
6 § ———————" £ 0
1 -12R3 + R1
0
3
2
1
1 - 12R2 + R3
-1 § ———————" £ 0
1
0
0
0
1
0
1
0
1
1
0
-2
3§
1
0
0
1
7
6§
3
3
1
0
2
-1 §
2
-1
1
0
2
-1 §
0
38. (x, y, z)=(z+2, –z-1, z) — the final matrix translates to x-z=2 and y+z=–1.
1 0
£ -2 1
2 1
-1
3
-1
2
1
12 2R1 + R2
-5 § ———————" £ 0
1 -2 2R1 + R3
3
0
0
1
1
2
1
1 -1 2R2 + R3
-1 § ———————" £ 0
-1
0
-1
1
1
0
1
0
39. (x, y, z)=(2-z, 1+z, z) — the final matrix translates to x+z=2 and y-z=1.
c
1
2
0
1
1
1
1 -2 2R1 + R2
2
1
d ———————" c
5
0
0
1
1
-1
2
d
1
40. (x, y, z)=(z+53, z-26, z) — the final matrix translates to x-z=53 and y-z=–26.
c
1
-3
2
-5
-3
8
132 R1 + R2
1
1
d ———————" c
-29
0
1
£3
2
41. No solution:
2
4
3
-3
-1
4
1
1 -32R1 + R2
5 § ———————" £ 0
1 -22R1 + R3
4
0
1
£2
2
42. (x, y)=(1, 2):
2
1
1
3
2
1 - 22R2 + R1
1
1
d ———————" c
0
-26
2
-2
-1
3
1
1 -22 R1 + R2
8 § ———————" £ 0
1 -2 2R1 + R3
6
0
0
1
1
4
11>2 2R2
-7 § —————" £ 0
0
-4
1
1
0
53
d
-26
-1
-1
2
-1
-1
4
-7>2 §
-4
3
1
1 -1 2R2 + R1
2 § ———————" £ 0
0
0
0
1
0
1
2§
0
43. (x, y, z)=(z+w+2, 2z-w-1, z, w) — the final matrix translates to x-z-w=2 and y-2z+w=–1.
1
£1
2
1
0
1
-3
-1
-4
1
0
1 -1 2R2 + R1
2 § ———————" £ 1
1 -2 2 R2 + R3
3
0
0
-1
-1
1
0
1
-2
-1
-2
1
-1
1
-1
1
1 -1 2R1 + R3
2 § ———————" £ 0
R12
-1
0
0
1
0
-1
-2
0
-1
1
0
2
-1 §
0
44. (x, y, z)=(z-w, w+3, z, w) — the final matrix translates to x-z+w=0 and y-w=3.
1
£2
1
45. c
2
1
5
46. £ 2
1
-1
-1
-2
-1
-2
-1
-3
1
1 -2 2 R1 + R2
-3 § ———————" £ 0
1 -1 2 R1 + R3
-6
0
-3
5 x
d
dc d = c
1
-2 y
-7
-3
1
1 x
2
-1 § £ y § = £ 3 §
1 z
-3
47. 3x- y=–1
2x+4y=3
48.
2
3
3
x-3z=3
2x-y+3z=–1
–2x+3y-4z=2
-1
1
-1
-1
0
0
2
-1
1
1
-3
11 2R2 + R1
3§ ———————" £ 0
11 2R2 + R3
0
-3
0
1
0
-1
0
0
1
-1
0
0
3§
0
Section 7.3
Multivariate Linear Systems and Row Operations
49. (x, y)=(–2, 3):
x
2
c d = c
y
4
- 3 -1 - 13
1
1
d c
d =
c
1
-5
14 -4
3 -13
-2
1 -28
dc
d =
c
d = c
d.
2
-5
14
42
3
50. (x, y)=(1, –1.5):
x
1
c d = c
y
3
2 -1 - 2
1 -4
d = - c
d c
9
-4
10 -3
x
2
51. (x, y, z)=(–2, –5, –7); £ y § = £ 1
z
3
x
1
52. (x, y, z)=(3, –0.5, 0.5); £ y § = £ 2
z
-3
1 -10
-2 - 2
1
dc
d = - c
d = c
d.
1
9
10
15
-1.5
-1
2
-2
1
-3 §
1
4
1
3
-2
1§
-5
-1
£
-1
-6
-2
9 § = £ -5 §
-3
-7
0
3
£
6 § = £ - 0.5 §
-13
0.5
-1
x
2
y
1
53. (x, y, z, w)=(–1, 2, –2, 3); ≥ ¥ = ≥
z
3
w
-2
-1
2
-1
3
1
-3
-1
1
1
1
¥
2
-3
2
x
3
y
54. (x, y, z, w)=(4, –2, 1, –3); ≥ ¥ = ≥
-2
z
4
w
1
2
1
-3
2
-1
0
2
0
-1
¥
-3
-5
-3
-1
12
2
≥
¥ = ≥
¥
3
-2
-3
3
-1
4
8
10
-2
≥
¥ = ≥
¥
-1
1
39
-3
55. (x, y, z)=(0, –10, 1): Solving up from the bottom gives z=1; then y-2=–12, so y=–10;
then 2x+10=10, so x=0.
2x-y=10
2x-y=10
2x-y=10
y-2z=–12
x-z=–1 1 2E2-E1 1 y-2z=–12
y+z=–9
y+z=–9 1 E3-E2 1
3z=3
56. (x, y, z)= 1 -2, 0, 0.5 2 : Solving up from the bottom gives z=0.5;
then y-(5.5)(0.5)=247-2.75, so y=0 ; then 1.25x+0.5=–2, so x=–2 .
1.25x+z=–2
1.25x+z=–2
y-5.5z=–2.75
y-5.5z=–2.75
3x-1.5y=–6
1 E3-2.4E1+1.5E2 1
–10.65z=–5.325
57. (x, y, z, w)=(3, 3, –2, 0): Solving up from the bottom gives w=0; then –z+0=2, so z=–2; then –3y+4=–5,
so y=3; then x+6-4=5, so x=3.
x+2y+2z+w=5
x+2y+2z+w=5
2x+y+2z=5 1 E2-2E1 1
–3y-2z-2w=–5
3x+3y+3z+2w=12 1 E3-3E1 1
–3y-3z-w=–3 1 E3-E2 1
x+z+w=1 1 E4-E1 1
–2y-z=–4 1 3E4-2E2 1
x+2y+2z+w=5
x+2y+2z+w=5
–3y-2z-2w=–5
–3y-2z-2w=–5
–z+w=2
–z+w=2
z+4w=–2 1 E4+E3 1
5w=0
58. (x, y, z, w)=(–1, 2, 4, –1): Solving up from the bottom gives w=–1; then –z+2=–2, so z=4; then –y+42=0, so y=2; then x-2-1=–4, so x=–1.
x-y+w=–4
x-y+w=–4
x-y+ w=–4
–2x+y+z=8
1 E2+2E1 1
–y+z+2w=0
–y+z+2w=0
2x-2y-z=–10 1 E3-2E1 1
–z-2w=–2
–z-2w=–2
–2x+z+w=5
1 E4+2E1 1 –2y+z+3w=–3 1 E4-2E2-E3 1
w=–1
59. (x, y, z)= a 2 -
1
3
1
z, - z - 4, z b : z can be anything; once z is chosen, we have 2y+z=–8, so y = - z - 4; then
2
2
2
3
1
x- a - z - 4 b + z = 6, so x=2- z
2
2
x-y+z=6
x+y+2z=–2 1 E2-E1 1
x-y+z=6
2y+z=–8
287
288
Chapter 7
Systems and Matrices
3
1
3
60. (x, y, z)= a z - 1, z - 2, z b : z can be anything; once z is chosen, we have 5y-3z=–10, so y = z - 2 ; then
5
5
5
3
1
x - 2 a z - 2 b + z = 3 , so x = z - 1 .
5
5
x-2y+z=3
x-2y+ z=3
2x+y-z=–4 1 E2-2E1 1
5y-3z=–10
61. (x, y, z, w)=(–1-2w, w+1, –w, w): w can be anything; once w is chosen, we have –z-w=0, so z=–w;
then y-w=1, so y=w+1; then x+(w+1)+(–w)+2w=0, so x=–1-2w.
2x+y+z+4w=–1 1 E1-2E3 1
–y-z=–1 1 E1+E2 1
–z- w=0
x+2y+z+w=1 1 E2-E3 1
y-w=1
y-w=1
x+y+z+2w=0
x+y+z+2w=0
x+y+z+2w=0
62. (x, y, z, w)=(w, 1-2w, –w-1, w): w can be anything; once w is chosen, we have –z-w=1, so z=–w-1;
then y+2w=1, so y=1-2w; then x+(1-2w)+2(–w-1)+3w=–1, so x=w.
2x+3y+3z+7w=0 1 E1-2E3 1
y-z+w=2 1 E1-E2 1
–z- w=1
x+2y+2z+5w=0 1 E2-E3 1
y+2w=1
y+2w=1
x+y+2z+3w=–1
x+y+2z+3w=–1
x+y+2z+3w=–1
63. (x, y, z, w)=(–w-2, 0.5-z, z, w): z and w can be anything; once they are chosen, we have –y-z=–0.5,
so y=0.5-z; then since y+z=0.5 we have x+0.5+w=–1.5, so x=–w-2.
2x+y+z+2w=–3.5 1 E1-2E2 1
–y-z=–0.5
x+y+z+w=–1.5
x+y+z+w=–1.5
64. (x, y, z, w)=(z-3w+1, 2w-2z+4, z, w): z and w can be anything; once they are chosen, we have
–y-2z+2w=–4, so y=2w-2z+4; then x+(2w-2z+4)+z+w=5, so x=z-3w+1.
2x+y+4w=6 1 E1-2E2 1
–y-2z+2w=–4
x+y+z+w=5
x+y+z+w=5
65. No solution: E1+E3 gives 2x+2y-z+5w=3, which contradicts E4.
66. (x, y, z, w)=(1, 1-w, 6w-2, w): Note first that E4 is the same as E1, so we ignore it. w can be anything, while x=1.
Once w is chosen, we have 1+y+w=2, so y=1-w; then 2(1-w)+z-4w=0, so z=6w-2.
x+y+w=2
x+y+w=2
x+y+w=2
x+4y+z-2w=3 1 E2-E1 1
3y+z-3w=1 1 E2-E1-E3
–x=–1
x+3y+z-3w=2 1 E3-E1 1
2y+z-4w=0
2y+z-4w=0
67. f(x)=2x2-3x-2: We have f(–1)=a(–1)2+b(–1)+c=a-b+c=3, f(1)=a+b+c=–3, and
f(2)=4a+2b+c=0. Solving this system gives (a, b, c)=(2, –3, –2).
a-b+c=3
a-b+c=3
a-b+c=3
a+b+c=–3 1 E2-E1 1
2b=–6
2b
=–6
4a+2b+c=0 1 E3-4E1 1
6b-3c=–12 1 E3-3E2 1
–3c=6
68. f(x)=3x3-x2+2x-5: We have f(–2)=–8a+4b-2c+d=–37, f(–1)=–a+b-c+d=–11,
f(0)=d=–5, and f(2)=8a+4b+2c+d=19. Solving this system gives (a, b, c, d)=(3, –1, 2, –5).
–8a+4b-2c+d=–37
–8a+4b-2c+d=–37 1 E1-8E2 1 –4b+6c-7d=51
–a+b-c+d=–11
–a +b-c+d=–11
–a+b-c+d=–11
d=–5
d=–5
d=–5
8a+4b+2c+d=19 1 E4-E1 1
8b+2d=–18
8b+2d=–18
69. f(x)=(–c-3)x2+x+c, for any c — or f(x)=ax2+x+(–a-3), for any a: We have f(–1)=a-b+c=–4
and f(1)=a+b+c=–2. Solving this system gives (a, b, c)=(–c-3, 1, c)=(a, 1, –a-3). Note that when c=–3
(or a=0), this is simply the line through (–1, –4) and (1, –2).
a-b+c=–4
a-b+c=–4
a+b+c=–2 1 E2-E1 1
2b=2
70. f(x)=(4-c)x3-x2+cx-1, for any c — or f(x)=ax3-x2+(4-a)x-1, for any a: We have f(–1)
=–a+b-c+d=–6, f(0)=d=–1, and f(1)=a+b+c+d=2. Solving this system gives (a, b, c, d)
=(4-c, –1, c, –1)=(a, –1, 4-a, –1). Note that when c=4 (or a=0), this is simply the parabola through the given
points.
–a+b-c+d=–6
–a+b-c+d=–6
d=–1
d=–1
a+b+c+d=2 1 E3+E1 1
2b+2d=–4
Section 7.3
71. In this problem, the graphs are representative of the population (in thousands) of the cities of Corpus Christi, TX,
and Garland, TX, for several years, where x is the number
of years past 1980.
(a) The following is a scatter plot of the Corpus Christi
data with the linear regression equation
y=2.0735x+234.0268 superimposed on it.
[–3, 30] by [0, 400]
(b) The following is a scatter plot of the Garland data with
the linear regression equation y=3.5302x+141.7246
superimposed on it.
Multivariate Linear Systems and Row Operations
289
72. In this problem, the graphs are representative of the population (in thousands) of the cities of Anaheim, CA, and
Anchorage, AK, for several years, where x is the number of
years past 1970.
(a) The following is a scatter plot of the Anaheim data
with the linear regression equation
y=5.1670x+166.2935 superimposed on it.
[–3, 35] by [0, 400]
(b) The following is a scatter plot of the Anchorage data
with the linear regression equation
y=6.3140x+78.3593 superimposed on it.
[–3, 30] by [0, 400]
[–3, 35] by [0, 400]
(c) Graphical solution: Graph the two linear equations
y=2.0735x+234.0268 and y=3.5302x+141.7246
on the same axis and find their point of intersection.
The two curves intersect at x≠63.4. So, the population of the two cities will be the same sometime in the
year 2043.
(c) Graphical solution: Graph the two linear equations
y=5.1670x+166.2935 and y=6.3140x+78.3593
on the same axis and find their point of intersection.
The two curves intersect at x≠76.7.
The population of the two cities will be the same sometime in the year 2046.
[–5, 100] by [0, 500]
[–5, 120] by [0, 800]
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
Algebraic solution:
Solve 2.0735x+234.0268=3.5302x+141.7246 for x.
2.0735x+234.0268=3.5302x+141.7246
1.4567x=92.3022
92.3022
x =
L 63.4
1.4567
The population of the two cities will be the same sometime
in the year 2043.
Another graphical solution would be to find where the
graph of the difference of the two curves is equal to 0.
Algebraic solution:
Solve 5.1670x+166.2935=6.3140x+78.3593 for x.
5.1670x+166.2935=6.3140x+78.3593
1.147x=87.9342
87.9342
x =
L 76.7
1.147
The population of the two cities will be the same sometime
in the year 2046.
73. (x, y, z)=(825, 410, 165), where x is the number of children, y is the number of adults, and z is the number of senior citizens.
x+y+z=1400
x+y+z=1400
25x+100y+75z=74,000 1 E2-75E1 1
–50x+25y=–31,000
x-y-z=250
1 E3+E1 1
2x=1650
Chapter 7
290
74. (x, y, z)= a
Systems and Matrices
160 320 400
,
,
b ≠(14.55, 29.09, 36.36) (all amounts in grams), where x is the amount of 22% alloy, y is the
11 11 11
amount of 30% alloy, and z is the amount of 42% alloy.
x+y+z=80
0.22x+0.30y+0.42z=27.2 1 50E2-11E1 1
2x-y=0
1
E3- 2E1 1
x+y+z=80
4y+10z=480
–3y-2z=–160 1 4E3+3E2 1
x+y+z=80
4y+10z=480
22z=800
75. (x, y, z)=(14,500, 5500, 60,000) (all amounts in dollars), where x is the amount invested in CDs, y is the amount in bonds,
and z is the amount in the growth fund.
x+y+z=80,000
x+y+z=80,000
0.067x+0.093y+0.156z=10,843 1 1000E2-67E1 1
26y+89z=5,483,000
3x+3y-z=0
1
E3-3E1
1
–4z=–240,000
76. (x, y, z)=(z-9000, 29,000-2z, z) (all amounts in dollars). The amounts cannot be determined: if z dollars are invested
at 10% (9000 z 14,500), then z-9000 dollars invested at 6% and 29,000-2z invested at 8% satisfy all conditions.
x+y+z=20,000
x+y+z=20,000
x+y+z=20,000
0.06x+0.08y+0.10z=1780 1
50E2
1 3x+4y+5z=89,000 1 E2-3E1 1
y+2z=29,000
–x+z=9000 1 E3+4E1 1 3x+4y+5z=89,000
1
(21,250,000-885z)≠
295
72,033.898-3z dollars must be invested in bonds, and x≠2z-22,033.898 dollars are invested in CDs. Since x 0, we see
that z 11016.95 (approximately); the minimum value of z requires that x=0 (this is logical, since if we wish to minimize z, we
should put the rest of our money in bonds, since bonds have a better return than CDs). Then y≠72,033.898-3z=38,983.05.
x+y+z=50,000
x+y+z=50,000
0.0575x+0.087y+0.146z=5000 1 10,000E2-575E1 1
295y+885z=21,250,000
77. (x, y, z)≠(0, 38,983.05, 11,016.95): If z dollars are invested in the growth fund, then y=
78. (x, y, z)=(0, 28.8, 11.2): If z liters of the 50% solution are used, then y=
8
1
(880-40z)= (22-z) liters of 25%
15
3
56
5
solution must be used, and x= z- liters of 10% solution are needed. Since x 0, we see that z 11.2 liters;
3
3
8
the minimum value of z requires that x=0. Then y= (22-z)=28.8 liters.
3
x+y+z=40
x+y +z=40
15y+40z=880
0.10x+0.25y+0.50z=12.8 1 100E2-10E1 1
79. 22 nickels, 35 dimes, and 17 quarters:
1
£5
1
1
1
10 25
-1
1
74
1
1 -5 2 R1 + R2
885 § ———————" £ 0
1 -1 2 R1 + R3
4
0
1
£0
0
1 1
5 20
-2
0
0
1
0
1
0
20
74
1
R23
515 § —————" £ 0
1 -1>22 R2
-70
0
39
1
11>202R3
"
35 § ————— £ 0
340
0
0
1
0
1
0
1
1
1
5
1
0
20
74
1 - 12R2 + R1
35 § ———————"
1 -5 2R2 + R3
515
39
1
1 -1 2R3 + R1
"
35 § ——————— £ 0
17
0
0
1
0
0
0
1
22
35 §
17
80. 27 one-dollar bills, 18 fives, and 6 tens:
1 1
£1 5
0 1
1
10
-3
51
1
1 -12 R1 + R2
177 § ———————" £ 0
1 -1 2 R3 + R1
0
0
0
4
1
4
51
1
1 -42 R3 + R2
9 126 § ———————" £ 0
R23
-3
0
0
1
£0
0
81. (x, p)= a
=
1 1
c
15 10
82. (x, p)= a
=
x
5
16 220
,
b: c d = c
p
- 10
3 3
1
- 1 100
80
1 16
dc
d =
c
d = c
d
5
20
15 1100
3 220
x
12
10
, 110 b : c d = c
p
- 24
3
1 1
c
36 24
1 -1 100
d c
d
1
20
1 -1 150
d c
d
1
30
1 120
- 1 150
1 10
dc
d =
c
d = c
d
12
30
36 3960
3 330
0
1
0
4
-3
1
0
1
0
4
-3
21
51
11>21 2 R3
0 § —————"
126
51
1
1 -4 2R3 + R1
"
0 § ——————— £ 0
13 2R3 + R2
6
0
0
1
0
0
0
1
27
18 §
6
83. Adding one row to another is the same as multiplying that
first row by 1 and then adding it to the other, so that it falls
into the category of the second type of elementary row
operations. Also, it corresponds to adding one equation to
another in the original system.
84. Subtracting one row from another is the same as multiplying that first row by –1 and then adding it to the other, so
that it falls into the category of the second type of elementary row operations. Also, it corresponds to subtracting one
equation from another.
Section 7.4
85. False. For a nonzero square matrix to have an inverse, the
determinant of the matrix must not be equal to zero.
86. False. The statement holds only for a system that has
1 0 1 0
exactly one solution. For example, £ 0 1 1 1 §
0 0 0 2
could be the reduced row echelon form for a system that
has no solution.
1
90. £ - 1
2
2
3
-1
-1
2
3
8
1
1 12 R1 + R2
" £0
3§
1 -2 2R1 + R3
-19
0
1 0 0
1 2 0
4
1 1>5 2 R2
" £0 1 0
£ 0 5 0 15 §
1 -2 2R2 + R1
0 0 1
0 0 1 -4
The answer is E.
2
5
-5
-1
1
5
Partial Fractions
291
87. 213 2 - 1 -12 12 2 = 8. The answer is D.
88. The augmented matrix has the variable coefficients in the
first three columns and the constants in the last column.
The answer is A.
89. Twice the first row was added to the second row. The
answer is D.
8 1 1 2R + R 1
2
3
" £0
11 §
11>62R3
-35
0
2
5
0
8
1 - 12R3 + R2
"
11 §
11 2R3 + R1
-4
-1
1
1
-2
3§
-4
91. (a) The planes can intersect at exactly one point
(b) At least two planes are parallel, or else the line of each
pair of intersecting planes is parallel to the third plane.
(c) Two or more planes can coincide, or else all three
planes can intersect along a single line.
92. Starting with any matrix in row echelon form, one can
perform the operation kRi+Rj, for any constant k, with
i>j, and obtain another matrix in row echelon form. As
1 1 1
1 2 2
a simple example, c
d and c
d are two
0 1 1
0 1 1
equivalent matrices (the second can be obtained from the
first via R2+R1), both of which are in row echelon form.
■ Section 7.4 Partial Fractions
Exploration 1
1. (a) 25-1=A1(5-5)+A2(5+3)
24=8 A2
3=A2
(b) –15-1=A1(–3-5)+A2(–3+3)
–16=–8 A1
2=A1
2. (a) –4+4+4=A1(2-2)2+2 A2(2-2)+2 A3
4=2 A3
2=A3
#
#
(b) 0+0+4=A1(0-2)2+0 A2(0-2)+0 A3
4=4 A1
1=A1
93. (a) C(x)=(x-3)(x-5)-(–1)(–2)
=x2-8x+13.
(b)
(c) Suppose x=3
–9+6+4=1 (3-2)2+2 3(3-2)+3A2
1=1+6+3A2
–6=3A2
–2=A2
#
[–1, 8.4] by [–3.1, 3.1]
(c) C(x)=0 when x=4 — 113 — approx. 2.27 and 5.73.
(d) det A=13, and the y-intercept is (0, 13). This is the
case because C(0)=(3)(5)-(1)(2)=det A.
Quick Review 7.4
1.
(e) a11+a22=3+5=8. The eigenvalues add to
(4- 113)+(4+ 113)=8, also.
94. (a) C(x)=(x-2)2-(–5)(–1)=x2-4x-1.
2.
(b)
3.
[–2.7, 6.7] by [–5.1, 1.1]
1 x - 3 2 + 2 1x - 1 2
1x - 1 2 1x - 3 2
3x - 5
= 2
x - 4x + 3
=
51 x + 1 2 - 21x + 42
1x + 4 2 1x + 1 2
3 1x - 12
= 2
x + 5x + 4
=
3x - 5
1x - 12 1x - 32
3x - 3
1x + 42 1x + 12
1x + 1 2 2 + 3x1x + 1 2 + x
x1x + 12
4x2 + 6x + 1
= 3
x + 2x2 + x
2
31x2 + 12 - 1x + 12
(c) C(x)=0 when 2 — 15 — approx. –0.24 and 4.24.
4.
(d) det A=–1, and the y-intercept is (0, –1). This is the
case because C(0)=(2)(2)-(–5)(–1)=det A.
5. –2 |
(e) a11+a22=2+2=4. The eigenvalues add to
(2- 15)+(2+ 15)=4, also.
#
1x + 12
2
3
3
f1x2
d1x2
2
–6
–6
0
= 3x2 - 2 +
=
=
4x2 + 6x + 1
x1x + 12 2
3x2 - x + 2
1x2 + 12 2
–2
0
–2
3
x - 2
7
4
3
Chapter 7
292
6.
Systems and Matrices
2x+1
4.
x +x-6R 2x +3x -14x-8
2
3
2
2x3+2x2-12x
x2- 2x-8
x2+ x-6
–3x-2
f1x2
3x + 2
d1x2
x2 + x - 6
7. Possible real rational zeros:
; 1, ; 2, ; 3, ; 4, ; 6, ; 12
;1
From a graph, x=–1 and x=3 seem reasonable:
–1 | 1
–2
1
–8
–12
1
–3
4
–12
1
–3
4
–12
0
–3 | 1
–3
4
–12
–3
0
–12
1
0
4
0
= 2x + 1 -
5.
6.
7.
x4 - 2x2 + x2 - 8x + 12 = 1x + 1 2 1x - 3 2 1x2 + 42
; 1, ; 2, ; 5, ; 10
From a
;1
graph, x=–1, x=–2 and x=5 seem reasonable:
–1 | 1
–1 –15
–23
–10
1
–2
–13
–10
1
–2 –13
–10
0
–2 | 1
–2 –13
–10
2
–8
–10
1
–4
–5
0
–5 | 1
–4
–5
–5
–5
1
1
0
x4 - x3 - 15x2 - 23x - 10 = 1x + 1 2 2 1x + 2 2 1x - 5 2
8. Possible real rational zeros:
x2 + 3x + 2
x2 + 3x + 2
=
3
3
1x - 12
1x - 12 3 1x2 + x + 1 2 3
A3
A1
A2
B1x + C1
+
+ 2
+
=
2
3
x - 1
1x - 12
1x - 12
x + x + 1
B3x + C3
B2x + C2
+ 2
+
1x2 + x + 12 2
1x + x + 12 3
-3
4
+
: x+22=A(x-2)+B(x+4)
x + 4
x - 2
=(A+B)x+(–2A+4B)
A
1 1 -1 1
-3
A + B = 1
d c d = c
d
1 c d = c
B
-2 4
22
4
-2A + 4B = 22
1
2
- : x-3=Ax+B(x+3)
x + 3
x
=(A+B)x+3B
A + B = 1
A
1 1 -1 1
2
1 c d = c
d c
d = c
d
B
0 3
-3
-1
0 + 3B = -3
2x - 1
3
+
: 3x2+2x+2
x + 1
1x2 + 12 2
=(Ax+B)(x2+1)+(Cx+D)
=Ax3+Bx2+(A+C)x+(B+D)
A
= 0
A
B
= 3
B
1 ≥ ¥
A
+ C
= 2
C
B
+ D = 2
D
2
-1
1
0
=≥
1
0
8.
In #9–10, equate coefficients.
9. A=3, B=–1, C=1
10. A=–2, B=2, C=–1, D=–5
0
1
0
1
0
0
1
0
0
0
¥
0
1
0
0
3
3
≥ ¥ = ≥
¥
2
2
2
-1
1
2
1
+ 2 : 4x+4
x
x + 2
x
=Ax(x+2)+B(x+2)+Cx2
=(A+C)x2+(2A+B)x+(2B)
A
+ C = 0
A
2A + B
= 4 1 £ B§
2B
= 4
C
-1
1
=£2
0
Section 7.4 Exercises
1.
A3
A1
A2
x2 - 7
=
+
+
x
x - 2
x + 2
x1 x2 - 42
2.
x4 + 3x2 - 1
2
1 x + x + 1 2 2 1x2 - x + 1 2
=
3.
B1x + C1
x2 + x + 1
+
x5 - 2x4 + x - 1
x3 1 x - 1 2 2 1 x2 + 9 2
=
B2x + C2
1 x2 + x + 1 2 2
9.
+
B3x + C3
x2 - x + 1
A3
A5
A1
A2
A4
B1x + C1
+ 2 + 3 +
+
+
2
x
x - 1
x
x
1x - 12
x2 + 9
0
1
2
1
0§
0
0
1
£4§ = £ 2§
4
-1
1
1
2
: x2-2x+1
+
+
x - 2
1x - 22 2
1x - 2 2 3
=A(x-2)2+B(x-2)+C
=Ax2+(–4A+B)x+(4A-2B+C)
A
= 1
1
0 0
1
-4A + B
= -2 1 £ -4
1 0 -2 §
4A - 2B + C = 1
4 -2 1
1
Using a grapher, we find that the reduced row echelon
form of the augmented matrix is:
1 0 0 1
A
1
£0 1 0 2§ 1 £ B§ = £2§
0 0 1 1
C
1
Section 7.4
10.
11.
-3x + 7
8x - 17
+ 2
:
x2 + 4
x + 9
5x3-10x2+5x-5
=(Ax+B)(x2+9) +(Cx+D)(x2+4)
=(A+C)x3+(B+D)x2 +(9A+4C)x
+(9B+4D)
A
+ C
=
5
B
+ D = -10
1
9A
+ 4C
=
5
9B
+ 4D = -5
1 0 1 0
5
0 1 0 1 - 10
≥
¥
9 0 4 0
5
0 9 0 4
-5
Using a grapher, we find that the reduced row echelon
form of the augmented matrix is:
1 0 0 0
-3
A
-3
0 1 0 0
7
B
7
≥
¥ 1 ≥ ¥ = ≥
¥
0 0 1 0
8
C
8
0 0 0 1 - 17
D
-17
x + 2
1
3x - 1
2
+
:
+ 2
x + 3
1 x + 32 2
x + 2
1 x2 + 2 2 2
5x5+22x4+36x3+53x2+71x+20
=A(x+3)(x2+2)2+B(x2+2)2
+(Cx+D)(x+3)2(x2+2)+(Ex+F)(x+3)2
=(A+C)x5+(3A+B+6C+D)x4
+(4A+11C+6D+E)x3+(12A+4B+12C
+11D+6E+F)x2+(4A+18C+12D+9E
+6F)x+(12A+4B+18D+9F)
A
+ C
= 5
3A + B + 6C
+ D
= 22
4A
+ 11C + 6D + E
= 36
1
12A + 4B + 12C + 11D + 6E + F = 53
4A
+ 18C + 12D + 9E + 6F = 71
12A + 4B
+ 18D
+ 9F = 20
1 0
1
0 0 0
5
3 1
6
1 0 0 22
4 0 11
6 1 0 36
≥12 4 12 11 6 1 53¥
4 0 18 12 9 6 71
12 4
0 18 0 9 20
Using a grapher, we find that the reduced row echelon
form of the augmented matrix is:
1 0 0 0 0 0
2
A
2
0 1 0 0 0 0 -1
B
-1
0 0 1 0 0 0
3
C
3
≥0 0 0 1 0 0 - 1¥ 1 ≥D¥ = ≥ -1¥
0 0 0 0 1 0
1
E
1
0 0 0 0 0 1
2
F
2
12.
-2
3
3
1
+
+
+
:
x - 1
x + 4
1 x - 12 2
1x + 422
–x3-6x2-5x+87=A(x-1)(x+4)2
+B(x+4)2+C(x-1)2(x+4)+D(x-1)2
=(A+C)x3+(7A+B+2C+D)x2
+(8A+8B -7C-2D)x
+(–16A+16B+4C+D)
Partial Fractions
293
A
+ C
= -1
7A
+ B + 2C + D = -6
1
8A + 8B - 7C - 2D = -5
- 16A + 16B + 4C + D = 87
1
0
1
0 -1
7
1
2
1 -6
≥
¥
8
8 -7 -2 -5
-16 16
4
1
87
Using a grapher, we find that the reduced row echelon
form of the augmented matrix is:
1 0 0 0 -2
A
-2
0 1 0 0
3
B
3
≥
¥ 1 ≥ ¥ = ≥
¥
0 0 1 0
1
C
1
0 0 0 1
3
D
3
A1
A2
2
13.
=
+
, so
1x - 5 2 1x - 3 2
x - 5
x - 3
2=A1(x-3)+A2(x-5). With x=5, we see that
2=2A1, so A1=1; with x=3 we have 2=–2A2, so
1
1
A2=–1:
.
x - 5
x - 3
A1
A2
4
=
+
14.
, so
1x + 3 2 1x + 7 2
x + 3
x + 7
4=A1(x+7)+A2(x+3). With x=–3, we see that
4=4A1, so A1=1; with x=–7 we have 4=–4A2,
1
1
so A2=–1:
.
x + 3
x + 7
15.
A1
A2
4
=
, so
+
x - 1
x + 1
x2 - 1
4=A1(x+1)+A2(x-1). With x=1, we see that
4=2A1, so A1=2; with x=–1 we have 4=–2A2, so
2
2
A2=–2:
.
x - 1
x + 1
A1
A2
6
=
+
, so
x - 3
x + 3
x - 9
6=A1(x+3)+A2(x-3). With x=3, we see that
6=6A1, so A1=1; with x=–3 we have 6=–6A2, so
1
1
A2=–1:
.
x - 3
x + 3
A1
A2
1
=
+
17. 2
, so 1=A1(x+2)+A2x.
x
x + 2
x + 2x
1
With x=0, we see that 1=2A1, so A1= ; with
2
1
x=–2, we have 1=–2A2, so A2= - :
2
1>2
1>2
1>2
1
=
.
x
x + 2
2x
x + 2
A1
A2
-6
=
+
18. 2
, so
x
x - 3
x - 3x
–6=A1(x-3)+A2x. With x=0, we see that
–6=–3A1, so A1=2; with x=3 we have –6=3A2,
so
-2
2
A2=–2:
+ .
x - 3
x
16.
2
Chapter 7
294
19.
Systems and Matrices
A1
A2
-x + 10
=
, so –x+10
+
x
3
x
+ 4
x + x - 12
x=0, we have 25=9A1; so A1=
2
7
have 7=3A3; so A3= ; with x=4, we have
3
28
25
1=A1+4A2+4A3= +4A2+ , so
9
3
7>3
25>9
25>9
25
+
.
A2= - :
9 1x - 32 2
x - 3
x
=A1(x+4)+A2(x-3). With x=3, we see that
7=7A1, so A1=1; with x=–4 we have 14=–7A2,
1
2
so A2=–2:
.
x - 3
x + 4
20.
A1
A2
7x - 7
=
+
, so 7x-7
x - 5
x + 2
x2 - 3x - 10
=A1(x+2)+A2(x-5). With x=5, we see that
28=7A1, so A1=4; with x=–2 we have –21=–7A2,
4
3
so A2=3:
.
+
x - 5
x + 2
21.
22.
A1
A2
x + 17
=
+
, so x+17
x + 3
2x - 1
2x2 + 5x - 3
=A1(2x-1)+A2(x+3). With x=–3, we see that
1
7
35
14=–7A1, so A1=–2; with x= we have
= A2,
2
2
2
-2
5
so A2=5:
+
.
x + 3
2x - 1
A1
A2
4x - 11
=
+
, so 4x-11
2
x
+
1
2x
- 3
2x - x - 3
=A1(2x-3)+A2(x+1). With x=–1, we see that
3
5
–15=–5A1, so A1=3; with x= we have - 5 = A2,
2
2
3
2
so A2=–2:
.
x + 1
2x - 3
B2x + C2
B1x + C1
2x2 + 5
+
=
, so 2x2+5
1 x2 + 1 2 2
x2 + 1
1 x2 + 1 2 2
=(B1x+C1)(x2+1)+B2x+C2. Expanding the
right side leaves 2x2+5=B1x3+C1x2
+(B1+B2)x+C1+C2; equating coefficients reveals
that B1=0, C1=2, B1+B2=0, and C1+C2=5.
3
2
This means that B2=0 and C2=3: 2
+
.
x + 1
1 x2 + 12 2
B2x + C2
B1x + C1
3x2 + 4
24. 2
+
, so 3x2+4
=
1x + 122
x2 + 1
1 x2 + 1 2 2
=(B1x+C1)(x2+1)+B2x+C2. Expanding the
right side leaves 3x2+4=B1x3+C1x2
+(B1+B2)x+C1+C2; equating coefficients reveals
that B1=0, C1=3, B1+B2=0, and C1+C2=4.
1
3
This means that B2=0 and C2=1: 2
+
.
x + 1
1 x2 + 12 2
23.
25. The denominator factors into x(x-1)2, so
A3
A1
A2
x2 - x + 2
. Then
=
+
+
3
2
x
x - 1
x - 2x + x
1x - 122
x2-x+2=A1(x-1)2+A2x(x-1)+A3x. With
x=0, we have 2=A1; with x=1, we have 2=A3; with
x=2, we have 4=A1+2A2+2A3=2+2A2+4,
2
1
2
so A2=–1: .
+
x
x - 1
1x - 1 2 2
26. The denominator factors into x(x-3)2, so
A3
A1
A2
-6x + 25
=
. Then
+
+
x
x - 3
x3 - 6x2 + 9x
1x - 322
–6x+25=A1(x-3)2+A2x(x-3)+A3x. With
25
; with x=3, we
9
27.
A3
A1
A2
3x2 - 4x + 3
. Then
=
+ 2 +
3
2
x
x - 3
x - 3x
x
3x2-4x+3=A1x(x-3)+A2(x-3)+A3x2.
With x=0, we have 3=–3A2, so A2=–1; with x=3,
we have 18=9A3, so A3=2; with x=1, we have
2=–2A1-2A2+A3=–2A1+2+2, so
1
2
1
A1=1: - 2 +
.
x
x - 3
x
A3
A1
A2
5x2 + 7x - 4
. Then
=
+ 2 +
3
2
x
x
+ 4
x + 4x
x
2
5x +7x-4=A1x(x+4)+A2(x+4)+A3x2.
With x=0, we have –4=4A2, so A2=–1; with
x=–4, we have 48=16A3, so A3=3; with x=1, we
have 8=5A1+5A2+A3=5A1-5+3, so
1
3
2
A1=2: - 2 +
.
x
x
+
4
x
B2x + C2
B1x + C1
2x3 + 4x - 1
+
=
29.
. Then
1x2 + 22 2
x2 + 2
1x2 + 22 2
2x3+4x-1=(B1x+C1)(x2+2)+B2x+C2.
Expanding the right side and equating coefficients reveals
that B1=2, C1=0, 2B1+B2=4, and 2C1+C2=–1.
This means than B2=0 and C2=–1:
1
2x
.
x2 + 2
1x2 + 22 2
B2x + C2
B1x + C1
3x3 + 6x - 1
30.
. Then
+
=
1x2 + 22 2
x2 + 2
1x2 + 22 2
3x3+6x-1=(B1x+C1)(x2+2)+B2x+C2.
Expanding the right side and equating coefficients reveals
that B1=3, C1=0, 2B1+B2=6, and 2C1+C2=–1.
This means than B2=0 and C2=–1:
3x
1
.
x2 + 2
1x2 + 22 2
28.
31. The denominator factors into (x-1)(x2+x+1), so
A
x2 + 3x + 2
Bx + C
=
+ 2
. Then
3
x
1
x - 1
x + x + 1
2
2
x +3x+2=A(x +x+1)+(Bx+C)(x-1).
With x=1, we have 6=3A, so A=2; with x=0,
2=A-C, so C=0. Finally, with x=2, we have
2
x
12=7A+2B, so B=–1:
.
- 2
x - 1
x + x + 1
32. The denominator factors into (x+1)(x2-x+1), so
A
2x2 - 4x + 3
Bx + C
=
+ 2
. Then
x + 1
x3 + 1
x - x + 1
2x2-4x+3=A(x2-x+1)+(Bx+C)(x+1).
With x=–1, we have 9=3A, so A=3; with x=0,
3=A+C, so C=0. Finally, with x=–2, we have
3
x
19=7A+2B, so B=–1:
.
- 2
x + 1
x - x + 1
Section 7.4
In #33–36, find the quotient and remainder via long division
or other methods (note in particular that if the degree of the
numerator and denominator are the same, the quotient is the
ratio of the leading coefficients). Use the usual methods to
find the partial fraction decomposition.
x + 5 r1 x2
x + 5
2x2 + x + 3
33.
= 2 + 2
;
= 2
2
x - 1
x - 1 h1x2
x - 1
A1
A2
=
, so x+5=A1(x+1)
+
x - 1
x + 1
+A2(x-1). With x=1 and x=–1 (respectively),
2
3
we find that A1=3 and A2=–2:
.
x - 1
x + 1
2x2 + x + 3
Graph of
:
x2 - 1
Graph of
Partial Fractions
295
3x2 + 2x
:
x2 - 4
[–4.7, 4.7] by [–10, 10]
Graph of y = 3:
[–4.7, 4.7] by [–10, 10]
Graph of
4
:
x - 2
[–4.7, 4.7] by [–10, 10]
Graph of y = 2:
[–4.7, 4.7] by [–10, 10]
Graph of [–4.7, 4.7] by [–10, 10]
Graph of
2
:
x + 2
3
:
x - 1
[–4.7, 4.7] by [–10, 10]
3
35.
[–4.7, 4.7] by [–10, 10]
Graph of -
2
:
x + 1
x - 2
x - 2 r1x2
x - 2
= x - 1 + 2
;
= 2
2
x + x
x + x h1 x2
x + x
A1
A2
=
, so x-2=A1x+A2(x+1). With
+
x + 1
x
x=–1 and x=0 (respectively), we find that A1=3
3
2
and A2=–2:
- .
x + 1
x
x3 - 2
Graph of y = 2
:
x + x
[–4.7, 4.7] by [–10, 10]
34.
3x2 + 2x
2x + 12 r1x2
2x + 12
= 3 + 2
;
= 2
2
x - 4
x - 4 h1 x2
x - 4
A1
A2
+
=
, so 2x+12=A1(x+2)
x - 2
x + 2
+A2(x-2). With x=2 and x=–2 (respectively),
4
2
we find that A1=4 and A2=–2:
.
x - 2
x + 2
[–4.7, 4.7] by [–10, 15]
Chapter 7
296
Systems and Matrices
Graph of y = x - 1:
Graph of y =
3
:
x - 1
[–4.7, 4.7] by [–10, 15]
Graph of y =
[–4.7, 4.7] by [–15, 10]
3
:
x + 1
2
Graph of y = - :
x
[–4.7, 4.7] by [–10, 15]
Graph of y = -
2
:
x
[–4.7, 4.7] by [–15, 10]
37. (c)
38. (f)
39. (d)
40. (b)
41. (a)
42. (e)
[–4.7, 4.7] by [–10, 15]
3
43.
x + 2
x + 2 r1x2
x + 2
= x + 1 + 2
;
= 2
36. 2
x - x
x - x h1 x2
x - x
A1
A2
+
=
, so x+2=A1x+A2(x-1). With
x - 1
x
x=1 and x=0 (respectively), we find that A1=3 and
3
2
A2=–2:
- (note the similarity to #35).
x - 1
x
x3 + 2
Graph of y = 2
x - x:
44.
[–4.7, 4.7] by [–15, 10]
Graph of y = x + 1:
[–4.7, 4.7] by [–15, 10]
-1
1
+
:
ax
a1 x - a2
1=A(x-a)+Bx=(A+B)x-aA
A + B = 0
-aA
= 1
1 1
1
Since A= - , - + B = 0, B =
a a
a
1
A
a
c d = ≥
¥
B
1
a
1
1
:
1b - 2 2 1x - 22
1b - 2 2 1x - b 2
–1=A(x-b)+B(x-2)=(A+B)x+(–bA-2B)
A + B = 0
1
1
0
1 c
d
- bA - 2B = -1
-b -2 - 1
1 1
0
1
1
0
£
1 § 1
d1
c
0 1
0 b - 2 -1
b - 2
1
1
1 0
b - 2
A
b - 2
¥ 1 c d = ≥
¥
≥
-1
B
-1
0 1
b - 2
b - 2
Section 7.5
45.
-3
3
:
+
1 b - a2 1x - a2
1 b - a2 1 x - b2
3=A(x-b)+B(x-a)=(A+B)x+(–bA-aB)
A + B = 0
1
1 0
1c
d 1
- bA - aB = 3
-b -a 3
c
1
b - a
1
0
0
1
≥
46.
0
1
3 §
1
b - a
-3
-3
b - a
A
b - a
¥ 1 c d = ≥
¥
3
B
3
b - a
b - a
1
0
0 1 1
d
£
3
0
1
-1
1
:
+
a1x + a2
a1x - a2
2=A(x-a)+B(x+a)=(A+B)x+(–aA+aB)
A + B = 0
1
1c
- aA + aB = 2
-a
c
1
0
1
2a
0d 1 1
£
2
0
1
0
-
0
1
≥
1
1
0
d 1
2
1
a
0
1
1§
a
47. True. The behavior of f(x) near x=3 is the same as the
1
1
behavior of y =
, and lim= - q.
xS3 x - 3
x - 3
48. True. The behavior of f(x) for |x| large is the same as the
behavior of y=–1, and lim 1 -1 2 = -1.
xSq
A1
A2
49. The denominator factor x2 calls for the terms
and 2
x
x
in the partial fraction decomposition, and the denominator
B1x + C1
x2 + 2
.
The answer is E.
54. One possible answer: For any two polynomials to be
equal over the entire range of real or complex numbers,
the coefficients of each power must be equal. (For example 2x3=2x2 only when x=0 and x=1: at all other
values of x, the functions are not equal).
b
has a greater effect on f(x) at x=1.
1x - 12 2
2
-1
+
56. Using partial fractions, f(x)=
while
x - 1
1x - 12 2
2
5
+
g(x)=
. Near x=1, the term
x - 1
1x - 1 2 2
-1
dominates f(x); at the same value of x, the term
1x - 1 2 2
5
dominates g(x). Near x=1, then, we expect
1x - 1 2 2
f(x) to approach –q and g(x) to approach +q.
■ Section 7.5 Systems of Inequalities
in Two Variables
Quick Review 7.5
1. x-intercept: (3, 0); y-intercept: (0, –2)
y
5
50. The denominator factor (x+3)2 calls for terms
A1
A2
in the partial fraction decomposition,
and
x + 3
1x + 322
2
5
x
2
and the denominator factor (x +4) calls for the terms
B2x + C2
B1x + C1
and 2
. The answer is C.
x2 + 4
1x + 422
51. The y-intercept is –1, and because the denominators are
both of degree 1, the expression changes sign at each
asymptote. The answer is B.
3
52. The y-intercept is , and the expression changes sign at
4
the x=1 asymptote but is negative on both sides of the
x=–2 asymptote. The answer is E.
53. (a) x=1: 1+4+1=A(1+1)+(B+C)(0)
6=2A
A=3
297
(b) x=i: –1+4i+1=3(–1+1)+(Bi+C)(i-1)
4i=(Bi+C)(i-1)
4i=–B-Bi+Ci-C
-B - C = 0
B + C = 0
1
1
- Bi + Ci = 4i
B - C = -4
1
1
0
1
1
0
1 1 0
c
d1c
d1 c
d1
1 - 1 -4
0 -2 -4
0 1 2
1 0 -2
B
-2
c
d1 c d = c
d
0 1
2
C
2
x=–i
–1-4i+1=A(–1-1)+(–Bi+C)(–i-1)
–4i=–B+Bi-Ci-C, which is the
=same as above.
B=–2, C=2
55. y=
1
1
a
a
A
¥ 1c d = ≥
¥
1
B
1
a
a
factor x2+2 calls for the term
Systems of Inequalities in Two Variables
2. x-intercept: (6, 0); y-intercept: (0, 3)
y
5
10
x
Chapter 7
298
Systems and Matrices
7.
3. x-intercept: (20, 0); y-intercept: (0, 50)
y
y
5
50
5
x
50
x
boundary line x=4 included
4. x-intercept: (30, 0); y-intercept: (0, –20)
8.
y
y
5
50
5
x
50
x
boundary line y=–3 included
For #5–9, a variety of methods could be used. One is shown.
4
5. c
1
6. c
7. c
= c
30
d
60
1
10
1
5
= c
70
d
20
4
10
1
4
1
1
180
d1£
90
0
1
1
1
5
9.
y
5
45
1
§1 c
60
0
30
x
d1c d
60
y
0
1
5
90
1
d1c
800
0
90
1
d1c
20
0
1
1
x
70
x
d1 c d
20
y
0
1
boundary line 2x+5y=7 included
10.
1
4
1
1
180
d1£
800
0
45
140
§1 c
1
0
0
1
y
5
10
d1
140
x
10
c d = c
d
y
140
5
8. c
1
8
1
2
6
1
d1c
24
0
1
1
6
1
d1c
4
0
0
1
2
x
2
d1c d = c d
4
y
4
9. c
1
2
1
8
6
1
d1 c
30
0
1
1
6
1
d1c
3
0
0
1
3
x
3
d1c d = c d
3
y
3
2
2
10. Use substitution: 2x + 3x = 4, 3x + 2x - 4 = 0,
- 2 ; 14 - 41 32 1 -4 2
,
x =
6
(x, y)≠(–1.54, 2.36) or (0.87, 0.75)
x
boundary line 3x-y=4 excluded
11.
y
9
Section 7.5 Exercises
1. Graph (c); boundary included
2. Graph (f); boundary excluded
3. Graph (b); boundary included
4. Graph (d); boundary excluded
5. Graph (e); boundary included
6. Graph (a); boundary excluded
5
x
boundary curve y=x2+1 excluded
Section 7.5
12.
17.
y
y
10
5
5
x
10
boundary curve y=x2-3 included
13.
Systems of Inequalities in Two Variables
Corner at (2, 3). Left boundary is excluded, the other
is included.
y
18.
5
x
y
10
5
x
10
x
boundary circle x2+y2=9 excluded
14.
Corner at (–3, 2). Boundaries included.
y
19.
5
y
16
5
x
10
x
boundary circle x2+y2=4 included
15.
Corners at about (–1.45, 0.10) and (3.45, 9.90).
Boundaries included.
y
8
20.
y
4
5
boundary curve y =
16.
3
x
x
ex + e -x
included
2
Corners at about (–3.48, –3.16) and (1.15, –1.62).
Boundaries excluded.
y
2
21.
y
5
2π
x
5
x
boundary curve y=sin x excluded
Corners at about (—1.25, 1.56) Boundaries included.
299
300
Chapter 7
22.
Systems and Matrices
27. x2 + y2 4
y -x2 + 1
y
5
5
x
Corners at about (—2.12, 2.12)
23.
y
90
90
x
For #29–30, first we must find the equations of the lines—then
the inequalities.
1 5 - 32
¢y
2
-1
-1
=
=
=
,y =
x + 5
29. line 1: m =
¢x
10 - 42
-4
2
2
1 0 - 32
¢y
-3
=
=
,
line 2: m =
¢x
16 - 42
2
-3
-3
1 x - 62, y =
x + 9
(y-0) =
2
2
line 3: x=0
line 4: y=0
-1
x + 5
y 2
-3
x + 9
y 2
x 0
y 0
11 - 6 2
¢y
-5
-5
=
=
,y =
x + 6
¢x
12 - 02
2
2
11 - 0 2
¢y
1
-1
=
=
, =
,
line 2:
¢x
12 - 52
-3
3
-1
5
-1
1 x - 52, y =
x +
(y-0) =
3
3
3
line 3: x=0
line 4: y=0
-5
x + 6
y 2
5
-1
x +
y 3
3
x 0
y 0
30. line 1:
Corners at (0, 40), (26.7, 26.7), (0, 0), and (40, 0).
Boundaries included.
24.
y
90
90
x
Corners at (6, 76.5), (32, 18), and (80, 0). Boundaries
included.
25.
y
9
28. x¤+y¤ 4
y 0
For #31–36, the feasible area, use your grapher to determine
the feasible area, and then solve for the corner points graphically or algebraically. Evaluate f(x) at the corner points to
determine maximum and minimum values.
31.
y
90
9
x
Corners at (0, 2), (0, 6), (2.18, 4.55), (4, 0), and (2, 0).
Boundaries included.
26.
y
80
80
x
Corners at (0, 30), (21, 21), and (30, 0). Boundaries
included.
90
x
Corner points: (0, 0)
(0, 80), the y-intercept of x+y=80
160 80
,
b , the intersection of x+y=80
a
3 3
and x-2y=0
160 80
,
b
1x, y2 10, 02 10, 802
a
3 3
880
f
0
240
L 293.33
3
160 80
, bd
fmin = 0 3at 10, 02 4; fmax L 293.33 c at a
3 3
Section 7.5
32.
35.
y
301
y
16
90
90
16
x
Corner points: (0, 0)
(90, 0), the x-intercept of x+y=90
45 135
b , the intersection of x+y=90
a ,
2 2
and 3x-y=0
45 135
1 x, y2 10, 02 1 90, 02 a ,
b
2 2
f
0
900
967.5
45 135
fmin = 0 3at 1 0, 0 2 4; fmax = 967.5 c at a ,
bd
2 2
33.
Systems of Inequalities in Two Variables
x
Corner points: (0, 12) y-intercept of 2x+y=12
(3, 6) intersection of 2x+y=12 and
4x+3y=30
(6, 2) intersection of 4x+3y=30 and
x+2y=10
(10, 0) x-intercept of x+2y=10
(x, y) (0, 12) (3, 6)
(6, 2) (10, 0)
`
f
36.
24
`
`
27
34
`
50
fmin = 24 3at 10, 122 4; fmax = none (region is unbounded)
y
16
y
90
16
90
x
Corner points: (0, 60) y-intercept of 5x+y=60
(6, 30) intersection of 5x+y=60 and
4x+6y=204
(48, 2) intersection of 4x+6y=204 and
x+6y=60
(60, 0) x-intercept of x+6y=60
(x, y) (0, 60) (6, 30) (48, 2) (60, 0)
` 240 ` 162 ` 344 ` 420
f
fmin=162 [at (6, 30)]; fmax=none (region is unbounded)
34.
y
20
40
x
Corner points: (16, 3) intersection of 3x+4y=60 and
x+8y=40
(4, 12) intersection of 3x+4y=60 and
11x+28y=380
(32, 1) intersection of x+8y=40 and
11x+28y=380
(x, y) (4, 12) (16, 3) (32, 1)
` 360 ` 315 ` 505
f
fmin = 315 3at 1 16, 32 4; fmax = 505 3 at 1 32, 1 2 4
x
Corner points: (0, 10) y-intercept of 3x+2y=20
(2, 7) intersection of 3x+2y=20 and
5x+6y=52
(8, 2) intersection of 5x+6y=52 and
2x+7y=30
(15, 0) x-intercept of 2x+7y=30
(x, y) (0, 10) (2, 7)
(8, 2) (15, 0)
`
f
50
`
`
41
34
`
45
fmin = 34 3at 18, 22 4; fmax = none (region is unbounded)
For #37–40, first set up the equations; then solve.
37. Let x=number of tons of ore R
y=number of tons of ore S
C=total cost=50x+60y, the objective function
80x+140y 4000 At least 4000 lb of mineral A
160x+50y 3200 At least 3200 lb of mineral B
x 0, y 0
The region of feasible points is the intersection of
80x+140y 4000 and 160x+50y 3200 in the first
quadrant. The region has three corner points: (0, 64),
(13.48, 20.87), and (50, 0). Cmin=$1,926.20 when 13.48
tons of ore R and 20.87 tons of ore S are processed.
y
80
80
x
302
Chapter 7
Systems and Matrices
38. Let x=number of units of food substance A
y=number of units of food substance B
C=total cost=1.40x+0.90y, the objective function
3x+2y 24 At least 24 units of carbohydrates
4x+y 16 At least 16 units of protein
x 0, y 0
The region of feasible points is the intersection of
3x+2y 24 and 4x+y 16 in the first quadrant.
The corner points are (0, 16), (1.6, 9.6), and (8, 0).
Cmin=$10.88 when 1.6 units of food substance A and
9.6 units of food substance B are purchased.
The region of feasible points is the intersection of
1
x+y 3000 and x-y 0 in the first quadrant. The
2
corners are (0, 3000), (2000, 1000) and (0, 0).
Pmax=$6,500 when 2000 units of product A and
1000 units of product B are produced.
y
3600
y
18
3600
x
41. False. The graph is a half-plane.
9
x
39. Let x=number of operations performed by Refinery 1
y=number of operations performed by Refinery 2
C=total cost=300x+600y, the objective function
x+y 100 At least 100 units of grade A
2x+4y 320 At least 320 units of grade B
x+4y 200 At least 200 units of grade C
x 0, y 0
The region of feasible points is the intersection of
x+y 100, 2x+4y 320, and x+4y 200 in the
first quadrant. The corners are (0, 100), (40, 60), (120, 20),
and (200, 0). Cmin=$48,000, which can be obtained by
using Refinery 1 to perform 40 operations and Refinery 2
to perform 60 operations, or using Refinery 1 to perform
120 operations and Refinery 2 to perform 20 operations, or
any other combination of x and y such that 2x+4y=320
with 40 x 120.
y
180
42. True. The half-plane determined by the inequality
2x-3y<5 is bounded by the graph of the equation
2x-3y=5, or equivalently, 3y=2x-5.
43. The graph of 3x+4y 5 is Regions I and II plus the
boundary. The graph of 2x-3y 4 is Regions I and IV
plus the boundary. And the intersection of the regions is
the graph of the system. The answer is A.
44. The graph of 3x+4y<5 is Regions III and IV without the
boundary. The graph of 2x-3y>4 is Regions II and III
without the boundary. And the intersection of the regions is
the graph of the system. The answer is C.
45. (3, 4) fails to satisfy x+3y 12. The answer is D.
46. At (3.6, 2.8), f=46. The answer is D.
47. (a) One possible answer: Two lines are parallel if they have
exactly the same slope. Let l1 be 5x+8y=a
a
-5
and l™ be 5x+8y=b. Then l1 becomes y=
+
8
8
-5
-5
b
and l™ becomes y= x + . Since Ml1=
8
8
8
=Ml2, the lines are parallel.
(b) One possible answer: If two lines are parallel, then a
line l2 going through the point (0, 10) will be further
away from the origin then a line l1 going through the
point (0, 5). In this case f1 could be expressed as
mx+5 and f2 could be expressed as mx+10. Thus,
l is moving further away from the origin as f increases.
200
x
40. Let x=units produced of product A
y=units produced of product B
P=total profit=2.25x+2.00y
x+y 3000 No more than 3000 units produced
1
y x
2
x 0, y 0
(c) One possible answer: The region is bounded and
includes all its boundary points.
48. Two parabolas can intersect at no points, exactly one point,
two points, or infinitely many points.
None: y1 = x2 and y2 = x2 + 1
One point: y1 = x2 and y2 = -x2
1
Two points: y1 = x2 and y2 = x2 + 4
4
Chapter 7 Review
303
51. 4x2 + 9y2 36
9y2 36 - 4x2
36 - 4x2
y2 9
49. 4x2 + 9y2=36
9y2=36 - 4x2
4
y2=4 - x2
9
4
x2
y1 = 4 - x2=2 1 B
9
B
9
4 2
x2
y2 =– 4 - x =–2 1 B
9
B
9
y1 36 - 4x2
B
9
y2 –
B
36 - 4x2
9
y3 x2 - 1
y
5
[–4, 4] by [–3, 3]
5
2
50. y =x2 - 4
y1 = 2x2 - 4
y2 =– 2x2 - 4
52.
x
y
5
[–4, 4] by [–3, 3]
5
–5
x
–5
■ Chapter 7 Review
1. (a) c
1 2
d
8 3
1 5 -1
2. (a) £ 3 3
1
-2 1
3
1 -1 2 13 2 +
3. AB= c
1 02 13 2 +
(b) c
-3 4
d
0 -3
6
3
1
0 § (b) £ -1
5
4
2 -7
1 4 2 10 2 1 -1 2 1 - 1 2 +
1 62 10 2
10 2 1 -1 2 +
4. AB is not possible; BA= £
5. AB=[(–1)(5)+(4)(2)
2 -6
(d)
d
-8 0
-1 -2
- 4 -6
2 -4
(c) £ - 2 -8
-5 -6 §
4
6 § (d)
1 -2
0
6 -4 -2
14 2 1 -22 1 -1 2 15 2 + 14 2 142
-3
-7
4 = c
16 2 1 -22
10 2 15 2 + 162 142
0 -12
1 -2 2 1 -1 2 + 13 2 132 + 11 2 14 2
12 2 1 -1 2 + 1 1 2 13 2 + 102 1 42
1 -1 2 1 -1 2 + 12 2 132 + 1 -3 2 142
c
-7 11
d
4 -6
8
5
-3
-2
£ -1
14 -12 - 15 §
4 -17
4
-3
11
d ; BA is not possible.
24
1 - 22 1 2 2 + 13 2 1 -1 2 + 11 2 13 2
15
12 2 12 2 + 11 2 1 -12 + 1 02 132 4 = £ 1
1 - 12 122 + 1 22 1 -1 2 + 1 -3 2 13 2
-5
-4
3§.
-13
(–1)(–3)+(4)(1)]=[3 7]; BA is not possible.
13 2 1 -1 2 + 1 - 4 2 1 02
11 2 1 -1 2 + 12 2 1 0 2
6. AB is not possible; BA= ≥
13 2 1 -1 2 + 11 2 1 0 2
112 1 - 1 2 + 11 2 102
102 12 2 + 112 11 2 + 1 02 1 -2 2
7. AB= £ 11 2 122 + 10 2 112 + 1 02 1 - 2 2
10 2 12 2 + 10 2 11 2 + 1 1 2 1 - 2 2
BA= £
(c) c
1 22 1 0 2 + 1 -3 2 1 1 2 + 14 2 10 2
1 12 1 0 2 + 1 22 1 1 2 + 1 -3 2 10 2
1 -2 2 1 0 2 + 1 1 2 11 2 + 1 -1 2 10 2
13 2 112 + 1 -4 2 11 2
-3
1 12 11 2 + 1 22 11 2
-1
¥=≥
1 32 11 2 + 11 2 11 2
-3
1 1 2 11 2 + 1 12 1 12
-1
-1
3
¥.
4
2
10 2 1 - 32 + 11 2 12 2 + 1 02 11 2 10 2 142 + 1 12 1 -3 2 + 102 1 -12
1
11 2 1 - 3 2 + 1 02 1 22 + 10 2 11 2 £ 11 2 14 2 + 10 2 1 -32 + 10 2 1 -1 2 § = £ 2
1 0 2 1 -3 2 + 10 2 12 2 + 1 12 1 12 10 2 142 + 1 02 1 -3 2 + 112 1 -1 2
-2
1 2 2 1 12 + 1 -3 2 10 2 + 142 102 122 10 2 + 1 -3 2 1 02 + 1 42 1 1 2
-3
1 1 2 1 12 + 12 2 102 + 1 - 32 102 11 2 102 + 1 22 10 2 + 1 -3 2 1 12 4 = £ 2
1 -2 2 11 2 + 11 2 10 2 + 1 - 12 102 1 -22 10 2 + 11 2 10 2 + 1 -12 1 1 2
1
2
-3
1
2
1
-2
-3
4§
-1
4
-3 §
-1
Chapter 7
304
Systems and Matrices
8. As in #7, the multiplication steps take up a lot of
space to write, but are easy to carry out, since A
contains only 0s and 1s. The intermediate steps
are not shown here, but note that the rows of
AB are a rearrangement of the rows of B
(specifically, rows 1 and 2, and rows 3 and 4, are
swapped), while the columns of BA are a
rearrangement of the columns of B (we swap
columns 1 and 2, and columns 3 and 4). The
nature of the rearrangement can be determined
by noting the locations of the 1s in A.
3
-2
AB = ≥
3
-1
0 2
1 0
-2 1
1 2
1
0
BA = ≥
1
-2
-2
3
-1
3
10. Carry out the multiplication of AB and BA and confirm
that both products equal I3.
11. Using a calculator:
-1
1
2 0 -1
- 2 -5
6 -1
0 -1
1
0
2 -1 1
2
≥
¥ =≥
¥
10 24 -27
4
2
0 1
2
-1
1 1
4
- 3 -7
8 -1
12. Using a calculator:
-1
-1
0 1
-0.4
£ 2 -1 1 § = £ -0.2
1
1 1
0.6
1
1
¥;
0
-1
0.2
-0.4
0.2
0.2
0.6 §
0.2
1
2
-2
-3
2
4 -1 †
0
1
-3
22
1
+0+(1)(–1)6 2
=(–2)(–1)4 2
4 -1
2
=–2(3-8)+(4-(–6))
=10+10
=20
13. det= †
1 0
1 2
¥
-1 2
0 1
-3 2
4
9. Carry out the multiplication of AB and BA and confirm
that both products equal I4.
-2
3
14. det= ∞
5
1
3
0
2
-1
= -3 c 3 1 -1 2 2 `
0 1
3
2 0
∞ =(3)(–1)3 † 2
-3 4
-1
2 3
-3
2
0 1
-2
-3 4 † +0+2(–1)5 † 5
2 3
1
4
2
` + 0 + 11 2 1 - 1 2 4 `
3
-1
3
2
-1
-3
2
` d -2 c -2 1 - 12 2 `
2
-1
1
4 † +0
3
4
5
` + 13 2 1 -12 3 `
3
1
4
5
` + 11 2 1 -1 2 4 `
3
1
2
`d
-1
=(–3)(3)(–9-8)+(–3)(1)(4-3)+(–2)(–2)(6+4)+(–2)(–3)(15-4)+(–2)(1)(–5-2)
=153-3+40+66+14=270
For #15–18, one possible sequence of row operations is shown.
1
15. £ 3
1
2
16. £ - 3
5
1
-1
2
1
£0
0
1
17. £ 2
1
1
18. £ - 2
2
2
1
1 - 32 R1 + R2
5 § ¬¬¬¬¬¬¬" £ 0
1 -1 2 R1 + R3
3
0
0
1
-1
2
3
2
1 1
1
11>2 2R1
- 2 1 § ————" £ -3
2 3
5
1 0
2 112R + R
2
3
-1 § ¬¬¬¬¬¬" £ 0 1
0 0
1
0
1
-1
0.5 0.5
-1 - 2
2
2
0
0.5
0
0
1
1
12 2 R2
- 0.5 2.5 § —————" £ 0
1 -1 2 R3
-1
3
0
0
1
0
3
3
4
1
1
1 -2 2 R1 + R2
- 2 § ———————" £ 0
1 -12 R1 + R3
6
0
2
-1
0
-2 0
5 3
4 1
4
1
1 2 2R1 + R2
"
-6 § ——————— £ 0
1 -2 2R1 + R3
9
0
0
-1
1
0.5
1
132 R1 + R2
1 § ———————" £ 0
1 -5 2R1 + R3
3
0
1
1
1 12R3 + R2
5 § ——————" £ 0
0
-3
0.5
0.5
-0.5
0
1
0
1
1
12 2R2 + R1
-4 § ——————" £ 0
132R3 + R2
5
0
3
-3
1
-2
1
0
2
-1 §
0
0
3
1
1
4
12 2R2 + R1
"
2 § ——————— £ 0
1 -32 R3 + R2
0
1
0
0
1
0
-1
0
0
1
0
0.5
-0.5
-0.5
0.5
11 2R3 + R1
2.5 § ——————"
11 2R2 + R3
0.5
1
2§
-3
-3
0
1
6
0
1
1
-7
1 -12 R2
11 § ——————" £ 0
5 13 2R3 + R1 0
0
1
0
0
0
1
8
-11 §
5
8
1 -6 2R3 + R1 1
—————
——" £ 0
-1 §
1
0
0
1
0
0
0
1
2
-1 §
1
For #19–22, use any of the methods of this chapter. Solving for x (or y) and substituting is probably easiest for these systems.
19. (x, y)=(1, 2): From E1, y=3x-1; substituting in E2 gives x+2(3x-1)=5. Then 7x=7, so x=1.
Finally, y=2.
20. (x, y)=(–3, –1): From E1, x=2y-1; substituting in E2 gives –2(2y-1)+y=5. Then –3y=3, so y=–1. Finally,
x=–3.
21. No solution: From E1, x=1-2y; substituting in E2 gives 4y-4=–2(1-2y), or 4y-4=4y-2
— which is impossible.
Chapter 7 Review
305
22. No solution: From E1, x=2y+9; substituting in E2
27
3
gives 3y- (2y+9)=–9, or - =–9 — which is not true.
2
2
23. (x, y, z, w)=(2-z-w, w+1, z, w): Note that the last equation in the triangular system is not useful. z and w can be
anything, then y=w+1 and x=2-z-w.
x+z+w=2
x+z+w=2
x +z+w=2
x+y+z=3 1 E2-E1 1
y-w=1
y-w=1
3x+2y+3z+w=8 1 E3-3E1 1
2y-2w=2 1 E3-2E2 1
0=0
24. (x, y, z, w)=(–w-2, –z-w, z, w): Note that the last equation in the triangular system is not useful. z and w can be anything, then y=–z-w and x=–w-2.
x+w=–2
x+w=–2
x+w=–2
x+y+z+2w=–2 1 E2-E1 1
y+z+w=0
y+z+w=0
–x-2y-2z-3w=2 1 E3+2E2 1
x+w=–2 1 E3-E1 1
0=0
25. No solution: E1 and E3 are inconsistent.
x+y-2z=2
3x-y+z=4
–2x-2y+4z=6 1 E3+2E1 1
x+y-2z=2
3x-y+z=4
0=10
1
3 7
5
26. (x, y, z)=a z+ , z+ , zb: Note that the last equation in the triangular system is not useful. z can be anything,
4
4 4
4
3
7
5
7
5
1
then y= z+ and x=2+2z-a z+ b= z+ .
4
4
4
4
4
4
x+y-2z=2
x+y-2z=2
–4y+7z=–5
3x-y+z=1
1 E2-3E1 1
–2x-2y+4z=–4 1 E3+2E1 1
0=0
27. (x, y, z, w)=(1-2z+w, 2+z-w, z, w): Note that the last two equations in the triangular system give no additional
information. z and w can be anything, then y=2+z-w and x=13-6(2+z-w)+4z-5w=1-2z+w.
–x-6y+4z-5w=–13
–x-6y+4z-5w=–13
–x-6y+4z-5w=–13
1
2x+y+3z-w=4
1 E2+2E1 1 –11y+11z-11w=–22 1 - E2 1
y-z+w=2
11
1
2x+2y+2z=6
1 E3+2E1 1 –10y+10z-10w=–20 1 - E3 1
y-z+w=2
10
1
3y-3z+3w=6
1 E4 1
y-z+w=2
–x-3y+z-2w=–7 1 E4-E1 1
3
28. (x, y, z, w)=(–w+2, –z-1, z, w): Note that the last two equations in the triangular system give no additional
information. z and w can be anything, then y=–z-1 and x=4+2(–z-1)+2z-w=2-w.
–x+2y+2z-w=–4
–x+2y+2z-w=–4
–x+2y+2z-w=–4
y+z=–1
y+z=–1
y+z=–1
1
–2x+2y+2z-2w=–6 1 E3-2E1 1
–2y-2z=2
1 - E3 1
y+z=–1
2
–x+3y+3z-w=–5 1 E4-E1 1
y+z=–1
y+z=–1
x
1
9
3
7
29. (x, y, z)= a , - , - b : £ y § = £ 1
4
4
4
z
2
2
-3
-3
x
1
1
5
5
30. (x, y, z)= a , - , - b : £ y § = £ 2
2
2
2
z
1
2
-1
1
1
2§
1
-1
£
-1
3
1
1§ =
£3
12
5
3
-1
-1
1§
-2
£
-2
1
1
1§ = £5
8
3
3
-5
-1
7
3
-1
1
7 -1
-1 § £ 1 § .
-5
5
1 -2
-3 § £ 1 § .
-5
3
31. There is no inverse, since the coefficient matrix, shown on the right, has determinant 0
(found with a calculator). Note that this does not necessarily mean there is no solution
— there may be infinitely many solutions. However, by other means one can determine
that there is no solution in this case.
x
1
13
8
1 22
2
y
b: ≥ ¥ = ≥
32. (x, y, z, w)= a , - , - ,
3
3
3 3
z
1
w
1
-2
1
-1
3
1
-1
2
-1
2
-1 -1
8
-1
1 -7
-1
¥ =
¥ ≥
≥
9 -2
-1
-1
4
1
11
2
2
≥
-1
1
-1
2
-2
-7
1
-1
1
-2
-2
4
5
-5
1
1
-1
1
-1
-1
¥
1
-1
2
5
-1
-1
¥
¥≥
1
-1
4
8
306
Chapter 7
Systems and Matrices
33. (x, y, z, w)=(2-w, z+3, z, w) — z and w can be anything:
1 - 22R1 + R2
1 2 -2 1
8
1
2 -2
c
d ¬¬¬¬¬¬¬" c
2 3 - 3 2 13
0 -1
1
122R2 + R1
8
1
d ¬¬¬¬¬¬¬" c
-3
0
1 -1 2R2
1
0
0
1
0
-1
1
0
2
d
3
1
0
0
2
3§
3
34. (x, y, z, w)=(2-w, z+3, z, w) — z and w can be anything. The final step, (–1)R2+R3, is not shown:
1
£2
1
2
7
3
-2
-7
-3
1
2
1
1
3
35. (x, y, z, w)=(–2, 1, 3, –1): ≥
2
3
1
0
≥
0
3
2
1
0
4
4
2
1
8
6
3
1
11
8
1
1 - 22 R1 + R2
25 § ¬¬¬¬¬¬¬" £ 0
1 -1 2 R1 + R3
11
0
2
4
4
5
4
8
7
10
0
1
0
0
0
0
1
0
0
0
0
-1
1
1 -2 2R1 + R2
———————" ≥ 0
1 - 2 2R4 + R1
0
0
0 0
1 0
0 1
0 0
0
1
-1
0
0
1
1
0
0
0
-1
1
0
1
2
-1
0
3
1
-1
0
1
1
1
1
0
0
8
1 0
11>3 2R2
9 § ¬¬¬¬¬¬¬" £ 0 1
1 -2 2R3 + R1
3
0 1
2
4
7
5
-2
-3
-6
-4
2
5
0
4
4
10
-1
8
6
14
-1
11
-2
1
1 -32R1 + R4
4 ———————
" ≥0
¥
1 -22R3 + R2
2
0
-5
0
1
-2
1 - 1 2R4
1 —————
" ≥0
¥
0
3
0
1
1
2
36. (x, y, z, w)=(1, –w-3, w+2, w): ≥
4
2
-2
-3
-1
6
1
+
R
1
-2
2R
1
3
11
3
¥ ———————" ≥
10
0
R24
15
3
6
1 0 0
1 -2 2R2 + R1
4 ———————
" ≥0 1 2
¥
1 -4 2R2 + R4
2
0 0 1
11
3 0 0
1
11 2 R4 + R2
———————" ≥ 0
1 12 R4 + R3
0
0
1
0
≥
0
0
6
11
11
14
2
3
1
0
1
0
0
0
0
1
0
0
0
0
1
0
1
0
0
6
1 -1 2R4 + R2
15
¥ ———————"
1 -12R3
-2
11
0
0
1
0
0
1
1
-1
-2
0
¥
2
1
-2
1
¥
3
-1
5
1
1 -4 2R1 + R3
7
2
¥ ———————" ≥
15
0
1 -1 2R2 + R4
9
0
1
1
1 -1 2R2 + R3
- 3 ———————
" ≥0
¥
0
-5
0
2
0
1
0
0
0
-1
-1
0
0
-1
1
0
1
1
-1
0
1
1
0
2
4
-1
1
-2
-3
2
-1
5
7
¥
-5
2
1
11 2R3 + R4
-3 ———————
"
¥
1 -12 R3
-2
2
1
-3
¥
2
0
37. (x, p)≠(7.57, 42.71): Solve 100-x2=20+3x to give x≠7.57 (the other solution, x≠–10.57, makes no sense in this
problem). Then p=20+3x≠42.71.
1 2
x =5+4x to give x≠13.91 (the other solution, x≠–53.91, makes no sense in this
10
problem). Then p=5+4x≠60.65.
38. (x, p)≠(13.91, 60.65): Solve 80-
39. (x,y)≠(0.14, –2.29)
[–5, 5] by [–5, 5]
40. (x, y)=(–1, –2.5) or (x, y)=(3, 1.5)
[–5, 5] by [–5, 5]
Chapter 7 Review
41. (x, y)=(–2, 1) or (x, y)=(2, 1)
[–5, 5] by [–5, 5]
42. (x, y)≠(–1.47, 1.35) or (x, y)≠(1.47, 1.35) or
(x, y)≠(0.76, –1.85) or (x, y)≠(–0.76, –1.85)
[–5, 5] by [–5, 5]
43. (x, y)≠(2.27, 1.53)
[–5, 5] by [–5, 5]
44. (x, y)≠(4.62, 2.22) or (x, y)≠(1.56, 1.14)
[–1, 5] by [–5, 5]
45. (a, b, c, d)= a
17
33
571 386
,,,
b
840
280
420 35
=(0.020…, –0.117…, –1.359…, 11.028…). In matrix
form, the system is as shown below. Use a calculator to
find the inverse matrix and multiply.
8
64
≥
216
729
4
16
36
81
2
4
6
9
1
a
8
1
b
5
¥≥ ¥ = ≥ ¥
1
c
3
1
d
4
46. (a, b, c, d, e)= a
19
29 59 505 68
,- ,
,
,- b
108
18 36 54
9
=(0.17592, –1.61, 1.638, 9.3518, –7.5). In matrix form,
the system is as shown below. Use a calculator to find the
inverse matrix and multiply.
16
1
E 81
256
2401
-8
1
27
64
343
4
1
9
16
49
-2
1
3
4
7
-4
a
1
2
b
1
1U E c U=E 6U
-2
d
1
8
e
1
307
A1
A2
3x - 2
, so 3x-2
=
+
x
+
1
x
- 4
x - 3x - 4
=A1(x-4)+A2(x+1). With x=–1, we see that
–5=–5A1, so A1=1; with x=4 we have 10=5A2,
2
1
so A2=2:
+
.
x + 1
x - 4
A1
A2
x - 16
=
+
48. 2
, so x-16
x + 2
x - 1
x + x - 2
=A1(x-1)+A2(x+2). With x=–2, we see that
–18=–3A1, so A1=6; with x=1 we have
6
5
.
–15=3A2, so A2=–5:
x + 2
x - 1
47.
2
49. The denominator factors into (x+2)(x+1)2, so
A3
A1
A2
3x + 5
=
+
+
.
3
2
x + 2
x + 1
x + 4x + 5x + 2
1x + 12 2
Then 3x+5=A1(x+1)2+A2(x+2)(x+1)
+A3(x+2). With x=–2, we have –1=A1; with
x=–1, we have 2=A3; with x=0, we have 5
=A1+2A2+2A3=–1+2A2+4, so that A2=1:
1
2
1
+
+
.
x + 2
x + 1
1x + 12 2
50. The denominator factors into (x-1)(x+2)2, so
31 3 + 2x + x2 2
A3
A1
A2
=
.
+
+
x - 1
x + 2
x3 + 3x2 - 4
1x + 22 2
2
2
Then 3x +6x+9=A1(x+2) +A2(x-1)(x+2)
+A3(x-1). With x=1, we have 18=9A1, so A1=2;
with x=–2, we have 9=–3A3, so A3=–3; with x=0,
we have 9=4A1-2A2-A3=8-2A2 +3,
2
1
3
+
.
so that A2=1:
x - 1
x + 2
1x + 2 2 2
51. The denominator factors into (x+1)(x2+1), so
5x2 - x - 2
A
Bx + C
=
+ 2
. Then
3
x + 1
x + x2 + x + 1
x + 1
5x2-x-2=A(x2+1)+(Bx+C)(x+1).
With x=–1, we have 4=2A1, so A=2; with x=0,
–2=A+C, so C=–4. Finally, with x=1 we have
2=2A+(B+C)(2)=4+2B-8, so that B=3:
2
3x - 4
+ 2
x + 1
x + 1
52. The denominator factors into (x+2)(x2+4), so
-x2 - 5x + 2
A
Bx + C
=
+ 2
. Then
3
x + 2
x + 2x2 + 4x + 8
x + 4
2
2
–x -5x+2=A(x +4)+(Bx+C)(x+2).
With x=–2, we have 8=8A1, so A=1, with x=0,
2=4A+2C, so C=–1. Finally, with x=1 we have
–4=5A+(B+C)(3)=5+3B-3, so that
2x + 1
1
B=–2:
- 2
x + 2
x + 4
53. (c)
54. (d)
55. (b)
56. (a)
308
Chapter 7
57.
Systems and Matrices
62. Corner points: approx. (–2.41, 3.20) and (2.91, 0.55).
Boundaries included.
y
5
y
10
5
x
5
58.
x
y
5
63. Corner points: approx. (–1.25, 1.56) and (1.25, 1.56).
Boundaries included.
y
5
x
5
5
59. Corner points: (0, 90), (90, 0), a
included.
x
360 360
,
b . Boundaries
13 13
y
64. No corner points. Boundaries included.
90
y
7
90
x
5
60. Corner points: (0, 3), (0, 7), a
Boundaries included.
x
30 70
, b , (3, 0), (5, 0).
13 13
65. Corner points: (0, 20), (25, 0), and (10, 6).
(x, y) (0, 20) (10, 6) (25, 0)
`
`
`
f
120
106
175
y
9
fmin=106 [at (10, 6)]; fmax=none (unbounded)
y
40
9
x
61. Corner points: approx. (0.92, 2.31) and (5.41, 3.80).
Boundaries excluded.
y
7
8
x
40
x
Chapter 7 Review
66. Corner points: (0, 30), (8, 10), and (24, 0).
(x, y) (0, 30) (8, 10) (24, 0)
`
`
`
f
150
138
264
309
(a) The following is a scatter plot of the data with the
linear regression equation y = 11.4428x + 116.681
superimposed on it.
fmin=138 [at (8, 10)]; fmax=none (unbounded)
y
40
[–5, 20] by [0, 400]
40
(b) The following is a scatter plot of the data with the
294.846
logistic regression equation y =
1 + 1.6278e - 0.1784x
superimposed on it.
x
67. Corner points: (4, 40), (10, 25), and (70, 10).
(x, y) (4, 40) (10, 25) (70, 10)
`
`
`
f
292
205
280
fmin=205 [at (10, 25)]; fmax=292 [at (4, 40)]
y
[–5, 20] by [0, 400]
60
120
x
68. Corner points: (0, 120), (120, 0), and (20, 30).
(x, y) (0, 120) (120, 0) (20, 30)
`
`
`
f
1680
1080
600
fmin=600 [at (20, 30)]; fmax=1680 [at (0, 120)]
(c) Graphical solution: The two regression models will
predict the same disbursement amounts when the
graph of their difference is 0. That will occur when the
graph crosses the x-axis. This difference function is
294.846
b
y = 11.4428x + 116.681 - a
1 + 1.6278e - 0.1784x
and it crosses the x-axis when x L 3.03 and
x L 10.15.
The disbursement amount of the two models will be
the same sometime in the years 1993 and 2000.
y
180
[0, 15] by [–10, 10]
180
69. (a) [1
(b) [1
cos 45 °
2] c
sin 45 °
2] c
cos 45 °
-sin 45 °
x
2.12
-sin 45 °
d
d L c
0.71
cos 45 °
-0.71
sin 45 °
d
d L c
cos 45 °
2.12
70. In this problem, the graphs are representative of the total
Medicare Disbursements (in billions of dollars) for several years, where x is the number of years past 1990.
Another graphical solution would be to find where
the graphs of the two curves intersect.
Algebraic solution: The algebraic solution of the
problem is not feasible.
(d) Both models appear to fit the data fairly well. The
logistic model should be used to make predictions
beyond 2000. The disbursements stabilize using the
logistic model but continue to rise according to the
linear model.
310
Chapter 7
Systems and Matrices
143
L 0.4917 and the ratio of females to the total
290.8
147.8
population is
L 0.5083. If we define Matrix A
290.8
as the population matrix for the states of California,
CA 35.5
Florida, and Rhode Island, we have A = FL £ 17.0 § .
RI 1.1
If we define Matrix B as the ratio of males and
females to the total population in 2003, we have
M
F
B=[0.4917 0.5083].
The product AB gives the estimate of males and
females in each of the three states in 2003.
M
F
35.5
CA 17.5 18.0
C = £ 17.0 § [0.4917 0.5083]= FL £ 8.4
8.6 §
1.1
RI 0.54 0.56
71. In this problem, the graphs are representative of the
population (in thousands) of the states of Hawaii and
Idaho for several years, where x is the number of years
past 1980.
(a) The following is a scatter plot of the Hawaii data with
the linear regression equation
y = 12.2614x + 979.5909 superimposed on it.
[–5, 30] by [0, 2000]
(b) The following is a scatter plot of the Idaho data with
the linear regression equation
y = 19.8270x + 893.9566 superimposed on it.
[–5, 30] by [0, 2000]
(c) Graphical solution: Graph the two linear equations
y = 12.2614x + 979.5909 and
y = 19.8270x + 893.9566 on the same axis and find
their point of intersection. The two curves intersect at
x L 11.3.
The population of the two states will be the same
sometime in the year 1991.
(b) The matrix for the percentages of the populations of
California, Florida, and Rhode Island under the age of
18 and age 65 or older is given as:
618 65
CA 26.5 10.6
FL £ 23.1 17.0 §
RI 22.8 14.0
(c) To change the matrix in (b) from percentages to decimals, multiply by the scalar 0.01 as follows:
618
65
26.5 10.6
CA 0.265 0.106
0.01 * £ 23.1 17.0 § = FL £ 0.231 0.170 §
22.8 14.0
RI 0.228 0.140
(d) The transpose of the matrix in (c) is
0.265 0.231 0.228
c
d.
0.106 0.170 0.140
Multiplying the transpose of the matrix in (c) by the
matrix in (a) gives the total number of males and
females who are under the age of 18 or are 65 or
older in all three states.
[–5, 30] by [0, 2000]
c
Another graphical solution would be to find where
the graph of the difference of the two curves is equal
to 0.
Algebraic solution:
Solve 12.2614x + 979.5909 = 19.8270x + 893.9566
for x.
12.2614x + 979.5909 = 19.8270x + 893.9566
7.5656x = 85.6343
85.6343
L 11.3
x =
7.5656
The population of the two states will be the same
sometime in the year 1991.
72. (a) According to data from the U. S. Census Bureau, there
were 143.0 million males and 147.8 million females in
2003. The ratio of males to the total population is
0.265
0.106
17.5
0.228
d £ 8.4
0.140
0.54
0.231
0.170
M
6 18 6.7
c
65 3.4
18.0
8.6 § =
0.56
F
6.9
d
3.4
(e) In 2003, there were about 6.7 million males under age
18 and about 3.4 million females 65 or older living in
the three states.
73. (a)
N= [200
400
600
250]
(b)
P= [$80
$120
$200
$300]
(c) NPT=[200
400
600
$ 80
$120
250] ≥
¥ = $259,000
$200
$300
Chapter 7 Review
74. (x, y)=(380, 72), where x is the number of students and
y is the number of nonstudents.
x+
y=452
0.75x+2.00y=429
One method to solve the system is to solve by elimination
as follows:
2x+2y=904
0.75x+2y=429
1.25x
Divide equation 2 by 0.025 to simplify:
x+y+z=650,000
–x
+z=160,000
2x
-z=0
Now add equation 2 to equation 3, replacing equation 3:
x+y+z=650,000
–x
+z=160,000
x
=160,000
Substitute x=160,000 into equation 2 to solve for
z: z=320,000. Substitute these values into equation 1
to solve for y: y=170,000.
=475
x=380
Substitute x=380 into x+y=452 to solve for y.
75. Let x be the number of vans, 8x+15y+22z 115
y be the number of small
3x+10y+20z 85
trucks, and z be the number
2x+ 6y+ 5z 35
of large trucks needed. The
requirements of the problem are summarized above
(along with the requirements that each of x, y, and z must
be a non-negative integer).
The methods of this chapter do not allow complete solution of this problem. Solving this system of inequalities
as if it were a system of equations gives (x, y, z)
=(1.77, 3.30, 2.34), which suggests the answer
(x, y, z)=(2, 4, 3); one can easily check that (x, y, z)
=(2, 4, 2) actually works, as does (1, 3, 3). The first of
these solutions requires 8 vehicles, while the second
requires only 7. There are a number of other seven-vehicle
answers (these can be found by trial and error): Use no
vans, anywhere from 0 to 5 small trucks, and the rest
should be large trucks — that is, (x, y, z) should be one of
(0, 0, 7), (0, 1, 6), (0, 2, 5), (0, 3, 4), (0, 4, 3), or (0, 5, 2).
76. (x, y)=(21,333.33, 16,666.67), where x is the amount
invested at 7.5% and y is the amount invested at 6%.
x+
y=38,000
0.075x+0.06y= 2,600
x + y = 38,000 1 x = 38,000 - y
0.075 138,000 - y2 + 0.06y = 2600
2850 - 0.075y + 0.06y = 2600
-0.015y = - 250
y = 16,666.67
Substitute y=16,666.67 into x+y=38,000 to solve
for x.
77. (x, y, z)=(160000, 170000, 320000), where x is the
amount borrowed at 4%, y is the amount borrowed at
6.5%, and z is the amount borrowed at 9%. Solve the
system below.
4x+
y+
z=650,000
0.04x+0.065y+0.09z= 46,250
z=
78. Sue: 9.3 hours (9 hours and
x+y+z=1/4
20 minutes), Esther: 12 hours,
x
+z=1/6
Murphy: 16.8 hours (16 hours 48
y+z=1/7
minutes). If x is the portion of the
room Sue completes in one hour, y is the portion that
Esther completes in one hour, and z is the portion that
Murphy completes in one hour, then
solving the system above gives (x, y, z)
3 1 5
1 1
1
=a , ,
b=a
, ,
b.
28 12 84
9.333 12 16.8
One method to solve the system is to find the row echelon form of the augmented matrix:
1
£1
0
1
0
1
1
1
1
1
1>4
R1 - R2
1>6 § —————" £ 0
0
1>7
1
R1 - R3
—————" £ 0
0
1
£0
0
0
1
0
0
0
1
0
1
1
0
0
1
0
One method to solve the system is to solve using
Gaussian elimination:
Multiply equation 1 by –0.065 and add the result to equation 2, replacing equation 2:
x+y+
z=650,000
–0.025x
+0.025z=4000
2x
z=0
1
1
1
1
0
1
1>4
1>12 §
1>7
3>28
R3 - R2
1>12 § —————"
1>7
3>28
1>12 §
5>84
79. Pipe A: 15 hours. Pipe B:
One method to solve the system is to solve by substitution
as follows:
02x-
311
60
≠5.45 hours (about
11
5 hours 27.3 minutes). Pipe C:
x+y+z=1/3
12 hours. If x is the portion of
x+y
=1/4
the pool that A can fill in one
y+z=1/3.75
hour, y is the portion that B fills
in one hour, and z is the portion that C fills in one hour,
then solving the system above gives
1 11 1
1x, y, z2 = a , ,
b
15 60 12
One method to solve the system is to use elimination:
Subtract equation 2 from equation 1:
x + y + z = 1>3
z = 1>12
y + z = 4>15 1 convert 1>3.75 to simpler form2
Subtract equation 2 from equation 3:
x + y + z = 1>3
z = 1>12
y = 11>60
Substitute the values for y and z into equation 1 to solve
for x: x=1/15.
312
Chapter 7
Systems and Matrices
80. B must be an n × n matrix. (There are n rows in B because
AB is defined, and n columns in B since BA is defined.)
has a greater slope. But since 2000, the female population
has always been greater. Since a span of only 15 years is
represented, the data are most likely not enough to create
a model for 100 or more years.
81. n=p — the number of columns in A is the same as the
number of rows in B.
Chapter 7 Project
4.
1. The graphs are representative of the male and female
population in the United States from 1990 to 2004, where
x is the number of years after 1990.
[–10, 120] by [30, 150]
412.574
1 + 10.956e 1-0.01539x2
315.829
Females: y L
1 + 9.031e 1-0.01831x2
The curves intersect at approximately (45, 64); this represents the time when the female population became
greater than the male population.
The curves intersect at approximately (159, 212); this represents the time when the male population will again
become greater the female population.
5. Males: y L
[–5, 15] by [120, 160]
The linear regression equation for the male population is
y L 1.7585x + 119.5765.
The linear regression equation for the female population
is y L 1.6173x + 126.4138.
2. The slope is the rate of change of people (in millions) per
year. The y-intercept is the number of people (either
males or females) in 1990.
3. Yes, the male population is predicted to eventually surpass
the female population, because the males’ regression line
7. Approximately
female
138.1
L 0.491 = 49.1% male and 50.9%
281.5
Section 8.1
Conic Sections and Parabolas
313
Chapter 8
Analytic Geometry in Two and Three Dimensions
■ Section 8.1 Conic Sections and Parabolas
2.
y
10
Exploration 1
y=4
1. From Figure 8.4, we see that the axis of the parabola is
x=0. Thus, we want to find the point along x=0 that
is equidistant from both (0, 1) and the line y=–1. Since
the axis is perpendicular to the directrix, the point on the
directrix closest to the parabola is (0, 1) and (0, –1), it
must be located at (0, 0).
10
F(2, –2)
x=2
2. Choose any point on the parabola (x, y). From figures 8.3
and 8.4, we see that the distance from (x, y) to the focus
is d1 = 21x - 02 + 1y - 1 2 = 2x + 1 y - 1 2
and the distance from (x, y) to the directrix is
d2 = 21x - x2 2 + 1y - 1 -1 2 2 2 = 2 1 y + 12 2.
Since d1 must equal d2, we have
2
2
2
2
The equation of the axis is x=2.
3.
y
10
y=4
d1 = 2x2 + 1y - 1 2 2 = 21 y + 12 2 = d2
x2+(y-1)2=(y+1)2
2
x +y2-2y+1=y2+2y+1
x2=4y
x2
= y or x2=4y
4
3. From the figure, we see that the first dashed line above
y=0 is y=1, and we assume that each subsequent
dashed line increases by y=1. Using the equation above,
x2
x2
x2
x2
x2
we solve b 1 = , 2 = , 3 = , 4 = , 5 = ,
4
4
4
4
4
x2
6 =
r to find:
4
5 1 -2 16, 62, 1 - 2 15, 5 2 , 1 -4, 4 2, 1 -2 13, 3 2,
V(2, 1)
10
x
F(2, –2)
x=2
4. Since the focus (h, k+p)=(2, –2) and the directrix
y=k-p=4, we have k+p=–2 and k-p=4.
Thus, k=1, p=–3. As a result, the focal length p is –3
and the focal width 4p is 12.
5. Since the focal width is 12, each endpoint of the chord is
6 units away from the focus (2, –2) along the line y=–2.
The endpoints of the chord, then, are (2-6, –2) and
(2+6, –2), or (–4,–2) and (8, –2).
1 -2 12, 2 2, 1 - 2, 1 2, 10, 0 2 , 12, 1 2 , 12 12, 22 , 1213, 3 2,
14, 42, 1 215, 5 2, 1 216, 6 2 6
y
10
y=4
Exploration 2
1.
x
V(2, 1)
y
10
10
F(2, –2)
A(–4, –2)
x
B(8, –2)
y=4
x=2
10
F(2, –2)
x
6.
y
10
y=4
V(2, 1)
10
A(–4, –2)
F(2, –2)
x=2
x
B(8, –2)
314
Chapter 8
Analytic Geometry in Two and Three Dimensions
7. Downward
2
8. h=2, p=–3, k=1, so (x-2) =–12(y-1)
Quick Review 8.1
1. 2 12 - 1 -1 2 2 2 + 1 5 - 3 2 2 = 19 + 4 = 113
2. 2 1a - 22 2 + 1b + 3 2 2
-3
-6
=
. Vertex: (–4, –1),
4
2
-5
1
-3
b , Directrix: y = -1 - a
b = ,
Focus: a -4,
2
2
2
4. k=–1, h=–4, p=
-3
Focal width: 2 4p 2 = 2 4 a
b2 = 6
2
5. k=0, h=0, 4p=
3. y2=4x, y=_2 1x
4. y2=5x, y=_ 15x
5. y+7=–(x2-2x), y+7-1=–(x-1)2,
y+6=–(x-1)2
9
3 2
6. y+5=2(x2+3x), y+5+ =2 a x + b
2
2
3 ¤
19
y+ =2 ¢x + ≤
2
2
7. Vertex: (1, 5). f(x) can be obtained from g(x) by
stretching x2 by 3, shifting up 5 units, and shifting right
1 unit.
-4
1
, so p= - . Vertex: (0, 0),
3
3
1
1
Focus: a 0, - b , Directrix: y = , Focal width:
3
3
2 4p 2 = 2 a -4 b 2 = 4
3
3
4
16
, so p = . Vertex: (0, 0),
5
5
4
4
Focus: a , 0 b , Directrix: x = - ,
5
5
6. k=0, h=0, 4p=
4
16
Focal width: 2 4p 2 = 2 4 a b 2 =
5
5
7. (c)
8. (b)
9. (a)
10. (d)
[–3, 4] by [–2, 20]
2
8. Vertex: (3, 19). f(x)=–2(x-3) +19. f(x) can be
obtained from g(x) by stretching x2 by 2, reflecting across
the x-axis, shifting up 19 units and shifting right 3 units.
For #11–30, recall that the standard form of the parabola is
dependent on the vertex (h, k), the focal length p, the focal
width @ [email protected] , and the direction that the parabola opens.
11. p=–3 and the parabola opens to the left, so
y2=–12x.
12. p=2 and the parabola opens upward, so x2=8y.
13. –p=4 (so p=–4) and the parabola opens downward,
so x2=–16y.
14. –p=–2 (so p=2) and the parabola opens to the right,
so y2=8x.
[–2, 7] by [–10, 20]
9. f(x)=a(x+1)2+3, so 1=a+3, a=–2,
f(x)=–2(x+1)2+3.
10. f(x)=a(x-2)2-5, so 13=9a-5, a=2,
f(x)=2(x-2)2-5
Section 8.1 Exercises
6
3
3
= . Vertex: (0, 0), Focus: a 0, b ,
4
2
2
3
3
Directrix: y= - , Focal width: 2 4p 2 = 2 4 # 2 = 6
2
2
1. k=0, h=0, p=
-8
2. k=0, h=0, p=
= - 2. Vertex: (0, 0),
4
Focus: (–2, 0), Directrix: x=2,
Focal width: @ [email protected] = @ 4 1 -2 2 @ = 8
4
3. k=2, h=–3, p= =1. Vertex: (–3, 2),
4
Focus: (–2, 2), Directrix: x=–3-1=–4,
Focal width: @ [email protected] = @ 4 11 2 @ = 4.
15. p=5 and the parabola opens upward, so x2=20y.
16. p=–4 and the parabola opens to the left, so y2=–16x.
17. h=0, k=0, @ [email protected] = 8 1 p = 2 (since it opens to the
right): (y-0)2=8(x-0); y2=8x.
18. h=0, k=0, @ [email protected] = 12 1 p = -3 (since it opens to
the left): (y-0)2=–12(x-0); y2=–12x
3
19. h=0, k=0, @ [email protected] = 6 1 p = - (since it opens
2
downward): (x-0)2=–6(y-0); x2=–6y
20. h=0, k=0, @ [email protected] = 3 1 p =
3
(since it opens upward):
4
(x-0)2=3(y-0); x2=3y
21. h=–4, k=–4, –2=–4+p, so p=2 and the
parabola opens to the right; (y+4)2=8(x+4)
22. h=–5, k=6, 6+p=3, so p=–3 and the parabola
opens downward; (x+5)2=–12(y-6)
Section 8.1
23. Parabola opens upward and vertex is halfway between
focus and directrix on x=h axis, so h=3 and
5
5
3
4 + 1
= ; 1 = - p, so p = .
k =
2
2
2
2
5
1x - 32 2 = 6 a y - b
2
24. Parabola opens to the left and vertex is halfway between
focus and directrix on y=k axis, so k=–3 and
2 + 5
7
7
3
h =
= ; 5 = - p, so p = - .
2
2
2
2
7
2
1y + 32 = -6 a x - b
2
Conic Sections and Parabolas
34.
y
10
2
35.
y
10
25. h=4, k=3; 6=4-p, so p=–2 and parabola opens
to the left. (y-3)2=–8(x-4)
26. h=3, k=5; 7=5-p, so p=–2 and the parabola
opens downward. (x-3)2=–8(y-5)
6
27. h=2, k=–1; @ 4p @ = 16 1 p = 4 (since it opens
upward): (x-2)2=16(y+1)
28. h=–3, k=3; @ 4p @ = 20 1 p = -5 (since it opens
downward): (x+3)2=–20(y-3)
29. h=–1, k=–4; @ [email protected] = 10 1 p = to the left): 1y + 4 2 2 = -101x + 1 2
30. h=2, k=3; @ 4p @ = 5 1 p =
31.
right): 1 y - 3 2 2 = 5 1x - 2 2
5
(since it opens
2
36.
10
5
(since it opens to the
4
20
y
37.
x
[–4, 4] by [–2, 18]
32.
y
38.
5
5
x
[–10, 10] by [–8, 2]
39.
33.
y
5
[–8, 2] by [–2, 2]
5
x
x
y
5
5
x
x
315
316
Chapter 8
Analytic Geometry in Two and Three Dimensions
40.
47.
[–13, 11] by [–10, 6]
[–2, 8] by [–3, 3]
48.
41.
[–10, 15] by [–3, 7]
[–20, 28] by [–10, 22]
49. Completing the square produces y-2=(x+1)2. The
vertex is (h, k)=(–1, 2), so the focus is
42.
(h, k+p)= a - 1, 2 +
9
1
b = a -1, b , and the
4
4
directrix is y=k-p=2 [–12, 8] by [–2, 13]
43.
50. Completing the square produces 2 a y The vertex is (h, k)= a 1,
(h, k+p)= a 1,
[–2, 6] by [–40, 5]
1
7
=
4
4
y=k-p=
7
b = 1x - 12 2.
6
7
b , so the focus is
6
7
1
5
+ b = a 1, b , and the directrix is
6
2
3
1
2
7
- = .
6
2
3
51. Completing the square produces 8(x-2)=(y-2)2.
The vertex is (h, k)=(2, 2) so the focus is
(h+p, k)=(2+2, 2)=(4, 2), and the directrix is
x=h-p=2-2=0.
44.
[–15, 5] by [–15, 5]
45.
[–22, 26] by [–19, 13]
46.
52. Completing the square produces
13
b =(y-1)2. The vertex is
-4 a x 4
13
(h, k)= a , 1 b so the focus is
4
9
13
(h+p, k)= a
- 1, 1 b = a , 1 b , and
4
4
17
13
the directrix is x=h-p=
.
+ 1 =
4
4
53. h=0, k=2, and the parabola opens to the left, so
(y-2)2=4p(x). Using (–6, –4), we find
36
(–4-2)2=4p(–6) 1 4p = = - 6. The equation
6
for the parabola is: (y-2)2=–6x
54. h=1, k=–3, and the parabola opens to the right, so
[–17, 7] by [–7, 9]
11
, 0 b , we find
2
2
11
(0-3)2=4p a
- 1 b 1 4p=9 # = 2. The equation
2
9
(y+3)2=4p(x-1). Using a
for the parabola is: (y+3)2=2(x-1).
55. h=2, k=–1 and the parabola opens down so
(x-2)2=4p(y+1). Using (0, –2), we find that
(0-2)2=4p(–2+1), so 4=–4p and p=–1.
The equation for the parabola is: (x-2)2=–4(y+1).
Section 8.1
56. h=–1, k=3 and the parabola opens up so
(x+1)2=4p(y-3). Using (3, 5), we find that
(3+1)2=4p(5-3), so 16=8p and p=2. The
equation for the parabola is (x+1)2=8(y-3)
57. One possible answer:
If p is replaced by –p in the proof, then the result is
x2=–4py, which is the correct result.
58. One possible answer:
Let P(x, y) be a point on the parabola with focus (p, 0)
and directrix x=–p. Then 21x - p2 2 + 1y - 0 2 2 =
distance from (x, y) to (p, 0) and
21x - 1 - p2 2 2 + 1 y - y2 2 = distance from (x, y) to
line x=–p. Because a point on a parabola is equidistant
from the focus and the directrix, we can equate these distances. After squaring both sides, we obtain
(x-p)2+(y-0)2=(x-(–p))2+(y-y)2
x2-2px+p2+y2=x2+2px+p2
y2=4px.
59. For the beam to run parallel to the axis of the mirror, the
filament should be placed at the focus. As with Example 6,
we must find p by using the fact that the points
(—3, 2) must lie on the parabola. Then,
(—3)2=4p(2)
9=8p
9
p= =1.125 cm
8
Because p=1.125 cm, the filament should be placed
1.125 cm from the vertex along the axis of the mirror.
60. For maximum efficiency, the receiving antenna should be
placed at the focus of the reflector. As with Example 6,
we know that the points (—2.5, 2) lie on the parabola.
Solving for p, we find
(—2.5)2=4p(2)
8p=6.25
p=0.78125 ft
The receiving antenna should be placed 0.78125 ft, or 9.375
inches, from the vertex along the axis of the reflector.
5
61. 4p=10, so p= and the focus is at (0, p)=(0, 2.5).
2
The electronic receiver is located 2.5 units from the vertex
along the axis of the parabolic microphone.
62. 4p=12, so p=3 and the focus is at (0, p)=(0, 3). The
light bulb should be placed 3 units from the vertex along
the axis of the headlight.
63. Consider the roadway to be the axis. Then, the vertex of
the parabola is (300, 10) and the points (0, 110) and
(600, 110) both lie on it. Using the standard formula,
(x-300)2=4p(y-10). Solving for 4p, we have
(600-300)2=4p(110-10), or 4p=900, so the
formula for the parabola is (x-300)2=900(y-10).
The length of each cable is the distance from the parabola to
the line y=0. After solving the equation of the parabola
1 2
2
for y (y=
x - x + 110), we determine that the
900
3
length of each cable is
2
1 2
2
x - x + 110 - 0 b =
1 x - x2 2 + a
900
3
B
1 2
2
x - x + 110. Starting at the leftmost tower, the
900
3
Conic Sections and Parabolas
317
lengths of the cables are:≠{79.44, 54.44, 35, 21.11, 12.78,
10, 12.78, 21.11, 35, 54.44, 79.44}.
64. Consider the x-axis as a line along the width of the road
and the y-axis as the line from the middle stripe of the
road to the middle of the bridge — the vertex of the
parabola. Since we want a minimum clearance of 16 feet
at each side of the road, we know that the points
(—15, 16) lie on the parabola. We also know that the
points (—30, 0) lie on the parabola and that the vertex
occurs at some height k along the line x=0, or (0, k).
From the standard formula, (x-0)2=4p(y-k), or
x2=4p(y-k). Using the points (15, 16), and (30, 0),
we have:
302=4p(0-k)
152=4p(16-k)
Solving these two equations gives 4p=–42.1875 and
k≠21.33. The maximum clearance must be at least
21.33 feet.
65. False. Every point on a parabola is the same distance
from its focus and its directrix.
66. False. The directrix of a parabola is perpendicular to the
parabola’s axis.
67. The word “oval” does not denote a mathematically
precise concept. The answer is D.
68. (0)2=4p(0) is true no matter what p is. The answer is D.
69. The focus of y2=4px is (p, 0). Here p=3, so the
answer is B.
70. The vertex of a parabola with equation
(y-k)2=4p(x-h) is (h, k). Here, k=3 and
h=–2. The answer is D.
71. (a)–(c)
y
slope = x – b
c–l
F(b, c)
P(x, y)
A(x, l)
y= l
x
Midpoint of AF
Max + b, l + c b
2
2
(d) As A moves, P traces out the curve of a parabola.
(e) With labels as shown, we can express the coordinates
of P using the point-slope equation of the line PM:
/ + c
x - b
x + b
y =
ax b
2
c - /
2
2
1 x - b2
/ + c
y =
2
21c - /2
/ + c
b = 1x - b2 2
2 1c - /2 a y 2
This is the equation of a parabola with vertex at
/ + c
/ + c
a b,
b and focus at a b,
+ p b where
2
2
c - /
.
p =
2
318
Chapter 8
Analytic Geometry in Two and Three Dimensions
72. (a)–(d)
(c)
y
Axis
5
y= n – 1
P(x, n – 1)
(0, 1)
5
y = –1
Generator
x
x2 + (y – 1)2 = n2
(e) A parabola with directrix y=–1 and focus at (0, 1)
has equation x2=4y. Since P is on the circle
x2+(y-1)2=n2 and on the line y=n-1, its
x-coordinate of P must be
x = 2n2 - 1 1n - 1 2 - 1 2 2 = 2n2 - 1n - 2 2 2.
Substituting 1 2n2 - 1n - 2 2 2, n-1) into x2=4y
Plane
(d)
2
shows that 1 2n2 - 1n - 2 2 2 2 =4(n-1) so P lies
on the parabola x2=4y.
73. (a)
Axis
Cylinder
(b)
1 2
x
4p
2
a
1
if and only if b =
. The parabola y= x2 and
4p
4p
a2
the line y=m(x-a)+ intersect in exactly
4p
a2
one point (namely the point a a,
b b if and only if
4p
1
a2
the quadratic equation x2 - mx + am = 0
4p
4p
has exactly one solution. This happens if and only if
the discriminant of the quadratic formula is zero.
1
a2
am
a2
1 -m2 2 - 4 a
b a am b = m2 +
4p
4p
p
4p2
2
a
a
= am b =0 if and only if m =
.
2p
2p
a
Substituting m =
and x=0 into the equation of the
2p
line gives the y-intercept
a
a2
a2
a2
a2
y =
10 - a2 +
= +
= - =–b.
2p
4p
2p
4p
4p
Single line
1 2
x is at (0, p) so any
4p
line with slope m that passes through the focus must
have equation y=mx+p.
The endpoints of a focal chord are the intersection
1 2
points of the parabola y =
x and the line
4p
y=mx+p.
1
Solving the equation x2 - mx - p = 0 using the
4p
quadratic formula, we have
1
m ;
m2 - 4 a
b 1 -p2
9
4p
x =
1
2a
b
4p
m ; 2m2 + 1
= 2p1m ; 2m2 + 12.
=
1
2p
75. (a) The focus of the parabola y =
Two parallel lines
Line
74. The point (a, b) is on the parabola y=
Generator
Circle
Plane
Section 8.2
(b) The y-coordinates of the endpoints of a focal chord
are
2
1
y =
12p 1m + 2m2 + 1 2 2 and
4p
2
1
12p 1m - 2m2 + 1 2 2
y =
4p
1
1 4p2 2 1m2 + 2m2m2 + 1 + 1 m2 + 1 2 2
4p
1
14p2 2 1m2 - 2m2m2 + 1 + 1m2 + 1 2 2
=
4p
319
2.
[–17.5, 12.5] by [–5, 15]
y2
x2
+
= 1, a parametric solution is
9
4
x=3 cos t and y=2 sin t.
y2
x2
Example 2: Since
+
= 1, a parametric solution is
13
4
y= 113 sin t and x=2 cos t.
1x - 322
1y + 1 2 2
Example 3: Since
+
= 1, a parametric
25
16
solution is x=5 cos t+3 and y=4 sin t-1.
3. Example 1: Since
= p 12m2 + 2m2m2 + 1 + 1 2
= p12m2 - 2m2m2 + 1 + 1 2
Using the distance formula for
12p1m - 2m2 + 12 , p12m2 - 2m 2m2 + 1 + 12 2
and 1 2p1 m + 2m2 + 12 ,
p1 2m2 + 2m2m2 + 1 + 1 2 2, we know that the
length of any focal chord is 21 x2 - x1 2 2 + 1y2 - y1 2 2
Ellipses
4.
= 314p2m2 + 1 2 2 + 14mp2m2 + 1 2 2
= 21 16m2p2 + 16p2 2 + 116m4p2 + 16m2p2 2
= 216m4p2 + 32m2p2 + 16p2
The quantity under the radical sign is smallest when
m=0. Thus the smallest focal chord has length
216p2 = @ [email protected] .
76. (a) For the parabola x2=4py, the axis and directrix
intersect at the point (0, –p). since the latus rectum is
perpendicular to the axis of symmetry, its slope is 0,
and from Exercise 65 we know the endpoints are
(–2p, p) and (2p, p). These points are symmetric
about the y-axis, so the distance from (–2p, p) to
(0, –p) equals the distance from (2p, p) to (0, –p).
The slope of the line joining (0, –p) and (2p, p) is
-p - p
= - 1 and the slope of the line joining
0 - 1 - 2p2
-p - p
(0, –p) and (2p, p) is
= 1. So the lines are
0 - 2p
perpendicular, and we know that the three points
form a right triangle.
(b) By Exercise 64, the line passing through (2p, p) and
(0, –p) must be tangent to the parabola; similarly for
(–2p, p) and (0, –p).
■ Section 8.2 Ellipses
Exploration 1
1. The equations x=–2+3 cos t and y=5+7 sin t can be
y - 5
x + 2
rewritten as cos t =
and sin t =
.
3
7
2
Substituting these into the identity cos t+sin2 t=1
1x + 22 2
1y - 522
yields the equation
+
= 1.
9
49
[–4, 4] by [–3, 3]
[–6, 6] by [–4, 4]
[–3, 9] by [–6, 4]
Answers may vary. In general, students should find that
the eccentricity is equal to the ratio of the distance
between foci over distance between vertices.
5. Example 1: The equations x=3 cos t, y=2 sin t can be
y
x
rewritten as cos t = , sin t = , which using
3
2
y2
x2
2
2
cos t+sin t=1 yield
+
= 1 or 4x2+9y2=36.
9
4
Example 2: The equations x=2 cos t, y= 113 sin t can
y
x
be rewritten as cos t = , sin t =
, which using
2
113
2
2
y
x
sin2 t+cos2 t=1 yield
+
= 1.
13
4
Example 3: By rewriting x=3+5 cos t,
y + 1
x - 3
y=–1+4 sin t as cos t =
, sin t =
and
5
4
using cos2 t+sin2 t=1, we obtain
1x - 3 2 2
1y + 12 2
+
= 1.
25
16
320
Chapter 8
Analytic Geometry in Two and Three Dimensions
Exploration 2
Answers will vary due to experimental error. The theoretical
answers are as follows.
2. a=9 cm, b= 180≠8.94 cm, c=1 cm, e=1/9≠0.11,
b/a≠0.99.
3. a=8 cm, b= 160≠7.75 cm, c=2 cm, e=1/4=0.25,
b/a≠0.97;
a=7 cm, b= 140≠6.32 cm, c=3 cm, e=3/7≠0.43,
b/a≠0.90;
a=6 cm, b= 120≠4.47 cm, c=4 cm, e=2/3≠0.67,
5. 3x+12=(10- 13x - 8)2
3x+12=100-20 13x - 8+3x-8
–80=–20 13x - 8
4= 13x - 8
16=3x-8
3x=24
x=8
6.
b/a≠0.75.
4. The ratio b/a decreases slowly as e=c/a increases rapidly.
The ratio b/a is the height-to-width ratio, which measures
the shape of the ellipse—when b/a is close to 1, the ellipse
is nearly circular; when b/a is close to 0, the ellipse is
elongated. The eccentricity ratio e=c/a measures how
off-center the foci are—when e is close to 0, the foci are
near the center of the ellipse; when e is close to 1, the foci
are far from the center and near the vertices of the ellipse.
The foci must be extremely off-center for the ellipse to be
significantly elongated.
5.
7. 6x2+12=(11- 26x2 + 1)2
6x2+12=121-22 26x2 + 1+6x2+1
–110=–22 26x2 + 1
2
6x +1=25
6x2-24=0
x2-4=0
x=2, x=–2
8.
[–0.3, 1.5] by [0, 1.2]
b
2a2 - c2
=
a
a
c2
B
a2
= 21 - e2
=
6x+12=(1+ 14x + 9)2
6x+12=(1+2 14x + 9+4x+9)
2x+2=2 14x + 9
x+1= 14x + 9
x2+2x+1=4x+9
x2-2x-8=0
(x-4)(x+2)=0
x=4
2x2+8=(8- 23x2 + 4)2
2x2+8=64-16 23x2 + 4+3x2+4
0=x2-16 23x2 + 4+60
x2+60=(16 23x2 + 4)2
x4+120x2+3600=256 (3x2+4)
x4-648x2+2576=0
x=2, x=–2
9. 2 a x -
3 2
15
3; 115
b - =0, so x=
2
2
2
1 -
10. 2(x+1)2-7=0, so x=–1 ;
7
C2
Section 8.2 Exercises
1. h=0, k=0, a=4, b= 17, so c= 116 - 7=3
Vertices: (4, 0), (–4, 0); Foci: (3, 0), (–3, 0)
2. h=0, k=0, a=5, b= 121, so c= 125 - 21=2
Vertices: (0, 5), (0, –5); Foci: (0, 2), (0, –2)
[–0.3, 1.5] by [0, 1.2]
Quick Review 8.2
1. 2 12 - 1 -3 2 2 2 + 1 4 - 1 -2 2 2 2 = 252 + 62 = 261
2. 2 1a - 1 -3 2 2 2 + 1 b - 1 -4 2 2 2
= 21a + 3 2 2 + 1b + 4 2 2
3. 4y2+9x2=36, 4y2=36-9x2,
y= ;
3
36 - 9x2
= ; 24 - x2
4
2
C
4. 25x2+36y2=900, 36y2=900-25x2,
y= ;
5
900 - 25x2
= ; 236 - x2
C
36
6
3. h=0, k=0, a=6, b=3 13, so c= 136 - 27=3
Vertices: (0, 6), (0, –6); Foci: (0, 3), (0, –3)
4. h=0, k=0, a= 111, b= 17, so c= 111 - 7=2
Vertices: ( 111, 0), ( - 111, 0); Foci: (2, 0), ( -2, 0)
y2
x2
+
= 1. h=0, k=0, a=2, b= 13, so
5.
4
3
c= 14 - 3=1
Vertices: (2, 0), (–2, 0); Foci: (1, 0), (–1, 0)
y2
x2
+
= 1. h=0, k=0, a=3, b=2, so
6.
9
4
c= 19 - 4 = 15.
Vertices: (0, 3), (0, –3); Foci: (0, 15), (0, – 15)
Section 8.2
7. (d)
16.
Ellipses
y
8. (c)
2
9. (a)
6
10. (b)
11.
x
y
10
10
12.
x
17.
y
[–9.4, 9.4] by [–6.2, 6.2]
10
y= ;
10
13.
x
2
2 -x2 + 36
3
18.
y
[–11.75, 11.75] by [–8.1, 8.1]
5
y= ;2 2-x2 + 16
19.
5
x
[–4.7, 4.7] by [–3.1, 3.1]
14.
y
y=1 ;
10
C
-
1x + 222
10
+
1
2
20.
10
x
[–9, 17] by [–6, 6]
15.
y= -4 ;
y
8
1
2- x2 + 8x + 112
16
y2
x2
+
= 1
4
9
y2
x2
22.
+
= 1
49
25
21.
4
x
23. c=2 and a=
y2
x2
+
= 1
25
21
10
=5, so b= 2a2 - c2 = 221 :
2
321
322
Chapter 8
24. c=3 and b=
Analytic Geometry in Two and Three Dimensions
10
=5, so a= 2b2 - c2 = 216 = 4 :
2
y2
x2
+
= 1
16
25
y2
x2
25.
+
= 1
16
25
y2
x2
+
= 1
26.
49
16
40. Center (–2, 1); Vertices: (–2, 1_5)=(–2, –4), (–2, 6);
Foci: (–2, 1_3)=(–2, –2), (–2, 4)
41.
[–8, 8] by [–6, 6]
y2
x2
+
= 1
16
36
y2
x2
+
= 1
28. b=2;
25
4
y2
x2
+
= 1
29. a=5;
25
16
y2
x2
+
= 1
30. a=13;
144
169
27. b=4;
x=2 cos t, y=5 sin t
42.
[–6, 10] by [–6, 5]
31. The center (h, k) is (1, 2) (the midpoint of the axes); a
and b are half the lengths of the axes (4 and 6,
1x - 122
1y - 2 2 2
+
= 1
respectively):
16
36
x= 130 cos t, y=2 15 sin t
43.
32. The center (h, k) is (–2, 2) (the midpoint of the axes); a
and b are half the lengths of the axes (2 and 5,
1x + 222
1y - 2 2 2
+
= 1
respectively):
4
25
33. The center (h, k) is (3, –4) (the midpoint of the major
axis); a=3, half the lengths of the major axis. Since
c=2 (half the distance between the foci),
1x - 3 2 2
1y + 4 2 2
+
= 1
b= 2a2 - c2 = 25 :
9
5
[–8, 2] by [0, 10]
x=2 13 cos t-3, y= 15 sin t+6
44.
34. The center (h, k) is (–2, 3) (the midpoint of the major
axis); b=4, half the lengths of the major axis. Since
c=2 (half the distance between the foci),
1x + 2 2 2
1y - 3 2 2
+
= 1
a= 2b2 - c2 = 212 :
12
16
35. The center (h, k) is (3, –2) (the midpoint of the major
axis); a and b are half the lengths of the axes (3 and 5,
respectively):
1x - 322
1y + 22 2
+
= 1
9
25
36. The center (h, k) is (–1, 2) (the midpoint of the major
axis); a and b are half the lengths of the axes (4 and 3,
1y - 2 2 2
1x + 122
+
= 1
respectively):
16
9
For #37–40, an ellipse with equation
1 x - h2 2
+
1y - k 2 2
a2
b2
has center (h, k), vertices (h _a, k), and foci (h_c, k)
where c = 2a2 - b2.
= 1
[–3, 7] by [–5, 3]
x= 16 cos(t)+2, y= 115 sin(t)-1
For #45–48, complete the squares in x and y, then put in standard form. (The first one is done in detail; the others just
show the final form.)
45. 9x2+4y2-18x+8y-23=0 can be rewritten as
9(x2-2x)+4(y2+2y)=23. This is equivalent to
9(x2-2x+1)+4(y2+2y+1)=23+9+4, or
9(x-1)2+4(y+1)2=36. Divide both sides by 36 to
1x - 12 2
1y + 12 2
obtain
+
= 1. Vertices: (1, –4) and
4
9
15
(1, 2) Foci: (1, –1 — 15). Eccentricity:
.
3
46.
39. Center (7, –3); Vertices: (7, –3_9)=(7, 6), (7, –12);
Foci: (7, –3_ 117)≠(7, 1.12), (7, –7.12)
5
+
1y + 32 2
3
= 1. Vertices: (2 ; 15, –3).
Foci: (2 ; 12, –3). Eccentricity:
37. Center (–1, 2); Vertices (–1_5, 2)=(–6, 2), (4, 2);
Foci (–1_3, 2)=(–4, 2), (2, 2)
38. Center (3, 5); Vertices: (3_ 111, 5)≠(6.32, 5), (–0.32, 5);
Foci=(3_2, 5)=(5, 5), (1, 5)
1x - 2 2 2
47.
1x + 3 2 2
16
+
1y - 12 2
9
12
2
=
C
5
15
= 1. Vertices: (–7, 1) and (1, 1).
Foci: (–3 ; 17, 1). Eccentricity:
17
4
Section 8.2
48. 1x - 42 2 +
1y + 8 2 2
4
= 1. Vertices: (4, –10) and (4, –6).
Foci: (4, –8 ; 13). Eccentricity:
13
2
49. The center (h, k) is (2, 3) (given); a and b are half the
lengths of the axes (4 and 3, respectively):
1y - 32 2
1x - 222
+
= 1
16
9
50. The center (h, k) is (–4, 2) (given); a and b are half the
lengths of the axes (4 and 3, respectively):
1x + 422
1y - 22 2
+
= 1
16
9
51. Consider Figure 8.15(b); call the point (0, c) F1, and the
point (0, –c) F2. By the definition of an ellipse, any point
P (located at (x, y)) satisfies the equation
PF + PF2 = 2a thus, 21x - 0 2 2 + 1 y - c2 2
+ 21x - 0 2 2 + 1 y + c2 2 = 2x2 + 1y - c2 2
2
2
+ 2x + 1 y + c2 = 2a
2
then 2x + 1 y - c2 2=2a- 2x2 + 1 y + c2 2
x2+(y-c)2=4a2-4a 2x2 + 1y + c2 2
+x2+(y+c)2
y2-2cy+c2=4a2-4a 2x2 + 1y + c2 2
+y2+2cy+c2
2
2
4a 2x + 1y + c2 =4a2+4cy
a 2x2 + 1y + c2 2=a2+cy
a2(x2+(y+c)2)=a4+2a2cy+c2y2
a2x2+(a2-c2)y2=a2(a2-c2)
a2x2+b2y2=a2 b2
y2
x2
=1
+
b2
a2
¡
53. Since the Moon is furthest from the Earth at 252,710
miles and closest at 221,463, we know that
2a=252,710+221.463, or a=237,086.5. Since
c+221,463=a, we know c=15,623.5 and
b= 2a2 - c2 = 2 1237,086.5 2 2 - 1 15,623.52 2
≠236,571.
15,623.5
c
From these, we calculate e= =
≠0.066.
237,086.5
a
The orbit of the Moon is very close to a circle, but still
takes the shape of an ellipse.
54. For Mercury, c=ea=(0.2056)(57.9)≠11.90 Gm and
its perihelion a-c=57.9-11.90≠46 Gm. Since the
diameter of the sin is 1.392 Gm. Mercury gets within
1.392
46≠45.3 Gm of the Sun’s surface.
2
323
55. For Saturn, c=ea=(0.0560)(1,427)≠79.9 Gm.
Saturn’s perihelion is a-c=1427-79.9≠1347 Gm
and its aphelion is a+c=1427+72.21≠1507 Gm.
56. Venus: c=ea=(0.0068)(108.2)≠0.74, so
b= 2 1108.2 2 2 - 10.74 2 2 L 108.2.
y2
x2
+
= 1
11,707.24
11,706.70
Mars: c=ea=(0.0934)(227.9)≠21.29, so
b= 2 1227.91 2 2 - 121.29 2 2 L 226.91
y2
x2
+
= 1
51,938
51,485
57. For sungrazers, a-c 6 1.5(1.392)=2.088. The
eccentricity of their ellipses is very close to 1.
58. a=
¡
c
52. Recall that e= means that c=ea, b= 2a2 - c2 and
a
a celestial object’s perihelion occurs at a-c for Pluto,
c=ea=(0.2484)(5900)≠1456.56, so its perhelion is
5900-1456.56=4,434.44 Gm. For Neptune,
c=ea=(0.0050)(4497)≠22.49, so its perihelion is
4497-22.49=4,474.51 Gm. As a result of its high by
eccentric orbit, Pluto comes over 40 Gm closer to the Sun
than Neptune.
Ellipses
36.18
9.12
, b=
, c= 2a2 - b2
2
2
9.12 2
36.18 2
b - a
b ≠17.51 Au
2
2
17.51
thus, e=
L 0.97
18.09
= a
C
59. a=8 and b=3.5, so c= 2a2 - b2 = 251.75. Foci at
(— 251.75, 0 2 L 1 ;7.19, 0 2 .
60. a=13 and b=5, so c= 2a2 - b2 = 12 Place the
source and the patient at opposite foci — 12 inches from
the center along the major axis.
61. Substitute y2=4-x2 into the first equation:
x2
4 - x2
+
=1
4
9
2
2
9x +4(4-x )=36
5x2=20
x2=4
x=—2, y=0
Solution: (–2, 0), (2, 0)
62. Substitute x=3y-3 into the first equation:
1 3y - 3 2 2
+ y2=1
9
y2-2y+1+y2=1
2y2-2y=0
2y(y-1)=0
y=0 or y=1
x=–3 x=0
Solution: (–3, 0), (0, 1)
63. (a)
[–4.7, 4.7] by [–3.1, 3.1]
Approximate solutions:
(—1.04, –0.86), (—1.37, 0.73)
; 294 - 2 1161 1 + 2161
,b,
8
16
; 294 + 2 1161 -1 + 2161
a
,
b
8
16
(b) a
324
Chapter 8
Analytic Geometry in Two and Three Dimensions
64. One possibility: a circle is perfectly “centric”: it is an
ellipse with both foci at the center. As the foci move off
the center and toward the vertices, the ellipse becomes
more eccentric as measured by the ratio=e=c/a. In
everyday life, we say a person is eccentric if he or she
deviates from the norm or central tendencies of behavior.
65. False. The distance is a-c=a(1-c/a)=a(1-e).
66. True, because a2=b2+c2 in any ellipse.
y2
x2
67. + =1, so c= 2a2 - b2= 222 - 12= 13. The
4
1
answer is C.
68. The focal axis runs horizontally through (2, 3). The
answer is C.
74. (a) The equations x(t)=5+3 sin a pt +
p
b and
2
p
b can be rewritten as
2
y
p
x - 5
p
sin a pt + b =
and cos a pt + b = .
2
3
2
3p
p
Substituting these into the identity cos2 a pt + b +
2
p
sin2 a pt + b =1 yields the equation
2
1x - 52 2
y2
+
= 1. This is the equation of an ellipse.
9
9p2
y(t)=3∏ cos a pt +
69. Completing the square produces
1 x - 4 2 2 1y - 3 2 2
+
=1. The answer is B.
4
9
70. The two foci are a distance 2c apart, and the sum of the
distances from each of the foci to a point on the ellipse is
2a. The answer is C.
71. (a) When a=b=r, A=pab=prr=pr2 and
213r + r2 1r + 3r2
P≠p(2r) a 3 b
r + r
4r
216r2
=2pr a 3 b =2pr a 3 b
2r
2r
=2pr (3-2)=2pr.
y2
x2
(b) One possibility:
+
= 1 with A=12p and
16
9
x2
P≠ 1 21 - 11952 p≠22.10, and
+ y2 = 1 with
100
A=10p and P≠ 1 33 - 14032p≠40.61.
[–8, 8] by [–10, 10]
(b) The pendulum begins its swing at t=0 so
p
x(0)=5+3 sin a b =8 ft, which is the maximum
2
distance away from the detector. When t=1,
p
x(1)=5+3 sin a p + b =2 ft, which is the
2
minimum distance from the detector. When t=3, the
pendulum is back to the 8-ft position.
As indicated in the table, the maximum velocity
(≠9.4 ft/sec) happens when the pendulum is at the
halfway position of 5 ft from the detector.
72. (a) Answers will vary. See Chapter III: The Harmony of
Worlds in Cosmos by Carl Sagan, Random House,
1980.
(b) Drawings will vary. Kepler’s Second Law states that as
a planet moves in its orbit around the sun, the line
segment from the sun to the planet sweeps out equal
areas in equal times.
p
73. (a) Graphing in parametric mode with Tstep =
.
24
[–4.7, 4.7] by [–3.1, 3.1]
(b) The equations x(t)=3+cos(2t-5) and y(t)=
–2 sin(2t-5) can be rewritten as cos (2t-5)=x-3
and sin(2t-5)= - y>2. Substituting these into the
identity cos2(2t-5)+sin2(2t-5)=1 yields the
equation y2>4 + 1x - 32 2 = 1. This is the equation of
an ellipse with x=3 as the focal axis. The center of
the ellipse is (3, 0) and the vertices are (3, 2) and
(3, –2). The length of the major axis is 4 and the
length of the minor axis is 2.
75. Write the equation in standard form by completing the
squares and then dividing by the constant on the righthand side.
D2
D2
E2
E2
Ax2+Dx+ +Cy2+Ey+
=
+
- F
4A
4C
4A
4C
D
D2
E
E2
x2 + x +
y2 + y +
2
A
C
4A
4C2
+
C
A
1
D2
E2
a
+
- Fb
=
AC 4A
4C
D 2
E 2
ax +
b
b
ay +
2A
2C
CD2 + AE2 - 4ACF
+
=
C
A
4A2C2
4A2C2
b*
a
CD2 + AE2 - 4ACF
D 2
E 2
b
b
ax +
ay +
£
2A
2C §
+
= 1
C
A
Section 8.3
4A2C‘ a x +
D 2
E 2
b
b
4AC2 a y +
2A
2C
+
= 1
CD2 + AE2 - 4ACF
CD2 + AE2 - 4ACF
Since AC 7 0, A Z 0 and C Z 0 (we are not dividing by
zero). Further, AC 7 0 1 4A2C 7 0 and 4AC2 7 0
(either A 7 0 and C 7 0, or A 6 0 and C 6 0), so the
equation represents an ellipse.
y - k 2
x - h 2
b + a
b = 0
76. Rewrite the equation to a
a
b
Since that a Z 0 and b Z 0 (otherwise the equation is not
defined) we see that the only values of x, y that satisfy the
equation are (x, y)=(h, k). In this case, the degenerate
ellipse is simply a single point (h, k). The semimajor and
semiminor axes both equal 0. See Figure 8.2.
1x - 3 2 2
25
-
1y + 12 2
16
Hyperbolas
325
= 1
[–7.4, 11.4] by [–6.2, 6.2]
1x + 2 2 2
9
-
1y - 52 2
49
= 1
■ Section 8.3 Hyperbolas
Exploration 1
1. The equations x=–1+3/cos t=–1+3 sec t and
y=1+2 tan t can be rewritten as
y - 1
x + 1
sec t=
and tan t =
. Substituting these
2
3
2
2
into the identity sec t-tan t=1 yields the equation
1x + 122
1y - 12 2
= 1.
9
4
2.
[–9.4, 9.4] by [–6.2, 6.2]
[–9.4, 9.4] by [–6.2, 6.2]
In Connected graphing mode, pseudo-asymptotes appear
because the grapher connects computed points by line
segments regardless of whether this makes sense. Using
Dot mode with a small Tstep will produce the best graphs.
3. Example 1: x=3>cos1 t 2, y = 2 tan 1 t2
Example 2: x=2 tan(t), y= 15/cos (t)
Example 3: x=3+5/cos(t), y=–1+4 tan(t)
Example 4: x=–2+3/cos(t), y=5+7 tan(t)
[–20, 18] by [–8, 18]
5. Example 1: The equations x = 3>cos t = 3 sec t,
y
x
y = 2 tan t can be rewritten as sec t = , tan t = , which
2
3
2
2
y
x
using the identity sec2 t-tan2 t=1 yield
= 1.
9
4
Example 2: The equations x=2 tan t, y = 15>cos t
y
x
,
= 15 sec t can be rewritten as tan t = , sec t =
2
15
y2
x2
which using sec2 t-tan2 t=1 yield
= 1.
5
4
Example 3: By rewriting x=3+5/cos t,
y + 1
x - 3
y=–1+4 tan t as sec t =
, tan t =
and
5
4
2
2
using sec t-tan t=1, we obtain
1y + 12 2
1x - 3 2 2
= 1.
25
16
Example 4: By rewriting x=–2+3/cos t,
y - 5
x + 2
y=5+7 tan t as sec t =
, tan t =
and using
3
7
2
1x + 22
1y - 52 2
sec2 t-tan2 t=1, we obtain
= 1.
9
49
4. 4x2-9y2=36
Quick Review 8.3
1. 2 1 -7 - 42 2 + 1 -8 - 1 -3 2 2 2
= 21 -112 2 + 1 -5 2 2 = 2146
[–9.4, 9.4] by [–6.2, 6.2]
2
2
y
x
= 1
5
4
[–9.4, 9.4] by [–6.2, 6.2]
2. 2 1b - a2 2 + 1c - 1 -32 2 2
= 2 1b - a2 2 + 1c + 32 2
3. 9y2-16x2=144
9y2=144+16x2
4
y= ; 29 + x2
3
4. 4x2-36y2=144
36y2=4x2-144
2
y= ; 2x2 - 36
6
1
y=; 2x2 - 36
3
Chapter 8
326
Analytic Geometry in Two and Three Dimensions
5. 13x + 12=10 + 13x - 8
3x+12=100+20 13x - 8+3x-8
–80=20 13x - 8
–4= 13x - 8
no solution
6.
14x + 12=1+ 1x + 8
4x+12=1+2 1x + 8+x+8
3x+3=2 1x + 8
9x2+18x+9=4x+32
9x2+14x-23=0
- 14 + 2196 - 4 19 2 1 -23 2
x=
18
- 14 ; 32
x=
18
23
23
x=1 or x=– . When x = - ,
9
9
14x + 12 - 1x + 8
16
49
4
7
= = -1
=
C9
C9
3
3
The only solution is x=1.
7. 26x2 + 12=1+ 26x2 + 1
6x2+12=1+2 26x2 + 1+6x2+1
10=2 26x2 + 1
25=6x2+1
6x2-24=0
x2-4=0
x=2, x=–2
22x2 + 12=–8+ 23x2 + 4
2x2+12=64-16 23x2 + 4+3x2+4
8.
x2+56=16 23x2 + 4
x +112x +3136=768x2+1024
x4-656x2+2112=0
x={25.55, –25.55} (the other
solutions are extraneous)
4
2
9. c=a+2, (a+2)2-a2=
a2+4a+4-a2=
7. (c)
8. (b)
9. (a)
10. (d)
11. Transverse axis from (–7, 0) to (7, 0); asymptotes:
5
y= ; x,
7
5
y= ; 2x2 - 49
7
y
15
20
x
12. Transverse axis from (0, –8) to (0, 8); asymptotes:
8
y= ; x,
5
8
y= ; 2x2 + 25
5
y
25
16a
,
3
16a
, 4a=12: a=3, c=5
3
10. c=a+1, (a+1)2-a2=
a2+2a+1-a2=
y2
x2
= 1; a=2, b= 13, c= 17;
4
3
Vertices: (—2, 0); Foci: (— 17, 0)
y2
x2
= 1; a=2, b=3, c= 113;
6.
4
9
Vertices: (—2, 0); Foci: (— 113, 0)
5.
20
x
25a
,
12
25a
: a=12, c=13
12
Section 8.3 Exercises
For #1–6, recall the Pythagorean relation that c2=a2+b2.
1. a=4, b= 17, c= 116 + 7 = 123;
Vertices: (—4, 0); Foci: (— 123, 0)
2. a=5, b= 121, c= 125 + 21 = 146;
Vertices: (0, —5); Foci: (0, — 146)
13. Transverse axis from (0, –5) to (0, 5); asymptotes:
5
y= ; x,
4
5
y= ; 2x2 + 16
4
y
15
3. a=6, b= 113, c= 136 + 13 = 7;
Vertices: (0, —6); Foci: (0, —7)
4. a=3, b=4, c= 19 + 16 = 5;
Vertices: (—3, 0); Foci: (—5, 0)
20
x
Section 8.3
14. Transverse axis from (–13, 0) to (13, 0); asymptotes:
12
y= ; x,
13
12
y= ; 2x2 - 169
13
Hyperbolas
327
18.
y
[–18.8, 18.8] by [–12.4, 12.4]
15
y= ;2 2x2 + 16
19.
20
x
[–9.4, 9.4] by [–6.2, 6.2]
15. The center (h, k) is (–3, 1). Since a2=16 and b2=4, we
have a=4 and b=2. The vertices are at (–3 — 4, 1) or
(–7, 1) and (1, 1).
3
y= ; 2x2 - 4
2
20.
y
4
3
x
[–9.4, 9.4] by [–6.2, 6.2]
y= ;
16. The center (h, k) is (1, –3). Since a2=2 and b2=4, we
have a= 12 and b=2. The vertices are at (1— 12, –3).
4
2x2 + 9
3
21.
y
3
6
x
[–9.4, 9.4] by [–3.2, 9.2]
y=3 ;
1
25x2 - 20
2
22.
17.
[–11.4, 7.4] by [–3.2, 9.2]
y=3 ;
[–18.8, 18.8] by [–12.4, 12.4]
2
y= ; 2x2 - 36
3
3
2x2 + 4x + 8
2
y2
x2
= 1
4
5
y2
x2
24. c=3 and b=2, so a= 2c2 - b2= 15:
= 1
4
5
23. c=3 and a=2, so b= 2c2 - a2= 15:
25. c=15 and b=4, so a= 2c2 - b2 = 2209 :
y2
x2
= 1
16
209
328
Chapter 8
Analytic Geometry in Two and Three Dimensions
26. c=5 and a=3/2, so b= 2c2 - a2 =
1
291 :
2
y2
y2
x2
x2
= 1 or
= 1
2.25
22.75
9>4
91>4
27. a=5 and c=ea=10, so b= 1100 - 25 = 5 15:
y2
x2
= 1
25
75
28. a=4 and c=ea=6, so b= 136 - 16 = 2 15:
y2
x2
= 1
16
20
29. b=5, a= 2c2 - b2 = 2169 - 25 = 12:
y2
x2
= 1
144
25
30. c=6, a=
c
= 3, b= 2c2 - a2 = 236 - 9 = 3 23:
e
y2
x2
= 1
9
27
31. The center (h, k) is (2, 1) (the midpoint of the transverse
axis endpoints); a=2, half the length of the transverse
axis. And b=3, half the length of the conjugate axis.
1 y - 12 2
1x - 22 2
= 1
4
9
32. The center (h, k) is (–1, 3) (the midpoint of the transverse
axis endpoints); a=6, half the length of the transverse
axis. And b=5, half the length of the conjugate axis.
1x + 122
1y - 32 2
= 1
36
25
33. The center (h, k) is (2, 3) (the midpoint of the transverse
axis); a=3, half the length of the transverse axis.
1x - 222
1y - 3 2 2
4
Since |b/a|= , b=4:
= 1
3
9
16
5
34. The center (h, k) is a-2, b , the midpoint of the
2
9
transverse axis); a= , half the length of the transverse
2
27
4
axis. Since |a/b|= , b= :
3
8
1y - 5>2 2 2
1x + 222
= 1
81>4
729>64
35. The center (h, k) is (–1, 2), the midpoint of the transverse axis. a=2, half the length of the transverse axis.
The center-to-focus distance is c=3, so b= 2c2 - a2
1 x + 12 2
1y - 2 2 2
= 15:
= 1
4
5
36. The center (h, k) is a-3, -
37. The center (h, k) is (–3, 6), the midpoint of the transverse axis. a=5, half the length of the transverse axis.
The center-to-focus distance c=ea
=2 # 5=10, so b= 2c2 - a2 = 2100 - 25 = 525
1y - 6 2 2
1x + 32 2
= 1
25
75
38. The center (h, k) is (1, –4), the midpoint of the transverse axis. c=6, the center-to-focus distance
c
6
a= = = 3, b= 2c2 - a2 = 236 - 9 = 227
e
2
1x - 1 2 2
1y + 42 2
= 1
9
27
For #39–42, a hyperbola with equation
1x - h2 2
1y - k 2 2
= 1 has center (h, k) vertices
a2
b2
(h_a, k), and foci (h_c, k) where c = 2a2 + b2.
1y - k2 2
1 x - h2 2
A hyperbola with equation
= 1 has
a2
b2
center (h, k), vertices (h, k_a), and foci (h, k_c) where
again c = 2a2 + b2.
39. Center (–1, 2); Vertices: (–1_12, 2)=(11, 2), (–13, 2);
Foci: (–1_13, 2)=(12, 2), (–14, 2)
40. Center (–4, –6); Vertices: (–4_ 112, –6);
Foci: (–4_5, –6)=(1, –6), (–9, –6)
41. Center (2, –3); Vertices: (2, –3_8)=(2, 5), (2, –11);
Foci: (2,–3_ 1145)
42. Center (–5, 1); Vertices: (–5, 1_5)=(–5, –4), (–5, 6);
Foci: (–5, 1_6)=(–5, –5), (–5, 7)
43.
[–14.1, 14.1] by [–9.3, 9.3]
y=5/cos t, x=2 tan t
44.
[–14.1, 14.1] by [–9.3, 9.3]
x= 130/cos t, y=2 15 tan t
45.
11
b , the midpoint of the
2
7
transverse axis. b= , half the length of the transverse
2
11
axis. The center-to-focus distance is c= , so
2
2
1y
+
5.5
2
1x
+ 322
a= 2c2 - b2 = 218 :
= 1
49>4
18
[–12.4, 6.4] by [–0.2, 12.2]
x=–3+213/cos t, y=6+ 15 tan t
Section 8.3
Hyperbolas
329
50.
46.
[–7.4, 11.4] by [–7.2, 5.2]
[–12.4, 6.4] by [–5.2, 7.2]
y=–1+ 115/cos t, x=2+ 16 tan t
47.
1y - 1 2 2
9
-
1x + 32 2
25
= 1. Vertices: (–3, –2) and
(–3, 4). Foci: (–3, 1 ; 134), e=
[–9.4, 9.4] by [–5.2, 7.2]
Divide the entire equation by 36. Vertices: (3, –2) and
113
(3, 4), Foci: (3, 1 ; 113), e=
.
3
48.
[–2.8, 6.8] by [–7.1, 0]
3
5
113
Vertices: a , -4 b and a , -4 b , Foci: a 2 ;
, -4 b
2
2
6
e=
21 1>4 2 + 11>9 2
1>2
9 + 4
113
=
.
B 36
3
= 2
For #49–50, complete the squares in x and y, then write the
equation in standard form. (The first one is done in detail; the
other shows just the final form.) As in the previous problems,
the values of h, k, a, and b can be “read” from the equation
1 y - k2 2
1x - h2 2
<
= 1. The asymptotes are
;
a2
b2
b
y-k= ; (x-h). If the x term is positive, the transverse
a
axis endpoints are (h ; a, k); otherwise the endpoints are
(h, k ; b).
49.
51. a=2, (h, k)=(0, 0) and the hyperbola opens to the left
y2
x2
4
9
and right, so
- 2 = 1. Using (3, 2): - 2 = 1,
4
4
b
b
5y2
16 x2
9b2-16=4b2, 5b2=16, b2= ;
= 1
5 4
16
52. a= 12, (h, k)=(0, 0) and the hyperbola opens upward
y2
x2
- 2 = 1. Using (2, –2):
and downward, so
2
b
y2
4
4
4
x2
2
- 2 = 1, 2 =1, b =4;
= 1
2
2
4
b
b
53. Consider Figure 8.24(b). Label (0, c) as point F1, label
(0, –c) as point F2 and consider any point P(x, y) along
the hyperbola. By definition, PF1-PF2=—2a, with
c 7 a 0
2 1x - 0 2 2 + 1 y - 1 -c2 2 2 - 2 1x - 02 2 + 1y - c2 2
=—2a
2x2 + 1y + c2 2= ;2a + 2x2 + 1y - c2 2
x2+y2+2cy+c2=4a2 ; 4a 2x2 + 1y - c2 2
+x2+y2-2cy+c2
2
2
;a2x + 1y - c2 =a2-cy
2
a (x2+y2-2cy+c2)=a4-2a2cy+c2y2
–a2x2+(c2-a2)y2=a2(c2-a2)
b2y2-a2x2=a2b2
y2
x2
=1
a2
b2
54. (a)
x2
- y2=0
4
x2
y2=
4
y= ;
x
2
y
4
[–9.4, 9.4] by [–6.2, 6.2]
2
9x -4y2-36x+8y-4=0 can be rewritten as
9(x2-4x)-4(y2-2y)=4. This is equivalent to
9(x2-4x+4)-4(y2-2y+1)=4+36-4, or
9(x-2)2-4(y-1)2=36. Divide both sides by 36 to
1 x - 22 2
1y - 12 2
obtain
= 1. Vertices: (0, 1) and
4
9
113
(4, 1). Foci: (2 ; 113, 1), e=
2
134
3
4
x
330
(b)
Chapter 8
Analytic Geometry in Two and Three Dimensions
y2
x2
=0
9
16
9x2
y2=
16
3x
y= ;
4
y
4
4
55.
x
c-a=120, b2=250a
c -a2=b2
(a+120)2-a2=250a
a2+240a+14,400-a2=250a
10a=14,400
a=1440 Gm
13
1560
a=1440 Gm, b=600 Gm, c=1560, e=
= .
12
1440
The Sun is centered at focus (c, 0)=(1560, 0).
2
56.
c-a=140, b2=405a
c -a2=b2
(a+140)2-a2=405a
a2+280a+19,600-a2=405a
125a=19,600
a=156.8
2
a=156.8 Gm, b=252 Gm, c=296.8 Gm, e=
hyperbola, c=3500 ft, 2a=(4 sec)(1100 ft/sec)=
4400 ft, and b≠ 235002 - 22002 = 100 2741 ft.
The two equations are therefore
1y - 20002 2
x2
= 1 and
2,790,000
11002
2
1 x - 3500 2
y2
= 1.
2
7,410,000
2200
The intersection of the upper branch of the first hyperbola
and the right branch of the second hyperbola (found
graphically) is approximately (11,714.3, 9792.5). The gun is
located about 11,714 ft (2.22 mi) east and 9793 ft
(1.85 mi) north of point B – a bearing and distance of
about 50.11 and 15,628.2 ft (2.89 mi), respectively.
y2
x2
59.
- =1
4
9
213
x y=–2
3
Solve the second equation for x and substitute into the
first equation.
213
x=
y-2
3
2
2
y
1 2 13
a
y - 2b = 1
4
3
9
2
y
813
1 4 2
a y y + 4b = 1
4 3
3
9
2 13
2 2
y y = 0
9
3
2
y1y - 3132 = 0
9
y=0 or y=313
53
.
28
The Sun is centered at focus (c, 0)=(297, 0).
57. The Princess Ann is located at the intersection of two
hyperbolas: one with foci O and R, and the other with foci
O and Q. For the first of these, the center is (0, 40), so the
center-to-focus distance is c=40 mi. The transverse axis
length is 2b=(323.27 Âsec)(980 ft/Âsec)=
316,804.6 ft≠60 mi. Then a≠ 2402 - 302 = 2700 mi.
For the other hyperbola, c=100 mi, 2a=(646.53 Âsec)
(980 ft/Âsec)=633599.4 ft≠120 mi, and
b≠ 21002 - 602 = 80 mi. The two equations are
therefore
1x - 100 2 2
1y - 402 2
y2
x2
= 1 and
= 1.
900
700
3600
6400
The intersection of the upper branch of the first hyperbola and the right branch of the second hyperbola (found
graphically) is approximately (886.67, 1045.83). The ship is
located about 887 miles east and 1046 miles north of point
O – a bearing and distance of about 40.29 and 1371.11
miles, respectively.
58. The gun is located at the intersection of two hyperbolas:
one with foci A and B, and the other with foci B and C.
For the first of these, the center is (0, 2000), so the centerto-focus distance is c=2000 mi. The transverse axis
length is 2b=(2 sec)(1100 ft/sec)=2200 ft. Then
a≠ 220002 - 11002 = 1002279 ft. For the other
[–9.4, 9.4] by [–6.2, 6.2]
Solutions: (–2, 0), (4, 313)
60. Add:
x2
- y2=1
4
x2+y2=9
5x2
=10
4
2
x =8
x= ;2 12
x2+y2=9
8+y2=9
y= ;1
[–9.4, 9.4] by [–6.2, 6.2]
There are four solutions: (—212, —1)
Section 8.3
61. (a)
There are four solutions: (— 2.13, — 1.81)
29
21
, ; 10
b.
B 641
B 641
62. One possibility: Escape speed is the minimum speed one
object needs to achieve in order to break away from the
gravity of another object. For example, for a NASA space
probe to break away from the Earth’s gravity is must
meet or exceed the escape speed for Earth
˜E= 12GM>r L 11,200 m/s. If this escape speed is
exceeded, the probe will follow a hyperbolic path.
63. True. The distance is c-a=a(c/a-1)=a(e-1).
64. True. For an ellipse, b¤+c¤=a¤.
y2
x2
= 1, so c= 14 + 1 and the foci are each 15
65.
4
1
units away horizontally from (0, 0). The answer is B.
66. The focal axis passes horizontally through the center,
(–5, 6). The answer is E.
67. Completing the square twice, and dividing to obtain 1 on
the right, turns the equation into
1 y + 32 2
1x - 22 2
= 1. The answer is B.
4
12
68. a=2, b= 23, and the slopes are —b/a. The answer is C.
69. (a–d)
y
5
(e) a=3, c=5, b=4;
x2/9-y2/16=1
331
70. Assume that the focus for the primary parabolic mirror
occurs at FP and the foci for the hyperbolic mirror occur
at FH and FH. Assume also that the x-axis extends from
the eye piece to the right most FH, and that the y-axis is
perpendicular through the x-axis 60 cm from the eye
piece. Then, the center (h, k) of the hyperbolic mirror is
(0, 0), the foci (—c, 0)=(—60, 0) and the vertices
(—a, 0)=(—40, 0).
Since a=40, c=60, b2=c2-a2=2000. The equation
y2
x2
for the hyperbolic mirror is
= 1.
1600
2000
[–9.4, 9.4] by [–6.2, 6.2]
(b) The exact solutions are a ;10
Hyperbolas
x
71. From Section 8.2, Question #75, we have
Ax2+Cy2+Dx+Ey+F=0 becomes
D 2
F 2
4AC2 a y +
b
b
4A2C a x +
2A
2C
+
= 1
CD2 + AE2 - 4ACF
CD2 + AE2 - 4ACF
Since AC 6 0 means that either (A 6 0 and C 7 0) or
(A 7 0 and C 6 0), either (4A2C 6 0 and 4AC2 7 0),
or (4A2C 7 0 and 4AC2 6 0). In the equation above,
that means that the + sign will become a (–) sign once
all the values A, B, C, D, E, and F are determined, which
is exactly the equation of the hyperbola. Note that if
A 7 0 and C 6 0, the equation becomes:
E 2
D 2
@ [email protected] a x +
4AC2 a y +
b
b
2C
2A
= 1
CD2 + AE2 - 4ACF
CD2 + AE2 - 4ACF
If A 6 0 and C 7 0, the equation becomes:
D 2
E 2
@ 4AC2 @ a y +
4A2C a x +
b
b
2A
2C
= 1
CD2 + AE2 - 4ACF
CD2 + AE2 - 4ACF
y - k 2
x - h 2
72. With a Z 0 and b Z 0, we have a
b = a
b .
a
b
y - k
x - h
Then a
b = a
b or
a
b
y - k
x - h
b = -a
b . Solving these two equations,
a
a
b
b
we find that y= ; 1x - h2 + k. The graph consists of
a
two intersecting slanted lines through (h, k). Its symmetry
is like that of a hyperbola. Figure 8.2 shows the relationship
between an ordinary hyperbola and two intersecting lines.
73. The asymptotes of the first hyperbola are
b
y= ; 1x - h2 + k and the asymptotes of the second
a
b
hyperbola are y= ; 1x - h2 + k; they are the same.
a
[Note that in the second equation, the standard usage of
a+b has been revised.] The conjugate axis for hyperbola
1 is 2b, which is the same as the transverse axis for hyperbola 2. The conjugate axis for hyperbola 2 is 2a, which is
the same as the transverse axis of hyperbola 1.
332
Chapter 8
Analytic Geometry in Two and Three Dimensions
y2
c2
- 2 =1
2
a
b
c2b2-a2y2=a2b2
a2y2=b2(c2-a2)
b2=c2-a2
b4
y2= 2
a
b2
y= ;
a
7
17
17 # 12
cos Å=
=
=
B8
2 12
2 12 12
114
cos Å=
4
74. When x=c,
One possible answer: Draw the points a c,
5
1
9. cos 2Å=1-2 sin2 Å= , –2 sin2 Å= - 1
6
6
1
1
1
sin2 Å= 1 sin Å=
1 sin Å=
12
B 12
112
b2
b and
a
-b2
b on a copy of figure 8.24(a). Clearly the points
a
b2
a c, ; b on the hyperbola are the endpoints of a
a
segment perpendicular to the x-axis through the focus
(c, 0). Since this is the definition of the focal width used
in the construction of a parabola, applying it to the hyperbola also makes sense.
a c,
75. The standard forms involved multiples of x, x2, y, and y2,
as well as constants; therefore they can be rewritten in the
general form Ax2+Cy2+Dx+Ey+F=0 (none of
the standard forms we have seen require a Bxy term). For
example, rewrite y=ax2 as ax2-y=0; this is the general form with A=a and E=–1, and all others 0.
y2
x2
Similarly, the hyperbola 2 - 2 = 1 can be put in
b
a
1
1
standard form with A= - 2 , C= 2 , F=–1, and
a
b
B=D=E=0.
■ Section 8.4 Translation and Rotation
of Axes
Quick Review 8.4
1. cos 2Å=
5
13
2. cos 2Å=
8
17
3. cos 2Å=
1
2
4. cos 2Å=
2
3
p
p
5. 2Å= , so Å=
2
4
p
1
p
6. 2Å=sin–1 a b = , so Å=
2
6
12
10. cos 2Å=1-2 sin2 Å=
sin Å=
45
8
8
, 2 sin2 Å= 1 sin2 Å=
1
53
53
106
2
153
Section 8.4 Exercises
1. Use the quadratic formula with a=1, b=10, and
c=x2-6x+18. Then b2-4ac=(10)2-4(x26x+18)=–4x2+24x+28=4(–x2+6x+7), and
y =
-10 ; 241 -x2 + 6x + 7 2
2
= -5 ; 2-x2 + 6x + 7
[–6.4, 12.4] by [–11.2, 1.2]
2. Use the quadratic formula with a=1, b=–2, and
c=4x2+24x+21. Then b2-4ac=
(–2)2-4(4x2+24x+21)=–16x2-96x-80=
16(–x2-6x-5), and
2 ; 2161 -x2 - 6x - 52
y =
2
=1 ; 22 -x2 - 6x - 5
[–9.4, 9.4] by [–6.2, 6.2]
3. Use the quadratic formula with a=1, b=–8, and
c=–8x+8. Then b2-4ac=
(–8)2-4(–8x+8)=32x+32=32(x+1), and
y =
8 ; 1321 x + 1 2
2
=4 ; 2 12x + 2
4
3
8
7. cos 2Å=2 cos2 Å-1= , 2 cos2 Å= , cos2 Å= ,
5
5
5
2
cos Å=
15
3
7
7
8. cos 2Å=2 cos2 Å-1= , 2 cos2 Å= , cos2 Å= ,
4
4
8
[–19.8, 17.8] by [–8.4, 16.4]
Section 8.4
Translation and Rotation of Axes
333
4. Use the quadratic formula with a=–4, b=–40, and
c=x2+6x+91. Then b2-4ac=
(–40)2-4(–4)(x2+6x+91)=16x2+96x+3056
=16(x2+6x+191), and
y =
40 ; 216 1 x2 + 6x + 191 2
-8
1
2
= -5 ; 2x + 6x + 191
2
[– 4.7, 4.7] by [–3.1, 3.1]
10. Use the quadratic formula with a=4, b=3x-10, and
c=–x2-5x-20. Then b2-4ac=(3x-10)2
–16(–x2-5x-20)=25x2 +20x+420, and
1
y = 310 - 3x ; 225x2 + 20x + 4204 .
8
[–37.6, 37.6] by [–24.8, 24.8]
5. –4xy+16=0 1 –4xy=–16 1 y=4/x
[–10, 10] by [–8, 8]
[–9.4, 9.4] by [–6.2, 6.2]
6. 2xy+6 1 2xy=–6 1 y=–3/x
11. Use the quadratic formula with a=8, b=4-4x, and
c=2x2-10x-13.
Then b2-4ac=(4-4x)2-32(2x2-10x-13)
=–48x2+288x+432=48(–x2+6x+9), and
y=
4x - 4 ; 248 1 -x2 + 6x + 92
16
1
1
1
= x - ; 231 -x2 + 6x + 92.
4
4
4
[–9.4, 9.4] by [–6.2, 6.2]
7. xy-y-8=0 1 y(x-1)=8 1 y=8/(x-1)
[–2, 8] by [–3, 3]
[–10, 12] by [–12, 12]
2
8. 2x -5xy+y=0 1 y(1-5x)=–2x2 1
y=2x2/(5x-1)
[–1, 1.4] by [–0.4, 0.8]
9. Use the quadratic formula with a=3, b=4-x, and
c=2x2-3x-6. Then b2-4ac
=(4-x)2-12(2x2-3x-6)=–23x2+28x+88,
x - 4 ; 2-23x2 + 28x + 88
and y =
6
12. Use the quadratic formula with a=2, b=6-4x, and
c=2x2-5x-15. Then b2-4ac=(6-4x)2
–8(2x2-5x-15)=156-8x=4(39-2x), and
4x - 6 ; 14139 - 2x2
y =
4
3
1
= x - ; 139 - 2x.
2
2
[–10, 30] by [–5, 20]
13. h=0, k=0 and the parabola opens downward, so
4py=x2 ). Using (2, –1): –4p=4, p=–1.
The standard form is x2=–4y.
14. h=0, k=0 and the parabola opens to the right, so
4px=y2 (p>0). Using (2, 4): 8p=16, p=2.
The standard form is y2=8x.
334
Chapter 8
Analytic Geometry in Two and Three Dimensions
15. h=0, k=0 and the hyperbola opens to the right and
left, so a=3, and b=4. The standard form is
y2
x2
= 1.
9
16
16. h=0, k=0, and the x-axis is the focal axis, so a=4
y2
x2
and b=3. The standard form is
+
= 1.
16
9
24. 2 a y -
7
b =(x-1)2, a parabola. The vertex is
6
7
(h, k)= a 1, b , so 2y=(x)2.
6
y'
8
For #17–20, recall that x=x-h and y=y-k.
17. (x, y)=(4, –1)
18. (x, y)=(2, 12)
19. (x, y)=(5, –3- 15 2
20. (x, y)= 1 -5 - 12, -1 2
21. 4(y2-2y)-9(x2+2x)=41, so
4(y-1)2-9(x+1)2=41+4-9=36. Then
1 y - 12 2
1x + 12 2
= 1. This is a hyperbola, with
9
4
a=3, b=2, and c= 113.
1 y¿ 2 2
1x¿ 2 2
= 1.
9
4
y'
8
4
x'
25. 9(x2-2x)+4(y2+4y)=11, so
9(x-1)2+4(y+2)2=11+9+16=36. Then
1x - 1 2 2
1y + 22 2
+
= 1.
4
9
This is an ellipse, with a=2, b=3, and c= 15.
Foci: 11, -2 ; 15 2. Center (1, –2), so
1x¿ 2 2
1 y¿ 2 2
+
= 1.
4
9
y'
4
5
x'
12
22. 2(x2+6x)+3(y2-8y)=–60, so
2(x+3)2+3(y-42)=–60+18+48=6. Then
1x + 3 2 2
1y - 42 2
+
= 1. This is an ellipse with
3
2
1 x¿ 2 2
1y¿ 2 2
+
= 1.
a= 13, b = 12, and c=1.
3
2
y'
3
x'
26. 16(x2-2x)-(y2+6y)=57, so 16(x-1)2(y+3)2=57+16-9=64.
1y + 3 2 2
1x - 122
= 1. This is a hyperbola, with
Then
4
64
a=2, b=8, and c= 168 = 2 117. Foci:
1x¿ 2 2
1y¿ 2 2
11 ; 2117, -3 2. Center (1, –3), so
= 1.
4
64
y'
4
x'
20
5
2
23. y-2=(x+1) , a parabola. The vertex is
(h, k)=(–1, 2), so y=(x)2.
y'
25
5
x'
x'
Section 8.4
27. 8(x-2)=(y-2)2, a parabola. The vertex is
(h, k)=(2, 2), so 8x=(y)2.
y'
Translation and Rotation of Axes
335
31. The horizontal distance from O to P is
x=h+x=x+h, and the vertical distance from O
to P is y=k+y=y+k.
32. Given x=x+h, subtract h from both sides: x-h=x
or x=x-h. And given y=y+k, subtract k from
both sides: y-k=y or y=y-k.
8
8
x'
For #33–36, recall that x=x cos Å+y sin Å and
y=–x sin Å+y cos Å.
33. (x, y)= a -2 cos
p
p
p
p
+ 5 sin , 2 sin + 5 cos b
4
4
4
4
3 12 712
,
b
=a
2
2
28. 2(x2-2x)+(y2-6y)=9, so 2(x-1)2+(y-3)2
1y - 322
1x - 1 2 2
=9+2+9=20. Then
+
= 1.
10
20
This is an ellipse, with
a= 110, b = 120 = 2 15, and c = 110.
1 x¿ 2 2
1y¿ 2 2
Foci: 11, 3 ; 1102. Center (1, 3), so
+
= 1
10
20
34. (x, y)= a 6 cos
p
p
p
p
- 3 sin , -6 sin - 3 cos b
3
3
3
3
6
3 13 -6 13
3
,
- b
2
2
2
2
6 - 313 -613 - 3
,
b L (0.40, –6.70)
=a
2
2
=a
35. Å≠1.06, (x, y)=(–5 cos (1.06)-4 sin (1.06),
5 sin (1.06)-4 cos (1.06))≠(–5.94, 2.38)
y'
8
6
x'
p
36. Å≠ , (x, y)
4
p
p
p
p
+ 3 cos b
= a 2 cos + 3 sin , -2 sin
4
4
4
4
=a
29. 2(x2+2x)-y2=–6, so 2(x+1)2-y2=–6+2
1x + 1 2 2
y2
=–4, Then
= 1. This is a hyperbola,
4
2
with a = 12, b = 2, and c = 16. Foci:
1y¿ 2 2
1x¿ 2 2
1 -1, ; 162. Center (–1, 0), so
= 1.
4
2
y
8
8
5 12 12
,
b
2
2
For #37–40, use the discriminant B2-4AC to determine the
A - C
type of conic. Then use the relationship of cot 2a =
B
to determine the angle of rotation.
p
37. B2-4AC=1>0, hyperbola; cot 2Å=0, so a = .
4
x¿ + y¿
x¿ - y¿
ba
b = 8,
Translating, a
12
12
2
2
1x¿ 2
1 y¿ 2
= 1, y¿ = ; 21x¿ 2 2 - 16
16
16
x
[–9.4, 9.4] by [–6.2, 6.2]
30. –4(x-3.25)=(y-1)2, a parabola. The vertex is
(h, k)=(3.25, 1), so –4x=(y)2
2
38. B -4AC=9>0, hyperbola; cot 2Å=0, so a =
Translating, 3 a
y'
5
1y¿ 2 2
10
2
-
1x¿ 2 2
10
x¿ - y¿
12
ba
x¿ + y¿
12
b + 15 = 0,
= 1, y¿ = ; 21x¿ 2 2 + 10
x'
[–9.4, 9.4] by [–6.2, 6.2]
p
.
4
Chapter 8
336
Analytic Geometry in Two and Three Dimensions
39. B2-4AC=3-4(2)(1)=–5<0, ellipse;
p
1
, a = . Translating,
cot 2a =
6
13
p
p
p
p
x = x¿ cos - y¿ sin , y = x¿ sin + y¿ cos , so the
6
6
6
6
5 1 x¿ 2 2
1y¿ 2 2
equation becomes
+
= 10,
2
2
2
2
1 x¿ 2
1 y¿ 2
+
= 1
4
20
[–28, 52] by [–15, 45]
1
cot 2Å= - , Å≠0.946≠54.22º
3
43. B2-4AC=16-4(1)(10)=–24<0; ellipse
44. B2-4AC=16-4(1)(0)=16>0; hyperbola
45. B2-4AC=36-4(9)(1)=0; parabola
46. B2-4AC=1-4(0)(3)=1>0; hyperbola
47. B2-4AC=16-4(8)(2)=–48<0; ellipse
[–9.4, 9.4] by [–6.2, 6.2]
1
40. B2-4AC=12-4(3)(1)=0, parabola; cot 2Å=
,
13
p
p
p
Å= . Translating, x=x cos
- y¿ sin ,
6
6
6
p
p
114
2
y=x sin
+ y¿ cos , so 4(x) =14, x¿ = ;
.
6
6
2
This is a degenerate form consisting of only two parallel
lines.
[–4.7, 4.7] by [–3.1, 3.1]
2
41. B -4AC=–176<0, ellipse. Use the quadratic
formula with a=9, b=–20x, and c=16x2-40. Then
b2-4ac=(–20x)2-4(9)(16x2-40)
=–176x2+1440=16(–11x2+90), and
20x ; 2161 -11x2 + 90 2
y =
18
=
10x ; 2 290 - 11x2
9
[–9.4, 9.4] by [–6.2, 6.2]
cot 2Å= -
7
, Å≠0.954≠54.65º
20
42. B2-4AC=4>0, hyperbola. Use the quadratic formula
with a=2, b=–6x+10, and c=4x2-3x-6. Then
b2-4ac=(–6x+10)2-4(2)(4x2-3x-6)
=4x2-96x+148=4(x2-24x+37), and
16x - 10 2 ; 24 1 x2 - 24x + 37 2
y =
4
2
13x - 52 ; 2x - 24x + 37
=
2
48. B2-4AC=144-4(3)(4)=96>0; hyperbola
49. B2-4AC=0-4(1)(–3)=12>0; hyperbola
50. B2-4AC=16-4(5)(3)=–44<0; ellipse
51. B2-4AC=4-4(4)(1)=–12<0; ellipse
52. B2-4AC=16-4(6)(9)=–200<0; ellipse
53. In the new coordinate system, the center (x, y)=
(0, 0), the vertices occur at (—3, 0) and the foci are
p
p
located at 1 ;312, 0 2. We use x = x¿ cos - y¿ sin ,
4
4
p
p
y=x sin +y cos to translate “back.” Under the
4
4
“old” coordinate system, the center (x, y)=(0, 0), the
3 12 3 12
3 12 312
vertices occured at a
,
b and a ,b.
2
2
2
2
and the foci are located at (3, 3) and (–3, –3).
54. (a) Reversing the translation and rotation of the
parabola, we see that the vertex in the (x, y)
21
coordinate system is V(0, 0), with h=
and
115,
315
. This means that the vertex of the
k =
10
parabola in the (x, y) coordinate system is
21
315
(x+h, y+k)= a 0 +
b . Since
,0 +
10
115
2
1
cos Å=
and sin Å=
, rotating “back” into the
15
15
(x, y) coordinate system gives
(x, y)=(x cos Å-y sin Å, x sin Å+y cos Å)
21 # 1
3 15 # 2 21 # 2
315 # 1
a
,
+
≤
10
15
10 15
15 15
15 15
=(3.6, 8.7).
(b) See (a).
55. Answers will vary. One possible answer: Using the
geometric relationships illustrated, it is clear that
p
x=x cos Å-y cos a - a b = x¿ cos a - y¿ sin a
2
p
and that y=x cos a - a b + y¿ cos a
2
=x sin Å+y cos Å.
Section 8.4
56. x=x cos Å+y sin Å
y=–x sin Å+y cos Å
x cos Å=x cos2 Å+y sin Å cos Å
y sin Å=–x sin2 Å+y cos Å sin Å
x cos Å-y sin Å=x cos2 Å+y sin Å cos Å+
x sin2 Å-y sin Å cos Å
x cos Å-y sin Å=x cos2 Å+x sin2 Å
x cos Å-y sin Å=x
Similarly,
x=cos Å+y sin Å
y=–x sin Å+y cos Å
x sin Å=x cos Å sin Å+y sin2 Å
y cos Å=–x sin Å cos Å+y cos2 Å
x sin Å+y cos Å=x cos Å sin Å+y sin2 Å
-x sin Å cos Å+y cos2 Å
x sin Å+y cos Å=y(sin2 Å+cos2 Å)
x sin Å+y cos Å=y
57. True. The Bxy term is missing and so the rotation angle Å
is zero.
58. True. Because the x2 and y2 terms have the same
coefficient (namely 1), completing the square to put the
equation in standard form will produce the same denominator under (y-k)2 as under (x-h)2.
59. Eliminating the cross-product term requires rotation, not
translation. The answer is B.
60. Moving the center or vertex to the origin is done through
translation, not rotation. The answer is C.
61. Completing the square twice, and dividing to obtain 1 on
the right, turns the equation into
1x - 122
1y + 22 2
+
= 1
16
9
The vertices lie 4 units to the left and right of center
(1, –2). The answer is A.
62. The equation is equivalent to y=4/x. The answer is E.
63. (a) The rotated axes pass through the old origin with
slopes of —1, so the equations are y=_x.
(b) The location of 1 x–, y– 2 = 10, 0 2 in the xy system can
be found by reversing the transformations. In the x¿y¿
system, 1x–, y– 2 = 1 0, 02 has coordinates
21 3 15
,
b .The coordinates of this point
1h, k2 = a
15 10
in the xy system are then given by the second set of
1
2
, sin Å =
:
rotation formulas; with cos Å =
15
15
1
3 15 2
18
21
a
b a
b =
x =
10
5
15 15
15
2
3 15 1
87
21
a
b +
a
b =
y =
10 15
10
15 15
The x–y– axes pass through the point (x, y)
2> 15
18 87
= 2
= a , b with slopes of
5 10
1> 15
1
and its negative reciprocal, - . Using this information
z
to write linear equations in point-slope form, and then
converting to slope-intercept form, we obtain
3
y = 2x +
2
1
21
y = - x +
2
2
Translation and Rotation of Axes
337
64. (a) If the translation on x=x-h and y=y-k is
applied to the equation, we have:
A(x)2+Bxy+C(y)2+Dx+Ey+F=0,
so A(x-h)2+B(x-h)(y-k)+C(y-k)2
+D(x-h)+E(y-k)+F=0, which becomes
Ax2 +Bxy+Cy2+(D-Bk-2Ah)x+
(E-2ck-Bh)y+(Ah2+Ck2-Ek-Dh)
+Bhk+F=0
The discriminants are exactly the same; the coefficients of the x2, xy, and y2 terms do not change (no
sign change).
(b) If the equation is multiplied by some constant k, we
have kAx2+kBxy+kCy2+kD+kE+kF=0,
so the discriminant of the new equation becomes
(kB)2-4(kA)(kC)=k2B2-4k2AC=k2(B24AC). Since k2>0 for k Z 0, no sign change occurs.
65. First, consider the linear terms:
Dx+Ey=D(xcos Å-ysin Å)
+E(xsin Å+ ycos Å)
=(D cos Å+E sin Å)x
+(E cos Å-D sin Å)y
This shows that Dx+Ey=Dx+Ey, where
D=D cos Å+E sin Å and E=E cos Å-D sin Å.
Now, consider the quadratic terms:
Ax2+Bxy+Cy2=A(xcos Å-ysin Å)2+
B(xcos Å-ysin Å)(xsin Å+ycos Å)+
C(xsin Å+ycos Å)2
=A(x2 cos2 Å-2xycos Å sin Å+y2 sin2 Å)
+B(x2 cos Å sin Å+xycos2 Å-xysin2 Å
-y2 sin Å cos Å)+C(x2 sin2 Å+2xysin Å cos Å
+y2 cos2 Å)
=(A cos2Å+B cos Å sin Å+C sin2 Å)x2
+[B(cos2 Å-sin2 Å)
+2(C-A)(sin Å cos Å)]xy
+(C cos2 Å-B cos Å sin Å+A sin2 Å)y2
=(A cos2 Å+B cos Å sin Å+C sin2 Å)x2
+[B cos 2Å+(C-A) sin 2Å]xy
+(C cos2Å-B cos Å sin Å+A sin2 Å)y2
This shows that
Ax2+Bxy+Cy2=Ax2+Bxy+Cy2, where
A=A cos2 Å+B cos Å sin Å+C sin2 Å,
B=B cos 2Å+(C-A) sin 2Å, and C=C cos2ÅB cos Å sin Å+A sin2 Å.
The results above imply that if the formulas for A, B,
C, D, and F are applied, then
Ax2+Bxy+Cy2+Dx+Ey+F=0 is
equivalent to Ax2+Bxy+Cy2+Dx+Ey+F=0.
Therefore, the formulas are correct.
66. This equation is simply a special case of the equation we
have used throughout the chapter, where B=0. The
discriminant B2-4AC, then, reduces simply to –4AC.
If –4AC>0, we have a hyperbola; –4AC=0, we have
a parabola; –4AC<0, we have a ellipse. More simply: a
hyperbola if AC<0; a parabola if AC=0; an ellipse if
AC>0.
338
Chapter 8
Analytic Geometry in Two and Three Dimensions
67. Making the substitutions x=xcos Å-ysin Å and
y=xsin Å+ycos Å, we find that:
Bxy=(B cos2 Å-B sin2 Å+2C sin Å cos Å
-2A sin Å cos Å)xy
Ax2=(A cos2 Å+B sin Å cos Å+C sin2 Å)(x)2
Cy2=(A sin2 Å+C cos2 Å-B cos Å sin Å)(y)2
B2-4AC=(B cos (2Å)-(A-C)sin (2Å))2
-4(A cos2 Å+B sin Å cos Å+C sin2 Å)
(A sin2 Å-B sin Å cos Å+C cos2 Å)
1
1
= B2 cos (4Å)+ B2+BC sin (4Å)-BA sin (4Å)
2
2
1 2
1 2
+ C - C cos (4Å)-CA+CA cos (4Å)
2
2
1 2
1
1
1
+ A - A2 cos (4Å)-4 a A cos (2Å)+ A
2
2
2
2
1
1
1
+ B sin (2Å)+ C- C cos (2Å) ≤
2
2
2
1
1
1
1
a A - A cos 1 2Å2 + C cos 12Å 2 + C
2
2
2
2
1
- B sin 12Å2 ≤
2
1
1
= B2 cos (4Å)+ B2+BC sin (4Å)-BA sin (4Å)
2
2
1
1
1
+ C2- C2 cos (4Å)-CA+CA cos (4Å)+ A2
2
2
2
1
- A2 cos (4Å)-BC sin (4Å)+BA sin (4Å)-3AC
2
1 2
1
1
1
1
- C - A2+ A2 cos (4Å)+ B2- B2 cos (4Å)
2
2
2
2
2
1 2
+ C cos (4Å)-AC cos (4Å)
2
=B2-4AC.
68. When the rotation is made to the (x, y) coordinate
system, the coefficients A, B, C, D, E, and F become:
A
B
A= (1+cos (2Å)+ sin (2Å)
2
2
C
+ (1-cos (2Å))
2
B=B cos (2Å)-(A-C) sin (2Å)
A
B
C= (1-cos (2Å))- sin (2Å)
2
2
C
+ (cos (2Å)+1)
2
D=D cos Å+E sin Å
E=–D sin Å+E cos Å
F=F
(a) Since F=F, F is invariant under rotation.
A
[1+cos (2Å)+1-cos (2Å)]
2
C
B
+ [sin (2Å)-sin (2Å)]+ [1-cos (2Å)
2
2
+cos (2Å)+1]=A+C, A+C is invariant
under rotation.
69. Intersecting lines: x2 + xy = 0 can be rewritten as x = 0
(the y-axis) and y = -x
[–4.7, 4.7] by [–3.1, 3.1]
A plane containing the axis of a cone intersects the cone.
Parallel lines: x2 = 4 can be rewritten as x = ;2 (a pair
of vertical lines)
[–4.7, 4.7] by [–3.1, 3.1]
A degenerate cone is created by a generator that is
parallel to the axis, producing a cylinder. A plane parallel
to a generator of the cylinder intersects the cylinder and
its interior.
One line: y2 = 0 can be rewritten as y = 0 (the x-axis).
[–4.7, 4.7] by [–3.1, 3.1]
A plane containing a generator of a cone intersects the
cone.
No graph: x2 = -1
[–4.7, 4.7] by [–3.1, 3.1]
A plane parallel to a generator of a cylinder fails to
intersect the cylinder.
Circle: x2 + y2 = 9
(b) Since A+C=
(c) Since D2+E2=(D cos Å+E sin Å)2
+(–D sin Å+E cos Å)2
=D2 cos2 Å+2DE cos Å sin Å+E2 sin2 Å
+D2 sin2 Å-2DE cos Å sin Å+E2 cos2 Å
=D2 (cos2 Å+sin2 Å)+E2 (sin2 Å+cos2 Å)
=D2+E2, D2+E2 is invariant under rotation.
[–4.7, 4.7] by [–3.1, 3.1]
A plane perpendicular to the axis of a cone intersects the
cone but not its vertex.
Section 8.5
Point: x2 + y2 = 0, the point (0, 0).
Polar Equations of Conics
Section 8.5 Exercises
1. r=
2
— a parabola.
1 - cos ¨
[–4.7, 4.7] by [–3.1, 3.1]
A plane perpendicular to the axis of a cone intersects the
vertex of the cone.
No graph: x2 + y2 = -1
[–10, 20] by [–10, 10]
2. r=
20
5
=
— a hyperbola.
4
+
5 cos ¨
5
1 + a b cos ¨
4
[–4.7, 4.7] by [–3.1, 3.1]
A degenerate cone is created by a generator that is perpendicular to the axis, producing a plane. A second plane
perpendicular to the axis of this degenerate cone fails to
intersect it.
■ Section 8.5 Polar Equations of Conics
Exploration 1
[–20, 40] by [–20, 20]
12
12
5
=
3. r=
— an ellipse.
5 + 3 sin ¨
3
1 + a b sin ¨
5
For e=0.7 and e=0.8, an ellipse; for e=1, a parabola; for
e=1.5 and e=3, a hyperbola.
[–7.5, 7.5] by [–7, 3]
[–12, 24] by [–12, 12]
4. r=
2
— a parabola.
1 + sin ¨
The five graphs all have a common focus, the pole (0, 0),
and a common directrix, the line x=3. As the eccentricity
e increases, the graphs move away from the focus and
toward the directrix.
Quick Review 8.5
1. r=–3
2. r=2
5p
7p
3. ¨=
or 6
6
4. ¨= -
5p
p
or
3
3
[–6, 6] by [–6, 2]
7
7
3
=
5. r=
— a hyperbola.
3 - 7 sin ¨
7
1 - a b sin ¨
3
5. h=0, k=0, 4p=16, so p=4
The focus is (0, 4) and the directrix is y=–4.
6. h=0, k=0, 4p=–12, so p=–3
The focus is (–3, 0) and the directrix is x=3.
7. a=3, b=2, c= 15; Foci: 1 ; 15, 02 ; Vertices: (—3, 0)
8. a=5, b=3, c=4; Foci: (0, —4); Vertices: (0, —5)
9. a=4, b=3, c=5; Foci: (—5, 0); Vertices: (—4, 0)
10. a=6, b=2, c=4 12; Foci: 10, ; 4 122 ;
Vertices: (0, —6)
[–5, 5] by [–4, 2]
339
340
Chapter 8
Analytic Geometry in Two and Three Dimensions
10
10
3
=
6. r=
— an ellipse.
3
2 cos ¨
2
1 - a b cos ¨
3
[–4, 11] by [–5, 5]
7. Parabola with e=1 and directrix x=2.
8. Hyperbola with e=2 and directrix x=3.
9. Divide numerator and denominator by 2.
Parabola with e=1 and directrix y= -
5
= -2.5.
2
10. Divide numerator and denominator by 4.
1
Ellipse with e= = 0.25 and directrix x=–2.
4
11. Divide numerator and denominator by 6.
5
Ellipse with e= and directrix y=4.
6
12. Divide numerator and denominator by 2.
7
Hyperbola with e= = 3.5 and directrix y=–6.
2
13. Divide numerator and denominator by 5.
2
Ellipse with e= = 0.4 and directrix x=3.
5
14. Divide numerator and denominator by 2.
5
Hyperbola with e= = 2.5 and directrix y=4.
2
15. (b) [–15, 5] by [–10, 10]
16. (d) [–5, 5] by [–3, 3]
17. (f) [–5, 5] by [–3, 3]
18. (e) [–5, 5] by [–3, 5]
19. (c) [–10, 10] by [–5, 10]
20. (a) [–3, 3] by [–6, 6]
For #21–28, one must solve two equations a=
ep
and
1 + e
ep
for e and p (given two constants a and b). The
1 - e
2ab
b - a
general solution to this is e=
and p=
.
b + a
b - a
b=
21. The directrix must be x=p 7 0, since the right majoraxis endpoint is closer to (0, 0) than the left one, so the
ep
equation has the form r=
. Then
1 + e cos ¨
ep
ep
ep
1.5=
and 6=
=
1 + e cos 0
1 + e
1 + e cos p
ep
3
=
(so a=1.5 and b=6). Therefore e= =0.6
1 - e
5
2.4
12
=
.
and p=4, so r=
1 + 13>5 2 cos ¨
5 + 3 cos ¨
22. The directrix must be x=–p 6 0, since the left majoraxis endpoint is closer to (0, 0) than the right one, so the
ep
equation has the form r=
. Then
1 - e cos ¨
ep
ep
ep
1.5=
=
and 1=
1 - e cos 0
1 - e
1 - e cos p
ep
1
=
(so a=1 and b=1.5). Therefore e= =0.2
1 + e
5
and p=6 (the directrix is x=–6) , so
1.2
6
r=
=
.
1 - 11>5 2 cos ¨
5 - cos ¨
23. The directrix must be y=p 7 0, since the upper majoraxis endpoint is closer to (0, 0) than the lower one, so the
ep
equation has the form r=
. Then
1 + e cos ¨
ep
ep
ep
1=
and 3=
=
1 + e sin1p>2 2
1 + e
1 + e sin13p>22
ep
1
=
(so a=1 and b=3). Therefore e= =0.5
1 - e
2
1.5
3
and p=3, so r=
=
.
1 + 11>22 sin ¨
2 + sin ¨
24. The directrix must be y=–p 6 0, since the lower majoraxis endpoint is closer to (0, 0) than the upper one, so the
ep
equation has the form r=
. Then
1 - e cos ¨
ep
ep
ep
3
3=
and =
=
1 - e sin 1p>22
1 - e
4
1 - e sin1 3p>2 2
ep
3
3
=
(so a= and b=3). Therefore e= =0.6
1 + e
4
5
and p=2 (the directrix is y=–2) , so
1.2
6
r=
=
.
1 - 13>5 2 sin ¨
5 - 3 sin ¨
25. The directrix must be x=p 7 0, since both transverseaxis endpoints have positive x coordinates, so the
ep
equation has the form r=
. Then
1 + e cos ¨
ep
ep
ep
=
3=
and –15=
1 + e cos 0
1 + e
1 + e cos p
ep
3
=
(so a=3 and b=–15). Therefore e=
1 - e
2
7.5
15
=
.
=1.5 and p=5, so r=
1 + 13>22 cos ¨
2 + 3 cos ¨
26. The directrix must be x=–p 6 0, since both transverseaxis endpoints have negative x coordinates, so the
ep
equation has the form r=
. Then
1 - e cos ¨
ep
ep
ep
–3=
and 1.5=
=
1 - e cos 0
1 - e
1 - e cos p
ep
=
(so a=1.5 and b=–3). Therefore e=3
1 + e
6
and p=2 (the directrix is x=–2), so r=
.
1 - 3 cos ¨
Section 8.5
27. The directrix must be y=p 7 0, since both transverseaxis endpoints have positive y coordinates, so the
ep
equation has the form r=
. Then 2.4
1 + e cos ¨
ep
ep
ep
=
and –12=
=
1 + e sin1p>2 2
1 + e
1 + e sin13p>2 2
ep
3
=
(so a=2.4 and b=–12). Therefore e=
1 - e
2
6
12
=1.5 and p=4, so r=
=
.
1 + 13>2 2 sin ¨
2 + 3 sin ¨
28. The directrix must be y=–p 6 0, since both transverseaxis endpoints have negative y coordinates, so the
ep
equation has the form r=
. Then
1 - e cos ¨
ep
ep
ep
–6=
and 2=
=
1 - e sin1p>2 2
1 - e
1 - e sin13p>2 2
ep
=
(so a=2 and b=–6). Therefore e=2
1 + e
6
and p=3 (the directrix is y=–3), so r=
.
1 - 2 sin ¨
Polar Equations of Conics
341
3p
p
b and a 1,
b , so 2a=12, a=6.
2
2
5
c=ae= # 6 = 5, so
6
are a 11,
b= 2a2 - c2 = 236 - 25 = 211.
[–11, 10] by [–2, 12]
5
e= , a=6, b= 111, c=5
6
24
6
1
=
, so e= . The vertices
4 + 2 sin ¨
1 + 11>2 2 sin ¨
2
p
3p
are (4, ) and (12,
), so 2a=16, a=8. c=ae
2
2
1
= # 8 = 4, so b= 2a2 - c2 = 264 - 16 = 4 23.
2
33. r=
29. The directrix must be x=p 7 0, so the equation has the
ep
ep
ep
form r=
. Then 0.75=
=
1 + e cos ¨
1 + e cos 0
1 + e
ep
ep
and 3=
=
(so a=0.75 and b=3).
1 + e cos p
1 - e
3
Therefore e= =0.6 and p=2, so
5
1.2
6
r=
=
.
1 + 13>52 cos ¨
5 + 3 cos ¨
30. Since this is a parabola, e=1, and with y=p 7 0 as the
p
directrix, the equation has the form r=
. Then
1 + sin ¨
p
p
1=
=
, p=2, and therefore
1 + sin1 p>2 2
1 + 1
2
r=
. Alternatively, for a parabola, the distance
1 + sin ¨
from the focus to the vertex is the same as the distance
from the vertex to the directrix (the same is true for all
points on the parabola). This distance is 1 unit, so we
again conclude that the directrix is y=2.
21
4.2
, so e=0.4. The vertices
=
5 - 2 cos ¨
1 - 0.4 cos ¨
are (7, 0) and (3, p), so 2a=10, a=5, c=ae
=(0.4)(5)=2, so b= 2a2 - c2 = 225 - 4 = 221.
31. r=
[–13, 14] by [–13, 5]
1
e= , a=8, b=413, c=4
2
16>5
16
3
34. r=
, so e= . The
=
5 + 3 cos ¨
1 + 1 3>5 2 cos ¨
5
vertices are (2, 0) and (8, p), so 2a=10, a=5, c=ae
3
=5 a b =3, so b= 2a2 - c2 = 225 - 9 = 4.
5
[–10, 5] by [–5, 5]
e=0.6, a=5, b=4, c=3
16>3
16
5
35. r=
, so e= . The
=
3 + 5 cos ¨
1 + 1 5>3 2 cos ¨
3
vertices are (2, 0) and (–8, p), so 2a=6, a=3, c=ae
5
= # 3 = 5 and b= 2c2 - a2 = 225 - 9 = 4.
3
[–6, 14] by [–7, 6]
e=0.4, a=5, b= 121, c=2
11>6
5
11
=
32. r=
, so e= . The vertices
6 - 5 sin ¨
1 - 1 5>6 2 sin ¨
6
[–3, 12] by [–5, 5]
5
e= , a=3, b=4, c=5
3
342
Chapter 8
Analytic Geometry in Two and Three Dimensions
12
p
, so e=5. The vertices are (–3, ) and
1 - 5 sin ¨
2
3p
1
1
5
a 2,
b , so 2a=1, a= . c=ae=5 # = and
2
2
2
2
25
1
2
16
b= 2c2 - a2 =
- =
= 16.
B4
4
2
36. r=
[–4, 4] by [–4, 0]
5
1
e=5, a= , b= 16, c=
2
2
4
2
1
so e= (an ellipse).
=
2 - sin ¨
1 - 1 1>2 2 sin ¨
2
p
4 3p
The vertices are a 4, b and a ,
b and the conic is
2
3 2
symmetric around x=0, so x=0 is the semi-major axis
1 8
16
8
4
and 2a= , so a= . c=ea= # = and
3
3
2 3
3
2
2
8
4
4
13
b= 2a2 - c2 =
a b - a b =
. The center
C 3
3
3
37. r=
(h, k)= a 0,
12
8
4
- b = a 0, b . The equation for the
3
3
3
The aphelion of Halley’s Comet is
18.09 1 1 - 0.972 2
r=
L 35.64 AU
1 - 0.97
42. Setting e=0.0461 and a=19.18,
19.18 1 1 - 0.04612 2
r=
1 + 0.0461 cos ¨
19.18 1 1 - 0.04612 2
Uranus’ perihelion is
L 18.30 AU
1 + 0.0461
2
19.18 11 - 0.0461 2
L 20.06 AU
Uranus’ aphelion is
1 - 0.0461
43. (a) The total radius of the orbit is
r=250+1740=1990 km=1,990,000 m. Then
v≠ 12,406,030≠1551 m/sec=1.551 km/sec.
(b) The circumference of one orbit is 2pr≠12503.5 km;
one orbit therefore takes about 8061 seconds, or
about 2 hr 14 min.
44. The total radius of the orbit is r=1000+2100=3100
miles. One mile is about 1.61 km, so r≠4991 km
= 4,991,000 m. Then v≠ 18,793,800≠2965 m/sec
=2.965 km/sec≠1.843 mi/sec.
45. True. For a circle, e=0. But when e=0, the equation
degenerates to r=0, which yields a single point, the pole.
46. True. For a parabola, e=1. But when e=1, the equation
degenerates to r=0, which yields a single point, the pole.
47. Conics are defined in terms of the ratio distance to focus :
distance to directrix. The answer is D.
ellipse is
4 2
4 2
ay - b
9ay - b
2
1x - 0 2
3
3
3x2
+
=
+ =1
2
2
64
16
8
413
a b
a
b
3
3
48. As the eccentricity increases beginning from zero, the
sequence of conics is circle (e=0), ellipse (e<1),
parabola (e=1), hyperbola (e>1). The answer is C.
6
, so e=2 (a hyperbola). The vertices are
1 + 2 cos ¨
(2, 0) and (–6, p) and the function is symmetric about
the x-axis, so the semi-major axis runs along x=0.
2a=4, a=2, so c=ea=2(2)=4 and
b= 2c2 - a2 = 216 - 4 = 2 13. The vertex
(h, k)=(4, 0). The equation of the hyperbola is
1x - 422
1y - 02 2
1x - 4 2 2
y2
=
= 1
2
2
4
12
2
1 2132
50. r = 1 + 2 cos ¨ is a limaçon curve. (See Section 6.5). The
answer is A.
38. r=
4
2
2
39. r=
, so e=1 and k= =2.
=
2 - 2 cos ¨
1 - cos ¨
e
Since k=2p, p=1 and 4p=4, the vertex (h, k)=
(–1, 0) and the parabola opens to the right, so the equation is y2=4(x+1).
12
4
4
=
, so e=1 and k= =4.
3 + 3 cos ¨
1 + cos ¨
e
Since k=2p, p=2 and 4p=8, the vertex (h, k)=
(2, 0) and the parabola opens to the left, so the equation
is y2=–8(x-2).
40. r=
41. Setting e=0.97 and a=18.09 AU,
18.09 1 1 - 0.972 2
r=
1 + 0.97 cos ¨
The perihelion of Halley’s Comet is
18.09 1 1 - 0.972 2
L 0.54 AU
r=
1 + 0.97
49. Conics written in polar form always have one focus at the
pole. The answer is B.
51. (a) When ¨=0, cos ¨=1, so 1+e cos ¨=1+e.
a 11 - e2 2
a11 - e2 2
Then
=
= a11 - e2
1 + e cos ¨
1 + e
Similarly, when ¨=∏, cos ¨=–1, so 1+e cos ¨=
a1 1 - e2 2
a1 1 - e2 2
1-e. Then
=
= a1 1 + e2
1 + e cos ¨
1 - e
(b) a(1-e)=a a 1 -
c
c
b =a-a # =a-c
a
a
c
c
a(1+e)=a a 1 + b =a+a # =a+c
a
a
(c) Planet
Perihelion (in Au)
Aphelion
Mercury
0.307
0.467
Venus
0.718
0.728
Earth
0.983
1.017
Mars
1.382
1.665
Jupiter
4.953
5.452
Saturn
9.020
10.090
(d) The difference is greatest for Saturn.
Section 8.5
52. e=0 yields a circle (degenerate ellipse); e=0.3
and e=0.7 yield ellipses; e=1.5 and e=3 yield hyperbolas. When e=1, we expect to obtain a parabola. But
a has no meaning for a parabola, because a is the centerto-vertex distance and a parabola has no center.
The equation
a1 1 - e2 2
yields no parabolas. When e=1, r=0.
r =
1 + e cos ¨
53. If r 6 0, then the point P can be expressed as the point
(r, ¨+p) then PF=r and PD=k-r cos ¨.
PF=ePD
r=e(k-r cos ¨)
ke
r=
1 + e cos ¨
Recall that P (r, ¨) can also be expressed as (–r, ¨-p)
then PD=–r and PF=–r cos (¨-p)-k
PD=ePF
–r=e[–r cos (¨-p)-k]
–r=–er cos (¨-p)-ek
–r=er cos ¨-ek
–r-er cos ¨=–ek
ke
r=
1 + e cos ¨
54. (a)
P(r, )
5
D
r
F
r cos 5
x
Conic section
Directrix
x = –k
PF=r and PD=k+r cos ¨, so PF=e PD
becomes
r=e(k+r cos ¨)
r-er cos ¨=ek
ke
r=
1 - e cos ¨
(b)
y
5
Directrix
y=k
D
k – r sin P(r, )
r
F
r sin 5
(c)
343
y
Conic section
5
r
r sin Focus at pole F
Directrix
y = –k
P(r, )
x
5
r sin + k
D
PF=r and PD=k+r sin ¨, so PF=e PD
becomes
r=e(k+r sin ¨)
r-er sin ¨=ek
ke
r=
1 - e sin ¨
16
. To transform
5 - 3 cos ¨
this to a Cartesian equation, rewrite the equation as
5r-3r cos ¨=16. Then use the substitutions
55. Consider the polar equation r=
r = 2x2 + y2 and x=r cos ¨ to obtain
5 2x2 + y2 - 3x = 16.
5 2x2 + y2 = 3x + 16; 25(x2+y2)=9x2+96x+256
25x2+25y2=9x2+96x+256
16x2-96x+25y2=256;
Completing the square on the x term gives
16(x2-6x+9)+25y2=256+144
16(x-3)2+25y2=400;
1x - 3 2 2
y2
The Cartesian equation is
+
= 1.
25
16
y
k + r cos Polar Equations of Conics
x
Conic section
PF=r and PD=k-r sin ¨, so PF=e PD
becomes
r=e(k-r sin ¨)
r+er sin ¨=ke
ke
r=
1 + e sin ¨
56. The focal width of a conic is the length of a chord through
a focus and perpendicular to the focal axis. If the conic is
ke
given by r=
, the endpoints of the chord
1 + e cos u
p
3p
occur when u =
and u =
. Thus, the points are
2
2
p
3p
a ke, b and a ke,
b and the length of the chord is
2
2
ke+ke=2ke.
The focal width of a conic is 2ke.
57. Apply the formula e # PD=PF to a hyperbola with one
focus at the pole and directrix x=–k, letting P be the
vertex closest to the pole. Then a+k=c+PD and
c
PF=c-a. Using e= , we have:
a
e # PD=PF
e(a+k-c)=c-a
e(a+k-ae)=ae-a
ae+ke-ae2=ae-a
ke-ae2=-a
ke=ae2-a
ke=a(e2-1)
ke
Thus, the equation r=
1 - e cos u
a 1e2 - 1 2
becomes r=
.
1 - e cos u
344
Chapter 8
Analytic Geometry in Two and Three Dimensions
58. (a) Let P(x, y) be a point on the ellipse. The horizontal
distance from P to the point Q(a2/c, y) on line L is
PQ=a2/c-x. The distance to the focus (c, 0) is
PF= 21x - c2 2 + y2 = 2x2 - 2cx + c2 + y2. To
confirm that PF/PQ=c/a, cross-multiply to get
a PF=c PQ; we need to confirm that
Section 8.6 Exercises
1.
z
8
a2x2 - 2cx + c2 + y2=a2-cx. Square both
sides: a2(x2-2cx+c2+y2)=a4-2a2cx+c2x2.
Substitute a2-b2 for c2, multiply out both sides, and
cancel out terms, leaving a2y2-a2b2=–b2x2. Since
P is on the ellipse, x2/a2+y2/b2=1, or equivalent
b2x2+a2y2=a2b2; this confirms the equality.
(b) According to the polar definition, the eccentricity is
the ratio PF/PQ, which we found to be c/a in (a).
y
8
(3, 4, 2)
8
x
a
=a2/c and ae=c; the disc>a
tance from F to L is a2/c-c=a/e-ea as desired.
(c) Since e=c/a, a/e=
59. (a) Let P(x, y) be a point on the hyperbola. The horizontal distance from P to the point Q(a2/c, y) on line L
is PQ=|a2/c-x|. The distance to the focus (c, 0) is
2.
PF= 21x - c2 2 + y2 = 2x2 - 2cx + c2 + y2. To
confirm that PF/PQ=c/a, cross-multiply to get
a PF=c PQ; we need to confirm that
z
8
a2x2 - 2cx + c2 + y2=|a2-cx|. Square both
sides: a2(x2-2cx+c2+y2)=a4-2a2cx+c2x2.
Substitute a2+b2 for c2, multiply out both sides, and
cancel out terms, leaving a2y2+a2b2=b2x2. Since P
is on the hyperbola, x2/a2-y2/b2=1, or equivalent
b2x2-a2y2=a2b2; this confirms the equality.
(2, –3, 6)
8
(b) According to the polar definition, the eccentricity is
the ratio PF/PQ, which we found to be c/a in (a).
8
x
a
=a2/c and ae=c; the
c>a
a2
a
distance from F to L is c- =ea- as desired.
c
e
(c) Since e=c/a, a/e=
3.
z
■ Section 8.6 Three-Dimensional Cartesian
Coordinates
Quick Review 8.6
y
1. 2 1x - 22 2 + 1y + 32 2
x + 2 y - 3
,
b
2. a
2
2
x
(1, –2, –4)
3. P lies on the circle of radius 5 centered at (2, –3)
4. |v|= 21 -4 2 2 + 15 2 2 = 141
4.
- 4, 5
v
-4
5
=
,
=
|v|
141
141 141
28, - 35
-7 # v
28 -35
=
,
6.
=
|v|
141
141 141
7. Circle of radius 5 centered at (–1, 5)
5.
8. A line of slope –2, passing through (2, –4)
2
z
2
9. (x+1) +(y-3) =4. Center: (–1, 3), radius: 2
10. –1-2, –4-5, =–3, –9
y
x
(–2, 3, –5)
y
Section 8.6
5. 213 - 1 -1 2 2 2 + 1 -4 - 2 2 2 + 16 - 5 2 2 = 153
Three-Dimensional Cartesian Coordinates
20.
z
6. 216 - 22 + 1 -3 - 1 -1 2 2 + 14 - 1 -82 2 = 2141
2
2
345
10
2
7. 21a - 12 2 + 1b - 1 -3 2 2 2 + 1 c - 2 2 2
= 21 a - 1 2 2 + 1b + 3 2 2 + 1c - 22 2
(0, 0, 6)
8. 21x - p2 2 + 1 y - q2 2 + 1 z - r2 2
9. a
11
3 - 1 -4 + 2 6 + 5
,
,
b = a 1, - 1, b
2
2
2
2
2 + 6 -1 - 3 -8 + 4
,
,
b = (4, –2, –2)
10. a
2
2
2
2x - 2 2y + 8 2z + 6
,
,
b = (x-1, y+4, z+3)
11. a
2
2
2
12. a
3a - a 3b - b 3c - c
,
,
b = (a, b, c)
2
2
2
2
2
(0, 3, 0)
y
6
5
x
21.
z
2
13. (x-5) +(y+1) +(z+2) =64
5
14. (x+1)2+(y-5)2+(z-8)2=5
15. (x-1)2+(y+3)2+(z-2)2=a
16. (x-p)2+(y-q)2+(z-r)2=36
17.
(0, –2, 0)
z
y
5
5
(6, 0, 0)
(0, 0, 3)
(0, 9, 0)
y
10
(9, 0, 0)
10
x
10
x
22.
z
5
18.
z
5
5
(3, 0, 0)
(0, 8, 0)
10
(0, 0, –4)
y
y
6
x
23. r+v=1, 0, –3+–3, 4, –5=–2, 4, –8
(8, 0, 0)
24. r-w=1, 0, –3-4, –3, 12=–3, 3, –15
10
x
25. v # w = –12-12-60=–84
19.
26. |w|= 242 + 1 -3 2 2 + 122 = 13
27. r # 1v + w2 = r # 1-3, 4, -5 + 4, - 3, 122
= 1, 0, -3 # 1, 1, 7 = 1 + 0 - 21 = - 20
z
6
28. r # v + r # w = (–3+0+15)+(4+0-36)=–20
(0, 0, 3)
29.
5
y
4, -3, 12
4
w
3 12
=
=
,- ,
2
2
2
|w|
13
13
13
24 + 1 -32 + 12
(3, 0, 0)
30. i # r = 1, 0, 0 # 1, 0, -3=1
6
x
32. 1r # v 2w = 1 1, 0, -3 # -3, 4, -524, - 3, 12
31. i # v, j # v, k # v=–3, 4, –5
=(–3+0+15)4, –3, 12=48, –36, 144
Chapter 8
346
Analytic Geometry in Two and Three Dimensions
33. The plane’s velocity relative to the air is
v1 = -200 cos 20 i + 200 sin 20 k
The air’s velocity relative to the ground is
v2 = -10 cos 45 i - 10 sin 45 j
Adding these two vectors and converting to decimal values rounded to two places produces the plane’s velocity
relative to the ground:
v = -195.01 i - 7.07 j + 68.40 k
1
1
46. Midpoint of AB: a - , 4, b . Direction vector:
2
2
1
1
1
5
- - 2, 4 - 1 -42, - 1 = - , 8, - ,
2
2
2
2
¡
OC=2, –4, 1, so a vector equation of the line is
5
1
r=2, –4, 1+t - , 8, 2
2
5
1
= 2 - t, -4 + 8t, 1 - t . This can be expressed in
2
2
5
1
parametric form: x=2- t, y=–4+8t, z=1- t.
2
2
34. The rocket’s velocity relative to the air is
v1 = 12,000 cos 80 i + 12,000 sin 80 k
The air’s velocity relative to the ground is
v2 = 8 cos 45 i + 8 sin 45 j
Adding these two vectors and converting to decimal values rounded to two places produces the rocket’s velocity
relative to the ground:
v = 2089.43 i + 5.66 j + 11,817.69 k
For #35–38, the vector form is r0 + tv with r0x0, y0, z0, and
the parametric form is x = x0 + ta, y = y0 + tb,
z = z0 + tc where v=a, b, c.
35. Vector form: r=2, –1, 5+t3, 2, –7; parametric form:
x = 2 + 3t, y = -1 + 2t, z = 5 - 7t
36. Vector form: r=–3, 8, –1+t–3, 5, 2; parametric
form: x = -3 - 3t, y = 8 + 5t, t = - 1 + 2t
37. Vector form: r=6, –9, 0+t1, 0, –4; parametric form:
x = 6 + t, y = -9, z = -4t
47. The length of AB=
2 10 - 1 -12 2 2 + 1 6 - 2 2 2 + 1 -3 - 4 2 2 = 166; the
length of BC=
2 12 - 0 2 2 + 1 -4 - 62 2 + 11 - 1 -3 2 2 2 = 2130; the
length of AC=
2 12 - 1 -12 2 2 + 1 -4 - 22 2 + 11 - 4 2 2 = 154. The
triangle ABC is scalene.
48. M=(1, 1, –1) (from #33). The midpoint of
3 3
AM= a 0, , b
2 2
49. (a)
z
5
38. Vector form: r=0, –1, 4+t0, 0, 1; parametric form:
x = 0, y = -1, z = 4 + t
39. Midpoint of BC: 1, 1, –1. Distance from A to midpoint
of BC:
2 1 -1 - 1 2 2 + 12 - 1 2 2 + 14 - 1 -1 2 2 2 = 130
5
5
x
40. 1-(–1), 1-2, –1-4=2, –1, –5
41. Direction vector: 0-(–1), 6-2, –3-4
¡
=1, 4, –7, OA=–1, 2, 4,
r=–1, 2, 4+t1, 4, –7
(b) the z-axis; a line through the origin in the direction k.
50. (a)
z
5
42. Direction vector: 2, –1, –5 (from #34). The vector
equation of the line is r=–1, 2, 4+t2, –1, –5.
43. Direction vector: 2-(–1), –4-2, 1-4
¡
5
=3, –6, –3, OA=–1, 2, 4, so a vector equation
of the line is r=–1, 2, 4+t3, –6, –3
=–1+3t, 2-6t, 4-3t. This can be expressed in
parametric form: x=–1+3t, y=2-6t, z=4-3t.
44. Direction vector: 2-0, –4-6, 1-(–3)
¡
=2, –10, 4, OB=0, 6, –3 so a vector equation of
the line is r=0, 6, –3+t2, –10, 4
=2t, 6-10t, –3+4t. This can be expressed in
parametric form: x=2t, y=6-10t, z=–3+4t.
1
5
45. Midpoint of AC: a , -1, b . Direction vector:
2
2
1
1
5
11
,
- 0, - 1 - 6, - 1 -3 2 = , -7,
2
2
2
2
y
y
5
x
(b) the intersection of the yz plane (at x=0) and xy plane
(at z=2); a line parallel to the y-axis through (0, 0, 2)
51. (a)
z
5
5
y
¡
OB=0, –6, –3, so a vector equation of the line is
11
1
r=0, 6, –3+t , - 7,
2
2
1
11
= t, 6 - 7t, - 3 +
t . This can be expressed in
2
2
1
11
parametric form: x= t, y=6-7t, z=–3+ t.
2
2
5
x
(b) the intersection of the xz plane (at y=0) and yz
plane (at x=–3); a line parallel to the z-axis
through (–3, 0, 0)
Section 8.6
52. (a)
Three-Dimensional Cartesian Coordinates 347
63. (a) Each cross-section is its own ellipse.
y2
z2
x = 0:
+
= 1, an ellipse centered at (0, 0)
4
16
z
5
5
(in the yz plane) of “width” 4 and “height” 8.
x2
z2
y = 0:
+
= 1, an ellipse centered at (0, 0)
9
16
(in the xz plane) of “width” 6 and “height” 8.
y2
x2
z = 0:
+
= 1, an ellipse centered at (0, 0)
9
4
(in the xy plane) of “width” 6 and “height” 4.
y
5
x
(b) the intersection of the xz plane (at y=1) and xy
plane (at z=3); a line through (0, 1, 3) parallel to
the x-axis
53. Direction vector: x2-x1, y2-y1, z2-z1,
¡
OP=x1, y2, z3, so a vector equation of the line is
r=x1+(x2-x1)t, y1+(y2-y1)t, z1+(z2-z1)t.
(b) Algebraically, z = 21 - x2 - y2 has only positive
values; 0 z 1 and the “bottom” half of the
sphere is never formed. The equation of the whole
sphere is x2+y2+z2=1.
(c)
z
54. Using the result from Exercise 49, the parametric
equations are x=x1+(x2-x1)t, y=y1+(y2-y1)t,
z=z1+(z2-z1)t.
55.
z
Q (x2, y2, z2)
R (x2, y2, z1)
P (x1, y1, z1)
2
y
x
By the Pythagorean Theorem,
d1 P, Q2 = 21d1 P, R2 2 2 + 1 d1R, Q2 2 2
= 41 21x1 - x2 2 2 + 1y1 - y2 2 2 2 2 + 1 ƒ z1 - z2 ƒ 2 2
= 21 x1 - x2 2 2 + 1 y1 - y2 2 2 + 1z1 - z2 2 2
56. Let u=x1, y1, z1. Then
u # u = x21 + y21 + z21
= 1 2x21 + y21 + z21 2 2
= @ [email protected] 2
57. True. This is the equation of a vertical elliptic cylinder.
The equation can be viewed as an equation in three variables, where the coefficient of z is zero.
58. False. Because the coefficient of t is always 0, the equations simplify to x = 1, y = 2, z = - 5; these represent
the point (1, 2,-5).
59. The general form for a first-degree equation in three variables is Ax + By + Cz + D = 0. The answer is B.
60. The equation for a plane is first-degree, or linear; there
are no squared terms. The answer is A.
61. The dot product of two vectors is a scalar. The answer is C.
62. The conversion to parametric form begins with
x = 2 + 1t, y = - 3 + 0t, z = 0 - 1t. The answer is E.
2
y
2
x
(d) A sphere is an ellipsoid in which all of the x=0,
y=0, and z=0 “slices” (i.e., the cross-sections of
the coordinate planes) are circles. Since a circle is a
degenerate ellipse, it follows that a sphere is a
degenerate ellipsoid.
64. (a) Since i points east and j points north, we determine
that the compass bearing ¨ is
22.63
0=90º-tan–1 a
b L 90 - 6.66 = 83.34°.
193.88
(Recall that tan ¨ refers to the x-axis (east) being
located at 0 °; if the y-axis (north) is 0 °, we must
adjust our calculations accordingly.)
(b) The speed along the ground is
2 1193.88 2 2 + 122.632 2 L 195.2 mph
(c) The tangent of the climb angle is the vertical speed
divided by the horizontal speed, so
125
¨ L tan-1
195.2
L 32.63°
(d) The overall speed is
2 1193.88 2 2 + 122.632 2 + 1 125 2 2≠231.8 mph
65. 2-3, –6+1, 1-4=–1, –5, –3
66. –2+6, 2-8, –12+1=4, –6, –11
67. iπj=1, 0, 0*0, 1, 0=0-0, 0-0, 1-0=
0, 0, 1=k
348
Chapter 8
Analytic Geometry in Two and Three Dimensions
68. u # (u*v)=
u1, u2, u3 # u2v3-u3v2, u3v1-u1v3, u1v2-u2v1
=(u1u2v3-u1u3v2)+(u2u3v1-u1u2v3)
+(u1u3v2-u2u3v1)
=0
v # (u*v)=
v1, v2, v3 # u2v3-u3v2, u3v1-u1v3, u1v2-u2v1
=(u2v1v3-u3v1v2)+(u3v1v2-u1v2v3)
+(u1v2v3-u2v1v3)
=0
So the angles between u and u*v, and v and u*v,
both have a cosine of zero by the theorem in Section 6.2.
It follows that the angles both measure 90 ° .
4. h=0, k=–2, 4p=16, so p=4.
Vertex: (0, –2), focus: (4, –2), directrix: x=–4,
focal width: 16
■ Chapter 8 Review
5. Ellipse. Center (0, 0). Vertices: (0, ;212). Foci: (0, ; 13)
since c= 18 - 5 = 13.
1. h=0, k=0, 4p=12, so p=3.
Vertex: (0, 0), focus: (3, 0), directrix: x=–3,
focal width: 12
y
10
8
x
y
7
y
10
7
9
x
x
6. Hyperbola. Center: (0, 0). Vertices: (0, ;4).
Foci: (0, ; 165) since c= 116 + 49 = 165.
2. h=0, k=0, 4p=–8, so p=–2.
Vertex: (0, 0), focus: (0, –2), directrix: y=2,
focal width: 8
y
10
y
2
10
10
x
x
7. Hyperbola. Center: (0, 0). Vertices: ( ;5, 0),
c= 2a2 + b2 = 225 + 36 = 261, so the foci are:
( ; 161, 0)
3. h=–2, k=1, 4p=–4, so p=–1.
Vertex: (–2, 1), focus: (–2, 0), directrix: y=2,
focal width: 4
y
10
y
2
8
x
10
x
Chapter 8
8. Hyperbola. Center: (0, 0). Vertices: ( ; 7, 0),
c= 2a + 6 = 249 + 9 = 258, so the foci are:
( ; 158, 0)
2
2
12. Ellipse. Center: (–6, 0). Vertices: (–6, ;6)
c= 2a2 - b2 = 236 - 20 = 4, so the foci are:
(–6, ; 4)
y
y
6
10
15
x
9. Hyperbola. Center: (–3, 5). Vertices: (–3 ; 312, 5),
c= 2a2 + b2 = 218 + 28 = 246, so the foci are:
(–3 ; 146, 5)
y
5
x
13. (b)
14. (g)
15. (h)
16. (e)
20
17. (f)
18. (d)
19. (c)
20. (a)
10
x
21. B2-4AC=0-4(1)(0)=0,
parabola (x2-6x+9)=y+3+9,
so (x-3)2=y+12
y
10. Hyperbola. Center: (7, 3). Vertices: (7, 3 ; 3)=(7, 0)
40
and (7, 6), c= 2a2 + b2 = 29 + 12 = 221, so the
foci are: (7, 3 ; 121)
y
13
10
17
x
x
22. B2-4AC=0-4(1)(3)=–12 6 0,
ellipse (x2+4x+4)+3y2=5+4,
1x + 22 2
y2
so
+
= 1
9
3
11. Ellipse. Center: (2, –1). Vertices: (2 ; 4, –1)=(6, –1)
and (–2, –1), c= 2a2 - b2 = 216 - 7 = 3, so the
foci are: (2 ; 3, –1)=(5, –1) and (–1, –1)
y
4
y
4
2
6
x
Review
x
349
Chapter 8
350
Analytic Geometry in Two and Three Dimensions
23. B2-4AC=0-4(1)(–1)=4 7 0,
hyperbola (x2-2x+1)-(y2-4y+4)
=1-4+6
1x - 122
1y - 22 2
= 1
3
3
27. B2-4AC=0-4(2)(–3)=24 7 0,
hyperbola 2(x2-6x+9)-3(y2+8y+16)
=18-48-60, so
1y + 4 2 2
1x - 32 2
= 1
30
45
y
y
7
10
–10
6
15
x
x
–15
2
24. B -4AC=0-4(1)(0)=0,
parabola (x2+2x+1)=–4y+7+1,
so (x+1)2=–4(y-2)
y
2
28. B2-4AC=0-4(12)(–4)=192>0,
hyperbola 12(x2-6x+9)-4(y2+4y+4)
=108-16-44, so
1x - 3 2 2
1y + 22 2
= 1
4
12
y
5
x
1
–2
25. B2-4AC=0-4(1)(0)=0,
parabola (y2-4y+4)=6x+13+4,
17
so (y-2)2=6 a x +
b
6
y
8
x
–5
29. By definition, every point P(x, y) that lies on the parabola
is equidistant from the focus to the directrix. The distance
between the focus and point P is:
21x - 0 2 2 + 1 y - p2 2 = 2x2 + 1y - p2 2, while the
distance between the point P and the line y=–p is:
11
10
x
26. B2-4AC=0-4(3)(0)=0,
parabola 3(x2-2x+1)=4y+9+3,
4
so (x-1)2= (y+3)
3
2 1x - x2 2 + 1y + p2 2 = 2 1 y + p2 2. Setting these
equal:
2x2 + 1y - p2 2=y+p
x2+(y-p)2=(y+p)2
x2+y2-2py+p2=y2+2py+p2
x2=4py
30. Let the point P1 x, y2 satisfy y2 = 4px. Then we have
y2 = 4px
x2 - 2px + p2 + y2 = x2 + 2px + p2
1x - p2 2 + y2 = 1x + p2 2 + 0
1x - p2 2 + 1y - 02 2 = 1 x - 1 -p2 2 2 + 1y - y2 2
2 1x - p2 2 + 1y - 02 2 = 2 1x - 1 -p2 2 2 + 1y - y2 2
y
9
5
x
distance from P1x, y2 to (p, 0)=distance from P1x, y2
to x = -p
Because P1x, y2 is equidistant from the point (p, 0) and
the line x = -p, by the definition of a parabola,
y2 = 4px is the equation of a parabola with focus (p, 0)
and directrix x = -p.
Chapter 8
31. Use the quadratic formula with a=6, b=–8x-5, and
c=3x2-5x+20. Then b2-4ac=(–8x-5)2
-24(3x2-5x+20)=–8x2+200x-455, and
1
y= B 8x + 5 ; 2-8x2 + 200x - 455R – an ellipse
12
[0, 25] by [0, 17]
32. Use the quadratic formula with a=6, b=–8x-5, and
c=10x2+8x-30. Then b2-4ac=(–8x-5)2
-24(10x2+8x-30)=–176x2-112x+745, and
1
y= B 8x + 5 ; 2-176x2 - 112x + 745R an ellipse
12
[–5, 5] by [–3, 3]
33. This is a linear equation in y:
(6-2x)y+(3x2-5x-10)=0. Subtract
3x2-5x-10 and divide by 6-2x, and we have
3x2 - 5x - 10
y=
— a hyperbola.
2x - 6
Review 351
35. Use the quadratic formula with a=–2, b=7x+20,
and c=–3x2-x-15. Then b2-4ac=(7x+20)2
+8(–3x2-x-15)=25x2+272x+280, and
1
y= B7x + 20 ; 225x2 + 272x + 280R a hyperbola.
4
[–24, 20] by [–20, 15]
36. Use the quadratic formula with a=–2, b=7x+3, and
c=–3x2-2x-10. Then b2-4ac=(7x+3)2
+8(–3x2-2x-10)=25x2+26x-71, and
1
y= B7x + 3 ; 225x2 + 26x - 71R a hyperbola.
4
[–15, 15] by [–15, 15]
37. h=0, k=0, p=2, and the parabola opens to the right
as y2=8x.
38. h=0, k=0, |4p|=12, and the parabola opens downward, so x2=–12y (p=–3).
39. h=–3, k=3, p=k-y=3-0=3 (since y=0 is
the directrix) the parabola opens upward, so
(x+3)2=12(y-3).
40. h=1, k=–2, p=2 (since the focal length is 2), and
the parabola opens to the left, so (y+2)2=–8(x-1).
[–8, 12] by [–5, 15]
34. Use the quadratic formula with a=–6, b=5x-17,
and c=10x+20. Then b2-4ac=(5x-17)2
+24(10x+20)=25x2+70x+769, and
y=
1
B 5x - 17 ; 225x2 + 70x + 769R a hyperbola.
12
41. h=0, k=0, c=12 and a=13, so b= 2a2 - c2
y2
x2
+
= 1
= 2169 - 144 = 5.
169
25
42. h=0, k=0, c=2 and a=6, so b= 2a2 - c2
y2
x2
+
= 1
= 236 - 4 = 412.
36
32
43. h=0, k=2, a=3, c=2-h (so c=2) and
b= 2a2 - c2= 29 - 4 = 25
1y - 2 2 2
x2
+
= 1
9
5
44. h=–3, k=–4, a=4, 0=–3_c, c=3,
[–15, 15] by [–10, 10]
b = 2a2 - c2 = 216 - 9 = 27, so
1x + 3 2 2
1y + 42 2
+
= 1
16
7
45. h=0, k=0, c=6, a=5,
b= 2c2 - a2 = 236 - 25 = 211, so
y2
x2
= 1
25
11
y2
b
x2
= 1
46. h=0, k=0, a=2, =2 (b=4), so
a
4
16
352
Chapter 8
Analytic Geometry in Two and Three Dimensions
47. h=2, k=1, a=3,
1x - 222
9
-
1y - 12 2
4
b
4
= (b= # 3 = 4), so
a
3
3
16
57.
= 1
48. h=–5, k=0, c-k=3 (c=3), a-k=2 (a=2),
b= 2c2 - a2 = 29 - 4 = 25, so
1x + 52 2
y2
= 1
4
5
y
y2
x2
x
49. =cos t and =sin t, so
+
= 1 — an ellipse.
5
2
25
4
y
y2
x
x2
+
= 1 — an ellipse.
50. =sin t and =cos t, so
4
6
16
36
51. x+2=cos t and y-4=sin t, so
(x+2)2+(y-4)2=1 — an ellipse (a circle).
y + 3
x - 5
52.
=cos t and
=sin t, so
3
3
1x - 522
1y + 32 2
+
= 1, or (x-5)2+(y+3)2=9
9
9
— an ellipse (a circle).
y
y2
x2
x
53. =sec t and =tan t, so
= 1 — a hyperbola.
3
5
9
25
y
y2
x
x2
= 1 — a hyperbola.
54. =sec t and =tan t, so
4
3
16
9
[–3, 3] by [–2, 2]
1
e= , so an ellipse. In polar coordinates the vertices are
3
(2, 0) and (1, p). Converting to Cartesian we have (2, 0)
3
1
1 3
and (–1, 0), so 2a=3, a= , c=ea= # = and
2
3 2
2
3
1
the center (h, k)= a 2 - , 0 b = a , 0 b (since it’s
2
2
symmetric about the polar x-axis). Solving for
b= 2a2 - c2 =
4ax 9
1 2
b
2
+
3 2
1 2
8
a b - a b =
= 22
B 2
2
B4
y2
= 1
2
58.
55.
[–2.3, 2.3] by [–2, 1]
1
e= , so an ellipse. In polar coordinates the vertices are
4
3 p
3p
a , b and a 1,
b . Converting to Cartesian we have
5 2
2
3
8
4
1
1 4
a 0, b and (0, –1), so 2a= , a= , c=ea= # = ,
5
5
5
4 5
5
3
4
-1
the center (h, k)= a 0, - b = a 0,
b (since it’s
5
5
5
symmetric on the y-axis). Solving for
[–8, 3] by [–10, 10]
Parabola with vertex at (2, 0), so h=2, k=0, e=1.
p
The graph crosses the y-axis, so (4, )=(0, 4) lies on the
2
parabola. Substituting (0, 4) into y2=4p(x-2) we have
16=4p(–2), p=–2. y2=–8(x-2).
56.
4 ¤
1 ¤
15
3
b¤=a¤-c¤= ¢ ≤ - ¢ ≤ = = .
5
5
25
5
1 2
25 a y + b
5
5x2
+
= 1
16
3
[–10, 10] by [–4, 10]
5
e=1, so a parabola. The vertex is (h, k)=(0, – ) and
2
the point (5, 0) lies on the curve. Substituting (5, 0) into
5
5
5
x2=4p a y + b , we have 25=4p a b , p= .
2
2
2
5
x2=10 a y + b
2
59.
[–8, 8] by [–11, 0]
7
e= , so a hyperbola. In polar coordinates the vertices
2
35 3p
p
are a -7, b and a ,
b . Converting to Cartesian we
2
9 2
14
- 35
28
have (0, –7) and a 0,
b , so 2a= , a= ,
9
9
9
Chapter 8
7 14
# = 49 the center (h, k)
2 9
9
- 35
14
-49
= a 0,
b = a 0,
b (since it’s symmetric
9
9
9
on the y-axis). Solving for
81 a y +
196
49 2
b
9
-
353
e=1, so this is a parabola. In polar coordinates, the ver1
tex is a , p b and the parabola crosses the y-axis at
2
p
a 1, b . Converting to Cartesian form, we have the
2
1
vertex (h, k)= a - , 0 b and a point on the parabola is
2
(0, 1). Since the parabola opens to the right,
1
y2=4p a x + b . Substituting (0, 1), we have 1=2p,
2
1
1
p= . y2=2 a x + b
2
2
c=ea=
b= 2c2 - a2 =
Review
49 2
14 2
21 15
715
b - a b =
=
B 9
9
9
3
a
9x2
= 1
245
60.
63. 2 13 - 1 -12 2 2 + 1 -2 - 02 2 + 1 -4 - 32 2
= 116 + 4 + 49 = 169
64. a
[–2, 6] by [–2, 3]
5
e= , so a hyperbola. In polar coordinates the vertices
2
15
are a , 0 b and (–5, p). Converting to Cartesian we
7
15
20
10
have a , 0 b and (5, 0), so 2a= , a= ,
7
7
7
25
5 # 10
c=ea=
the center (h, k)
=
2 7
7
5 + 15>7
25
=a
, 0b = a
, 0 b (since it’s symmetric on
2
7
the y-axis). Solving for b¤=a¤-c¤
25 ¤
10 ¤
75
525
=¢ ≤ -¢ ≤ =
=
7
7
49
7
25 2
49 a x b
7y2
7
= 1
100
75
61.
3 - 1 -2 + 0 -4 + 3
1
,
,
b = a 1, -1, - b
2
2
2
2
65. v+w=–3, 1, –2+3, –4, 0=0, –3, –2
66. v-w=–3, 1, –2-3, –4, 0=–6, 5, –2
67. v # w=–3, 1, –2 # 3, –4, 0=–9-4+0=–13
68. |v|= 2 1 -32 2 + 12 + 1 -22 2 = 29 + 1 + 4 = 214
69.
w
@ [email protected]
=
3, -4, 0
23 + 1 -4 2 + 0
2
2
=
2
3
4
5 , - 5 , 0
70. (v # w)(v+w)=–13 0, –3, –2=0, 39, 26
71. (x+1)2+y2+(z-3)2=16
¡
72. The direction vector PQ is 3-(–1), –2-0,
–4-3=4, –2, –7. Since the line l through P in the
¡
direction of PQ is l=(–1, 0, 3)+t (4, –2, –7), the
parametric equations are: x=–1+4t, y=–2t,
z=3-7t.
73. The direction vector is –3, 1, –2 so the vector equation
of a line in the direction of v through P is
r=–1, 0, 3+t–3, 1 –2
¡
1
74. The mid-point M of PQ is: a 1, - 1, - b (from Exercise
2
[–20, 4] by [–8, 8]
e=1, so a parabola. In polar coordinates, the vertex is
p
(1, 0) and the parabola crosses the y-axis at a 2, b .
2
Converting to Cartesian form, we have the vertex
(h, k)=(1, 0) and a point on the parabola is (0, 2). Since
the parabola opens to the left, y2=4p(x-1).
Substituting (0, 2), we have 4=–4p, p=–1
y2=–4(x-1)
62.
1
. The direction vector is
2
w=3, –4, 0, so a vector equation of the line is
1
v= 1+3t, –1-4t, – . This can be expresses in
2
1
parametric form: x=1+3t, y=–1-4t, z= - .
2
¡
#64) so OM= 1, –1, –
75. 4p=18, so p=4.5; the focus is at (0, 4.5).
76. 4p=15, so p=3.75; the focus is at (3.75, 0).
77. (a) The “shark” should aim for the other spot on the
table, since a ball that passes through one focus will
end up passing through the other focus if nothing gets
in the way.
(b) Let a=3, b=2, and c= 15. Then the foci are at
(– 15, 0) and ( 15, 0). These are the points at which
to aim.
[–1.7, 7.7] by [–3.1, 3.1]
354
Chapter 8
Analytic Geometry in Two and Three Dimensions
78. The total radius of the orbit is r=0.500+6380
=6380.5 km, or 6,380,500 m.
(a) v=7908 m/sec=7.908 km/sec
(b) The circumference of the one orbit is 2pr≠40,090
km; one orbit therefore takes about 5070 seconds, or
about 1 hr 25 min.
79. The major axis length is 18,000 km, plus 170 km, plus the
diameter of the earth, so a≠15,465 km=15,465,000 m.
At apogee, r=18,000+6380=24,380 km, so v≠2633
m/sec. At perigee, r=6380+170=6550 km, so
v≠9800 m/sec.
80. Kepler’s third law: T 2=a3, T is in Earth years and a is in
AU.
2>3
409 days
b
L 1.08 AU=161 Gm
a=T2/3= a
365.2 days>year
c=ae=(161 Gm)(0.83)≠134 Gm
perihelion: a-c=161 Gm-134 Gm=27 Gm
aphelion: a+c=161 Gm+134 Gm=295 Gm
Chapter 8 Project
Answers are based on the sample data provided in the table.
1.
[0.4, 0.75] by [–0.7, 0.7]
2. The endpoints of the major and minor axes lie at approximately (0.438, 0), (0.700, 0), (0.569, 0.640) and
(0.569, –0.640). The ellipse is taller than it is wide, even
though the reverse appears to be true on the graphing
calculator screen. The semimajor axis length is 0.640, and
the semiminor axis length is 10.700 - 0.438 2>2 = 0.131.
The equation is
1y - 0 2 2
1x - 0.5692 2
+
= 1
10.640 2 2
10.1312 2
3. With respect to the graph of the ellipse, the point (h, k)
represents the center of the ellipse. The value a is the
length of the semimajor axis, and b is the length of the
semiminor axis.
4. Physically, h=0.569 m is the pendulum’s average distance from the CBR, and k=0 m/sec is the pendulum’s
average velocity. The value a=0.64 m/sec is the maximum velocity, and b=0.131 m is the maximum displacement of the pendulum from its average position.
5. The parametric equations for the sample data set (using
sinusoidal regression) are
x1T L 0.131 sin14.80T + 2.102 + 0.569 and
y1T L 0.639 sin14.80T - 2.65 2.
[–0.1, 1.4] by [–1, 1]
[0.4, 0.75] by [–0.7, 0.7]
Section 9.1
Basic Combinatorics
355
Chapter 9
Discrete Mathematics
■ Section 9.1 Basic Combinatorics
1. Six: ABC, ACB, BAC, BCA, CAB, CBA.
6. There are 3 # 4 = 12 possible
routes. In the tree diagram,
B1 represents the first road
from town A to town B, etc.
2. Approximately 1 person out of 6, which would mean
10 people out of 60.
7. 9!=362,880
(ALGORITHM)
Exploration 1
3. No. If they all looked the same, we would expect approximately 10 people to get the order right simply by chance.
The fact that this did not happen leads us to reject the
“look-alike” conclusion.
4. It is likely that the salesman rigged the test to mislead the
office workers. He might have put the copy from the more
expensive machine on high-quality bond paper to make it
look more like an original, or he might have put a tiny ink
smudge on the original to make it look like a copy. You can
offer your own alternate scenarios.
Quick Review 9.1
1. 52
2. 13
3. 6
4. 11
5. 10
6. 4
7. 11
8. 4
9. 64
10. 13
Section 9.1 Exercises
1. There are three possibilities for who stands on the left,
and then two remaining possibilities for who stands in the
middle, and then one remaining possibility for who stands
on the right: 3 # 2 # 1 = 6.
8. 22 # 21 # 20
C1
B1
C1
A
B2
C3
C1
B3
10. There are 11 letters, where A
appears 3 times and O and T each appear 2 times.
The number of distinguishable permutations is
11!
= 1,663,200.
3!2!2!
C2
C3
C4
11. The number of ways to fill 3 distinguishable offices from a
13!
pool of 13 candidates is 13P3 =
= 1716.
10!
12. The number of ways to select and prioritize 6 out of 12
12!
projects is 12P6 =
= 665,280.
6!
13. 4 # 3 # 2 # 1 = 24
14. 13 # 2 # 1 2 112 = 6
15.
16.
6!
6 # 5 # 4!
=
= 30
16 - 22 !
4!
9!
9 # 8 # 7!
=
= 72
19 - 22 !
7!
10!
10 # 9 # 8 # 7!
=
= 120
7!110 - 72 !
7! # 3 # 2 # 1
2. Any of the four jobs could be ranked most important, and
then any of the remaining three jobs could be ranked second, and so on: 4 # 3 # 2 # 1 = 24.
18.
10 # 9 # 8 # 7!
10 # 9 # 8
10!
=
=
= 120
#
3!110 - 32 !
3! 7!
3#2#1
3. Any of the five books could be placed on the left, and
then any of the four remaining books could be placed
next to it, and so on: 5 # 4 # 3 # 2 # 1 = 120.
20. permutations
5. There are 3 # 4 = 12 possible pairs: K1Q1, K1Q2, K1Q3,
K1Q4, K2Q1, K2Q2, K2Q3, K2Q4, K3Q1, K3Q2, K3Q3, and
K3Q4.
C2
C4
17.
4. Any of the five dogs could be awarded 1st place, and then
any of the remaining four dogs could be awarded 2nd
place, and so on: 5 # 4 # 3 # 2 # 1 = 120.
C3
C4
= 9240
9. There are 11 letters, where
S and I each appear 4 times
and P appears 2 times. The
number of distinguishable
permutations is
11!
= 34,650.
4!4!2!
C2
19. combinations
21. combinations
22. permutations (different roles)
23. There are 10 choices for the first character, 9 for the second, 26 for the third, then 25, then 8, then 7, then 6:
10 # 9 # 26 # 25 # 8 # 7 # 6 = 19,656,000.
24. There are 36 choices for each character: 365=60,466,176.
25. There are 6 possibilities for the red die, and 6 for the
green die: 6 # 6 = 36.
26. There are 2 possibilities for each flip: 210=1024.
356
Chapter 9
27.
25C3
=
28.
52C5
=
29.
48C3
=
Discrete Mathematics
25!
25!
=
= 2300
3!125 - 3 2 !
3!22!
52!
52!
=
= 2,598,960
5!152 - 5 2 !
5!47!
48!
48!
=
= 17,296
3!148 - 3 2 !
3!45!
30. Choose 7 positions from the 20:
20!
20!
=
= 77,520
20C7 =
7!120 - 7 2 !
7!13!
31. Choose A♠ and K♠, and 11 cards from the other 50:
50!
50!
#
#
=
2C2 50C11 = 1 50C11 =
11!1 50 - 11 2 !
11!39!
=37,353,738,800
32. 8C3 =
8!
8!
=
= 56
3!18 - 3 2!
3!5!
33. We either have 3, 2, or 1 student(s) nominated:
6C3+6C2+6C1=20+15+6=41
34. We either have 3, 2, or 1 appetizer(s) represented:
5C3+5C2+5C1=10+10+5=25
35. Each of the 5 dice have 6 possible outcomes: 65=7776
36.
20C8
=
20!
20!
=
= 125, 970
8!120 - 8 2 !
8!12!
37. 29-1=511
38. 3*4*3*26=2304
39. Since each topping can be included or left off, the total
number of possibilities with n toppings is 2n. Since
211=2048 is less than 4000 but 212=4096 is greater
than 4000, Luigi offers at least 12 toppings.
40. There are 2n subsets, of which 2n-2 are proper subsets.
41. 210=1024
42. 510=9,765,625
(d) Number of 5-digit numbers that can be formed using
only the digits 1 and 2
(e) Number of possible pizzas that can be ordered at a
place that offers 3 different sizes and up to 10
different toppings.
50. Counting the number of ways to choose the 2 eggs you
are going to have for breakfast is equivalent to counting
the number of ways to choose the ten eggs you are not
going to have for breakfast.
51. (a) Twelve
(b) Every 0 represents a factor of 10, or a factor of
5 multiplied by a factor of 2. In the product
50 # 49 # 48 # . . . # 2 # 1, the factors 5, 10, 15, 20, 30, 35, 40,
and 45 each contain 5 as a factor once, and 25 and 50
each contain 5 twice, for a total of twelve occurrences.
Since there are 47 factors of 2 to pair up with the
twelve factors of 5, 10 is a factor of 50! twelve times.
52. (a) Each combination of the n vertices taken 2 at a time
determines a segment that is either an edge or a diagonal. There are nC2 such combinations.
(b) Subtracting the n edges from the answers in (a),
n!
we find that nC2 - n =
- n
2!1n - 2 2!
n1n - 1 2
2n
n2 - 3n
=
=
.
2
2
2
53. In the nth week, 5n copies of the letter are sent. In the last
week of the year, that’s 552≠2.22*1036 copies of the
letter. This exceeds the population of the world, which is
about 6*109, so someone (several people, actually) has
had to receive a second copy of the letter.
54. Six. No matter where the first person sits, there are the
3!=6 ways to sit the others in different positions relative
to the first person.
n
n!
n!
n!
n
=
=
= a b
43. True. a b =
a
a!1 n - a2!
a!b!
1n - b2!b!
b
55. Three. This is equivalent to the round table problem
(Exercise 42), except that the necklace can be turned
upside-down. Thus, each different necklace accounts for
two of the six different orderings.
6
45. There are a b = 15 different combinations of vegeta2
bles. The total number of entreé-vegetable-dessert
variations is 4 # 15 # 6 = 360. The answer is D.
56. The chart on the left is more reasonable. Each pair of
actresses will require about the same amount of time to
interview. If we make a chart showing n (the number of
actresses) and nC2 (the number of pairings), we can see
that chart 1 allows approximately 3 minutes per pair
throughout, while chart 2 allows less and less time per
pair as n gets larger.
5
5
44. False. For example, a b = 10 is greater than a b = 5.
2
4
46.
10P5
= 30, 240. The answer is D.
47. nPn =
n!
= n! The answer is B.
1 n - n2!
48. There are as many ways to vote as there are subsets of a
set with 5 members. That is, there are 25 ways to fill out
the ballot. The answer is C.
49. Answers will vary. Here are some possible answers:
(a) Number of 3-card hands that can be dealt from a
deck of 52 cards
(b) Number of ways to choose 3 chocolates from a box
of 12 chocolates
(c) Number of ways to choose a starting soccer team
from a roster of 25 players (where position matters)
Number
n
Number of
pairs nC2
Time per Pair Time per Pair
Chart1
Chart 2
3
3
6
15
3
2
9
36
3.06
1.67
12
66
3.03
1.52
15
105
3.05
1.43
3.33
3.33
Section 9.2
57. There are 52C13=635,013,559,600 distinct bridge hands.
Every day has 60 # 60 # 24 = 86,400 seconds; a year has
365.24 days, which is 31,556,736 seconds. Therefore it will
635,013,559,600
take about
L 20,123 years. (Using 365 days
31,556,736
per year, the computation gives about 20,136 years.)
58. Each team can choose 5 players in 12C5=792 ways, so
there are 7922=627,264 ways total.
■ Section 9.2 The Binomial Theorem
Exploration 1
3!
3!
= 1, 3C1 =
= 3,
0!3!
1!2!
3!
3!
= 3, 3C3 =
= 1. These are (in order) the
3C2 =
2!1!
3!0!
coefficients in the expansion of (a+b)3.
1. 3C0 =
2. {1 4 6 4 1}. These are (in order) the coefficients in the
expansion of (a+b)4.
3. {1 5 10 10 5 1}. These are (in order) the coefficients in
the expansion of (a+b)5.
Quick Review 9.2
1. x2+2xy+y2
2. a2+2ab+b2
3. 25x2-10xy+y2
The Binomial Theorem
4. 1x + y 2 10 = a
10 10 0
10
10
b x y + a b x9y1 + a b x8y2
0
1
2
10 7 3
10 6 4
+ a bx y + a bx y
3
4
10 5 5
10 4 6
+ a bx y + a bx y
5
6
10 3 7
10 2 8
+ a bx y + a bx y
7
8
10 1 9
10 0 10
+ a bx y + a bx y
9
10
10
9
= x + 10x y + 45x8y2 + 120x7y3 + 210x6y4
+ 252x5y5 + 210x4y6 + 120x3y7
+ 45x2y8 + 10xy9 + y10
5. Use the entries in row 3 as coefficients:
(x+y)3=x3+3x2y+3xy2+y3
6. Use the entries in row 5 as coefficients:
(x+y)5=x5+5x4y+10x3y2+10x2y3
+5xy4+y5
7. Use the entries in row 8 as coefficients:
(p+q)8=p8+8p7q+28p6q2+56p5q3+70p4q4
+56p3q5+28p2q6+8pq7+q8
8. Use the entries in row 9 as coefficients:
(p+q)9=p9+9p8q+36p7q2+84p6q3+126p5q4
+126p4q5+84p3q6+36p2q7+9pq8+q9
9
9!
9#8
= # = 36
9. a b =
2
2!7!
2 1
10. a
15
15!
15 # 14 # 13 # 12
b =
=
= 1365
11
11!4!
4#3#2#1
11. a
166
166!
b =
= 1
166
166!0!
12. a
166
166!
= 1
b =
0!166!
0
10. 64m3+144m2n+108mn2+27n3
13. ¢
14
14
≤ = ¢ ≤ = 364
3
11
Section 9.2 Exercises
14. ¢
13
13
≤ = ¢ ≤ = 1287
8
5
4. a2-6ab+9b2
5. 9s2+12st+4t2
6. 9p2-24pq+16q2
7. u3+3u2v+3uv2+v3
8. b3-3b2c+3bc2-c3
9. 8x3-36x2y+54xy2-27y3
4
4
4
4
1. 1a + b2 4 = a b a4b0 + a b a3b1 + a b a2b2 + a b a1b3
0
1
2
3
4 0 4
+ a ba b
4
4
3
= a + 4a b + 6a2b2 + 4ab3 + b4
6
6
6
6
2. 1a + b2 6 = a b a6b0 + a b a5b1 + a b a4b2 + a ≤a3b3
0
1
2
3
6 2 4
6 1 5
6 0 6
+ a ba b + a ba b + a ba b
4
5
6
6
5
4 2
3 3
2 4
= a + 6a b + 15a b + 20a b + 15a b
+ 6ab5 + b6
7
7
7
7
3. 1 x + y2 7 = a b x7y0 + a b x6y1 + a b x5y2 + a b x4y3
0
1
2
3
7 3 4
7 2 5
+ a bx y + a bx y
4
5
7 1 6
7
+ a b x y + a b x0y7
6
7
= x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4
+ 21x2y5 + 7xy6 + y7
357
15. 1 -22 8 ¢
12
12
≤ = 1 -2 2 8 ¢ ≤ =126,720
8
4
16. 1 -32 4 ¢
11
11
≤ = 1 -3 2 4 ¢ ≤ =26,730
4
7
17. f(x)=(x-2)5
=x5+5x4(–2)+10x3(–2)2+10x2(–2)3
+5x(–2)4+(–2)5
5
=x -10x4+40x3-80x2+80x-32
18. g(x)=(x+3)6
=x6+6x5 # 3 + 15x4 # 32 + 20x3 # 33
+15x2 # 34 + 6x # 35 + 36
6
=x +18x5+135x4+540x3+1215x2+1458x
+729
19. h(x)=(2x-1)7
=(2x)7+7(2x)6(–1)+21(2x)5(–1)2
+35(2x)4(–1)3+35(2x)3(–1)4
+21(2x)2(–1)5+7(2x)(–1)6+(–1)7
=128x7-448x6+672x5-560x4+280x3
-84x2+14x-1
358
Chapter 9
Discrete Mathematics
20. f(x)=(3x+4)5
=(3x)5+5(3x)4 # 4 + 1013x2 3 # 42
+10(3x)2 # 43 + 5 13x2 # 44 + 45
5
=243x +1620x4+4320x3+5760x2+3840x
+1024
21. (2x+y)4=(2x)4+4(2x)3y+6(2x)2y2+4(2x)y3+y4
=16x4+32x3y+24x2y2+8xy3+y4
22. (2y-3x)5=(2y)5+5(2y)4(–3x)+10(2y)3(–3x)2
+10(2y)2(–3x)3+5(2y)(–3x)4
+(–3x)5
5
=32y -240y4x+720y3x2-1080y2x3
+810yx4-243x5
23. 1 1x - 1y2 6 = 1 1x2 6 + 6 1 1x2 5 1 - 1y2 + 151 1x2 4
# 1 - 1y2 2 + 20 1 1x2 3 1 - 1y2 3
+ 15 1 1x2 2 1 - 1y2 4
+ 6 1 1x2 1 - 1y2 5 + 1 - 1y2 6
3
=x -6x5/2y1/2+15x2y-20x3/2y3/2
+15xy2-6x1/2y5/2+y3
24. 1 1x + 132 4 = 1 1x2 4 + 41 1x2 3 1 132 + 6 1 1x2 2
# 1 132 2 + 41 1x2 1 132 3 + 1 132 4
= x2 + 4x13x + 18x + 12 13x + 9
25. (x–2+3)5=(x–2)5+5(x–2)4 # 3 + 10 1x-2 2 3 # 32
+ 10 1x-2 2 2 # 33 + 5 1 x-2 2 # 34 + 35
-10
= x
+ 15x-8+90x–6+270x–4+405x–2
+243
26. (a-b–3)7=a7+7a6(–b–3)+21a5(–b–3)2
+35a4(–b–3)3+35a3(–b–3)4
+21a2(–b–3)5+7a(–b–3)6+(–b–3)7
7
=a -7a6b–3+21a5b–6-35a4b–9
+35a3b–12-21a2b–15+7ab–18-b–21
27. Answers will vary.
28. Answers will vary.
n!
n
n!
29. a b =
= n =
1!1n - 1 2 !
1 n - 1 2 !1!
1
n!
n
=
= a
b
1 n - 1 2!3n - 1 n - 1 2 4!
n - 1
32. (a) Any pair (n, m) of nonnegative integers — except for
(1, 1) — provides a counterexample. For example,
n=2 and m=3: (2+3)!=5!=120, but
2!+3!=2+6=8.
(b) Any pair (n, m) of nonnegative integers — except for
(0, 0) or any pair (1, m) or (n, 1) — provides a counterexample. For example, n=2 and m=3:
12 # 3 2! = 6! = 720, but, 2! # 3! = 2 # 6 = 12.
1n + 12 !
n
n + 1
n!
33. a b + a
b =
+
2
2
2!1n - 22!
2!1n - 12!
n1n - 1 2
1n + 12 1n2
=
+
2
2
=
n2 - n + n2 + n
= n2
2
n
n + 1
n!
≤ + ¢
≤ =
n - 2
n - 1
1n - 22!3n - 1n - 22 4!
1n + 12 !
+
1n - 12!3 1n + 12 - 1n - 1 2 4!
1n + 12 !
n!
=
+
1n - 22!2!
1n - 1 2!2!
n1n - 1 2
1n + 12n
=
+
2
2
34. ¢
=
n2 - n + n2 + n
=n2
2
35. True. The signs of the coefficients are determined by the
powers of the (–y) terms, which alternate between odd
and even.
36. True. In fact, the sum of every row is a power of 2.
37. The fifth term of the expansion is
8
a b 12x2 4 1 12 4 = 1120x4. The answer is C.
4
38. The two smallest numbers in row 10 are 1 and 10. The
answer is B.
39. The sum of the coefficients of (3x-2y)10 is the same as
the value of (3x-2y)10 when x=1 and y=1.
The answer is A.
n!
n
n!
30. a b =
=
r!1n - r2!
1n - r2!r!
r
n!
n
=
= a
b
1 n - r2!3 n - 1n - r2 4!
n - r
40. The even-numbered terms in the two expressions are
opposite-signed and cancel out, while the odd-numbered
terms are identical and add together. The answer is D.
31. ¢
41. (a) 1, 3, 6, 10, 15, 21, 28, 36, 45, 55.
n - 1
n - 1
≤ + ¢
≤
r - 1
r
1n - 12!
1n - 1 2!
=
+
1r - 1 2!3 1 n - 1 2 - 1r - 1 2 4 !
r!1n - 1 - r2!
1n - 1 2!1 n - r2
r1n - 1 2 !
+
=
r1r - 1 2 !1n - r2 !
r!1n - r2 1 n - r - 1 2!
1 n - r2 1n - 1 2 !
r1n - 1 2 !
+
=
r!1n - r2 !
r!1n - r2!
1r + n - r2 1 n - 12 !
=
r!1n - r2!
n!
=
r!1n - r2 !
n
= ¢ ≤
r
(b) They appear diagonally down the triangle, starting
with either of the 1’s in row 2.
(c) Since n and n+1 represent the sides of the given
rectangle, then n(n+1) represents its area. The triangular number is 1/2 of the given area. Therefore, the
n1n + 1 2
triangular number is
.
2
Section 9.3
(d) From part (c), we observe that the nth triangular
n1 n + 1 2
number can be written as
. We know that
2
n
binomial coefficients are the values of ¢ r ≤ for
r=0, 1, 2, 3, . . . , n. We can show that
n + 1
n1 n + 1 2
= ¢ 2 ≤ as follows:
2
n1 n + 1 2
=
2
=
=
1 n + 1 2n1 n - 12 !
2 1n - 1 2!
1n + 1 2!
n
n
n
n
= ¢ ≤1n + ¢ ≤1n - 12 + ¢ ≤1n - 222 + p + ¢ ≤2n
n
0
1
2
n
n
n
n
= ¢ ≤ + 2¢ ≤ + 4¢ ≤ + p + 2n ¢ ≤
0
1
2
n
■ Section 9.3 Probability
Exploration 1
1.
Antibodies
present
p = 0.006
2!1 1n + 1 2 - 22 !
n + 1
≤
2
p = 0.994
Antibodies
absent
So, to find the fourth triangular number, for example,
=
5 # 4 # 3!
4 + 1
5
5!
≤=¢ ≤=
=
2
2
2!15 - 2 2!
2!3!
2.
p = 0.997
5#4
=10.
2
p = 0.006
p = 0.003
–
p = 0.015
+
p = 0.985
–
(b) 1
(c) No. (They all appear in order down the second
diagonal.)
p = 0.994
Antibodies
absent
(d) 0. (See Exercise 44 for a proof.)
(e) All are divisible by p.
(f) Rows that are powers of 2: 2, 4, 8, 16, etc.
(g) Rows that are 1 less than a power of 2: 1, 3, 7, 15, etc.
3.
p = 0.997
(h) For any prime numbered row, or row where the first
element is a prime number, all the numbers in that
row (excluding the 1’s) are divisible by the prime. For
example, in the seventh row (1 7 21 35 35 21
7 1) 7, 21, and 35 are all divisible by 7.
n
n
n
= a b 1n10 + a b 1n-111 + a b 1n-212
0
1
2
p = 0.006
n
n
n
n
= ¢ ≤1n + ¢ ≤1n - 1 1 -1 2 + ¢ ≤1n - 2 1 -1 2 2
0
1
2
n
+ p + ¢ ≤ (–1)n
n
n
n
n
n
= ¢ ≤ - ¢ ≤ + ¢ ≤ + p + 1 - 12 n ¢ ≤
0
1
2
n
p = 0.003
– p = 0.00002
p = 0.015
+ p = 0.01491
p = 0.985
– p = 0.97909
p = 0.994
Antibodies
absent
4. P(±)=0.00598+0.01491=0.02089
P1antibody present and + 2
P1 + 2
0.00598
L 0.286 (A little more than 1 chance in 4.)
0.02089
5. P(antibody present |±)=
n
+ p + a b 101n
n
n
n
n
n
= a b + a b + a b + p + a b
0
1
2
n
+ p = 0.00598
Antibodies
present
43. The sum of the entries in the nth row equals the sum of
the coefficients in the expansion of (x+y)n. But this
sum, in turn, is equal to the value of (x+y)n when x=1
and y=1:
2 n = 11 + 1 2 n
+
Antibodies
present
42. (a) 2. (Every other number appears at least twice.)
44. 0 = 11 - 12
359
45. 3n=(1+2)n
2!1 n - 1 2 !
1n + 12!
= ¢
compute ¢
Probability
Quick Review 9.3
1. 2
2. 6
3. 23=8
4. 63=216
5.
52C5=2,598,960
6.
10C2=45
7. 5!=120
360
Chapter 9
Discrete Mathematics
8. 5P3=60
5!
C
1
3!2!
5 3
9.
=
=
10!
12
10C3
3!7!
5!
C
2!3!
2
5 2
10.
=
=
10!
9
10C2
2!8!
16. P[not (B or T)]=1-P(B or T)=1-(0.3+0.1)
=0.6
17. P(B1 and B2)=P(B1) # P(B2)=(0.3)(0.3)=0.09
18. P(O1 and O2)=P(O1) # P(O2)=(0.1)(0.1)=0.01
19. P[(R1 and G2) or (G1 and R2)]=P(R1) # P(G2)
+P(G1) # P(R2)=(0.2)(0.2)+(0.2)(0.2)=0.08
20. P(B1 and Y2)=P(B1) # P(Y2)=(0.3)(0.2)=0.06
21. P (neither is yellow)=P(not Y1 and not Y2)
=P(not Y1) # P(not Y2)=(0.8)(0.8)=0.64
Section 9.3 Exercises
For #1–8, consider ordered pairs (a, b) where a is the value of
the red die and b is the value of the green die.
1
4
1. E={(3, 6), (4, 5), (5, 4), (6, 3)}: P(E)=
=
36
9
2. E={both dice even}, {both dice odd}:
3#3 + 3#3
18
1
P(E)=
=
=
36
36
2
3. E={(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2),
(5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}:
15
5
P(E)=
=
36
12
4. E={(1, 1), (1, 2), . . ., (6, 2), (6, 3)}
30
5
=
P(E)=
36
6
5. P(E)=
3#3
1
=
36
4
6. P(E)=
3#3
1
=
36
4
7. E={(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2),
(3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
15
5
=
P(E)=
36
12
22. P(not R1 and not O2)=P(not R1) # P(not O2)
=(0.8)(0.9)=0.72
23. There are 24C6=134,596 possible hands; of these, only 1
1
consists of all spades, so the probability is
.
134,596
24. Of the 24C6=134,596 possible hands, one consists of all
spades, one consists of all clubs, one consists of all hearts,
and one consists of all diamonds, so the probability is
4
1
=
.
134,596
33,649
25. Of the 24C6=134,596 possible hands, there are
#
4C4 20C2=190 hands with all the aces, so the probability
190
5
=
.
is
134,596
3542
26. There are 2C2 # 22C4 = 7315 ways to get both black jacks
and 4 “other” cards. Similarly, there are 7315 ways to get
both red jacks. These two numbers together count twice
the 2C2 # 2C2 # 20C2 = 190 ways to get all four jacks.
Therefore, altogether we have 2 # 7315 - 190 = 14,440
distinct ways to have both bowers, so the probability is
14,440
190
=
.
134,596
1771
27. (a)
0.3
8. E={(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}:
8
2
=
P(E)=
36
9
9. (a) No. 0.25+0.20+0.35+0.30=1.1. The numbers
do not add up to 1.
(b) There is a problem with Alrik’s reasoning. Since the
gerbil must always be in exactly one of the four
rooms, the proportions must add up to 1, just like a
probability function.
10. Since 4+3+2+1=10, we can divide each number
in the ratio by 10 and get the proportions relative to the
whole. The table then becomes
Compartment
A
B
C
D
Proportion
0.4
0.3
0.2
0.1
Yes, this is a valid probability function.
A
(b) 0.3
(c) 0.2
(d) 0.2
(e) Yes. P(A |B)=
P1A and B2
P1 B2
=
0.3
= 0.6 = P1A2
0.5
=
0.2
= 0.5 Z P1A2
0.4
28. (a)
0.5
0.2
A
0.2
B
0.1
(b) 0.5
(c) 0.2
12. P(R or G or O)=P(R)+P(G)+P(O)
=0.2+0.1+0.1=0.4
13. P(R)=0.2
(d) 0.1
15. P[not (O or Y)]=1-P(O or Y)
=1-(0.2+0.1)=0.7
B
0.2
11. P(B or T)=P(B)+P(T)=0.3+0.1=0.4
14. P(not R)=1-P(R)=1-0.2=0.8
0.2
0.3
(e) No. P(A |B)=
P1A and B2
P1B2
29. P(John will practice)=(0.6)(0.8)+(0.4)(0.4)=0.64
Section 9.3
1
= 0.20
5
(b) P(meatloaf and peas are served)=(0.20)(0.70)
=0.14
(c) P(peas are served)=(0.20)(0.70)+(0.80)(0.30)
=0.38
30. (a) P(meatloaf is served)=
31. If all precalculus students were put on a single list and a
name then randomly chosen, the probability P(from
Mr. Abel’s class | girl) would be 12 | 22=6 | 11. But when
one of the two classes is selected at random, and then a
student from this class is selected,
P(from Mr. Abel’s class | girl)
P1girl from Mr. Abel’s class2
=
P1girl2
12
1
a ba b
2
20
=
1
12
1
10
a ba b + a ba b
2
20
2
25
3
=
5
32. Within each box, any of the coins is equally likely to be
chosen and either side is equally likely to be shown. But a
head in the 2-coin box is more likely to be displayed than
a head in the 3-coin box.
P1H from 2-coin box2
P(from 2-coin box | H)=
P1 H2
3
1
a ba b
2
4
=
1
3
1
3
a ba b + a ba b
2
4
2
6
3
=
5
190
19
20C2
33.
=
=
300
30
25C2
34. P(none are defective)=4C0 # (0.037)0(0.963)4
=(0.963)4≠0.860
35. (a) P(cardiovascular disease or cancer)=0.45+0.22
=0.67
(b) P(other cause of death)=1-0.67=0.33
1 5
1
36. P(Yahtzee)=6 # a b =
6
1296
38. P(at least one false positive)
=1-P(no false positives)=1-(0.993)60≠0.344.
172
86
=
254
127
(b) P(went to graduate school)=
(b)
(c)
14C8
20C8
14C5
=
14C6
3003
77
=
125,970
3230
# 6C3
20C8
361
=
40,040
308
=
125,970
969
# 6C2 + 14C7 # 6C1 +
14C8
# 6C0
20C8
45,045 + 20,592 + 3003
68,640
176
=
=
=
125,970
125,970
323
41.
1
1
=
C
36
9 2
42. This cannot be true. Let A be the event that it is cloudy
all day, B be the event that there is at least 1 hour of sunshine, and C be the event that there is some sunshine, but
less than 1 hour. Then A, B, and C are mutually exclusive
events, so P(A or B or C)=P(A)+P(B)+P(C)
=0.22+0.78+P(C)=1+P(C). Then it must be
the case that P(C)=0. This is absurd; there must be
some probability of having more than 0 but less than
1 hour of sunshine.
1 10
1
1
43. P(HTTTTTTTTT)= a b = 10 =
2
1024
2
1 10
1
44. P(THTTTTHTTT)= a b =
2
1024
1 10
1
45. P(HHHHHHHHHH)= a b =
2
1024
1 10
10
5
=
46. P(9 H and 1 T)=10C1 # a b =
2
1024
512
1 10
45
47. P(2 H and 8 T)=10C2 # a b =
2
1024
1 10
120
15
=
48. P(3 H and 7 T)=10C3 # a b =
2
1024
128
# a1b
10
49. P (at least one H)=1-P(0 H)=1 -
10C0
2
1
1023
=
=1 1024
1024
50. P (at least two H)=1-P(0 H)-P(1 H)=
1 10
1 10
11
1013
1 - 10C0 # a b - 10C1 # a b = 1 =
2
2
1024
1024
51. False. A sample space consists of outcomes, which are not
necessarily equally likely.
37. The sum of the probabilities is greater than 1 – an impossibility, since the events are mutually exclusive.
39. (a) P(a woman)=
40. (a)
Probability
52. False. All probabilities are between 0 and 1, inclusive.
53. Of the 36 different, equally-likely ways the dice can land,
4 ways have a total of 5. So the probability is 4/36=1/9.
The answer is D.
54. A probability must always be between 0 and 1, inclusive.
The answer is E.
91
124 + 58
=
254
127
(c) P(a woman who went to graduate school)
62
124
=
=
254
127
55. P(B and A)=P(B) P(A|B), and for independent
events, P(B and A)=P(B)P(A). It follows that the
answer is A.
56. A specific sequence of one “heads” and two “tails” has
probability (1/2)3=1/8. There are three such sequences.
The answer is C.
362
Chapter 9
57. (a)
Discrete Mathematics
Type of Bagel
Probability
Plain
0.37
Onion
0.12
Rye
0.11
Cinnamon Raisin
0.25
Sourdough
0.15
(b) (0.37)(0.37)(0.37)≠0.051
(c) No. They are more apt to share bagel preferences if
they arrive at the store together.
58. (a) P(at least one king)=1-P(no kings)
18,472
48C5
=1=
L 0.34
C
54,145
52 5
(b) The number of ways to choose, e.g., 3 fives and 2 jacks
is 4C3 # 4C2. There are 13P2=13 # 12 different combinations of cards that can make up the full house, so
13 # 12 # 4C3 # 4C2
6
P(full house)=
=
L 0.0014
C
4165
52 5
59. (a) We expect (8)(0.23)=1.84 (about 2) to be married.
(b) Yes, this would be an unusual sample.
(c) P(5 or more are married)=8C5 # (0.23)5(0.77)3
+8C6 # (0.23)6(0.77)2+8C7 # (0.23)7(0.77)1
+8C8 # (0.23)8(0.77)0≠0.01913=1.913%.
60. (a) 32C17 # (0.75)17(0.25)15≠0.00396=0.396%
17
(b) a 32Ck # 10.75 2 k 1 0.252 15 - k L 0.00596 = 0.596%
k=0
(c) The university’s graduation rate seems to be exaggerated; at least, this particular class did not fare as well
as the university claims.
61. (a) $1.50
10
L 0.0000010676
62. (a)
9,366,819
Value
–10
Probability
9,366,809
9,366,819
3,999,990
10
9,366,819
10
+ 1 - 102
9,366,819
10
a1 b L -5.73
9,366,819
(c) 3,999,990 #
#
(d) In the long run, Gladys is losing $5.73 every time she
buys the ten tickets. Given the low probability of a
positive payoff, she stands to lose a lot of money if she
does this often.
■ Section 9.4 Sequences
Quick Review 9.4
1. 3+(5-1)4=3+16=19
2.
4.
5 11 - 43 2
11 - 42
5. a10 =
=
- 315
= 105
-3
10
11
6. a10=5+(10-1)3=32
7. a10=5 # 29=2560
4
1 9
4
1
1
b =
8. a10 = a b a b = a b a
3
2
3
512
384
9. a10=32-17=15
10. a10 =
102
100
25
=
=
1024
256
210
Section 9.4 Exercises
For #1–4, substitute n=1, n=2, . . . , n=6, and n=100.
3 4 5 6 7 101
1. 2, , , , , ;
2 3 4 5 6 100
2.
4
4 2 4 1 2
, 1, , , , ;
3
5 3 7 2 51
3. 0, 6, 24, 60, 120, 210; 999,900
4. –4, –6, –6, –4, 0, 6; 9500
For #5–10, use previously computed values of the sequence to
find the next term in the sequence.
5. 8, 4, 0, –4; –20
6. –3, 7, 17, 27; 67
7. 2, 6, 18, 54; 4374
8. 0.75, –1.5, 3, –6; –96
9. 2, –1, 1, 0; 3
10. –2, 3, 1, 4; 23
2
4
6
4
1
=
(b) 3 # + 1 - 1 2 # = 6
6
6
6
3
(b)
3. 5 # 42 = 80
5
5
[(6+(5-1)4]= (22)=55
2
2
11. lim n2 = q, so the sequence diverges.
nSq
1
= 0, so the sequence converges to 0.
2n
1
1 1 1 1
13. , , , , . . . , 2 , . . .
1 4 9 16
n
1
lim
= 0, so the sequence converges to 0.
nSq n2
14. lim 13n - 12 = q, so the sequence diverges.
12. lim
nSq
nSq
15. Since the degree of the numerator is the same as the
degree of the denominator, the limit is the ratio of the
3n - 1
= -1. The
leading coefficients. Thus lim
nSq 2 - 3n
sequence converges to –1.
16. Since the degree of the numerator is the same as the
degree of the denominator, the limit is the ratio of the
2n - 1
= 2. The sequence
leading coefficients. Thus lim
nSq n + 1
converges to 2.
1 n
17. lim 10.52 n = lim a b = 0, so the sequence converges
nSq
nSq 2
to 0.
3 n
18. lim 11.52 n = lim a b = q, so the sequence diverges.
nSq
nSq 2
Section 9.4
Sequences
19. a1 = 1 and an + 1 = an + 3 for n 1 yields 1, 4, 7,
. . . , (3n-2), . . .
lim 13n - 2 2 = q, so the sequence diverges.
nSq
un
1
1 1
20. u1 = 1 and un + 1 =
for n 1 yields 1, , , . . . , n - 1 , . . .
3
3 9
3
1
= 0, so the sequence converges to 0.
lim
nSq 3 n - 1
For #21–24, subtract the first term from the second to find the
common difference d. Use the formula an=a1+(n-1)d
with n=10 to find the tenth term. The recursive rule for the
nth term is an=an-1+d, and the explicit rule is the one
given above.
31. a2=3=a1 # r1 and a8=192=a1 # r7, so
a8/a2=64=r6. Therefore r=—2, so a1=3/(—2)
3
3
5 ; and an= - # 1 -22 n - 1 = 3 # 1 -2 2 n - 2 or
2
2
3 # n-1
an= 2
= 3 # 2n - 2.
2
21. (a) d=4
33.
363
30. a5=–5=a1+4d and a9=–17=a1+8d, so
a9-a5=–12=4d. Therefore d=–3, so
a1=–5-4d=7 and an=an-1-3 for n 2.
32. a3=–75=a1 # r2 and a6=–9375=a1 # r5, so
a6/a3=125=r3. Therefore r=5, so a1=–75/52=–3
and an=–3 # 5n-1.
(b) a10=6+9(4)=42
(c) Recursive rule: a1=6; an=an-1+4 for n 2
(d) Explicit rule: an=6+4(n-1)
22. (a) d=5
(b) a10=–4+9(5)=41
(c) Recursive rule: a1=–4; an=an-1+5 for n 2
[0, 5] by [–2, 5]
34.
(d) Explicit rule: an=–4+5(n-1)
23. (a) d=3
(b) a10=–5+9(3)=22
(c) Recursive rule: a1=–5; an=an-1+3 for n 2
(d) Explicit rule: an=–5+3(n-1)
[0, 10] by [–3, 1]
24. (a) d=11
(b) a10=–7+9(11)=92
35.
(c) Recursive rule: a1=–7; an=an-1+11 for n 2
(d) Explicit rule: an=–7+11(n-1)
For #25–28, divide the second term by the first to find the
common ratio r. Use the formula an=a1 # rn-1 with n=8 to
find the eighth term. The recursive rule for the nth term is
an=an-1 # r, and the explicit rule is the one given above.
[0, 10] by [–10, 100]
36.
25. (a) r=3
(b) a8=2 # 3 =4374
7
(c) Recursive rule: a1=2; an=3an-1 for n 2
(d) Explicit rule: an=2 # 3n-1
26. (a) r=2
(b) a8=3 # 27=384
(c) Recursive rule: a1=3; an=2an-1 for n 2
(d) Explicit rule: an=3 # 2n-1
27. (a) r=–2
(b) a8=(–2)7=–128
(c) Recursive rule: a1=1; an=–2an-1 for n 2
(d) Explicit rule: an=(–2)n-1
28. (a) r=–1
(b) a8=–2 # (–1)7=2
(c) Recursive rule: a1=–2; an=–1an-1=–an-1
for n 2
(d) Explicit rule: an=–2 # (–1)n-1=2 # (–1)n
29. a4=–8=a1+3d and a7=4=a1+6d, so
a7-a4=12=3d. Therefore d=4, so
a1=–8-3d=–20 and an=an-1+4 for n 2.
[0, 10] by [0, 25]
37. The height (in cm) will be an arithmetic sequence with
common difference d=2.3 cm, so the height in week n is
700+2.3(n-1): 700, 702.3, 704.6, 706.9, . . . , 815, 817.3.
38. The first column is an arithmetic sequence with common
difference d=14. The second column is a geometric
1
sequence with common ratio r= .
2
Time
(billions of years)
0
14
28
42
56
Mass
(g)
16
8
4
2
1
364
Chapter 9
Discrete Mathematics
39. The numbers of seats in each row form a finite arithmetic
sequence with a1=7, d=2, and n=25. The total
number of seats is
25
3 21 7 2 + 125 - 1 2 1 2 2 4 = 775 .
2
40. The numbers of tiles in each row form a finite arithmetic
sequence with a1=15, an=30, and n=16. The total
number of tiles is
15 + 30
16 a
b = 360 .
2
41. The ten-digit numbers will vary; thus the sequences will
vary. The end result will, however, be the same. Each limit
will be 9. One example is:
Five random digits: 1, 4, 6, 8, 9
Five random digits: 2, 3, 4, 5, 6
List: 1, 2, 3, 4, 4, 5, 6, 6, 8, 9
Ten-digit number: 2, 416, 345, 689
Ten-digit number: 9, 643, 128, 564
a1=positive difference of the ten-digit numbers
=7, 226, 782, 875
an+1= sum of the digits of an, so
a2=sum of the digits of a1=54
a3=sum of the digits of a2=9
All successive sums of digits will be 9, so the sequence
converges and the limit is 9.
42. Everyone should end up at the word “all”.
43. True. Since two successive terms are negative, the common
ratio r must be positive, and so the sign of the first term
determines the sign of every number in the sequence.
44. False. For example, consider the sequence 5, 1, –3, –7, . . . .
45. a1=2 and a2=8 implies d=8-2=6
c=a1-d so c=2-6=–4
a4=6 # 4+(–4)=20
The answer is A.
(b) Notice that after n-2 months, there are an – 1 pairs,
of which an – 2 (the number of pairs present one
month earlier) are fertile. Therefore, after n-1
months, the number of pairs will be an=an – 1
+an – 2: to last month’s total, we add the number of
new pairs born. Thus a4=3, a5=5, a6=8,
a7=13, a8=21, a9=34, a10=55, a11=89,
a12=144, a13=233.
(c) Since a1 is the initial number of pairs, and a2 is the
number of pairs after one month, we see that a13 is the
number of pairs after 12 months.
50. Use a calculator: a1=1, a2=1, a3=2, a4=3, a5=5,
a6=8, a7=13. These are the first seven terms of the
Fibonacci sequence.
51. (a) For a polygon with n sides, let A be the vertex in
quadrant I at the top of the vertical segment, and let
B be the point on the x-axis directly below A.
Together with (0, 0), these two points form a right triangle; the acute angle at the origin has measure
2p
p
¨ =
= , since there are 2n such triangles making
2n
n
up the polygon. The length of the side opposite this
angle is
p
sin ¨ = sin , and there are 2n such sides making up
n
an
p
the perimeter of the polygon, so sin
=
, or
n
2n
an=2n sin(∏/n).
(b) a10≠6.1803, a100≠6.2822, a1000≠6.2832,
a10,000≠6.2832≠2∏. It appears that an S 2∏ as
n S q , which makes sense since the perimeter of
the polygon should approach the circumference of
the circle.
y
46. lim 1n = lim n1>2 = q, so the sequence diverges.
nSq
nSq
The answer is B.
a2
6
= = 3
47. r =
a1
2
a6 = a1r5 = 2 # 35 = 486 and a2 = 6, so
a6
486
=
= 81.
a2
6
The answer is E.
48. The geometric sequence will be defined by an+1=an÷3
for n 1 and a1 Z 0.
a1
a2 =
3
a1>3
a2
a1
=
=
a3 =
3
3
9
a1>9
a3
a1
a4 =
=
=
3
3
27
a1
an = n - 1 , which represents a geometric sequence.
3
The answer is C.
49. (a) a1=1 because there is initially one male-female pair
(this is the number of pairs after 0 months). a2=1
because after one month, the original pair has only
just become fertile. a3=2 because after two months,
the original pair produces a new male-female pair.
1
θ
an
2n
x
52. P1=525,000; Pn=1.0175Pn-1, n 2
53. The difference of successive terms in {log (an)} will be of
an + 1
the form log (an1)-log (an)=log a
b . Since {an} is
an
an + 1
an + 1
b congeometric,
is constant. This makes log a
an
an
stant, so {log (an)} is a sequence with a constant difference
(arithmetic).
54. The ratios of successive terms in {10bn} will be of the form
10bn±1÷10bn=10bn±1 – bn. Since {bn} is arithmetic, bn1-bn
is constant. This makes 10bn ± 1-bn constant, so {10bn} is a
sequence with a common ratio (geometric).
Section 9.5
55. a1=[1 1], a2=[1 2], a3=[2 3], a4=[3 5],
a5=[5 8], a6=[8 13], a7=[13 21]. The entries in
the terms of this sequence are successive pairs of terms
from the Fibonacci sequence.
1 d
d
0 1
#
a2=a1 r=[1 a+d]
a3=a2 # r=[1 d+a+d]=[1 a+2d]
a4=a3 # r=[1 d+a+2d]=[1 a+3d]
an=[1 a+(n-1)d].
So, the second entries of this geometric sequence of
matrices form an arithmetic sequence with the first term a
and common difference d.
56. a1=[1
a] r = c
5. a1=1 and a2=2 yields r =
Exploration 1
1. 3 + 6 + 9 + 12 + 15 = 45
2. 52 + 62 + 72 + 82 = 25 + 36 + 49 + 64 = 174
3. cos(0)+cos(∏)+. . .+cos(11∏)+cos(12∏)
=1-1+1+. . .-1+1=1
4. sin(0)+sin(∏)+. . .+sin(k∏)
+. . .=0+0+. . .+0+. . .=0
5.
3
3
3
3
1
+
+
+ ... +
+ . . .=
10
100
1,000
1,000,000
3
Exploration 2
3+. . .+99+100
2. 100+ 99+ 98+. . .+ 2+ 1
3. 101+101+101+. . .+101+101
1.
1+
2+
4. 100(101)=10,100
5. The sum in 4 involves two copies of the same progression,
so it doubles the sum of the progression. The answer that
Gauss gave was 5050.
365
2
= 2
1
a10=1 # 29=512
a10=512
6. a4=1 and a4=a1 # r 3; a6=2 and a6=a1 # r 5
a1 # r5
2
=
1
a1 # r3
r 2=2 1 r = 12
1 = a1 1 122 3; a 1 =
a10
a10
■ Section 9.5 Series
Series
1
1
=
3
1 122
212
1
1612
=
1 122 9 =
= 8
212
2 12
= 8
7. a7=5 and r=–2 1 5=a1(–2)6
5
5
-2560
a1 =
; a10 =
1 -22 9 =
= -40
64
24
24
a10 = -40
8. a8=10 and a8=a1 # r7; a12=40 1 a12=a1 # r11
a1 # r11
40
=
7
#
10
a1 r
r4 = 4; so r = 142 1>4
10 = a1 A 14 2 1>4 B 7;
10 1>4 9
14 2 =
4 7>4
= 20
a10 =
a10
10
4 7>4
10 14 9>4 2
a1 =
4 7>4
= 10 14 2>4 2 = 10 # 2 = 20
5
9. a n2 = 1 + 4 + 9 + 16 + 25 = 55
n=1
5
10. a 12n - 12 = 1 + 3 + 5 + 7 + 9 = 25
n=1
Section 9.5 Exercises
11
1. a 16k - 13 2
k=1
Quick Review 9.5
1. a1=4; d=2 so a10=a1+(n-1)d
a10=4+(10-1)2=4+18=22
a10=22
2. a1=3; a2=1 so d=1-3=–2
a10=a1+(n-1)d
a10=3+(10-1)(–2)=3-18=–15
a10=–15
3. a3=6 and a8=21
a3=a1+2d and a8=a1+7d
(a1+7d)-(a1+2d)=21-6 so 5d=15 1 d=3.
6=a1+2(3) so a1=0
a10=0+9(3)=27
a10=27
4. a5=3, and an+1=an+5 for n 1 1
a6=3+5=8
a5=3 and a6=8 1 d=5
a5=a1+4d so 3=a1+4(5) 1 a1=–17
a10=–17+9(5)=28
a10=28
10
2. a 13k - 1 2
k=1
n +1
3. a k2
k=1
n+1
4. a k3
k=1
q
5. a 61 -22 k
k=0
q
6. a 51 -32 k
k=0
For #7–12, use one of the formulas Sn=n a
a1 + an
b or
2
n
Sn= [2a1+(n-1)d]. In most cases, the first of these is
2
easier (since the last term an is given); note that
an - a1
n=
+ 1.
d
7. 6 # a
-7 + 13
b = 6 # 3 = 18
2
366
Chapter 9
8. 6 # a
Discrete Mathematics
-8 + 27
b = 3 # 19 = 57
2
24. (a) The first six partial sums are {–2, 0, –2, 0, –2, 0}. The
numbers approach no limit. The series is divergent.
(b) The first six partial sums are {1, 0.3, 0.23, 0.223,
0.2223, 0.22223}. The numbers appear to be approaching a limit of 0.2 = 2>9. The series is convergent.
9. 80 # a
1 + 80
b = 40 # 81 = 3240
2
10. 35 # a
2 + 70
b = 35 # 36 = 1260
2
25. r =
11. 13 # a
117 + 33
b = 13 # 75 = 975
2
26. r =
12. a
111 + 27
b = 29 # 69 = 2001
2
For #13–16, use the formula Sn=
n=1+log|r| 2
311 - 2 2
14.
51 1 - 3 2
15.
423 1 - 1 1>6 2 9 4
16.
42 3 1 - 1 -1>62 10 4
1 - 2
a1 11 - rn 2
1 - r
28. r=3, so it diverges.
. Note that
29. r =
30. r =
= 24,573
= 147,620
1 - 11>6 2
1 - 1 -1>6 2
= 50.4 1 1 - 6-9 2 L 50.4
= 36 11 - 6-10 2 = 36 - 6-8 L 36
n
For #17–22, use one of the formulas Sn= [2a1+(n-1)d]
2
a1 1 1 - rn 2
.
or Sn=
1 - r
17. Arithmetic with d=3:
10 #
[2 3+(10-1)(3)]
2
=5 # 31=155
9
18. Arithmetic with d=–6: [2 # 14+(9-1)(–6)]
2
=9 # (–10)=–90
12
1 4 3 1 - 1 - 1>2 2 4
2
1 - 1 - 1>2 2
19. Geometric with r= - :
=
8
# 11 - 2-12 2 L 2.666
3
11
1 6 31 - 1 -1>2 2 4
20. Geometric with r= - :
2
1 - 1 -1>22
=4 # 11 + 2-11 2 L 4.002
14
693
14
707
=
+
=
99
99
99
99
32. 5 +
93
31
196
= 5 +
=
99
33
33
33. -17 -
17,251
268
= 999
999
34. - 12 -
876
292
4288
= -12 = 999
333
333
35. (a) The ratio of any two successive account balances is
r=1.1. That is,
$22,000
$24,200
$26,620
$29,282
=
=
=
= 1.1.
$20,000
$22,000
$24,200
$26,620
(b) Each year, the balance is 1.1 times as large as the year
before. So, n years after the balance is $20,000, it will
be $20,000 (1.1)n.
(c) The sum of the eleven terms of the geometric
$20,000 11 - 1.111 2
= $370,623.34
sequence is
1 - 1.1
36. (a) The difference of any two successive account balances
is d=$2016. That is $20,016-$18,000
=$22,032-$20,016=$24,048-$22,032
=$26,064-$24,048=$2016.
(b) Each year, the balance is $2016 more than the year
before. So, n years after the balance is $18,000, it will
be $18,000+$2016n.
(c) The sum of the eleven terms of the arithmetic
sequence is
11
321$18,0002 + 110 2 1$2016 2 4 = $308,880
2
- 23 1 - 1 -12 2 8 4
37. (a) The first term, 120(1+0.07/12)0, simplifies to 120.
The common ratio of terms, r, equals 1+0.07/12.
1 - 1 -12 2
2 #
8
=11 - 12 2 = 66,151,030
13
22. Geometric with r= -12 :
10>3
2
= 10.
, so it converges to S=
3
1 - 12>32
- 13 1 - 1 -11 2 9 4
1 - 1 -11 2
1
9
= - # 11 + 11 2 = -196,495,641
12
21. Geometric with r= -11 :
3>4
1
= 1.
, so it converges to S=
4
1 - 11>42
31. 7 +
10
1 - 3
4
1
, so it converges to S=
= 6.
3
1 - 11>32
27. r=2, so it diverges.
ln @ an>a1 @
an
2 = 1 +
.
a1
ln @ [email protected]
13.
13
6
1
, so it converges to S=
= 12.
2
1 - 11>22
23. (a) The first six partial sums are {0.3, 0.33, 0.333, 0.3333,
0.33333, 0.333333}. The numbers appear to be approaching a limit of 0.3 = 1>3. The series is convergent.
(b) The first six partial sums are {1, –1, 2, –2, 3, –3}. The
numbers approach no limit. The series is divergent.
(b) The sum of the 120 terms is
120 31 - 11 + 0.07>122 120 4
1 - 1 1 + 0.07>122
= $20,770.18
38. (a) The first term, 100(1+0.08/12)0, simplifies to 100.
The common ratio of terms, r, equals 1+0.08/12.
(b) The sum of the 120 terms is
100 31 - 11 + 0.08>122 120 4
1 - 11 + 0.08>122
= $18,294.60 .
Section 9.5
39. 2+2 # [2(0.9)+(2(0.9)2+2(0.9)3+ ... +2(0.9)9]
3.6 1 1 - 0.9 2
9
9
=2+ a 4 10.9 2 k = 2 +
=
1 - 0.9
38-36(0.99)≠24.05 m: The ball travels down 2 m.
Between the first and second bounces, it travels 2(0.9) m
up, then the same distance back down. Between the
second and third bounces, it travels 2(0.9)2 m up, then the
same distance back down, etc.
k=1
40. f(x) does have a horizontal asymptote to the left (since
as x S – q , f(x) S –40), but as x S q , f(x) S q . The
given series is geometric with r=1.05; the sum of the first
n terms in the series is f(n). Observing that f(x) S q as
x S q is confirmation that the geometric series diverges.
41. False. The series might diverge. For example, examine the
series 1+2+3+4+5 where all of the terms are
positive. Consider the limit of the sequence of partial
sums. The first five partial sums are {1, 3, 6, 10, 15}. These
numbers increase without bound and do not approach a
limit. Therefore, the series diverges and has no sum.
42. False. Justifications will vary. One example is to examine
q
n=1
a 1 -n2.
q
an
and
Series
367
43. 3–1+3–2+3–3+. . .+3–n+. . .=
1
1
1
1
1
1
+
+
+
+
+ ... + n + ...
3
9
27
81
243
3
1 4 13 40 121
The first five partial sums are b , , , ,
r. These
3 9 27 81 243
appear to be approaching a limit of 1/2, which would
suggest that the series converges to 1/2. The answer is A.
q
44. If a xn = 4,
then x=0.8.
n=1
q
n
a 0.8 = 0.8+0.64+0.512+0.4096+0.32768
n=1
+0.262144+. . .
The first six partial sums are {0.8, 1.44, 1.952, 2.3616,
2.68928, 2.951424}. It appears from this sequence of
partial sums that the series is converging. If the sequence
of partial sums were extended to the 40th partial sum, you
would see that the series converges to 4. The answer is D.
45. The common ratio is 0.75/3=0.25, so the sum of the
infinite series is 3/(1-0.25)=4. The answer is C.
46. The sum is an infinite geometric series with 0 r 0 = 5>3 7 1.
The answer is E.
n=1
Both of these diverge, but a 1n + 1 -n2 2 = a 0 = 0.
q
q
n=1
n=1
So the sum of the two divergent series converges.
47. (a) Heartland: 19,237,759 people.
Southeast: 42,614,977 people.
(b) Heartland: 517,825 mi2.
Southeast: 348,999 mi2.
19,237,759
L 37.15 people/mi2.
(c) Heartland:
517,825
Southeast:
42,614,977
L 122.11 people/mi2.
348,999
Heartland:
Southeast:
Iowa
≠52.00
Kansas
≠32.68
Minnesota ≠58.29
Alabama
Arkansas
≠86.01
≠50.26
Florida
≠272.53
≠138.97
Missouri
≠80.28
Georgia
Nebraska
≠22.12
Louisiana
N. Dakota
≠9.08
Mississippi
≠59.65
S. Dakota
≠9.79
S.Carolina
≠128.95
Total
≠829.96
Average
≠118.57
Total
≠264.24
Average
≠37.75
≠93.59
(d) The table is shown on the right; the answer differs
because the overall population density
a population
population
1
is generally not the same as the average of the population densities, a a
b . The larger states
n
area
area
a
within each group have a greater effect on the overall mean density. In a similar way, if a student’s grades are based on a 100point test and four 10-point quizzes, her overall average grade depends more on the test grade than on the four quiz grades.
Chapter 9
368
Discrete Mathematics
■ Section 9.6 Mathematical Induction
8
48. a 1k2 - 2 2
k=1
Exploration 1
n
49. The table suggests that Sn = a Fk = Fn + 2 - 1.
k=1
Sn
Fn+2-1
1. Start with the rightmost peg if n is odd and the middle
peg if n is even. From that point on, the first move for
moving any smaller stack to a destination peg should be
directly to the destination peg if the smaller stack’s size n
is odd and to the other available peg if n is even. The fact
that the winning strategy follows such predictable rules is
what makes it so interesting to students of computer programming.
n
Fn
1
1
1
1
2
1
2
2
3
2
4
4
4
3
7
7
5
5
12
12
6
8
20
20
7
13
33
33
1. 43, 47, 53, 61, 71, 83, 97, 113, 131, 151. Yes.
8
21
54
54
2. 173, 197, 223, 251, 281, 313, 347, 383, 421, 461. Yes.
9
34
88
88
3. 503, 547, 593, 641, 691, 743, 797, 853, 911, 971. Yes.
Inductive thinking might lead to the conjecture that
n¤+n+41 is prime for all n, but we have no proof
as yet!
50. The nth triangular number is simply the sum of the first n
consecutive positive integers:
n1n + 1 2
1 + n
1+2+3+. . .+n=n a
b =
2
2
51. Algebraically: Tn1+Tn=
1n - 1 2n
2
+
n1n + 1 2
Exploration 2
4. The next 9 numbers are all prime, but 40¤+40+41
is not. Quite obviously, neither is the number
41¤+41+41.
2
Quick Review 9.6
n2 - n + n2 + n
=
= n2.
2
1. n2+5n
Geometrically: the array of black dots in the figure
represents Tn=1+2+3+. . .+n (that is, there are
Tn dots in the array). The array of gray dots represents
Tn1=1+2+3+. . .+(n-1). The two triangular
arrays fit together to form an n*n square array, which
has n2 dots.
2. n2-n-6
3. k3+3k2+2k
4. (n+3)(n-1)
5. (k+1)3
6. (n-1)3
7. f(1)=1+4=5, f(t)=t+4,
f(t+1)=t+1+4=t+5
n–1
1
k
1
= , f1k2 =
,
1 + 1
2
k + 1
k + 1
k + 1
f(k+1)=
=
k + 1 + 1
k + 2
8. f(1)=
n–1
n
2#1
1
= ,
+ 1
2
21k + 12
2k
2k + 2
P1 k2 =
=
=
3k + 1
31 k + 1 2 + 1
3k + 4
9. P(1)=
n
n
1
52. If a ln n for all n, then the sum diverges since as
k=1 k
n S q , ln n S q .
[–1, 10] by [–2, 4]
3#1
10. P(1)=2(1)2-1-3=–2, P(k)=2k2-k-3,
P(k+1)=2(k+1)2-(k+1)-3=2k2+3k-2
Section 9.6 Exercises
1. Pn: 2+4+6+. . .+2n=n2+n.
P1 is true: 2(1)=12+1.
Now assume Pk is true: 2+4+6+…+2k
=k2+k. Add 2(k+1) to both sides:
2+4+6+. . .+2k+2(k+1)
=k2+k+2(k+1)=k2+3k+2
=k2+2k+1+k+1=(k+1)2+(k+1), so
Pk+1 is true. Therefore, Pn is true for all n 1.
Section 9.6
2. Pn: 8+10+12+. . .+(2n+6)=n2+7n.
P1 is true: 2(1)+6=12+7 # 1.
Now assume Pk is true:
8+10+12+. . .+(2k+6)=k2+7k.
Add 2(k+1)+6=2k+8 to both sides:
8+10+12+. . .+(2k+6)+[2(k+1)+6]
=k2+7k+2k+8=(k2+2k+1)+7k+7
=(k+1)2+7(k+1), so Pk+1 is true.
Therefore, Pn is true for all n 1.
3. Pn: 6+10+14+. . .+(4n+2)=n(2n+4).
P1 is true: 4(1)+2=1(2+4).
Now assume Pk is true:
6+10+14+. . .+(4k+2)=k(2k+4).
Add 4(k+1)+2=4k+6 to both sides:
6+10+14+. . .+(4k+2)+[4(k+1)+2]
=k(2k+4)+4k+6=2k2+8k+6
=(k+1)(2k+6)=(k+1)[2(k+1)+4], so Pk+1
is true. Therefore, Pn is true for all n 1.
4. Pn: 14+18+22+. . .+(4n+10)=2n(n+6).
P1 is true: 4(1)+10=2 # 1(1+6). Now assume Pk is
true:
14+18+22+. . .+(4k+10)=2k(k+6). Add
4(k+1)+10=4k+14 to both sides:
14+18+22+. . .+(4k+10)+[4(k+1)+10]
=2k(k+6)+(4k+14)
=2(k2+8k+7)=2(k+1)(k+7)
=2(k+1)(k+1+6), so Pk+1 is true. Therefore, Pn
is true for all n 1.
5. Pn: 5n-2. P1 is true: a1=5 # 1-2=3.
Now assume Pk is true: ak=5k-2.
To get ak+1, add 5 to ak; that is,
ak+1=(5k-2)+5=5(k+1)-2 This shows that
Pk+1 is true. Therefore, Pn is true for all n 1.
6. Pn: an=2n+5. P1 is true: a1=2 # 1+5=7.
Now assume Pk is true: ak=2k+5.
To get ak+1, add 2 to ak; that is,
ak+1=(2k+5)+2=2(k+1)+5 This shows that
Pk+1 is true. Therefore, Pn is true for all n 1.
7. Pn: an=2 # 3n-1.
P1 is true: a1=2 # 31-1=2 # 30=2.
Now assume Pk is true: ak=2 # 3k-1.
To get ak+1, multiply ak by 3; that is,
ak+1=3 # 2 # 3k - 1=2 # 3k=2 # 3(k+1)-1. This shows that
Pk+1 is true. Therefore, Pn is true for all n 1.
8. Pn: an=3 # 5n-1.
P1 is true: a1=3 # 51-1=3 # 50=3.
Now assume Pk is true: ak=3 # 5k-1.
To get ak+1, multiply ak by 5; that is,
ak+1=5 # 3 # 5k-1=3 # 5k=3 # 5(k+1)-1. This shows that
Pk+1 is true. Therefore, Pn is true for all n 1.
9. P1: 1=
111 + 12
2
.
Pk : 1 + 2 + . . . + k =
k1k + 12
2
Pk+1: 1+2+. . .+k+(k+1)=
1 k + 1 2 1k + 22
2
.
10. P1: (2(1)-1)2=
Mathematical Induction
1 12 - 12 12 + 12
3
Pk: 12+32+. . .+(2k-1)2=
.
k 12k - 1 2 12k + 12
3
Pk+1: 12+32+. . .+(2k-1)2+(2k+1)2
1k + 12 12k + 12 12k + 3 2
=
.
3
1
1
=
.
1#2
1 + 1
1
1
1
k
=
Pk :
+
+ ... +
.
1#2
2#3
k 1k + 12
k + 1
1
1
1
Pk+1:
+ # + ... +
1#2
2 3
k1k + 12
1
k + 1
=
.
+
1k + 1 2 1k + 22
k + 2
11. P1 :
12. P1: 14=
1 11 + 12 12 + 1 2 1 3 + 3 - 1 2
#
30
Pk: 14+24+. . .+k4
k1 k + 1 2 12k + 1 2 13k2 + 3k - 12
=
.
30
4
4
4
4
Pk+1: 1 +2 +. . .+k +(k+1)
1k + 1 2 1k + 22 12k + 3 2 13k2 + 9k + 52
=
.
30
13. Pn: 1+5+9+. . .+(4n-3)=n(2n-1).
P1 is true: 4(1)-3=1 # (2 # 1-1).
Now assume Pk is true:
1+5+9+. . .+(4k-3)=k(2k-1).
Add 4(k+1)-3=4k+1 to both sides:
1+5+9+. . .+(4k-3)+[4(k+1)-3]
=k(2k-1)=4k+1=2k2+3k+1
=(k+1)(2k+1)=(k+1)[2(k+1)-1],
so Pk+1 is true.
Therefore, Pn is true for all n 1.
14. Pn: 1+2+22+. . .+2n-1=2n-1.
P1 is true: 21-1=21-1.
Now assume Pk is true:
1+2+22+. . .+2k-1=2k-1.
Add 2k to both sides,
1+2+22+. . .+2k-1+2k
=2k-1+2k=2 # 2k-1=2k+1-1, so
Pk+1 is true.
Therefore, Pn is true for all n 1.
1
1
n
1
+ # +. . .+
=
.
1#2
2 3
n1 n + 12
n + 1
1
1
P1 is true:
=
.
1#2
1 + 1
Now assume Pk is true:
1
1
1
+ # + ... +
.
1#2
2 3
k1k + 12
1
Add
to both sides:
1k + 12 1k + 2 2
1
1
1
1
+ # +. . .+
+
1#2
2 3
k1 k + 12
1k + 1 2 1k + 22
k1k + 2 2 + 1
1
k
=
=
+
k + 1
1 k + 1 2 1k + 22
1 k + 12 1 k + 22
1k + 12 1k + 12
k + 1
k + 1
=
=
,
=
1k + 12 1k + 22
k + 2
1k + 12 + 1
so Pk+1 is true.
Therefore, Pn is true for all n 1.
15. Pn:
369
.
370
Chapter 9
Discrete Mathematics
1
1
1
n
=
+ # +. . .+
.
1#3
3 5
12n - 1 2 12n + 1 2
2n + 1
1
1
P1 is true: it says that
= #
.
1#3
2 1 + 1
Now assume Pk is true:
1
1
k
1
+
+ ... +
=
.
1#3
3#5
12k - 1 2 12k + 1 2
2k + 1
1
Add
3 2 1 k + 12 - 1 4 32 1 k + 1 2 + 14
1
=
to both sides, and we have
12k + 12 12k + 3 2
1
1
1
+ # +. . .+
#
1 3
3 5
12k - 1 2 12k + 1 2
1
+
32 1k + 12 - 1 4 3 21k + 1 2 + 14
k1 2k + 3 2 + 1
k
1
=
+
=
...
2k + 1
1 2k + 1 2 12k + 3 2
12k + 1 2 12k + 32
12k + 12 1k + 12
k + 1
=
=
,
12k + 12 12k + 3 2
2 1k + 1 2 + 1
so Pk+1 is true.
Therefore, Pn is true for all n 1.
16. Pn:
17. Pn: 2n 2n.
P1 is true: 21 2 # 1 (in fact, they are equal). Now assume
Pk is true: 2k 2k.
Then 2k+1=2 # 2k 2 # 2k
=2 # (k+k) 2(k+1), so Pk+1 is true. Therefore,
Pn is true for all n 1.
18. Pn: 3n 3n.
P1 is true: 31 3 # 1 (in fact, they are equal). Now assume
Pk is true: 3k 3k.
Then 3k+1=3 # 3k 3 # 3k
=3 # (k+2k) 3(k+1), so Pk+1 is true.
Therefore, Pn is true for all n 1.
19. Pn: 3 is a factor of n3+2n.
P1 is true: 3 is a factor of 13+2 # 1=3.
Now assume Pk is true: 3 is a factor of k3+2k.
Then (k+1)3+2(k+1)
=(k3+3k2+3k+1)+(2k+2)
=(k3+2k)+3(k2+k+1).
Since 3 is a factor of both terms, it is a factor of the sum,
so Pk+1 is true. Therefore, Pn is true for all n 1.
20. Pn: 6 is a factor of 7n-1.
P1 is true: 6 is a factor of 71-1=6.
Now assume Pk is true, so that 6 is a factor of
7k-1=6.
Then 7k+1-1=7 # 7k-1=7(7k-1)+6. Since 6 is
a factor of both terms of this sum, it is a factor of the sum,
so Pk+1 is true.
Therefore, Pn is true for all positive integers n.
21. Pn: The sum of the first n terms of a geometric sequence
with first term a1 and common ratio r Z 1 is
P1 is true: a1=
a1 11 - r1 2
.
1 - r
Now assume Pk is true so that
a1 11 - rk 2
.
a1+a1r+p=a1rk-1=
11 - r2
a1 11 - rn 2
1 - r
.
Add a1rk to both sides: a1+a1r+p=a1rk-1
+a1rk=
=
a1 11 - rk 2
+ a1rk
11 - r2
a1 11 - rk 2 + ark 11 - r2
1 - r
a1 - a1rk - a1rk + 1
a1 - a1rk + 1
=
=
,
1 - r
1 - r
so Pk+1 is true. Therefore, Pn is true for all positive
integers n.
n
22. Pn: Sn= [2a1+(n-1)d].
2
First note that an=a1+(n-1)d. P1 is true:
1
1
S1= [2a1+(1-1)d]= (2a1)=a1.
2
2
k
Now assume Pk is true: Sk= [2a1+(k-1)d].
2
Add ak+1=a1+kd to both sides, and observe that
Sk+ak+1=Sk+1.
Then we have
k
Sk+1= [2a1+(k-1)d]+a1+kd
2
1
=ka1+ k(k-1)d+a1+kd
2
1
=(k+1)a1+ k(k+1)d
2
k + 1
=
[2a1+(k+1-1)d].
2
Therefore, Pk+1 is true, so Pn is true for all n 1.
n
n1n + 12
23. Pn: a k =
2
k=1
.
1
P1 is true: a k = 1 =
k=1
1#2
.
2
k
Now assume Pk is true: a i =
k1k + 12
.
2
Add (k+1) to both sides, and we have
k+1
k 1k + 1 2
+ 1k + 1 2
ai =
2
i=1
21k + 1 2
1k + 12 1k + 2 2
k1 k + 1 2
+
=
=
2
2
2
1k + 1 2 1k + 1 + 12
, so Pk+1 is true.
=
2
Therefore, Pn is true for all n 1.
i=1
n
24. Pn: a k3 =
n2 1n + 1 2 2
. P1 is true: 13 =
1222
.
4
4
Now assume Pk is true so that
k2 1k + 1 2 2
13+23+p+k3=
. Add (k+1)3 to both
4
sides and we have
13+23+p+k3+(k+1)3
k2 1k + 12 2 + 41k + 12 3
k2 1k + 12 2
=
+(k+1)3=
4
4
1k + 12 2 1 1k + 12 + 12 2
1k + 1 2 2 1k2 + 4k + 42
=
=
4
4
so Pk+1 is true. Therefore, Pn is true for all positive
integers.
k=1
Section 9.6
1500 2 1501 2
500
25. Use the formula in 23: a k =
2
k=1
250
26. Use the formula in Example 2: a k2 =
= 125,250
1250 2 12512 1501 2
6
k=1
=5,239,625
n
n
3
27. Use the formula in 23: a k = a k - a k
k=4
=
k=1
k=1
1n - 32 1n + 42
n2 + n - 12
3#4
=
=
2
2
2
n1n + 12
2
1752 2 1762 2
75
28. Use the formula in 24: a k3 =
4
k=1
=8,122,500
35
29. Use the formula in 14: a 2k - 1
k=1
=235-1≠3.44*1010
1152 2 1162 2
15
30. Use the formula in 24: a k3 =
4
k=1
=14,400
n
n
n
n
31. a (k2-3k+4)= a k2 - a 3k + a 4
k=1
=
n1n + 12 12n + 12
6
n1n - 3n + 82
k=1
k=1
- 3 c
k=1
n1 n + 12
2
d + 4n
2
=
3
n
n
n
n
k=1
k=1
32. a 12k2 + 5k - 2 2 = a 2k2 + a 5k - a 2 = 2
k=1
k=1
= 2c
n1n + 1 2 1 2n + 1 2
6
n14n + 21n + 52
d +5 c
2
=
=
6
n
n
n1n + 1 2
d - 2n
2
n 1n + 5 2 1 4n + 1 2
6
n2 1 n + 1 2 2
n
33. a 1k3 - 1 2 = a k3 - a 1 =
k=1
=
n1n3 + 2n2 + n - 4 2
k=1
4
n
k=1
=
n 1n + 1 2
2
4
n1n - 1 2 1 n2 + 3n + 42
4
n
n
n
k=1
k=1
k=1
2
n1 n + 12
+ 4c
d - 5n
4
2
n1n - 1 2 1 n2 + 3n + 122
n1n3 + 2n2 + 9n - 12 2
=
=
4
4
=
371
40. The anchor step, proving P1, comes first. The answer is A.
41. Mathematical induction could be used, but the formula
for a finite arithmetic sequence with a1=1, d=2 would
also work. The answer is B.
42. The first two partial sums are 1 and 9. That eliminates all
answers except C. Mathematical induction can be used to
show directly that C is the correct answer.
43. Pn: 2 is a factor of (n+1)(n+2). P1 is true because 2
is a factor of (2)(3). Now assume Pk is true so that 2 is a
factor of (k+1)(k+2). Then
[(k+1)+1][(k+2)+2]
=(k+2)(k+3)=k2+5k+6
=k2+3k+2+2k+4
=(k+1)(k+2)+2(x+2). Since 2 is a factor of
both terms of this sum, it is a factor of the sum, and so Pk+1
is true. Therefore, Pn is true for all positive integers n.
44. Pn: 6 is a factor of n(n+1)(n+2). P1 is true because 6
is a factor of (1)(2)(3). Now assume Pk is true so that 6 is
a factor of k(k+1)(k+2). Then
(k+1)[(k+1)+1][(k+1)+2]
=k(k+1)(k+2) +3(k+1)(k+2). Since 2 is a
factor of (k+1)(k+2), 6 is a factor of both terms of
the sum and thus of the sum itself, and so Pk+1 is true.
45. Given any two consecutive integers, one of them must be
even. Therefore, their product is even. Since n+1 and
n+2 are consecutive integers, their product is even.
Therefore, 2 is a factor of (n+1)(n+2).
46. Given any three consecutive integers, one of them must
be a multiple of 3, and at least one of them must be even.
Therefore, their product is a multiple of 6. Since n, n+1
and n+2 are three consecutive integers, 6 is a factor of
n(n+1)(n+2).
n
-n
34. a 1k3 + 4k - 5 2 = a k3 + a 4k- a 5
k=1
Mathematical Induction
35. The inductive step does not work for 2 people. Sending
them alternately out of the room leaves 1 person (and
one blood type) each time, but we cannot conclude that
their blood types will match each other.
36. The number k is a fixed number for which the statement
Pk is known to be true. Once the anchor is established, we
can assume that such a number k exists. We can not
assume that Pn is true, because n is not fixed.
37. False. Mathematical induction is used to show that a statement Pn is true for all positive integers.
2
38. True. (1+1) =4=4(1). Pn is false, however, for all
other values of n.
39. The inductive step assumes that the statement is true for
some positive integer k. The answer is E.
47. Pn: Fn+2-1= a Fk. P1 is true since
k=1
F1+2=1=F3-1=2-1=1, which equals
1
a Fk = 1. Now assume that Pk is true:
k=1
k
Fk + 2 - 1 = a Fi. Then F(k+1)+2-1
i=1
=Fk+3-1=Fk+1+Fk+2-1
k
=(Fk+2-1)+Fk+1= a a Fi b + Fk + 1
i=1
k+1
= a Fi, so Pk+1 is true. Therefore, Pn is true
i=1
for all n 1.
48. Pn: an<2. P1 is easy: a1= 22<2. Now assume that Pk
is true: ak<2. Note that
ak+1= 22 + ak , so that
a2k+1<2+ak<2+2=4; therefore ak+1<2,
so Pk+1 is true. Therefore, Pn is true for all n 1.
49. Pn: a-1 is a factor of an-1. P1 is true because a-1
is a factor of a-1. Now assume Pk is true so that a-1
is a factor of ak-1. Then ak+1-1=a # ak-1
=a(ak-1)+(a-1). Since a-1 is a factor of both
terms in the sum, it is a factor of the sum, and so Pk+1 is
true. Therefore, Pn is true for all positive integers n.
372
Chapter 9
Discrete Mathematics
50. Let P(a)=an-1. Since P(1)=1n-1=0, the Factor
Theorem for polynomials allows us to conclude that
a-1 is a factor of P.
51. Pn: 3n-4 n for n 2. P2 is true since
3 # 2 - 4 2. Now assume that Pk is true:
3k-4 2. Then 3(k+1)-4=3k+3-4
=(3k-4)+3k k+3 k+1, so Pk+1 is true.
Therefore, Pn is true for all n 2.
52. Pn: 2n n2 for n 4. P4 is true since 24 42. Now assume
that Pk is true: 2k k2. Then
2k+1=2 # 2 k 2 # k2 2 # k2 k2 + 2k + 1
=(k+1)2. The inequality 2k2 k2+2k+1, or
equivalently, k2 2k+1, is true for all k 4 because
k2=k # k 4k=2k+2k>2k+1.) Thus Pk+1 is
true, so Pn is true for all n 4.
53. Use P3 as the anchor and obtain the inductive step by
representing any n-gon as the union of a triangle and an
(n-1)-gon.
2. Maris
985
643
863
93
Aaron
0
1 023
2 04679
3 0244899
4 00444457
5
1 6
Except for Maris’s one record-breaking year, his home
run output falls well short of Aaron’s.
3.
Males
6 0 3
6 7 8 8 8
7 1 2 2 3 3 3
7
This stemplot shows the life expectancies of males in the
nations of South America are clustered near 70, with two
lower values clustered near 60.
4.
Females
6 5 8
7 1 2
7 5 6 7 7 9 9
8 0 0
This stemplot shows the life expectancies of females in
the nations of South America are clustered in the high
70’s and at 80, with four lower values in the high 60’s and
low 70’s.
■ Section 9.7 Statistics and Data (Graphical)
Exploration 1
1. We observe that the numbers seem to be centered a bit
below 13. We would need to take into account the
different state populations (not given in the table) in
order to compute the national average exactly; but, just
for the record, it was about 12.5 percent.
5.
2. We observe in the stemplot that 3 states have
percentages above 15.
3. We observe in the stemplot that the bottom five states are
all below 10%. Returning to the table, we pick these out
as Alaska, Colorado, Georgia, Texas, and Utah.
4. The low outlier is Alaska, where older people would be less
willing or able to cope with the harsh winter conditions.
The high outlier is Florida, where the mild weather and
abundant retirement communities attract older residents.
Males
Females
3 0 6
8 8 8 7 6 5 8
3 3 3 2 2 1 7 1 2
7 5 6 7 7 9 9
8 0 0
This stemplot shows that the life expectancies of the
women in the nations of South America are about
5–6 years higher than that of the men in the nations of
South America.
6. 9200
6. Female-Male Difference
3 5
4 7 8
5 2 4 8
6 3 5 7
7 7 8
8 1
This stemplot of differences shows that the women in the
nations of South America have higher life expectancies
than the men, by about 5–6 years.
7. 235 thousand
7.
Quick Review 9.7
1. ≠15.48%
2. ≠20.94%
3. ≠14.44%
4. ≠27.22%
5. ≠1723
8. 238 million
9. 1 million
10. 1 billion
Section 9.7 Exercises
1. 0 5 8 9
1 346
2 368
3 39
4
5
6 1
61 is an outlier.
Life
expectancy
(years)
Frequency
60.0–64.9
2
65.0–69.9
4
70.0–74.9
6
Section 9.7
8.
Life
expectancy
(years)
Frequency
60.0–64.9
1
65.0–69.9
1
70.0–74.9
2
75.0–79.9
8
Statistics and Data (Graphical)
373
15.
[–1.5, 17] by [–2, 80]
16.
9.
[–1, 25] by [–5, 50]
[50, 80] by [–1, 9]
17.
10.
[1965, 2008] by [–1000, 11000]
The top male’s earnings appear to be growing exponentially, with unusually high earnings in 1999 and 2000.
Since the graph shows the earnings of only the top player
(as opposed to a mean or median for all players), it can
behave strangely if the top player has a very good year —
as Tiger Woods did in 1999 and 2000.
[50, 80] by [–1, 9]
11.
18.
[0, 60] by [–1, 5]
12.
[1965, 2008] by [–500, 3400]
The top female’s earnings appear to be growing faster
than linearly, but with more fluctuation since 1990. Since
the graph shows the earnings of only the top player
(as opposed to a mean or median for all players), it is
strongly affected by how well the top player does in a
given year. The top player in 2002 and 2003, Annika
Sorenstam, played in fewer tournaments in 2003 and
hence won less money.
[0, 60] by [–1, 5]
13.
19.
[–1, 25] by [–5, 60]
14.
[1965, 2008] by [–1000, 11000]
[–1, 20] by [–5, 60]
After approaching parity in 1985, the top PGA player’s
earnings have grown much faster than the top LPGA
player’s earnings, even if the unusually good years for
Tiger Woods (1999 and 2000) are not considered part of
the trend.
374
Chapter 9
Discrete Mathematics
20. The 1985 male total was unusually low in the context of
the rest of the time plot. If it had been in line with the
earlier and later data, the difference between the PGA
and LPGA growth rates would not have been that large.
(The unusual PGA point in 1985, oddly enough, belonged
to Curtis Strange.)
(b)
21.
Interval
6.0–6.9
7.0–7.9
8.0–8.9
9.0–9.9
10.0–10.9
11.0–11.9
12.0–12.9
Frequency
4
4
8
13
8
4
3
(c)
[–1, 25] by [–5, 60]
The two home run hitters enjoyed similar success, with Mays
enjoying a bit of an edge in the earlier and later years of his
career, and Mantle enjoying an edge in the middle years.
22.
[5, 14] by [–2, 15]
(d) The data are not categorical.
25.
[–1, 25] by [–5, 80]
The two home run hitters were comparable for the first
seven years of their careers.
23. (a)
Stem Leaf
28
2
29 † 3 7
30
31
67
32
78
33
5558
34
288
35
334
36
37
37 †
38
5
∞
(b) Interval
25.0–29.9
30.0–34.9
35.0–39.9
Frequency
3
11
6
(c)
[1890, 2010] by [–4, 40]
=CA
+=NY
■=TX
26.
[1890, 2010] by [–4, 40]
=PA
+=IL
■=FL
27. False. The empty branches are important for visualizing
the true distribution of the data values.
28. False. They are outliers only if they are significantly
higher or lower than the other numbers in the data set.
29. A time plot uses a continuous line. The answer is C.
30. Back-to-back stemplots are designed for comparing data
sets. The answer is B.
31. The histogram suggests data values clustered near an
upper limit — such as the maximum possible score on an
easy test. The answer is A.
32. 45° is 1/8, or 12.5%, of 360°. The answer is B.
[20, 45] by [–1, 13]
(d) Time is not a variable in the data.
24. (a) Stem
6
7
8
9
10
11
12
Leaf
2399
7999
02377899
0001123334567
23366679
0246
059
33. Answers will vary. Possible outliers could be the pulse
rates of long-distance runners and swimmers, which are
often unusually low. Students who have had to run to
class from across campus might have pulse rates that
are unusually high.
34. Answers will vary. Female heights typically have a
distribution that is uniformly lower than male heights,
but the difference might not be apparent from a stem
plot, especially if the sample is small.
Section 9.8
35.
Statistics and Data (Algebraic)
375
p
g(t)≠6.5 +15.5 sin c 1t 3.52 d or
6
p
6.5+15.5 sin c 1t 4 2 d .
6
[0, 13] by [–15, 40]
36. Using the form f(t)=k+a sin[b(t-h)], we have
b = p>6 for both graphs since the period is 12. a is half
the difference between the extremes, and k is the average
of the extremes. Finally, h is the offset to the first time this
average occurs.
The maxima of both graphs occur in July (t=7), while
the minima occur in January (t=1), so we should choose
b=3.5 or 4 (if the latter is halfway between the
maximum and the minimum, but the former gives a
better match for some values of t). Thus, for the high
p
temperature, use f(t)≠17+15 sin c 1t 3.52 d or
6
■ Section 9.8 Statistics and Data (Algebraic)
Exploration 1
1. In Figure 9.19 (a) the extreme values will cause the range
to be big, but the compact distribution of the rest of the
data indicate a small interquartile range. The data in
Figure 9.19 (b) exhibit high variability.
2. Figure 9.20 (b) has a longer “tail” to the right (skewed
right), so the values in the tail pull the mean to the right
of the (resistant) median. Figure 9.20 (c) is skewed left, so
the values in the tail pull the mean to the left of the
median. Figure 9.20 (a) is symmetric about a vertical line,
so the median and the mean are close together.
p
17+15 sin c 1t 4 2 d , and for the low temperature, use
6
Quick Review 9.8
7
1. a xi = x1 + x2 + x3 + x4 + x5 + x6 + x7
i=1
5
2. a 1xi - x2 = 1x1 - x2 + 1x2 - x2 + 1x3 - x2 + 1x4 - x2 + 1x5 - x2 =x1+x2+x3+x4+x5-5x.
i=1
Note that, since x =
3.
4.
1 5
xi, this simplifies to 0.
5 ia
=1
1 7
1
xi = (x1+x2+x3+x4+x5+x6+x7)
7 ia
7
=1
1 5
1
1
1 5
1xi - x2 = (x1+x2+x3+x4+x5-5x)= (x1+x2+x3+x4+x5)-x. Note that, since x = a xi, this
a
5 i=1
5
5
5 i=1
simplifies to 0.
5. The expression at the end of the first line is a simple expansion of the sum (and is a reasonable answer to the given question).
By expanding further, we can also arrive at the final expression below, which is somewhat simpler.
1 5
1
1 xi - x2 2= 3 1x1 - x2 2 + 1x2 - x2 2 + . . . + 1 x5 - x2 2 4
5 ia
5
=1
1
= 3 1 x21 - 2x1x + x2 2 + 1 x22 - 2x2x + x2 2 + . . . + 1x25 - 2x5x + x2 2 4
5
1
= 3x21 + x22 + . . . + x25 - 2x1x1 + x2 + . . . + x5 2 4 + x2
5
1
1
= 1x21 + x22 + . . . + x25 2 - 2x2 + x2 = 1x21 + x22 + . . . + x25 2 - x2
5
5
6. The square root does not allow for further simplication. The final answer is the square root of the expression from #5: either
1 2
1
3 1 x1 - x2 2 + 1x2 - x2 2 + . . . + 1 x5 - x2 2 4 or
1x + x22 + . . . + x25 2 - x 2.
C5 1
C5
8
7.
a xifi
i=1
1 50
9.
1 xi - x2 2
50 ia
=1
10
8. a 1 xi - x2 2
i=1
10.
1 7
1 xi - x2 2
C 7 ia
=1
376
Chapter 9
Discrete Mathematics
Section 9.8 Exercises
1. (a) Statistic. The number characterizes a set of known data values.
(b) Parameter. The number describes an entire population, on the basis of some statistical inference.
(c) Statistic. The number is calculated from information about all rats in a small, experimental population.
2. (a) Mode. No quantitative measure beyond what is “most common” is implied.
(b) Mean. A pitcher’s earned run average (ERA) is total earned runs divided by number of nine-inning blocks pitched.
(c) Median. The middle height is considered average.
1
134
= 26.8
3. x= (12+23+15+48+36)=
5
5
1
57
= 9.5
4. x= 1 4 + 8 + 11 + 6 + 21 + 7 2 =
6
6
1
360.7
3607
=
L 60.12
5. x= (32.4+48.1+85.3+67.2+72.4+55.3)=
6
6
60
1
992
198.4
=
L 28.3
6. x= (27.4+3.1+9.7+32.3+12.8+39.4+73.7)=
7
7
35
1
7. x= (1.5+0.5+4.8+7.3+6.3+3.0)=3.9 million, or 3,900,000
6
1
8. x= (29.8+12.9+11.4+18.0+11.9+17.0)≠16.83 million, or about 16,830,000
6
1
137
L 15.2 satellites
9. x= (0+0+1+2+61+33+26+13+1)=
9
9
1
1
10. x= (30,065,000+13,209,000+44,579,000+. . .+17,819,000)= 1148,196,000 2 = 21,171,000 km2
7
7
11. There are 9 data values, which is an odd number, so the median is the middle data value when they are arranged in order. In
order, the data values are {0, 0, 1, 1, 2, 8, 21, 28, 30}. The median is 2.
12. There are 7 data values, which is an odd number, so the median is the middle data value when they are arranged in order. In
order, the data values are {7687, 9938, 13,209, 17,819, 24,256, 30,065, 44,579} in thousands of km2. The median is 17,819,000 km2.
13. Mays:
536
660
= 30 home runs/year. Mantle:
L 29.8 home runs/year. Mays had the greater production rate.
22
18
14. State College:
12
15
= 2.4 houses/day. College Station:
L 2.14 houses/day. The State College workers were faster.
5
7
15. No computations needed — Hip-Hop House: 1147 skirts in 4 weeks. What-Next Fashions: 1516 skirts in 4 weeks. What-Next
Fashions had the greater production rate.
311,000,000,000
L $352.
882,575,000
218,000,000,000
L $2485. Mexico has the higher PCI.
Mexico:
87,715,000
16. India:
17. For {79.5, 82.1, 82.3, 82.9, 84.0, 84.5, 84.8, 85.4, 85.7, 85.7, 86.0, 86.4, 87.0, 87.5, 87.7, 88.0, 88.0, 88.2, 88.2, 88.5, 89.2, 89.2, 89.8, 89.8,
87.7 + 88.0
= 87.85, and there is no mode.
91.1, 91.3, 91.6, 91.8, 94.1, 94.4}, the median is
2
18. x =
5118,5222 + 4 127,102 2 + 3 131,286 2 + 21 20,7322 + 1 118,8752
18,522 + 27,102 + 31,286 + 20,732 + 18,875
L 3.05
19. x =
5198792 + 4151192 + 3 16143 2 + 2 12616 2 + 1 130272
9879 + 5119 + 6143 + 2616 + 3027
L 3.61
1
221
L 18.42°C
(2+5+12+20+. . .+10+3)=
12
12
1 2 2 131 2 + 1 52 128 2 + 1 12 2 131 2 + 120 2 130 2 + … + 1102 1302 + 13 2 131 2
6748
L 18.49°C
(b) Weighted: x =
=
31 + 28 + 31 + 30 + … + 30 + 31
365
20. (a) Non-weighted: x =
(c) The weighted average is the better indicator.
Section 9.8
Statistics and Data (Algebraic)
377
77
1
(–9-7-1+7+. . .-1-7)=
L 6.42°C
12
12
1 -9 2 1 31 2 + 1 -7 2 128 2 + 1 -1 2 131 2 + 17 2 130 2 + … + 1 -12 1 302 + 1 -7 2 131 2
2370
(b) Weighted: x =
=
L 6.49°C
31 + 28 + 31 + 30 + … + 30 + 31
365
21. (a) Non-weighted: x =
(c) The weighted average is the better indicator.
22. The ordered data for Mark McGwire is {3, 9, 9, 22, 29, 32,
32, 33, 39, 39, 42, 49, 52, 58, 65, 70} so the median is
33 + 39
= 36,
2
22 + 29
49 + 52
Q3=
= 50.5, Q1=
= 25.5.
2
2
Five-number summary: {3, 25.5, 36, 50.5, 70}
Range: 70-3=67
IQR: 50.5-25.5=25
No outliers
The ordered data for Barry Bonds is:
{16, 19, 24, 25, 25, 33, 33, 34, 34, 37, 37, 40, 42, 46, 49, 73} so
34 + 34
the median is
= 34,
2
40 + 42
25 + 25
Q3=
= 41, and Q1=
= 25
2
2
Five-number summary: {16, 25, 34, 41, 73}
Range: 73-16=57
IQR: 41-25=16
Outlier: 73
23. The ordered data for Willie Mays is: {4, 6, 8, 13, 18, 20, 22,
23, 28, 29, 29, 34, 35, 36, 37, 38, 40, 41, 47, 49, 51, 52} so the
median is
29 + 34
= 31.5, Q3=40, and Q1=20
2
Five-number summary: {4, 20, 31.5, 40, 52}
Range: 52-4=48
IQR: 40-20=20
No outliers
The ordered data for Mickey Mantle is:
{13, 15, 18, 19, 21, 22, 23, 23, 27, 30, 31, 34, 35, 37, 40, 42,
52, 54} so the median is
27 + 30
= 28.5, Q3 = 37, and Q1 = 21
2
Five-number summary: {13, 21, 28.5, 37, 54}
Range: 54-13=41
IQR: 37-21=16
No outliers
24. While a stemplot is not needed to answer this question,
the sorted stemplot below is more compact than a sorted
list of the 44 numbers. The underlined numbers are the
ones used for the five-number summary, which is
Min
Q1
Median
Q3
Max
8.5
9.25
10.6
12.9
8.3 + 8.7
9.2 + 9.3
10.6 + 10.6
=
=
=
2
2
2
The range is 12.9-6.2=6.7 and
IQR =10.6-8.5=2.1 . There are no outliers, since
none of the numbers fall below Q1-1.5*IQR=5.35
or above Q3+1.5*IQG=13.75.
Stem Leaf
6 2399
7 7999
8 ∞ 02377899
9 0001123334567
10 2 3 3 6 6 6 7 9
11 P 0 2 4 6
12 0 0 5 9
25. The sorted list is {28.2, 29.3, 29.7, 31.6, 31.7, 32.7, 32.8, 33.5,
33.5, 33.5, 33.8, 34.2, 34.8, 34.8, 35.3, 35.4, 36.7, 37.3, 38.5}. Since
there are 19 numbers, the median is the 10th, Q1 is the
5th, and Q3 is the 15th; no additional computations are
needed. All these numbers are underlined above, and are
summarized below:
Min
Q1
Median
Q3
Max
28.2
31.7
33.5
35.3
38.5
The range is 38.5-28.2=10.3, and
IQR =35.3-31.7=3.6. There are no outliers, since
none of the numbers fall below Q1-1.5* IQR=26.3
or above Q3+1.5* IQR=40.7.
Note that some computer software may return
31.7 + 32.7
34.8 + 35.3
Q1=32.2=
and Q3=35.05=
2
2
(The TI calculator which computes the five-number
summary produces the results shown in the table).
26. (a)
[–4, 80] by [–1, 3]
(b)
[–4, 80] by [–1, 3]
27. (a)
6.2
[–3, 80] by [–1, 2]
(b)
[–3, 80] by [–1, 2]
Chapter 9
378
Discrete Mathematics
28. (a)
32. Although Mays’ best year was not as good as Mantle’s
best year, Mays was in some ways a better home-run
hitter: his median season was 31.5 home runs, compared
to 28.5 for Mantle. However, Mays had three extremely
bad years when he hit fewer than 10 home runs; these
make him look worse by comparison.
For #33–38, the best way to do the computation is with the statistics features of a calcuator.
[0, 1000] by [–1, 3]
(b)
33. Í≠9.08, Í2=82.5
34. Í≠23.99, Í2=575.64
35. Í≠186.36; Í2≠34,828.12
36. Í≠126.84; Í2≠16,088.08
37. Í≠1.53, Í2≠2.34
[0, 1000] by [–1, 3]
29. 12 of the 44 numbers are more than 10.5:
38. Í≠2.60, Í2≠6.77
12
3
=
44
11
30. All but 4 of the 44 numbers are more than 7.5:
40
10
=
44
11
31. The five-number summaries are:
Mays: 4, 20, 31.5, 40, 52 (top of graph)
Mantle: 13, 21, 28.5, 37, 54
(bottom of graph)
39. No. An outlier would need to be less than
25.5-1.5(25)=–12 or greater than
50.5+1.5(25)=88.
40. The standard deviation of a set can never be negative,
since it is the (positive) square root of the variance. It is
possible for the standard deviation of a set to be zero, but
all the numbers in the set would have to be the same.
41. (a) 68%
(a) Mays’s data set has the greater range.
(b) 2.5%
(b) Mays’s data set also has the greater IQR.
(c) A parameter, since it applies to the entire population.
42. (a) 16%
(b) 32.7
(c) No. The mean would have to be weighted according to
the number of people in each state who took the ACT.
43. False. The median is a resistant measure. The mean is
strongly affected by outliers.
[0, 60] by [0, 3]
44. True. The box extends from the first quartile, Q1, to the
third quartile, Q3, and Q3-Q1 is the interquartile range.
45. The plot of an ideal normal distribution is a symmetric
“bell curve.” The answer is A.
46. x =
10 13 2 + 913 2 + 8 15 2 + 7 16 2 + 6 14 2 + 51 32 + 4 1 12 + 3102 + 2 10 2 + 1 1 0 2
25
= 7.28
The answer is B.
47. The total number of points from all 30 exams combined is
30*81.3=2439. Adding 9 more points and recalculating produces a new mean of (2439+9)/30=81.6. The
median will be unaffected by an adjustment in the top
score. The answer is B.
48. In a normal distribution, 95% of the data values lie within
2 standard deviations of the mean. The answer is C.
49. There are many possible answers; examples are given.
(a) {2, 2, 2, 3, 6, 8, 20} — mode=2, median=3, and
43
x= ≠6.14.
7
(b) {1, 2, 3, 4, 6, 48, 48} — median=4, x=16, and
mode=48
1
(c) {–20, 1, 1, 1, 2, 3, 4, 5, 6} — x= , mode=1, and
3
median=2.
50. There are many possible answers; examples are given.
(a) {2, 4, 6, 8} — Í≠2.24 and IQR=7-3=4.
(b) {1, 5, 5, 6, 6, 9} — IQR=6-5=1, and Í≠2.36.
(c) {3, 3, 3, 9, 9, 9} — IQR=9-3=6 and
range=9-3=6.
51. No: (xi-x) (max-min)2=(range)2, so that
1 n
1 n
Í2= a 1xi - x)2 a (range)2=(range)2. Then
n i=1
n i=1
Í= 2Í2 21range2 2 = range.
52. The mode does not have to be a numerical value because
it is the data value that occurs most often; both the mean
and the median are for numerical data only, since they
both involve calculations of numerical data values.
Chapter 9
53. There are many possible answers; example data sets are
given.
Review
379
(c) {1, 1, 2, 6, 7} — range=7-1=6 and
2*IQR=2(6-1)=10.
(a) {1, 1, 2, 6, 7} — median=2 and x=3.4.
[–1, 12] by [–1, 5]
[–1, 10] by [–1, 5]
54. One possible answer: {1, 2, 3, 4, 5, 6, 6, 6, 30}.
(b) {1, 6, 6, 6, 6, 10} — 2*IQR=2(6-6)=0 and
range=10-1=9.
[–1, 12] by [–1, 5]
55. For women living in South American nations, the mean life expectancy is
179.72 1 39.1 2 + 167.9 2 1 8.72 + . . . + 179.2 2 1 3.42 + 1 77.32 1 25.02
27,825.56
x =
=
L 75.9 years.
.
.
.
39.1 + 8.7 +
+ 3.4 + 25.0
366.4
56. For men living in South American nations, the mean life expectancy is
172.02 1 39.1 2 + 162.5 2 1 8.72 + . . . + 172.7 2 1 3.42 + 1 71.02 1 25.02
25,160.64
x =
L 68.7 years.
=
39.1 + 8.7 + . . . + 3.4 + 25.0
366.4
12. Choose a king, then 8 more cards from the other 44:
# #
#
4C4 4C1 44C8=4C1 44C8=708,930,508 hands
57. Since Í=0.05 mm, we have 2Í=0.1 mm, so 95% of the
ball bearings will be acceptable. Therefore, 5% will be
rejected.
13. 5C2+5C3+5C4+5C5=25-5C0-5C1=26 outcomes
58. Use Â=12.08 and Í=0.04.
14.
Then Â-2Í=12.00 and Â+2Í=12.16, so 95% of
the cans contain 12 to 12.16 oz of cola, 2.5% contain less
than 12 oz, and 2.5% contain more than 12 oz. Therefore,
2.5% of the cans contain less than the advertised amount.
■ Chapter 9 Review
1. a
2. a
12
12!
12!
b =
=
= 792
5
5!1 12 - 5 2 !
5!7!
789
789!
789!
b =
=
= 310,866
787
787!1 789 - 787 2 !
787!2!
3.
18C12
=
4.
35C28
=
18!
18!
=
= 18,564
12!1 18 - 12 2!
12!6!
35!
35!
=
= 6,724,520
28!1 35 - 28 2!
28!7!
12!
12!
=
= 3,991,680
5. 12P7 =
112 - 72!
5!
6.
15P8
=
15!
15!
=
= 259,459,200
115 - 82!
7!
7. 26 # 364=43,670,016 code words
8. 3+(3 # 4)=15 trips
9.
10.
26P2
# 10P4+10P3 # 26P3=14,508,000 license plates
45C3=14,190
committees
11. Choose 10 more cards from the other 49:
#
3C3 49C10=49C10=8,217,822,536 hands
21C2
# 14C2=19,110 committees
15. 5P1+5P2+5P3+5P4+5P5=325
16. 24=16 (This includes the possibility that he has no coins
in his pocket.)
17. (a) There are 7 letters, all different. The number of distinguishable permutations is 7!=5040. (GERMANY
can be rearranged to spell MEG RYAN.)
(b) There are 13 letters, where E, R, and S each appear
twice. The number of distinguishable permutations is
13!
= 778,377,600
2!2!2!
(PRESBYTERIANS can be rearranged to spell
BRITNEY SPEARS.)
18. (a) There are 7 letters, all different. The number of distinguishable permutations is 7!=5040.
(b) There are 11 letters, where A appears 3 times and L,
S, and E each appear 2 times. The number of distinguishable permutations is
11!
= 831,600
3!2!2!2!
19. (2x+y)5=(2x)5+5(2x)4y+10(2x)3y2
+10(2x)2y3+5(2x)y4+y5
=32x5+80x4y+80x3y2+40x2y3
+10xy4+y5
380
Chapter 9
Discrete Mathematics
20. (4a-3b)7=(4a)7+7(4a)6(–3b)+21(4a)5(–3b)2
+35(4a)4(–3b)3+35(4a)3(–3b)4
+21(4a)2(–3b)5+7(4a)(–3b)6
+(–3b)7
=16,384a7-86,016a6b+193,536a5b2
-241,920a4b3+181,440a3b4
-81,648a2b5+20,412ab6-2187b7
21. (3x2+y3)5=(3x2)5+5(3x2)4(y3)
+10(3x2)3(y3)2+10(3x2)2(y3)3
+5(3x2)(y3)4+(y3)5
10
=243x +405x8y3+270x6y6
+90x4y9+15x2y12+y15
1 6
22. a 1 + b =1+6(x–1)+15(x–1)2+20(x–1)3
x
+15(x–1)4+6(x–1)5+(x–1)6
=1+6x–1+15x–2+20x–3
+15x–4+6x–5+x–6
23. (2a3-b2)9=(2a3)9+9(2a3)8(–b2)+36(2a3)7(–b2)2
+84(2a3)6(–b2)3+126(2a3)5(–b2)4
+126(2a3)4(–b2)5+84(2a3)3(–b2)6
+36(2a3)2(–b2)7+9(2a3)(–b2)8
+(–b2)9
27
=512a -2304a24b2+4608a21b4
-5376a18b6+4032a15b8
-2016a12b10+672a9b12
-144a6b14+18a3b16-b18
24. (x–2+y–1)4=(x–2)4+4(x–2)3(y–1)
+6(x–2)2(y–1)2+4(x–2)(y–1)3
+(y–1)4
=x–8+4x–6y–1+6x–4y–2
+4x–2y–3+y–4
25. a
11
11!8
11 # 10 # 9 # 8
b 112 8 1 -22 3 = = = - 1320
8
8!3!
3#2#1
8
8!4
8#7#4
=
= 112
26. a b 122 2 1 1 2 6 =
2
2!6!
2#1
40. We can redefine the words “success” and “failure” to mean
the opposite of what they meant before. In this sense, the
experiment is symmetrical, because P(S)=P(F).
41. P(SF)=(0.4)(0.6)=0.24
42. P(SFS)=(0.4)(0.6)(0.4)=0.096
43. P(at least 1 success)=1-P(no successes)
=1-(0.6)2=0.64
44. Successes are less likely than failures, so the two are not
interchangeable.
45. (a) P(brand A)=0.5
(b) P(cashews from brand A)=(0.5)(0.3)=0.15
(c) P(cashew)=(0.5)(0.3)+(0.5)(0.4)=0.35
(d) P(brand A/cashew)=
0.15
L 0.43
0.35
46. (a) P(track wet and Mudder Earth wins)
=(0.80)(0.70)=0.56
(b) P(track dry and Mudder Earth wins)
=(0.20)(0.40)=0.08
(c) 0.56+0.08=0.64
(d) P(track wet/Mudder Earth wins)=
For #47–48, substitute n=1, n=2, . . . , n=6, and n=40.
47. 0, 1, 2, 3, 4, 5; 39
4
16
16 64
48. –1, , –2, , - , ;≠2.68 1010
3
5
3 7
For #49–54, use previously computed values of the sequence
to find the next term in the sequence.
49. –1, 2, 5, 8, 11, 14; 32
50. 5, 10, 20, 40, 80, 160; 10,240
51. –5, –3.5, –2, –0.5, 1, 2.5; 11.5
27. {1, 2, 3, 4, 5, 6}
1 1 1 1
1
52. 3, 1, , , , ; 3 - 10 =
3 9 27 81
59,049
28. {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), ... ,
(6, 6)}
53. –3, 1, –2, –1, –3, –4; –76
29. {13, 16, 31, 36, 61, 63}
30. {Defective, Nondefective}
31. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
32. {HHT, HTH, THH, TTH, THT, HTT}
33. {HHH, TTT}
1 3
34. P(at least one head)=1-P(no heads)=1- a b
2
7
=
8
6
1
1
1
35. P(HHTHTT)= a b = 6 =
2
64
2
1 5
10
5
36. P(2 H and 3 T)=5C2 # a b = 5 =
2
16
2
4
1
4
1
37. P(1 H and 3 T)=4C1 # a b = 4 =
2
4
2
38. P(10 nondefectives in a row)=(0.997)10≠0.97
1
4
1
39. P(3 successes and 1 failure)=4C1 # a b 4 = 4 = = 0.25
2
4
2
0.56
= 0.875
0.64
54. –3, 2, –1, 1, 0, 1; 13
For #55–62, check for common difference or ratios between
successive terms.
55. Arithmetic with d=–2.5;
an=12+(–2.5)(n-1)=14.5-2.5n
56. Arithmetic with d=4;
an=–5+4(n-1)=4n-9
57. Geometric with r=1.2;
an=10 # (1.2)n1
58. Geometric with r=–2;
1
1
an= # 1 -2 2 n - 1 = - 1 - 22 n
8
16
59. Arithmetic with d=4.5;
an=–11+4.5(n-1)=4.5n-15.5
1
60. Geometric with r= ;
4
1 n-1
1 n
bn=7 # a b
= 28 # a b
4
4
Chapter 9
61. an=a1rn–1, so –192=a1r3 and 196,608=a1r8. Then
-192
r5=–1024, so r=–4, and a1=
= 3;
1 - 42 3
n–1
an=3(–4)
62. an=a1+(n-1)d, so 14=a1+2d, and
–3.5=a1+7d. Then 5d=–17.5, so
d=–3.5, and a1=14-2(–3.5)=21;
an=21-3.5(n-1)=24.5-3.5n.
a1 + an
b or
For #63–66, use one of the formulas Sn=n a
2
n
Sn= [2a1+(n-1)d]. In most cases, the first of these is
2
easier (since the last term an is given); note that
an - a1
n=
+ 1.
d
-11 + 10
b = 4 # 1 -1 2 = -4
63. 8 # a
2
64. 7 # a
13 - 11
b = 7
2
65. 27 # a
2.5 - 75.5
1
b = # 27 # 1 - 732 = -985.5
2
2
66. 31 # a
-5 + 55
b = 31 # 25 = 775
2
For #67–70, use the formula Sn=
n=1+log|r| 2
ln @ an>a1 @
an
2 =1+
.
a1
ln @ [email protected]
41 1 - 1 -1>2 2 6 2
68.
1 - 1 -1>2 2
- 3 11 - 11>3 2 5 2
70.
11 1 - 1 -22 14 2
67.
a1 11 - rn 2
=
1 - r
. Note that
21
8
121
= 1 - 1 1>3 2
27
10
21 1 - 3 2
= 59,048
69.
1 - 3
1 - 1 -2 2
= - 5461
1
71. Geometric with r= :
3
21871 1 - 1 1>3 2 10 2
29,524
=
= 3280.4
S10=
1 - 11>3 2
9
72. Arithmetic with d=–3:
10
S10= [2(94)+9(–3)]=5 # 161=805
2
73.
Review
74.
[0, 16] by [–10, 460]
75. With a1=$150, r=1+0.08/12, and n=120, the sum
becomes
$150 31 - 1 1 + 0.08>12 2 120 4
= $27,441.91
1 - 11 + 0.08>122
76. The payment amount P must be such that
0.08 0
0.08 1
Pa1 +
b + Pa1 +
b + ...
12
12
0.08 119
b
$30,000
12
Using the formula for the sum of a finite geometric series,
P 31 - 11 + 0.08>122 120 4
$30,000
1 - 11 + 0.08>12 2
-0.08>12
or P $30,000
1 - 11 + 0.08>122 120
L $163.983
L $163.99 rounded up
+ Pa1 +
3
3
77. Converges: geometric with a1= and r= , so
2
4
3>2
3>2
S=
=
= 6
1 - 13>4 2
1>4
1
2
78. Converges: geometric with a1= - and r= - , so
3
3
-2>3
-2>3
1
S=
=
= 1 - 1 -1>32
4>3
2
4
79. Diverges: geometric with r= 3
6
80. Diverges: geometric with r=
5
81. Converges: geometric with a1=1.5 and r=0.5, so
1.5
1.5
S=
=
= 3
1 - 0.5
0.5
82. Diverges; geometric with r=1.2
21
21
83. a 3 -8 + 51k - 1 2 4 = a 1 5k - 13 2
k=1
10
10
k=1
k=1
q
k=1
q
k=0
q
k=1
84. a 41 -22 k - 1 = a 1 - 22 k + 1
85. a 12k + 1 2 2 or a 1 2k - 1 2 2
q
1 k
1 k-1
86. a a b or a a b
k=0 2
k=1 2
n
n
n
k=1
k=1
87. a 13k + 1 2 = 3 a k + a 1
k=1
[0, 15] by [0, 2]
381
n1n + 1 2
n1 3n + 52
3n2 + 5n
+ n =
=
=3 #
2
2
2
n
n
n1n
+
12
12n
+
1
2
88. a 3k2 = 3 a k2 = 3 #
6
k=1
k=1
n1n + 1 2 12n + 12
=
2
Chapter 9
382
Discrete Mathematics
25
89. a 1k2 - 3k + 42 =
k=1
25 # 26 # 51
25 # 26
- 3#
6
2
+4 # 25 = 4650
175
90. a 13k2 - 5k + 1 2 = 3
# 175 # 176 # 351
k=1
6
175 # 176
- 5#
+ 175 = 5,328,575
2
n1n + 1 2
n1 n + 1 2 1n + 2 2
=
.
91. Pn: 1+3+6+. . .+
2
6
111 + 12 11 + 22
1 11 + 1 2
=
.
P1 is true:
2
6
k 1k + 1 2
Now assume Pk is true: 1+3+6+. . .+
2
1k + 1 2 1k + 22
k1 k + 12 1 k + 2 2
=
. Add
to both sides:
6
2
1k + 12 1k + 2 2
k 1k + 1 2
1+3+6+. . .+
+
2
2
1 k + 12 1k + 2 2
k1 k + 12 1k + 2 2
+
=
6
2
=(k+1)(k+2) a
Since both terms are divisible by 3, so is the sum, so Pk1
is true. Therefore, Pn is true for all n 1.
95. (a) 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
=(k+1)(k+2) a
=
6
,
so Pk+1 is true. Therefore, Pn is true for all n 1.
92. Pn: 1 # 2+2 # 3+3 # 4+. . .+n(n+1)
n1n + 12 1n + 2 2
=
. P1 is true:
3
1 11 + 1 2 1 1 + 2 2
.
1(1+1)=
3
Now assume Pk is true: 1 # 2+2 # 3+3 # 4+. . . +k(k+1)
k1 k + 12 1k + 2 2
=
.
3
Add (k+1)(k+2) to both sides:
1 # 2+2 # 3+. . . +k(k+1)+(k+1)(k+2)
=
k1 k + 12 1k + 2 2
3
=(k+1)(k+2) a
+(k+1)(k+2)
k
+ 1b
3
=(k+1)(k+2) a
=
k + 3
b
3
1k + 12 1 1k + 1 2 + 1 2 1 1k + 1 2 + 2 2
;
3
so Pk is true. Therefore, Pn is true for all n 1.
93. Pn: 2n–1 n!. P1 is true: it says that 21–1 1! (they are
equal). Now assume Pk is true: 2k–1 k!. Then
2k1–1=2 # 2k–1 2 # k! (k+1)k!=(k+1)!, so
Pk1 is true. Therefore, Pn is true for all n 1.
94. Pn: n3+2n is divisible by 3. P1 is true because
13+2 # 1=3 is divisible by 3. Now assume Pk is true:
k3+2k is divisible by 3. Then note that
(k+1)3+2(k+1)=(k3+3k2+3k+1)
+(2k+2)=(k3+2k)+3(k2+k+1).
9
4
(b)
k
1
+ b
6
2
k + 3
b
6
1k + 12 1 1k + 1 2 + 1 2 1 1k + 1 2 + 2 2
12
67
45577
024677
56
16778
48
14
06
Price
Frequency
90,000–99,999
100,000–109,999
110,000–119,999
120,000–129,999
130,000–139,999
140,000–149,999
150,000–159,999
160,000–169,999
170,000–179,999
210,000–219,999
230,000–239,999
2
2
5
6
2
5
2
2
2
1
1
(c)
[8, 24] by [–1, 7]
96. (a) 1
1
2
2
3
4
4
5
5
6
6
7
11111223344
5999
000
689
3
8
1
1
(b) Number of Visitors
10,000,000–14,999,999
15,000,000–19,999,999
20,000,000–24,999,999
25,000,000–29,999,999
30,000,000–34,999,999
35,000,000–39,999,999
40,000,000–44,999,999
45,000,000–49,999,999
55,000,000–59,999,999
60,000,000–64,999,999
65,000,000–69,999,999
70,000,000–74,999,999
Frequency
11
4
3
3
1
0
0
0
1
1
0
1
Chapter 9
(c)
Review
383
(c)
[0.5, 8] by [–1, 12]
97. (a) 12
13
14
15
16
17
18
19
20
21
22
23
[0, 6500] by [–1, 6]
0044
112679
0348
6
3
79
0
017
2
99. The ordered data is {9.1, 9.2, 10.6, 10.7, 11.4, 11.5, 11.5,
11.7, 11.7, 12, 12.2, 12.4, 12.6, 12.7, 12.7, 13.5, 13.6, 14.1,
14.6, 14.7, 14.7, 14.8, 15.4, 15.8, 16.1, 16.4, 17.0, 17.6, 21.9,
23.4} so the
median is
0
(b) Length (in seconds)
Frequency
120—129
130—139
140—149
150—159
160—169
170—179
180—189
190—199
200—209
210—219
220—229
230—239
4
6
4
1
1
2
1
3
1
0
0
1
(c)
[120, 240] by [0, 7]
98. (a) For the stemplot (but not for the other frequency
table or histogram), round to nearest hundred yards
first:
Stem Leaf
1 1
2 34578
3 13556
4 02377
5 06
(b) Yardage
1000–1999
2000–2999
3000–3999
4000–4999
5000–5999
Frequency
1
5
6
3
2
12.7 + 13.5
= 13.1, Q3=15.4, and Q1=11.7.
2
Five-number summary: {9.1, 11.7, 13.1, 15.4, 23.4}
Range: 23.4-9.1=14.3 ($143,000)
IQR: 15.4-11.7=3.7 ($37,000)
Í≠3.19, Í2≠10.14
Outliers: 21.9 and 23.4 are greater than
15.4+1.5(3.7)=20.95.
100. The ordered data is {1.1, 1.1, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3, 1.3,
1.4, 1.4, 1.5, 1.9, 1.9, 1.9, 2,0, 2.0, 2.0, 2.6, 2.8, 2.9, 3.3, 5.8,
6.1, 7.1} so the median is 1.9,
2.6 + 2.8
Q3=
= 2.7 and Q1=1.2.
2
Five-number summary: {1.1, 1.2, 1.9, 2.7, 7.1}
Range: 7.1-1.1=6.0 (6 million)
IQR: 2.7-1.2=1.5 (1.5 million)
Í≠1.63, Í2≠2.64
Outliers: 5.8, 6.1, and 7.1 are greater than
2.7+1.5(1.5)=4.95.
101. The ordered data is {120, 120, 124, 124, 131, 131, 132, 136,
137, 139, 140, 143, 144, 148, 156, 163, 177, 179, 180, 190,
191, 197, 202, 230} so the median is
143 + 144
179 + 180
= 143.5, Q3=
= 179.5, and
2
2
131 + 132
Q1 =
= 131.5.
2
Five-number summary: {120, 131.5, 143.5, 179.5, 230}
Range: 230-120=110
IQR: 179.5-131.5=48
Í=29.9, Í2=891.4
No Outliers.
102. The ordered data is: {1112, 2327, 2382, 2521, 2709, 2806,
3127, 3338, 3485, 3489, 3631, 3959, 4228, 4264, 4689, 4690,
5000, 5648},
3485 + 3489
so the median is
= 3487,
2
Q3=4264 and Q1=2709.
Five-number summary: {1112, 2709, 3487, 4264, 5648}
Range: 5648-1112=4536
IQR: 4264-2709=1555
≠1095, 2≠1,199,223
No Outliers.
384
Chapter 9
Discrete Mathematics
107.
103.(a)
[8, 24] by [–1, 1]
(b)
[8, 4] by [–1, 1]
104.(a)
4 0 0 12 4
9 2 1 13 1 6 7
8 4 3 0 14
15 6
3 16
7 17 9
18 0
19 0 1 7
20 2
21
22
23 0
The songs released in the earlier years tended to be
shorter.
108. Earlier years are in the upper box plot. The range and
interquartile range are both greater in the lower graph,
which shows the times for later years.
[0, 8] by [–1, 1]
(b)
[100, 250] by [–5, 10]
109.
[0, 8] by [–1, 1]
105.(a)
[–1, 25] by [100, 250]
Again, the data demonstrates that songs appearing later
tended to be longer in length.
110.
[100, 250] by [–1, 1]
(b)
[0, 7] by [50, 250]
The average times are {130.5, 132.75, 157, 142.5,
168.75, 202} The trend is clearly increasing overall, with
less fluctuation than the time plot for Exercise 105.
[100, 250] by [–1, 1]
106.(a)
111. 1 9 36 84 126 84 36 9 1
1n - k2!
n!
1n - k 2! 3 1n - k2 - j4 !
n!
=
1 n - k - j2!
n!
=
= nPk + j
3 n - 1 k + j2 4!
112. nPkn–kPj=
[0, 6000] by [–1, 1]
(b)
113. (a) P(no defective bats)=(0.98)4≠0.922
(b) P(one defective bat)=4C1 # (0.98)3(0.02)
=4(0.98)3(0.02)≠0.075
114. (a) P(no defective light bulbs)=(0.9996)10≠0.996
(b) P(two defective light bulbs)=10C2 #
(0.9996)8(0.0004)2 ≠7.18 * 10–6
[0, 6000] by [–1, 1]
Chapter 9
Chapter 9 Project
Answers are based on the sample data shown in the table.
1. Stem Leaf
5
5 9
6 1123334444
6 566678899
7 00111223
7 5
The average is about 66 or 67 inches.
Review
385
6. The stem and leaf plot puts the data in order.
Minimum value: 59
Maximum value: 75
66 + 67
Median:
= 66.5
2
Q1: 64
Q3: 70
The five-number summary is {59, 64, 66.5, 70, 75}.
7.
2. A large number of students are between 63 and 64 inches
and also between 69 and 72 inches.
Height
Frequency
59–60
61–62
63–64
65–66
67–68
69–70
71–72
73–74
75–76
1
3
7
4
3
5
5
1
1
[56, 78] by [–1, 7]
The boxplot visually represents the five-number summary.
The whisker-to-whisker size of the boxplot represents the
range of the data, while the width of the box represents
the interquartile range.
8. Mean=67.5; median=67; The new five-number summary is {59, 64, 67, 71, 86}.
3.
[56, 88] by [–1, 7]
[59, 78] by [–1, 7]
Again, the average appears to be about 66 or 67 inches.
Since the data are not broken out by gender, one can only
speculate about average heights for males and females
separately. Possibly the two peaks within the distribution
represent an average height of 63–64 inches for females
and 70–71 inches for males.
4. Mean=66.9 in.; median=66.5 in.; mode=64 in.
The mean and median both appear to be good measures
of the average, but the mode is too low. Still, the mode
might well be similar in other classes.
5. The data set is well distributed and probably does not
have outliers.
The minimum and first quartile are unaffected, but the
median, third quartile, and maximum are shifted upon to
varying degrees.
9. The new student’s height, 86 inches, lies 15 inches away
from Q3, and that is more than 1.5(Q3-Q1)=10.5.
The height of the new student should probably be tossed
out during prediction calculations.
10. Mean=68.9; median=68; The new five-number
summary is {59, 64, 68, 71, 86}.
All three of the new students are outliers by the
1.5*IQR test. Their heights should be left out during
prediction-making.
386
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
Chapter 10
An Introduction to Calculus: Limits, Derivatives, and Integrals
■ Section 10.1 Limits and Motion: The
Tangent Problem
3. s'(4)=lim
hS0
=lim
Exploration 1
¢s
3 ft
=
= 3 ft>sec
¢t
1 sec
hS0
4. s'(2)=lim
h
2
2
h + 2 + 1
3
=lim
hS0
h
6 - 2 1h + 32 1
#
=lim
hS0
31h + 32
h
-2h #
1
=lim
hS0
h 3 1h + 3 2
-2
2
=lim
= hS0 3 1h + 32
9
Quick Review 10.1
-1 - 3
-4
4
=
= 5 - 1 -2 2
7
7
3 - 1 -1 2
4
2
= =
2. m=
3 - 1 -3 2
6
3
1. m=
4. m=
5. s'(2)=lim
hS0
-1 - 6
-7
7
7
=
= - , y - 6 = - 1x - 12
4 - 1
3
3
3
5. y - 4 =
=lim
7.
4 + 4h + h2 - 4
4h + h2
=
= h + 4
h
h
8.
9 + 6h + h2 + 3 + h - 12
h2 + 7h
=
= h + 7
h
h
Section 10.1 Exercises
1. vave
¢s
21 miles
=
=
= 12 mi per hour
¢t
1.75 hours
2. vave =
¢s
540 km
=
= 120 km per hour
¢t
4.5 hours
h
a1h + 2 2 2 + 5 - 14a + 52
h
2
ah + 4ah
=lim
hS0
h
=lim (ah+4a)=4a
hS0
6. s'(1)=lim
s11 + h2 - s112
h
1h + 2 - 12
=lim
hS0
h
1h + 2 - 12 1h + 2 + 12
#
=lim
hS0
h
1h + 2 + 12
h + 2 - 2
=lim
hS0 h1 1h + 2 + 122
h
1
1
=lim #
=lim
hS0 h 1h + 2 + 12
hS0 1h + 2 + 12
1
=
2 12
hS0
1
1
2 - 1 2 + h2 1
2 + h
2
#
=
9.
h
212 + h2
h
1
-h #
1
=
=h 2 12 + h2
21 h + 22
1
1
x - 1x + h2 1
x + h
x
#
=
10.
h
x1x + h2
h
1
-h
1
#
=
=h x1 x + h2
x1 x + h2
s12 + h2 - s122
hS0
3
1x - 12
4
4
6. y - 4 = - 1 x - 12
3
s12 + h2 - s122
hS0
4. As the slope of the line joining (a, s(a)) and (b, s(b))
3
3
1 x + 2 2 or y = x + 6
2
2
h
hS0
3. They are the same.
3. y - 3 =
h
31h + 42 - 5 - 7
=lim 3=3
4 - 1
3
= = 3
1. m =
2 - 1
1
2. vave =
s14 + h2 - s142
7. Try
f11 2 - f1 02
1 - 0
=
8. Try
f12 2 - f1 12
3 - 2
= 1
1
2 - 1
=
1 - 2
= -1
1
9. No tangent
10. No tangent
Section 10.1
11.
Limits and Motion: The Tangent Problem
(c)
387
y
19
[–7, 9] by [–1, 9]
m=4
4
12.
18. (a) m=lim
hS0
=lim
x
f12 + h2 - f122
h
2 1h + 2 2 - 1 h + 2 2 2 - 0
h
2h + 4 - h2 - 4h - 4
=lim
= lim 1 -h - 22
hS0
hS0
h
=–2
hS0
[–10, 5] by [–7, 3]
m=4
13.
(b) Since (2, f(2))=(2, 0) the equation of the tangent
line is y=–2(x-2).
(c)
y
8
[–10, 11] by [–12, 2]
m=12
14.
3
19. (a) m=lim
hS0
=lim
[–2∏, 2∏] by [–3, 3]
m=–2
15. (a) f'(0)=lim
hS0
=lim
h
3 + 48 10 + h2 - 16 10 + h2 2 - 3
h
170 10 + h2 - 161 0 + h2 2 - 0
hS0
(b) The initial velocity of the rock is f'(0)=170 ft/sec.
hS0
=lim
f1 -1 + h2 - f1 - 12
h
21h - 1 2 2 - 2
=lim
h
=lim 12h - 42 = -4
hS0
y
5
f1 0 + h2 - f10 2
hS0
h
=lim 1170 - 16h2 = 170
17. (a) m=lim
hS0
(b) Since (2, f(2))=(2, –3) the equation of the tangent
line is y+3=1(x-2), or y=x-5.
(c)
(b) The initial velocity of the rock is f'(0)=48 ft/sec.
=lim
h
2 1h + 2 2 2 - 7 1h + 2 2 + 3 - 1 - 32
h
2h2 + 8h + 8 - 7h - 14 + 6
=lim
hS0
h
=lim 12h + 1 2 = 1
f1 0 + h2 - f10 2
h
48h - 16h2
= 48
=lim
hS0
h
hS0
f12 + h2 - f122
hS0
hS0
16. (a) f'(0)=lim
x
hS0
2h2 - 4h + 2 - 2
h
hS0
(b) Since (–1, f(–1))=(–1, 2) the equation of the
tangent line is y-2=–4(x+1).
4
x
388
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
20. (a) m=lim
f11 + h2 - f11 2
h
1
1
3 - 1 h + 32 1
h + 1 + 2
3
#
=lim
= lim
hS0
hS0
h
31h + 32
h
-h #
1
-1
1
=lim
= lim
= hS0 h
hS0 3 1h + 3 2
31h + 32
9
hS0
1
(b) Since (1, f(1))= a 1, b the equation of the tangent
3
1
1
line is y - = - 1x - 1 2 .
3
9
(c)
y
3
25. lim
f1 -2 + h2 - f1 - 22
hS0
=lim
h
3h2 - 12h + 12 - 12
=lim
=lim 13h - 122 = -12
hS0
hS0
h
hS0
26. lim
f11 + h2 - f11 2
hS0
=lim
h
h2 + 2h + 1 - 3h - 3 + 2
=lim
= lim 1 h - 12
hS0
hS0
h
=–1
f1 -2 + h2 - f1 - 22
hS0
x
h
1h + 12 2 - 31 h + 12 + 1 - 1 -1 2
hS0
27. lim
3
h
31h - 22 2 + 2 - 114 2
=lim
hS0
@ [email protected]
h
#
h
When h>0,
= lim
hS0
ƒh - 2 + 2ƒ - 0
h
|h|
= 1 while when h<0,
h
|h|
= -1. The limit does not exist. The derivative does
h
not exist.
1
1
h - 1 + 2
1
28. lim
=lim
hS0
hS0
h
h
1 - 1h + 1 2 1
-h
1
# =lim #
=lim
hS0
hS0 h
h + 1
h
h + 1
1
=lim =–1
hS0
h + 1
21.
f1 -1 + h2 - f1 - 12
29. f'(x)=lim
2 - 31x + h2 - 12 - 3x2
30. f'(x)=lim
12 - 31 x + h2 2 2 - 12 - 3x2 2
[–5, 5] by [–1, 5]
At x=–2: m=–1, at x=2: m=1, at x=0,
m does not exist.
22.
hS0
h
2 - 3x2 - 6xh - 3h2 - 2 + 3x2
=lim
hS0
h
-6xh - 3h2
=lim
=lim 1 -6x - 3h 2 = -6x
hS0
hS0
h
[–3, 3] by [–2, 2]
At x=–2: m=0.5, at x=2: m=0.1, at x=0,
m=0.5.
23. lim
f12 + h2 - f1 2 2
24. lim
f12 + h2 - f1 2 2
h
2 - 3x - 3h - 2 + 3x
-3h
=lim
=lim
= -3
hS0
hS0
h
h
hS0
1 - 12 + h2 2 - 11 - 4 2
=lim
hS0
h
h
2
- h - 4h - 4 + 4
=lim
=lim 1 -h - 4 2 = -4
hS0
hS0
h
hS0
31. f'(x)
=lim
31x + h2 2 + 2 1x + h2 - 1 - 13x2 + 2x - 12
h
3x2 + 6xh + 3h2 + 2x + 2h - 1 - 3x2 - 2x + 1
=lim
hS0
h
6xh + 3h2 + 2h
=lim
= lim 16x + 3h + 22 = 6x + 2
hS0
hS0
h
hS0
h
hS0
=lim
2 12 + h2 +
hS0
4 + 2h +
=lim
hS0
1
1 2 + h2 2 - 4 - 2
2
h
1 2
h + 2h + 2 - 6
2
h
1
=lim a h + 4 b = 4
hS0
2
32. f'(x)=lim
1
1
1x + h2 - 2
x - 2
hS0
h
1x - 22 - 1 x + h - 22 1
#
=lim
hS0
1x + h - 2 2 1x - 22
h
-h
1
#
=lim
hS0
h 1x + h - 22 1x - 2 2
1
1
=lim
= hS0 1x + h - 22 1x - 2 2
1x - 2 2 2
Section 10.1
3.2
0.6
7.3
Between 0.8 and 0.9 seconds:
0.9
33. (a) Between 0.5 and 0.6 seconds:
-
Limits and Motion: The Tangent Problem
389
2.3
= 9 ft/sec
0.5
5.8
= 15 ft/sec
0.8
(b) f(x)=8.94x2+0.05x+0.01, x=time in seconds
[–0.1, 1] by [–0.1, 8]
(c) f(2)≠35.9 ft
34. (a)
15.76 - 21.24
= - 27.4 ft/sec
1.0 - 0.8
(b) s(t)=–16.01t2+1.43t+30.35
[0, 1.5] by [0, 31]
s1t + h 2 - s1t 2
(c) s'(t)=lim
h
3 -16.0151 t + h2 2 + 1.431 t + h2 + 30.35 4 - 1 -16.015t2 + 1.43t + 30.352
hS0
=lim
h
hS0
- 32.03th - 16.015h2 + 1.43h
=lim
hS0
h
=lim(–32.03t-16.015h+16.015)
hS0
=–32.03t+1.43;
s(1)=–32.03(1)+1.43=–30.6
At t=1, the velocity is about –30.60 ft/sec.
35. (a)
36. (a)
y
y
10
9
5
5
x
x
(b) Since the graph of the function does not have a definable slope at x = 2, the derivative of f does not exist
at x=2. The function is not continuous at x=2.
(c) Derivatives do not exist at points where functions
have discontinuities.
(b) From the graph of the function, it appears that the
derivative may exist at x=2. Using the first definition of the derivative and taking secant lines on the
left of x=2 (so that f(x)=1+(x-2)2), we have
f1x2 - f12 2
1 + 1x - 22 2 - 1
lim
= lim
xS2
xS2
x - 2
x - 2
2
1x - 22
=lim
=lim 1 -|x - 2| 2 =0.
xS2 x - 2
xS2
390
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
Now, taking secant lines on the right of x=2 (so that
f(x)=1-(x-2)2), we have
f1 x2 - f1 22
1 - 1x - 2 2 2 - 1
lim
= lim
xS2
xS2
x - 2
x - 2
- 1x - 222
=lim
=lim (–|x-2|)=0.
xS2
xS2
x - 2
Since the limits are the same, f(x) exists at x=2
and f(2)=0.
37. (a)
39. Answers will vary. One possibility:
y
10
5
–1
x
y
3
–10
40. Answers will vary. One possibility:
y
5
x
(b) Since the graph of the function does not have a definable slope at x=2, the derivative of f does not exist
at x=2. The function is not continuous at x=2.
(c) Derivatives do not exist at points where functions
have discontinuities.
38. (a)
5
5
–1
x
–5
41. Answers will vary. One possibility:
y
y
2
5
π
x
5
–1
x
–5
(b) From the graph of the function, it appears that the
derivative may exist at x=0.
sin h
- 1
f1 0 + h2 - f10 2
h
f(0)=lim
=lim
hS0
hS0
h
h
sin h - h
=lim
hS0
h2
This limit cannot be found using algebraic techniques.
The table of values below suggests that this limit
equals 0.
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
`
`
`
`
`
`
`
sin x - x
x2
0.01666
0.00167
0.00017
–0.00017
–0.00167
–0.01666
The graph supports this since it appears that there is a
horizontal tangent line at x=0. Thus, f'(0)=0.
42. Answers will vary. One possibility:
y
5
5
–1
x
–5
43. Since f(x)=ax+b is a linear function, the rate of
change for any x is exactly the slope of the line. No calculations are necessary since it is known that the slope
a=f'(x).
44. f'(0)=lim
xS0
=lim
@ [email protected]
f1x2 - f102
x - 0
= lim
xS0
@ [email protected] - @ 0 @
x
. Looking at secant lines, we see that this limit
x
does not exist. If the secant line is to the left of x=0, it
will have slope m=–1, while if it is to the right of
x=0, it will have slope m=1. At x=0, the graph of
the function does not have a definable slope.
xS0
Section 10.2
45. False. The instantaneous velocity is a limit of average
velocities. It is nonzero when the ball is moving.
46. True. Both the derivative and the slope equal
f1x2 - f1 a2
lim
.
xSa
x - a
391
55. (a) The average velocity is
16 13 2 2 - 16 10 2 2
^s
=
= 48 ft/sec.
^t
3 - 0
(b) The instantaneous velocity is
161 3 + h2 2 - 144
96h + h2
lim
= lim
hS0
hS0
h
h
=lim (96+h)=96 ft/sec
47. For Y1=x2+3x-4, at x=0 the calculator shows
dy/dx=3. The answer is D.
48. For Y1=5x-3x2, at x=2 the calculator shows
dy>dx = - 7. The answer is A.
Limits and Motion: The Area Problem
hS0
56. (a) g=
49. For Y1=x3, at x=2 the calculator shows dy/dx=12.
The answer is C.
y
t2
=
125
= 5 m/sec2
25
(b) Average speed:
1
, at x=1 the calculator shows
x - 3
dy/dx=–0.25. The answer is A.
50. For Y1 =
^x
125
=
= 25 m/sec
^t
5
(c) Since y=5t 2, the instantaneous speed at t=5 is
51 5 + h2 2 - 5 15 2 2
50h + 5h2
=lim
lim
hS0
hS0
h
h
=lim 150 + 5h2
51. (a)
hS0
=50 m/sec
y
57.
1
[–4.7, 4.7] by [–3.1, 3.1]
No, there is no derivative because the graph has
a corner at x=0.
10
x
(b) No
52. (a)
y
58.
10
[–4.7, 4.7] by [–3.1, 3.1]
No, there is no derivative because the graph has a
cusp (“spike”) at x=0.
5
x
(b) Yes, the tangent line is x=0.
53. (a)
■ Section 10.2 Limits and Motion: The Area
Problem
[–4.7, 4.7] by [–3.1, 3.1]
No, there is no derivative because the graph has a
vertical tangent (no slope) at x=0.
(b) Yes, the tangent line is x=0.
Exploration 1
1. The total amount of water remains 1 gallon. Each of the
1 gal
10 teacups holds
= 0.1 gallon of water.
10
2. The total amount of water remains 1 gallon. Each of the
1 gal
100 teacups holds
= 0.01 gallon of water.
100
54. (a)
[–4.7, 4.7] by [–3.1, 3.1]
Yes, there is a derivative because the graph has a
nonvertical tangent line at x=0.
(b) Yes, the tangent line is y=x.
3. The total amount of water remains 1 gallon. Each of the
1 gal
1,000,000,000 teacups holds
= 0.000 000 001
1,000,000,000
gallon of water.
4. The total amount of water remains 1 gallon. Each of the
teacups holds an amount of water that is less than what
was in each of the 1 billion teacups in step 3. Thus each
teacup holds about 0 gallons of water.
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
392
Quick Review 10.2
1.
5
1 1 9
25 9 49
81 25
, , , 2, , , , 8, ,
8 2 8
8 2 8
8 2
81 25 121 9 169 49 225
289 81
2.
, ,
, ,
, ,
, 4,
,
64 16 64 4 64 16 64
64 16
3.
1
65
[2+3+4+5+6+7+8+9+10+11]=
2
2
4.
2+3
+
4 +. . .+ n
+(n+1)
± (n+1)+(n)
+(n-1)+. . .+ 3
+2
(n+3)+(n+3)+(n+3)+. . .+(n+3)+(n+3)
n
Thus 2 a 1 k + 1 2 =n(n+3), and
k=1
n
1
a 1k + 1 2 = 2 n(n+3).
k=1
5.
6.
1
505
[4+9+. . .+121]=
2
2
1
[1+4+9+. . .+(n-1)2+n2]
2
n1n + 1 2 12n + 1 2
1 n1n + 1 2 12n + 1 2
d=
= c
2
6
12
#
8. a 1 f1k2 =f(1)+f(2)+f(3)+f(4)+f(5)
k=1
1
1
=1+3+4 +4+0=12 (answers will vary)
2
2
5
#
9. a 1 f1k2 =f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)
k=1
=3.5+5.25+2.75+0.25+1.25=13 (answers will
vary)
5
#
10. a 1 f1k2 =f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)
k=1
=3+1.5+1.75+3.25+5=14.5 (answers will vary)
8
11. a 110 - x2i 2 Δxi
i=1
= 19 + 9.75 + 10 + 9.75 + 9 + 7.75 + 6 + 3.75 2 10.5 2
= 32.5 square units
8
12. a 110 - x2i 2 Δxi
i=1
= 19.75 + 10 + 9.75 + 9 + 7.75 + 6 + 3.75 + 1 2 10.5 2
= 28.5 square units
7. (57 mph)(4 hours)=228 miles
13. c 0,
3
1
1
3
d , c , 1 d , c 1, d , c , 2 d
2
2
2
2
8. a
14. c 0,
7
1
1 1
1 3
3
5
5 3
3 7
d , c , d , c , d , c , 1 d , c 1, d , c , d , c , d , c , 2 d
4
4 2
2 4
4
4
4 2
2 4
4
15. c 1,
7
3
3
5
5
7
d , c , 2 d , c 2, d , c , 3 d , c 3, d , c , 4 d
2
2
2
2
2
2
16. c 1,
3
3
5
5
7
7
9
9
d , c , 2 d , c 2, d , c , 3 d , c 3, d , c , 4 d , c 4, d , c , 5 d
2
2
2
2
2
2
2
2
5 gal
min
b (120 min)=600 gallons
200 ft3
60 minutes
60 seconds
b (6 hours) a
ba
b
sec
hour
minute
=4,320,000 ft3
9. a
10. a
560 people
mi2
b (35,000 mi )=19,600,000 people
2
Section 10.2 Exercises
1. Let the line y=65 represent the situation. The area
under the line is the distance traveled, a rectangle,
(65)(3)=195 miles.
For #17–20, the intervals are of width 1, so the area of each rectangle is 1 f(k)=f(k).
#
17. (a)
y
18
2. Let the line y=15 represent the situation. The area
under the line is the number of gallons pumped, a
rectangle, (15)(30)=450 gallons.
3. Let the line y=150 represent the situation. The area
under the line is the total number of cubic feet of water
pumped, a rectangle, (150)(3600)=540,000 ft3.
4. Let the line y=650 represent the situation. The area
under the line is the total population, a rectangle,
(650)(20)=13,000 people.
(b)
1
2
3
4
5
1
2
3
4
5
x
y
18
Δs #
Δ t = 1640 km>h2 13.4 h2 = 2176 km
5. Δ s =
Δt
Δs
# Δ t = 124 mi>h2 a 4 5 hb = 116 mi
Δt
6
6. Δ s =
5
#
7. a 1 f1k2 =f(1)+f(2)+f(3)+f(4)+f(5)
k=1
1
1
1
3
=3 +4 +3 +1 +0=13 (answers will vary)
2
4
2
4
x
RRAM: f(1)+f(2)+f(3)+f(4)
=1+4+9+16=30
Section 10.2
(c)
Limits and Motion: The Area Problem
19. (a) y
y
18
5
1
2
3
4
5
x
LRAM: f(0)+f(1)+f(2)+f(3)
=0+1+4+9=14
2
3
4
5
1
2
3
4
5
x
(b) y
5
14 + 30
(d) Average:
=22
2
18. (a)
1
y
40
x
RRAM: f(1)+f(2)+f(3)+f(4)
=3+4+3+0=10
1 2
(b)
3
4
5 6
7
x
(c)
y
5
y
40
1
1 2
3
4
5 6
7
x
RRAM: f(1)+f(2)+. . .+f(6)
=3+6+11+18+27+38=103
(c)
3
4
5
x
LRAM: f(0)+f(1)+f(2)+f(3)
=0+3+4+3=10
(d) Average:
20. (a)
y
2
10 + 10
=10
2
y
30
40
1 2
3
4
5 6
7
x
LRAM: f(0)+f(1)+. . .+f(5)
=2+3+6+11+18+27=67
(d) Average:
67 + 103
=85
2
(b)
1
2
3
4
1
2
3
4
x
y
30
x
RRAM: f(1)+f(2)+f(3)=1+8+27=36
393
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
394
(c)
4
y
25.
30
1
2
3
4
(x+3) dx=16.5 (Trapezoid with bases 4 and 7 and
31
altitude 3)
x
[–1, 6] by [–1, 12]
LRAM: f(0)+f(1)+f(2)=0+1+8=9
(d) Average:
9 + 36
45
=
2
2
4
26.
7
21.
(3x-2) dx=16.5 (Trapezoid with bases 1 and 10
31
and altitude 3)
5 dx=20 (Rectangle with base 4 and height 5)
33
[–1, 6] by [–1, 12]
2
[–1, 10] by [–1, 7]
27.
4
22.
3-1
3-2
24 - x2 dx=2∏ (Semicircle with radius 2)
6 dx=30 (Rectangle with base 5 and height 6)
[–2, 2] by [–0.5, 2]
6
[–2, 10] by [–1, 7]
28.
5
23.
30
236 - x2 dx=9∏ (Quarter circle with radius 6)
3x dx=37.5 (Triangle with base 5 and altitude 15)
30
[–1, 8] by [–0.5, 6]
∏
[–1, 6] by [–1, 20]
7
24.
29.
30
sin x dx=2 (One arch of sine curve)
0.5x dx=12 (Trapezoid with bases of 0.5 and 3.5 and
31
height 6)
[–2∏, 2∏] by [–3, 3]
[–1, 8] by [–1, 5]
Section 10.2
∏
30.
30
rectangle with base ∏ and height 2)
35.
2 sin x dx=4 (Rectangles in sum are twice as tall,
30
yielding twice the sum)
[–2∏, 2∏] by [–3, 3]
2+∏
2∏
sin (x-2) dx=2 (One arch of sine curve trans-
32
lated 2 units right)
36.
x
sin a b dx=4 (Rectangles in sum are twice as
2
30
wide, yielding twice the sum)
[–2∏, 2∏] by [–3, 3]
[–2∏, 2∏] by [–3, 3]
∏>2
32.
395
∏
(sin x+2) dx=2+2∏ (Arch of sine curve plus
[–2∏, 2∏] by [–3, 3]
31.
Limits and Motion: The Area Problem
2∏
cos x dx=2 (One arch of cosine curve, which is
3-∏>2
sine curve translated ∏/2 units)
37.
|sin x| dx=4 (Two arches of the sine curve)
30
[–2∏, 2∏] by [–3, 3]
[–2∏, 2∏] by [–3, 3]
38.
∏>2
33.
sin x dx=1 (Half-arch of sine curve)
30
3∏>2
|cos x| dx=5 (Two-and-a-half arches of the cosine
3-∏
curve)
[–2∏, 2∏] by [–3, 3]
∏>2
34.
cos x dx=1 (Half-arch of cosine curve, congruent
30
to half-arch of sine curve)
[–2∏, 2∏] by [–3, 3]
39. The graph of f(x)=kx+3 is a line. If k is a number
between 0 and 4, the integral is the area of a trapezoid
with bases of 0k+3=3 and 4k+3 and height of
1
4-0=4. The area is 142 13 + 4k + 32 = 214k + 62
2
4
=8k+12, so
[–2∏, 2∏] by [–3, 3]
30
1 kx + 32 dx = 8k + 12.
40. The graph of f(x)=4x+3 is a line. The integral is the
area of a trapezoid with bases of 4 # 0 + 3 = 3 and
4k+3 and height of k-0=k. The area is
1
1
k 13 + 4k + 32 = k14k + 62 =2k2+3k, so
2
2
k
30
14x + 3 2dx = 2k2 + 3k.
Chapter 10 An Introduction to Calculus: Limits, Derivatives, and Integrals
396
41. The graph of f(x)=3x+k is a line. The integral is the
area of a trapezoid with bases of 3 # 0 + k = k and
3 # 4 + k = 12 + k and height of 4-0=4. The area is
1
142 1 k + 12 + k 2 =2(12+2k)=24+4k, so
2
4
30
1 3x + k2dx = 24 + 4k.
42. The graph of f(x)=4x+3 is a line. The integral is the
area of a trapezoid with bases of 4k+3 and
4 # 4 + 3 = 19 and height of 4-k. The area is
1
1
14 - k 2 14k + 3 + 19 2 = 14 - k2 122 + 4k2
2
2
=(4-k)(11+2k)=44-2k2-3k
43. Since g(x)=–f(x), we consider g to be symmetric with
f about the x-axis. For every value of x in the interval,
|f(x)| is the distance to the x-axis and similarly, |g(x)| is
the distance to the x-axis; f(x) and g(x) are equidistant
from the x-axis. As a result, the area under f(x) must be
exactly equal to the area above g(x).
44. The graph of f(x)= 216 - x2 is the top half of a circle
of radius 4. The area of the graph from x=0 to x=4 is
1
the area of of the entire circle. Thus the desired area is
4
1 # 2
1
1p 4 2 = 116p2 = 4p.
4
4
45. The distance traveled will be the same as the area under
the velocity graph, v(t)=32t, over the interval [0, 2]. That
triangular region has an area of A=(1/2)(2)(64)=64.
The ball falls 64 feet during the first 2 seconds.
46. The distance traveled will be the same as the area under
the velocity graph, v(t)=6t, over the interval [0, 7]. That
triangular region has an area of A=(1/2)(7)(42)=147.
The car travels 147 feet in the first 7 seconds.
(b) The rocket reaches its maximum height when the
velocity function is zero; this is the point where the
rocket changes direction and starts its descent. Solving
for t when 170-32t=0, t≠5.31 sec.
(c) The distance the rocket has traveled is the area under
the curve, a triangle with base 5.3125 and height 170
1
thus, d= (170)(5.3125)≠451.6 ft.
2
49. (a)
[0, 2] by [–50, 0]
(b) Each RRAM rectangle will have width 0.2. The heights
(using the absolute value of the velocity) are 5.05, 11.43,
17.46, 24.21, 30.62, 37.06, and 43.47. The height of the
building is approximately 0.2[5.05+11.43+17.46+
24.21+30.62+37.06+43.47]=33.86 feet.
50. Work is defined as force times distance. The work done in
moving the barrel 35 feet is the area under the curve created by the given data points, assuming the barrel weighs
approximately 550 lbs after being moved 35 feet. In this
case, the area under the curve is the sum of a rectangle of
width 35 and height 550 and a triangle of base 35 and
height (1250-550)=700. The total work performed is
1
(35)(550)+ (35)(700)=31,500 ft-pounds.
2
51. True. The exact area under a curve is given by the limit as
n approaches infinity. This is true whether LRAM or
RRAM is used.
52. False. The statement lim f1x2 = L means that f(x) gets
47. (a)
xSq
arbitrarily close to L as x gets arbitrarily large.
53. Since y = 21x represents a vertical stretch, by a factor of
2, of y = 1x, the area under the curve between x=0
and x=9 is doubled. The answer is A.
[0, 3] by [0, 50]
(b) The ball reaches its maximum height when the velocity
function is zero; this is the point where the ball
changes direction and starts its descent. Solving for
t when 48-32t=0, we find t=1.5 sec.
(c) The distance the ball has traveled is the area under
the curve, a triangle with base 1.5 and height 48 thus,
d=0.5(1.5)(48)=36 units.
48. (a)
[0, 8] by [0, 180]
54. Since y = 1x + 5 represents a vertical shift, by 5 units
upward, of y = 1x, the area is inc