Calculus Regina Summer Review Packet

Calculus Regina Summer Review Packet
ReginaSummer
MathReview
Forstudentswhowill
betaking
HSCalculus
Completedreviewpacketdue
thefirstdayofclasses
Calculus
SummerReviewPacket
WelcometoCalculus!
Inthefollowingpages,youwillfindreviewmaterialsthatwillprepareyouforCalculus.Pleasetakethe
exercisesseriouslyasthiswillallowustohitthegroundrunninginthefall.Therearecertainmathskills
thathavebeentaughttoyouoverthepreviousyearsthatarenecessarytobesuccessfulincalculus.If
youdonotfullyunderstandthetopicsinthispacket,itispossiblethatyouwillgetcalculusproblems
wronginthefuture,notbecauseyoudonotunderstandthecalculusconcept,butbecauseyoudonot
understandthealgebraortrigonometrybehindtheproblem.
Inshort:Don’tfakeyourwaythroughthispacket.Youwillneedtounderstandthispacketverywellto
succeedinCalculusthisyear.Don’twaituntilthelastminutetodoeverythinginthepacketbecauseyou
mayrunoutoftimeorfeeltoorushed.Likewise,donotdoalltheproblemsimmediatelyandforgetto
reviewbeforeyoucomebacktoschool.
Ifyoufeelthatyouneedhelpwithsomeofthesetopics,thebestresourcemaybetheinternet.There
are great websites where you can just type in the topics to get help. Using youtube.com and
khanacademy.org not only will help you on this packet this summer, but will be referenced often
throughouttheupcomingschoolyear.IcanalsosendaPDFversionofChapter1ofourCalculustextbook
ifyouemailmearequest.Aftercheckingwiththesereferences,youmayalsoe-mailme.Pleasegiveme
atleasttwodaystorespond(thisismysummertoo,afterall!).
Finally,whencompletingthispacket,pleaseshowallofyourworkinaconcisemanner.Yourworkmust
bewellorganized,neat,andproperlylabeled.IfIcannoteasilyfindandreadeachquestionwiththework,
Iwillassumetheproblemsarenotdoneorareincorrect.Donotrelyonacalculatortodoallofthework
foryou—approximatelyhalfofourworkwillbedonewithoutacalculatorthisyear.
Thispacketduethefirstdayofclass.TherewillbeaBasicSkillsAssessmentthefirstweekofclassto
proveyouhavethemathematicalcapabilitiesofbeinganactiveparticipantinCalculusthisyear.
Materialsforthisclassaresuggestedasthefollowing:
“ GraphingCalculator(Required,TI-83/TI-84familiesstronglypreferred)
“ Loose-leafPaper
“ 3-RingBinder(atleast1”)
“ Pencils(nopensallowed)
“ SummerReviewPacket
Haveagreatsummer!
This packet contains the topics that you have learned in your previous courses that are
most important to calculus. Make sure that you are familiar with all of the topics listed
below when you enter Calculus.
I. Using Your Graphing Calculator
A graphing calculator is required for this course and is not provided by the school. It is
recommended that you own a TI-83 or TI-84 (any edition) as this is the calculator that will be
used by the instructor on a regular basis.
“ Be able to graph functions on your calculator
“ Be able to find key points on your graph like maximum values, minimum values, points
of intersection, etc.
“ Do not depend on the calculator for graphs that should be memorized. (See the section
on “Parent Functions”.)
“ Do not depend on your calculator for simple algebra!
II. Interval and Set Notation
“ You will be expected to be able to read and write using different notation
Ø {0,1} is the set of 0 and 1
Ø {x│x>1} is the set of all x’s that are greater than 1
Ø {x│1≤x≤7} is the set of all x’s that are greater than or equal to 1 and less than or
equal to 7
Ø [0,5] is the set of numbers from 0 to 5, including both 0 and 5
Ø [0,5) is the set of numbers from 0 to 5, which includes 0 and excludes 5
Ø (0,∞) is the set of numbers from 0 to infinity, not including 0
III. Cartesian Coordinate System
“
“
“
“
Know the quadrants, axes, and how to graph
Know the different forms of the equation of a line, especially Point-Slope Form
Slope (incredibly important in calculus)
Parallel and perpendicular lines
IV. Algebra Basics
“ You must be able to factor polynomials (GCF, by grouping or ac-Box method, trinomials,
difference of squares, and sum & difference of cubes)
“ You must know how to add, subtract, multiply, and divide rational expressions
“ You must know how to work with complex fractions
Function Basics
We use functions almost every day in calculus. It is critical that you are able to identify and do
the following:
“
“
“
“
“
“
“
“
“
“
“
“
“
Identify functions from equations, graphs, sets
Use function notation
Find the domain and range of a function
Identify functions as even, odd or neither. You should also be able to identify what lines
of symmetry go with even and odd functions
Perform operations on functions
Compose functions
Graph and write the equations of piecewise functions
Identify discontinuities
Find inverse functions algebraically and graphically
Use long and synthetic division
Find the left hand and right hand behavior of a graph
Find the zeros of a polynomial function algebraically and graphically
Find vertical, horizontal, and slant asymptotes and any holes in the graph of a function
V. Logarithmic and Exponential Functions
“
“
“
“
Know the properties of exponents
Know the properties of logarithms
Natural logs will be used a lot in Calculus, so be familiar with them
Solve exponential and logarithmic equations
VI. Trigonometry
The Pre-Calculus course that you just took was your formal trigonometry course. However, we
use the trigonometric functions and a few of the identities regularly in Calculus. The following
are the most important:
“ Calculus is always in radians, never degrees
“ Know the special right triangles
“ You need to know, without a calculator, sinθ, cosθ, tanθ, cscθ, secθ, and cosθfor the
following values
- - - Ø 0, , , ,
. / 0 1
“ Be able to identify the graph of sine, cosine, and tangent
“ The Pythagorean Identities and Double Angle Identities are the most used
“ Be able to work with arcsin, arcos, and arctan (sin23 , cos 23 , tan23 )
*Summer Review Packet is from mastermathmentor.com and used with permission.
A. Functions
The lifeblood of precalculus is functions. A function is a set of points ( x, y ) such that for every x, there is one
and only one y. In short, in a function, the x-values cannot repeat while the y-values can. In AB Calculus, all of
your graphs will come from functions.
The notation for functions is either “ y = ” or “ f ( x ) = ”. In the f ( x ) notation, we are stating a rule to find y
given a value of x.
1. If f ( x ) = x 2 − 5x + 8 , find a) f ( −6 )
⎛ 3⎞
b) f ⎜ ⎟
⎝ 2⎠
2
a) f ( −6 ) = ( −6 ) − 5 ( −6 ) + 8
2
36 + 30 + 8
74
f ( x + h) − f ( x )
h
f ( x + h) − f ( x )
c)
h
c)
⎛ 3⎞ ⎛ 3⎞
⎛ 3⎞
b) f ⎜ ⎟ = ⎜ ⎟ − 5 ⎜ ⎟ + 8
⎝ 2⎠ ⎝ 2⎠
⎝ 2⎠
9 15
− +8
4 2
11
4
( x + h)2 − 5( x + h) + 8 − ( x 2 − 5x + 8)
h
x + 2xh + h − 5x − 5h + 8 − x 2 + 5x − 8
5
2
h + 2xh − 5h h ( h + 2x − 5)
=
= h + 2x − 5
h
h
2
2
Functions do not always use the variable x. In calculus, other variables are used liberally.
2. If A ( r ) = π r 2 , find a) A (3)
A (3) = 9π
b) A ( 2s )
c) A ( r +1) − A ( r )
A ( 2s ) = π ( 2s ) = 4π s
2
A ( r +1) − A ( r ) = π ( r +1) − π r 2
2
2
π ( 2r +1)
One concept that comes up in AP calculus is composition of functions. The format of a composition of
functions is: plug a value into one function, determine an answer, and plug that answer into a second function.
3. If f ( x ) = x 2 − x +1 and g ( x ) = 2x −1 , a) find f ( g ( −1)) b) find g ( f ( −1)) c) show that f ( g ( x )) ≠ g ( f ( x ))
f ( g ( x )) = f ( 2x −1) = ( 2x −1) − ( 2x −1) +1
2
g ( −1) = 2 ( −1) −1 = −3
f ( −3) = 9 + 3 +1 = 13
f ( −1) = 1 +1 +1 = 3
g (3) = 2 (3) −1 = 5
= 4x 2 − 4x +1 − 2x +1 +1 = 4x 2 − 6x + 3
(
) (
)
g ( f ( x )) = g x 2 − x +1 = 2 x 2 − x +1 −1
= 2x 2 − 2x +1
Finally, expect to use piecewise functions. A piecewise function gives different rules, based on the value of x.
⎧ x 2 − 3, x ≥ 0
4. If f ( x ) = ⎨
, find a) f ( 5)
b) f ( 2 ) − f ( −1)
c) f ( f (1))
⎩2x +1, x < 0
f ( 5) = 25 − 3 = 22
www.mastermathmentor.com
f ( 2 ) − f ( −1) = 1 − ( −1) = 2
7
f (1) = −2, f ( −2 ) = −3
RU Ready?
A. Function Assignment
• If f ( x ) = 4x − x 2 , find
1. f ( 4) − f ( −4)
4
• If V ( r ) = π r 3 , find
3
⎛ 3⎞
4. V ⎜ ⎟
⎝ 4⎠
2.
⎛ 3⎞
f⎜ ⎟
⎝ 2⎠
5. V ( r +1) − V ( r −1)
• If f ( x ) and g ( x ) are given in the graph below, find
7. ( f − g) (3)
⎧ x + 2 − 2, x ≥ 2
⎪
• If f ( x ) = ⎨ x 2 −1, 0 ≤ x < 2 , find
⎪ −x, x < 0
⎩
10. f ( 0 ) − f ( 2 )
www.mastermathmentor.com
f ( x + h) − f ( x − h)
2h
6.
V ( 2r )
V (r )
• If f ( x ) = x 2 − 5x + 3 and g ( x ) = 1 − 2x , find
8. f ( g (3))
11.
3.
9. f ( g ( x ))
5 − f ( −4)
8
12. f ( f (3))
RU Ready?
B. Domain and Range
First, since questions in calculus usually ask about behavior of functions in intervals, understand that intervals
can be written with a description in terms of <, ≤, >, ≥ or by using interval notation.
Description Interval
notation
x>a
( a, ∞)
x≥a
[ a, ∞)
x<a
( −∞, a)
Description
x≤a
a<x<b
a≤x≤b
Interval
notation
( −∞, a ]
Description
Interval
notation
[ a, b)
a≤x<b
( a, b) - open interval
[ a, b ] - closed interval
a<x≤b
All real numbers
( a, b ]
( −∞, ∞)
If a solution is in one interval or the other, interval notation will use the connector ∪ . So x ≤ 2 or x > 6 would
be written ( −∞, 2 ] ∪ ( 6, ∞ ) in interval notation. Solutions in intervals are usually written in the easiest way to
define it. For instance, saying that x < 0 or x > 0 or ( −∞, 0 ) ∪ ( 0, ∞ ) is best expressed as x ≠ 0 .
The domain of a function is the set of allowable x-values. The domain of a function f is ( −∞, ∞ ) except for
values of x which create a zero in the denominator, an even root of a negative number or a logarithm of a nonpositive number. The domain of a p( x) where a is a positive constant and p ( x ) is a polynomial is ( −∞, ∞ ) .
• Find the domain of the following functions using interval notation:
6
1. f ( x ) = x 2 − 4x + 4
2. y =
x−6
x≠6
( −∞, ∞)
2x
x − 2x − 3
x ≠ −1, x ≠ 3
3. y =
2
x 2 + 4x + 6
4. y = x + 5
5. y = x + 5
6. y =
2x + 4
( −∞, ∞)
( −2, ∞)
[ −5, ∞)
The range of a function is the set of allowable y-values. Finding the range of functions algebraically isn’t as
easy (it really is a calculus problem), but visually, it is the [lowest possible y-value, highest possible y-value].
Finding the range of some functions are fairly simple to find if you realize that the range of y = x 2 is [ 0, ∞ ) as
3
any positive number squared is positive. Also the range of y = x is also positive as the domain is [ 0, ∞ ) and
the square root of any positive number is positive. The range of y = a x where a is a positive constant is ( 0, ∞ )
as constants to powers must be positive.
• Find the range of the following functions using interval notation:
1
7. y = 1 − x 2
8. y = 2
x
( −∞,1]
( 0, ∞)
9. y = x − 8 + 2
[ 2, ∞)
• Find the domain and range of the following functions using interval notation.
10.
www.mastermathmentor.com
Domain: ( −∞, ∞ )
11.
Range: [ −0.5, 2.5]
9
Domain: ( 0, 4)
Range: [ 0, 4)
RU Ready?
B. Domain and Range Assignment
• Find the domain of the following functions using interval notation:
1. f ( x ) = 3
4. y =
2. y = x − x + x
3
x−4
x 2 −16
2
5. f ( x ) =
1
4x − 4x − 3
2
7. f (t ) = t 3 +1
8. f ( x ) = 5 x 2 − x − 2
10. y = log ( x −10 )
11. y =
2x +14
x 2 − 49
x3 − x2 + x
3. y =
x
6. y = 2x − 9
9. y = 5 x
12. y =
2
−4 x−2
5− x
log x
Find the range of the following functions.
13. y = x 4 + x 2 −1
14. y = 100 x
15. y = x 2 +1 +1
Find the domain and range of the following functions using interval notation.
16.
www.mastermathmentor.com
17.
18.
10
RU Ready?
C. Graphs of Common Functions
No practice problems for this section.
There are certain graphs that occur all the time in calculus and students should know the general shape of them,
where they hit the x-axis (zeros) and y-axis (y-intercept), as well as the domain and range. There are no
assignment problems for this section other than students memorizing the shape of all of these functions. In
section 5, we will talk about transforming these graphs.
Function: y = a
Function: y = x
Function: y = x 2
Function: y = x 3
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Range: [ a, a ]
Range: ( −∞, ∞ )
Range: [ 0, ∞ )
Range: ( −∞, ∞ )
Domain: [ 0, ∞ )
Domain: ( −∞, ∞ )
1
x
Domain: x ≠ 0
Range: y ≠ 0
Function: y = e x
Function: y = e − x
Function: y = sin x
Function: y = cos x
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Domain: ( −∞, ∞ )
Function: y = x
Range: [ 0, ∞ )
Range: ( 0, ∞ )
www.mastermathmentor.com
Function: y =
Function: y = x
Range: [ 0, ∞ )
Range: ( 0, ∞ )
Range: [ −1,1]
11
Function: y = ln x
Domain: ( 0, ∞ )
Range: ( −∞, ∞ )
Range: [ −1,1]
RU Ready?
E. Transformation of Graphs
A curve in the form y = f ( x ) , which is one of the basic common functions from section C can be transformed
in a variety of ways. The shape of the resulting curve stays the same but zeros and y-intercepts might change
and the graph could be reversed. The table below describes transformations to a general function y = f ( x ) with
the parabolic function f ( x ) = x 2 as an example.
Notation
How f ( x ) changes
Example with f ( x ) = x 2
f ( x) + a
Moves graph up a units
f ( x) − a
Moves graph down a units
f ( x + a)
Moves graph a units left
f ( x − a)
Moves graph a units right
a ⋅ f ( x)
a > 1: Vertical Stretch
a ⋅ f ( x)
0 < a < 1: Vertical shrink
f ( ax )
a >1: Horizontal compress
(same effect as vertical stretch)
f ( ax )
0 < a < 1: Horizontal elongated
(same effect as vertical shrink)
− f ( x)
Reflection across x-axis
f ( −x )
Reflection across y-axis
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15
RU Ready?
E. Transformation of Graphs Assignment
• Sketch the following equations:
1. y = −x 2
2. y = 2x 2
3. y = ( x − 2 )
4. y = 2 − x
5. y = x +1 +1
6. y = 4x
7. y = x +1 − 3
8. y = −2 x −1 + 4
9. y = −
10. y = 2 x − 2
11. y = −2 x+2
12. y = 2 −2 x
13. y =
1
x−2
www.mastermathmentor.com
14. y =
−2
x +1
16
15. y =
2
x
−1
2
1
−3
( x + 2 )2
RU Ready?
F. Special Factorization
While factoring skills were more important in the days when A topics were specifically tested, students still
must know how to factor. The special forms that occur most regularly are:
(
)
Common factor: x 3 + x 2 + x = x x 2 + x +1
(
)(
Difference of squares: x 2 − y 2 = ( x + y ) ( x − y ) or x 2n − y 2n = x n + y n x n − y n
Perfect squares: x 2 + 2xy + y 2 = ( x + y )
Perfect squares: x 2 − 2xy + y 2 = ( x − y )
(
)
2
2
)
Sum of cubes: x 3 + y 3 = ( x + y ) x 2 − xy + y 2 - Trinomial unfactorable
(
)
Difference of cubes: x − y = ( x − y ) x + xy + y 2 - Trinomial unfactorable
3
3
2
Grouping: xy + xb + ay + ab = x ( y + b) + a ( y + b) = ( x + a ) ( y + b)
The term “factoring” usually means that coefficients are rational numbers. For instance, x 2 − 2 could
technically be factored as x + 2 x − 2 but since 2 is not rational, we say that x 2 − 2 is not factorable.
(
)(
)
It is important to know that x 2 + y 2 is unfactorable.
• Completely factor the following expressions.
1. 4a 2 + 2a
2. x 2 +16x + 64
( x + 8)2
2a ( 2a +1)
4. 5x 4 − 5y 4
5( x 2 + y2 )( x + y)( x − y)
7. 2x 2 − 40x + 200
2 ( x −10 )
10. x 4 +11x 2 − 80
( x + 4) ( x − 4) x 2 + 5
(
13. x 3 − x 2 + 3x − 3
x 2 ( x −1) + 3( x −1)
( x −1) ( x 2 + 3)
www.mastermathmentor.com
4 ( x + 4) ( x − 4)
5. 16x 2 − 8x +1
( 4x −1)2
)
6. 9a 4 − a 2 b 2
a 2 (3a + b) (3a − b)
8. x 3 − 8
9. 8x 3 + 27y 3
11. x 4 −10x 2 + 9
( x +1) ( x −1) ( x + 3) ( x − 3)
12. 36x 2 − 64
4 (3x + 4) (3x − 4)
14. x 3 + 5x 2 − 4x − 20
15. 9 − x 2 + 2xy + y 2
( x − 2) ( x 2 + 2x + 4)
2
3. 4x 2 − 64
x 2 ( x + 5) − 4 ( x + 5)
( x + 5) ( x − 2) ( x + 2)
17
( 2x + 3y) ( 4x 2 − 6xy + 9y 2 )
(
9 − ( x + y)
)
2
(3 + x + y ) (3 − x − y )
RU Ready?
F. Special Factorization - Assignment
• Completely factor the following expressions
1. x 3 − 25x
2. 30x − 9x 2 − 25
3. 3x 3 − 5x 2 + 2x
4. 3x 8 − 3
5. 16x 4 − 24x 2 y + 9y 2
6. 9a 4 − a 2 b 2
7. 4x 4 + 7x 2 − 36
8. 250x 3 −128
9.
10. x 5 +17x 3 +16x
11. 144 + 32x 2 − x 4
12. 16x 4a − y8a
13. x 3 − xy 2 + x 2 y − y 3
14. x 6 − 9x 4 − 81x 2 + 729
15. x 2 − 8xy +16y 2 − 25
16. x 5 + x 3 + x 2 +1
17. x 6 −1
18. x 6 +1
www.mastermathmentor.com
18
8x 3 64
+
125 y 3
RU Ready?
G. Linear Functions
Probably the most important concept from precalculus that is required for differential calculus is that of linear
functions. The formulas you need to know backwards and forwards are:
Slope: Given two points ( x1, y1 ) and ( x2 , y2 ) , the slope of the line passing through the points can be written as:
rise Δy y2 − y1
.
m=
=
=
run Δx x2 − x1
Slope intercept form: the equation of a line with slope m and y-intercept b is given by y = mx + b .
Point-slope form: the equation of a line passing through the points ( x1, y1 ) and slope m is given by
y − y1 = m ( x − x1 ) . While you might have preferred the simplicity of the y = mx + b form in your algebra
course, the y − y1 = m ( x − x1 ) form is far more useful in calculus.
x y
+ =1.
a b
General form: Ax + By + C = 0 where A, B and C are integers. While your algebra teacher might have required
your changing the equation y −1 = 2 ( x − 5) to general form 2x − y − 9 = 0 , you will find that on the AP
calculus test, it is sufficient to leave equations for a lines in point-slope form and it is recommended not to
waste time changing it unless you are specifically told to do so.
Parallel lines Two distinct lines are parallel if they have the same slope: m1 = m2 .
Normal lines: Two lines are normal (perpendicular) if their slopes are negative reciprocals: m1 ⋅ m2 = −1.
Horizontal lines have slope zero. Vertical lines have no slope (slope is undefined).
Intercept form: the equation of a line with x-intercept a and y-intercept b is given by
1. Find the equation of the line in slope-intercept form, with the given slope, passing through the given point.
2
⎛ 1 3⎞
a. m = −4, (1, 2 )
b. m = , ( −5,1)
c. m = 0, ⎜ − , ⎟
⎝ 2 4⎠
3
y − 2 = −4 ( x −1) ⇒ y = −4x + 6
y −1 =
2
2x 7
( x − 5) ⇒ y = −
3
3 3
y=−
3
4
2. Find the equation of the line in slope-intercept form, passing through the following points.
⎛3 ⎞
a. ( 4, 5) and ( −2, −1)
b. ( 0, −3) and ( −5, 3)
c. ⎜ , −1⎟ and
⎝4 ⎠
5 +1
m=
=1
4+2
y − 5 = x − 4 ⇒ y = x +1
3 + 3 −6
=
−5 − 0 5
−6
−6
y+3=
x⇒y=
x−3
5
5
m=
⎛ 1⎞
⎜⎝ 1, ⎟⎠
2
⎛ 1 +1⎞ ⎛ 4 ⎞ 2 + 4
m =⎜ 2 3⎟⎜ ⎟ =
=6
⎝1− 4 ⎠ ⎝ 4⎠ 4 − 3
y−
1
11
= 6 ( x −1) ⇒ y = 6x −
2
2
3. Write equations of the line through the given point a) parallel and b) normal to the given line.
⎛2 ⎞
a. ( 4, 7), 4x − 2y = 1
b. ⎜ ,1⎟ , x + 5y = 2
⎝3 ⎠
y = 2x −
1
⇒m=2
2
a ) y − 7 = 2 ( x − 4) www.mastermathmentor.com
−1
−1
x+2⇒m=
5
5
−1 ⎛
2⎞
a ) y −1 = ⎜ x − ⎟ ⎝
5
3⎠
y=
b) y − 7 =
−1
( x − 4)
2
2⎞
⎛
b) y −1 = 5 ⎜ x − ⎟
⎝
3⎠
19
G. Linear Functions - Assignment
1. Find the equation of the line in slope-intercept form, with the given slope, passing through the given point.
2 ⎛
1⎞
−1
a. m = −7, ( −3, −7)
b. m = , ( 2, −8)
c. m = , ⎜ −6, ⎟
3 ⎝
3⎠
2
2. Find the equation of the line in slope-intercept form, passing through the following points.
2⎞
⎛
⎛1 ⎞
a. ( −3, 6 ) and ( −1, 2 )
b. ( −7,1) and (3, −4)
c. ⎜ −2, ⎟ and ⎜ ,1⎟
⎝
⎝2 ⎠
3⎠
3. Write equations of the line through the given point a) parallel and b) normal to the given line.
a. ( 5, −3), x + y = 4
b. ( −6, 2 ), 5x + 2y = 7
c. ( −3, −4), y = −2
4. Find an equation of the line containing ( 4, −2 ) and parallel to the line containing ( −1, 4) and ( 2, 3) . Put your
answer in general form.
5. Find k if the lines 3x − 5y = 9 and 2x + ky = 11 are a) parallel and b) perpendicular.
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20
H. Solving Quadratic Equations
Solving quadratics in the form of ax 2 + bx + c = 0 usually show up on the AP exam in the form of expressions
that can easily be factored. But occasionally, you will be required to use the quadratic formula. When you have
a quadratic equation, factor it, set each factor equal to zero and solve. If the quadratic equation doesn’t factor or
if factoring is too time-consuming, use the quadratic formula:
x=
−b ± b 2 − 4ac
. The discriminant b 2 − 4ac will tell you how many solutions the quadratic has:
2a
⎧> 0, 2 real solutions ( if a perfect square, the solutions are rational)
⎪
b − 4ac ⎨ = 0, 1 real solution
⎪< 0, 0 real solutions ( or 2 imaginary solutions, but AP calculus does not deal with imaginaries)
⎩
2
1. Solve for x.
a. x 2 + 3x + 2 = 0
( x + 2) ( x +1) = 0
b. x 2 −10x + 25 = 0
c. x 2 − 64 = 0
( x − 8) ( x + 8) = 0
x=5
x = 8, x = −8
e. 12x 2 + 23x = −10
( 4x + 5) (3x + 2) = 0
f. 48x − 64x 2 = 9
(8x − 3)2 = 0
5
2
x = − ,x = −
4
3
x=
g. x 2 + 5x = 2
−5 ± 25 + 8
x=
2
−5 ± 33
x=
2
h. 8x − 3x 2 = 2
8 ± 64 − 24
x=
6
8 ± 2 10 4 ± 10
x=
=
6
3
i. 6x 2 + 5x + 3 = 0
j. x 3 − 3x 2 + 3x − 9 = 0
k.
x = −2, x = −1
d. 2x 2 + 9x = 18
( 2x − 3) ( x + 6) = 0
3
x = , x = −6
2
x 2 ( x − 3) − 3( x − 3) = 0
( x − 3) ( x
2
)
−3 = 0
x = 3, x = ± 3
( x − 5)2 = 0
x 5 −3
− =
3 2 x
⎛ x 5 −3 ⎞
6x ⎜ − = ⎟
⎝3 2 x ⎠
2x 2 −15x +18 = 0
( 2x − 3) ( x − 6) = 0
3
8
−5 ± 25 − 72 −5 ± −47
=
12
12
No real solutions
x=
l. x 4 − 7x 2 − 8 = 0
(x
2
)(
)
− 8 x 2 +1 = 0
x = ± 8 = ±2 2
3
x = , x = 6
2
2. If y = 5x 2 − 3x + k , for what values of k will the quadratic have two real solutions?
9
( −3)2 − 4 ( 5) k > 0 ⇒ 9 − 20k > 0 ⇒ k <
20
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21
H. Solving Quadratic Equations Assignment
1. Solve for x.
1
=0
4
a. x 2 + 7x −18 = 0
b. x 2 + x +
d. 12x 2 − 5x = 2
e. 20x 2 − 56x +15 = 0
f. 81x 2 + 72x +16 = 0
g. x 2 +10x = 7
h. 3x − 4x 2 = −5
i. 7x 2 − 7x + 2 = 0
k. x 3 − 5x 2 + 5x − 25 = 0
l. 2x 4 −15x 3 +18x 2 = 0
j. x +
1 17
=
x 4
c. 2x 2 − 72 = 0
2. If y = x 2 + kx − k , for what values of k will the quadratic have two real solutions?
3. Find the domain of y =
www.mastermathmentor.com
2x −1
.
6x − 5x − 6
2
22
I. Asymptotes
p( x)
possibly have vertical asymptotes, lines that the graph of the curve
q ( x)
approach but never cross. To find the vertical asymptotes, factor out any common factors of numerator and
denominator, reduce if possible, and then set the denominator equal to zero and solve.
Rational functions in the form of y =
Horizontal asymptotes are lines that the graph of the function approaches when x gets very large or very small.
While you learn how to find these in calculus, a rule of thumb is that if the highest power of x is in the
denominator, the horizontal asymptote is the line y = 0. If the highest power of x is both in numerator and
highest degree coefficient in numerator
denominator, the horizontal asymptote will be the line y =
. If the
highest degree coefficient in denominator
highest power of x is in the numerator, there is no horizontal asymptote, but a slant asymptote which is not used
in calculus.
1) Find any vertical and horizontal asymptotes for the graph of y =
y=
−x 2
.
x2 − x − 6
−x 2
−x 2
=
x 2 − x − 6 ( x − 3) ( x + 2 )
Vertical asymptotes: x − 3 = 0 ⇒ x = 3 and x + 2 = 0 ⇒ x = −2
Horizontal asymptotes: Since the highest power of x is 2 in both numerator and
denominator, there is a horizontal asymptote at y = −1 .
This is confirmed by the graph to the right. Note that the curve actually crosses its horizontal asymptote on the
left side of the graph.
2) Find any vertical and horizontal asymptotes for the graph of y =
y=
3( x +1)
3x + 3
3
=
=
x − 2x − 3 ( x − 3) ( x +1) x − 3
3x + 3
.
x − 2x − 3
2
2
Vertical asymptotes: x − 3 = 0 ⇒ x = 3 . Note that since the ( x +1) cancels, there
is no vertical asymptote at x = 1, but a hole (sometimes called a removable discontinuity) in the graph.
Horizontal asymptotes: Since there the highest power of x is in the denominator, there is a horizontal asymptote
at y = 0 (the x-axis). This is confirmed by the graph to the right.
2x 2 − 4x
3) Find any vertical and horizontal asymptotes for the graph of y = 2
.
x +4
2x 2 − 4x 2x ( x − 2 )
y= 2
= 2
x +4
x +4
Vertical asymptotes: None. The denominator doesn’t factor and setting it equal
to zero has no solutions.
Horizontal asymptotes: Since the highest power of x is 2 in both numerator and
denominator, there is a horizontal asymptote at y = 2. This is confirmed by the graph to the right.
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23
I. Asymptotes - Assignment
• Find any vertical and horizontal asymptotes and if present, the location of holes, for the graph of
1. y =
x −1
x+5
2. y =
2x 2 + 6x
4. y = 2
x + 5x + 6
7. y =
4 + 3x − x 2
3x 2
10. y =
13. y =
x3
x2 + 4
1
x
−
x x+2
8
x2
x
5. y = 2
x − 25
8. y =
5x +1
2
x − x −1
11. y =
x 3 + 4x
x 3 − 2x 2 + 4x − 8
3. y =
2x +16
x +8
x2 − 5
6. y = 2
2x −12
9. y =
1 − x − 5x 2
x 2 + x +1
12. y =
10x + 20
x − 2x 2 − 4x + 8
3
( hint: express with a common denominator )
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24
J. Negative and Fractional Exponents
In calculus, you will be required to perform algebraic manipulations with negative exponents as well as
1
fractional exponents. You should know the definition of a negative exponent: x −n = n , x ≠ 0. Note that
x
negative powers do not make expressions negative; they create fractions. Typically expressions in multiplechoice answers are written with positive exponents and students are required to eliminate negative exponents.
Fractional exponents create roots. The definition of x1 2 = x and x a b = b x a =
( x) .
a
b
( )( )
As a reminder: when we multiply, we add exponents: x a x b = x a+b .
When we divide, we subtract exponents:
xa
= x a−b , x ≠ 0
b
x
When we raise powers, we multiply exponents:
(x )
a b
= x ab
In your algebra course, leaving an answer with a radical in the denominator was probably not allowed. You had
x
1
⎛ 1 ⎞⎛ x⎞
to rationalize the denominator:
changed to ⎜
. In calculus, you will find that it is not
=
⎟
⎜
⎟
⎝ x⎠⎝ x⎠
x
x
necessary to rationalize and it is recommended that you not take the time to do so.
• Simplify and write with positive exponents. Note: # 12 involves complex fractions, covered in section K.
−2
⎛ −3 ⎞
3 −2
−2
1. −8x
2. −5x
3. ⎜ 4 ⎟
⎝x ⎠
(
−8
x2
(
( −5)
4. 36x10
)
12
12
−x
(
5. 27x 3
(27x )
(
)
2
)
−2
1
=
x − 2x 3 2 + x 2
−2
10.
(8x )−5 3 (8)
3
)
−16
−16
1
=− 53
53 =
53
3(32 ) x
6x
3(8x )
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(x )
4 −2
6. 16x −2
1
9x 2
=
=
(
−2 3
(
8. 4x 2 −12x + 9
1
12
⎡( 2x − 3)2 ⎤
⎣
⎦
11.
( −3)−2
1
1
x =
2 6 =
( −5) x 25x 6
−6
3 23
7. x1 2 − x
(x
−2
1
6x 5
1
)
1
( −3)2 x −8
34
8
x4 3
⎛1
⎞
⎛1
⎞
9. x1 3 ⎜ x −1 2 ⎟ + x1 2 +1 ⎜ x −1 3 ⎟
⎝2
⎠
⎝3
⎠
( )
−1 2
1
=
2x − 3
x8
9
)
16 3 4 x −4 3 =
)
=
(
)
x1 3 x1 2 +1
1
x1 2 +1
+
= 16 +
2x1 2
3x1 3
2x
3x1 3
( x + 4)1 2
( x − 4)−1 2
12.
( x + 4)1 2 ( x − 4)1 2 = ( x 2 −16 )
12
(x
−1
+ y −1
)
−1
⎛
⎞
⎜ 1 ⎟ ⎛ xy ⎞
xy
⎜ 1 1⎟⎜ ⎟ =
⎜ + ⎟ ⎝ xy ⎠ y + x
⎝ x y⎠
25
J. Negative and Fractional Exponents - Assignment
Simplify and write with positive exponents.
1. −12 2 x −5
⎛ −4 ⎞
4. ⎜ 4 ⎟
⎝x ⎠
(
−3
7. 121x 8
⎛ 5x 3 ⎞
5. ⎜ 2 ⎟
⎝ y ⎠
)
( )
12
10. ( x + y )
(
(
2. −12x 5
8. 8x 2
(
−2
3. 4x −1
−3
) (32x )
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−1
(
)
(
−4 3
−2
9. −32x −5
)
11. x 3 + 3x 2 + 3x +1
−3 4
)
6. x 3 −1
(
−2
1
13. 16x 2
4
)
( x −1)
14.
( x +1)
2
−1 2
2
12
−2 3
)
−3 5
(
12. x x1 2 − x
15.
(x
−2
+ 2 −2
)
−2
)
−1
26
O. Solving Inequalities
You may think that solving inequalities are just a matter of replacing the equal sign with an inequality sign. In
reality, they can be more difficult and are fraught with dangers. And in calculus, inequalities show up more
frequently than solving equations. Solving inequalities are a simple matter if they are based on linear equations.
They are solved exactly like linear equations, remembering that if you multiply or divide both sides by a
negative number, the direction of the inequality sign must be reversed.
However, if the inequality is more complex than a linear function, it is advised to bring all terms to one side.
Pretend for a moment it is an equation and solve. Then create a number line which determines whether the
transformed inequality is positive or negative in the intervals created on the number line and choose the correct
intervals according to the inequality, paying attention to whether the zeroes are included or not.
If the inequality involves an absolute value, create two equations, replacing the absolute value with a positive
parentheses and a negative parentheses and the inequality sign with an equal sign. Solve each, placing each
solution on your number line. Then determine which intervals satisfy the original inequality.
If the inequality involves a rational function, set both numerator and denominator equal to zero, which will give
you the values you need for your number line. Determine whether the inequality is positive or negative in the
intervals created on the number line and choose the correct intervals according to the inequality, paying
attention to whether the endpoints are included or not.
• Solve the following inequalities.
2. 1 −
1. 2x − 8 ≤ 6x + 2
−10 ≤ 4x
or
3. −7 ≤ 6x −1 < 11
2 − 3x > 2x −10
12
12 > 5x ⇒ x <
5
−4x ≤ 10
−5
≤x
2
3x
> x−5
2
−5
x≥
2
4. 2x −1 ≤ x + 4
−6 ≤ 6x ≤ 12
−1 ≤ x ≤ 2
5. x 2 − 3x > 18
2x −1 − x − 4 ≤ 0
x 2 − 3x −18 > 0 ⇒ ( x + 3) ( x − 6 ) > 0
2x −1 − x − 4 = 0 −2x +1 − x − 4 = 0
x=5
x = −1
For ( x + 3) ( x − 6 ) = 0, x = −3, x = 6
+ + + + + + +0 − − − − − − − − − 0 + + + + + + +
-1
++++++0−−−−−−−0+++++++
-3
6
5
So −1 ≤ x ≤ 5 or [ −1, 5]
6.
So x < −3 or x > 6
2x − 7
≤1
x−5
2x − 7
2x − 7 x − 5
x−2
−1 = 0 ⇒
−
<0⇒
<0
x−5
x−5 x−5
x−5
++++++0−−−−−−−∞+++++++
2
7. Find the domain of
32 − 2x 2
2 ( 4 + x )( 4 − x ) ≥ 0
−−−−−−0+++++++0−−−−−−−
-4
5
4
So -4 ≤ x ≤ −4 or [ −4, 4 ]
So 2 ≤ x < 5 or [ 2, 5)
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( −∞, −3) ∪ ( 6, ∞)
or
35
RU Ready?
O. Solving Inequalities - Assignment
• Solve the following inequalities.
1. 5 ( x − 3) ≤ 8 ( x + 5)
2. 4 −
5x
1⎞
⎛
> − ⎜ 2x + ⎟
⎝
3
2⎠
3.
3
1
> x +1 >
4
2
5.
( x + 2)2 < 25
6. x 3 < 4x 2
7.
5
1
≥
x−6 x+2
8. Find the domain of
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4. x + 7 ≥ 5 − 3x
36
x2 − x − 6
x−4
RU Ready?
Q. Right Angle Trigonometry
Trigonometry is an integral part of AP calculus. Students must know the basic trig function definitions in terms
of opposite, adjacent and hypotenuse as well as the definitions if the angle is in standard position.
Given a right triangle with one of the angles named θ , and the sides of the triangle relative to θ named opposite
(y), adjacent (x) , and hypotenuse (r) we define the 6 trig functions to be:
opposite
y
=
hypotenuse r
adjacent
x
cosθ =
=
hypotenuse r
opposite y
tan θ =
=
adjacent x
hyptotenuse
=
opposite
hyptotenuse
secθ =
=
adjacent
adjacent x
cot θ =
=
opposite y
sin θ =
cscθ =
r
y
r
x
The Pythagorean theorem ties these variables together: x 2 + y 2 = r 2 . Students
should recognize right triangles with integer sides: 3-4-5, 5-12-13, 8-15-17,
7-24-25. Also any multiples of these sides are also sides of a right triangle.
Since r is the largest side of a right triangle, it can be shown that the range of
sin θ and cosθ is [ −1,1], the range of cscθ and secθ is ( −∞, −1] ∪ [1, ∞ )
and the range of tan θ and cot θ is ( −∞, ∞ ).
Also vital to master is the signs of the trig functions in the four quadrants. A good way to remember this is
A – S – T – C where All trig functions are positive in the 1st quadrant, Sin is positive in the 2nd quadrant, Tan is
positive in the 3rd quadrant and Cos is positive in the 4th quadrant.
1. Let be a point on the terminal side of θ . Find the 6 trig functions of θ . (Answers need not be rationalized).
a) P (−8,6)
b. P (1, 3)
c. P − 10, − 6
(
x = −8, y = 6, r = 10
3
5
sin θ =
cscθ =
5
3
4
5
cosθ = −
secθ = −
5
4
3
4
tan θ = −
cot θ = −
4
3
2.
2
If cosθ = , θ in quadrant IV,
3
find sin θ and tanθ
x = 2, r = 3, y = − 5
sin θ = −
5
5
, tan θ = −
3
2
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x = − 10, y = − 6, r = 4
x = 1, y = 3, r = 10
3
10
1
cosθ =
10
sin θ =
3.
10
3
sin θ = −
secθ = 10
cosθ = −
cscθ =
tan θ = 3
cot θ =
1
3
If secθ = 3
find sin θ and tanθ
θ is in quadrant I or IV
x = 1, y = ± 2, r = 3
sin θ = ±
2
, tanθ = ± 2
3
39
)
tan θ =
6
4
10
4
3
5
cscθ = −
4
6
secθ = −
4
10
cot θ =
5
3
4. Is 3cosθ + 4 = 2 possible?
3cosθ = −2
2
cosθ = − which is possible.
3
RU Ready?
Q. Right Angle Trigonometry - Assignment
1. Let be a point on the terminal side of θ . Find the 6 trig functions of θ . (Answers need not be rationalized).
a) P (15,8)
12
, θ in quadrant III,
2.
5
find sin θ and cosθ
If tan θ =
(
b. P ( −2, 3)
c. P −2 5, − 5
6
If cscθ = , θ in quadrant II,
3.
5
find cosθ and tanθ
)
−2 10
3
find sin θ and cosθ
4. cot θ =
5. Find the quadrants where the following is true: Explain your reasoning.
a. sin θ > 0 and cosθ < 0
b. cscθ < 0 and cotθ > 0
c. all functions are negative
6. Which of the following is possible? Explain your reasoning.
a. 5sin θ = −2
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b. 3sin α + 4 cos β = 8
40
c. 8tan θ + 22 = 85
RU Ready?
R. Special Angles
Students must be able to find trig functions of quadrant angles ( 0, 90°,180°, 270°) and special angles, those
based on the 30° − 60° − 90° and 45° − 45° − 90° triangles.
First, for most calculus problems, angles are given and found in radians. Students must know how to convert
degrees to radians and vice-versa. The relationship is 2π radians = 360° or π radians = 180° . Angles are
π
assumed to be in radians so when an angle of
is given, it is in radians. However, a student should be able to
3
180°
picture this angle as
= 60° . It may be easier to think of angles in degrees than radians, but realize that
3
⎛ 1⎞ π
unless specified, angle measurement must be written in radians. For instance, sin −1 ⎜ ⎟ = .
⎝ 2⎠ 6
π
3π ⎞
⎛
The trig functions of quadrant angles ⎜ 0, 90°,180°, 270° or 0, , π , ⎟ can quickly be found. Choose a point
⎝
2
2⎠
along the angle and realize that r is the distance from the origin to that point and always positive. Then use the
definitions of the trig functions.
θ
0
point
x
y
r
sin θ
cosθ
tan θ
(1, 0)
1
0
1
0
1
0
π
or 90°
2
π or 180°
( 0,1)
0
1
1
1
0
does not
exist
1
does not
exist
0
( −1, 0)
-1
0
1
0
-1
0
does not
exist
-1
does not
exist
( 0, −1)
0
-1
1
-1
0
Does not
exist
-1
does not
exist
0
3π
or 270°
2
cscθ
does not
exist
secθ
1
cot θ
does not
exist
If you picture the graphs of y = sin x and y = cos x as shown to
the right, you need not memorize the table. You must know
these graphs backwards and forwards.
• Without looking at the table, find the value of
a. 5cos180° − 4sin 270°
5 ( −1) − 4 ( −1)
−5 + 4 = −1
π
⎛
⎞
8sin − 6 tan π
⎜
⎟
2
b. ⎜
3π ⎟
⎜⎝ 5sec π − csc ⎟⎠
2
2
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⎡ 8 (1) − 6 ( 0 ) ⎤ ⎛ 8 ⎞ 2
⎢
⎥ =⎜ ⎟ =4
⎣ 5 ( −1) − ( −1) ⎦ ⎝ −4 ⎠
2
41
RU Ready?
Because over half of the AP exam does not use a calculator, you must be able to determine trig functions of
⎛π π π⎞
special angles. You must know the relationship of sides in both 30° − 60° − 90° ⎜ , , ⎟
⎝ 6 3 2⎠
⎛π π π⎞
and 45° − 45° − 90° ⎜ , , ⎟ triangles.
⎝ 4 4 2⎠
⎛π π π⎞
In a 30° − 60° − 90° ⎜ , , ⎟ triangle,
⎝ 6 3 2⎠
⎛π π π⎞
In a 45° − 45° − 90° ⎜ , , ⎟ triangle,
⎝ 4 4 2⎠
the ratio of sides is 1 − 3 − 2.
the ratio of sides is 1 −1 − 2.
θ
⎛
30° ⎜or
⎝
⎛
45° ⎜or
⎝
⎛
60° ⎜or
⎝
!⎞
⎟
6⎠
!⎞
⎟
4⎠
!⎞
⎟
3⎠
sin θ
1
2
2
2
3
2
cosθ
3
2
2
2
1
2
tan θ
3
3
1
3
⎛π⎞
⎛π⎞
Special angles are any multiple of 30° ⎜ ⎟ or 45° ⎜ ⎟ . To find trig functions of any of these angles, draw
⎝ 6⎠
⎝ 2⎠
them and find the reference angle (the angle created with the x-axis). Although most problems in calculus will
use radians, you might think easier using degrees. This will create one of the triangles above and trig functions
can be found, remembering to include the sign based on the quadrant of the angle. Finally, if an angle is outside
the range of 0° to 360° ( 0 to 2π ) , you can always add or subtract 360° ( 2π ) to find trig functions of that angle.
These angles are called co-terminal angles. It should be pointed out that 390° ≠ 30° but sin390° = sin 30°.
• Find the exact value of the following
7π ⎞
⎛
b. ⎜ 2 cos π − 5tan ⎟
⎝
4 ⎠
a. 4sin120° − 8cos570°
Subtract 360° from 570°
( 2 cos180° − 5tan 315°)2
4sin120° − 8cos210°
120° is in quadrant II with reference angle 60°.
180° is a quadrant angle
315° is in quadrant III with reference angle 45°
210° is in quadrant III with reference angle 30°.
⎡⎣ 2 ( −1) − 5 ( −1)⎤⎦ = 9
2
⎛ 3⎞
⎛ − 3⎞
4 ⎜ ⎟ − 8⎜
=6 3
⎝ 2 ⎠
⎝ 2 ⎟⎠
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2
42
RU Ready?
R. Special Angles – Assignment
• Evaluate each of the following without looking at a chart.
1. sin 2 120° + cos2 120°
2. 2 tan 2 300° + 3sin 2 150° − cos2 180°
3. cot 2 135° − sin 210° + 5cos2 225
4. cot ( −30°) + tan ( 600°) − csc ( −450°)
2π
3π ⎞
⎛
− tan ⎟
5. ⎜ cos
⎝
3
4⎠
5π ⎞ ⎛ 11π
5π ⎞
⎛ 11π
− tan ⎟ ⎜ sin
+ tan ⎟
6. ⎜ sin
⎝
6
6 ⎠⎝
6
6 ⎠
2
• Determine whether each of the following statements are true or false.
7. sin
π
π
⎛π π⎞
+ sin = sin ⎜ + ⎟
⎝ 6 3⎠
6
3
3π ⎞ ⎛
3π ⎞
⎛ 3π
+ sin ⎟ ⎜ 1 + cos ⎟ > 0
9. 2 ⎜
⎝ 2
2 ⎠⎝
2⎠
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5π
5π
+1
cos
3
3
=
8.
5π
2 5π
tan
sec
−1
3
3
cos
10.
cos3
44
4π
4π
+ sin
3
3 >0
2 4π
cos
3
RU Ready?
T. Solving Trig Equations and Inequalities
Trig equations are equations using trig functions. Typically they have many (or infinite) number of solutions so
usually they are solved within a specific domain. Without calculators, answers are either quadrant angles or
special angles, and again, they must be expressed in radians.
For trig inequalities, set both numerator and denominator equal to zero and solve. Make a sign chart with all
these values included and examine the sign of the expression in the intervals. Basic knowledge of the sine and
cosine curve is invaluable from section R is invaluable.
• Solve for x on [0,2π )
2. tan x + sin 2 x = 2 − cos2 x
1. x cos x = 3cos x
Do not divide by cos x as you will lose solutions
tan x + sin 2 x + cos2 x = 2
cos x ( x − 3) = 0
cos x = 0
tan x +1 = 2
x−3= 0
You must work in radians.
tan x = 1
π 5π
x= ,
4 4
Two answers as tangent is positive
Saying x = 90° makes no sense.
in quadrants I and III.
x=
π 3π
,
2 2
x=3
3. 3tan 2 x −1 = 0
4. 3cos x = 2sin 2 x
(
3cos x = 2 1 − cos2 x
3tan 2 x = 1
1
tan 2 x =
3
1
3
tan x = ±
=±
3
3
π 5π 7π 11π
x= , , ,
6 6 6 6
7. Solve for x on [ 0, 2π ) :
)
2 cos x + 3cos x − 2 = 0
2
( 2 cos x −1) (cos x + 2) = 0
cos x = −2
2 cos x = 1
No solution
1
cos x = 2
π 5π
x= ,
3 3
2 cos x +1
>0
sin 2 x
−1
2π 4π
+ + + + + + +0 − − − − − − − ∞ − − − − − − − 0 + + + + + + +
⇒x=
,
2
3 3
2π
4π
0
π
2π
2
sin x = 0 ⇒ x = 0, π
3
3
2 cos x = −1 ⇒ cos x =
⎡ 2π ⎞ ⎛ 4π
⎞
Answer: ⎢ 0, ⎟ ∪ ⎜ , 2π ⎟
⎠
⎣ 3 ⎠ ⎝ 3
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T. Solving Trig Equations and Inequalities - Assignment
• Solve for x on [ 0, 2π )
1. sin 2 x = sin x
2. 3tan 3 x = tan x
3. sin 2 x = 3cos2 x
4. cos x + sin x tan x = 2
5. sin x = cos x
6. 2 cos2 x + sin x −1 = 0
7. Solve for x on [ 0, 2π ) :
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x −π
<0
cos2 x
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U. Graphical Solutions to Equations and Inequalities
You have a shiny new calculator. So when are we going to use it? So far, no mention has been made of it. Yet,
the calculator is a tool that is required in the AP calculus exam. For about 25% of the exam, a calculator is
permitted. So it is vital you know how to use it.
There are several settings on the calculator you should make. First, so you don’t get into
rounding difficulties, it is suggested that you set your calculator to three decimal places.
That is a standard in AP calculus so it is best to get into the habit. To do so, press MODE
and on the 2nd line, take it off FLOAT and change it to 3. And second, set your calculator to
radian mode from the MODE screen. There may be times you might want to work in
degrees but it is best to work in radians.
You must know how to graph functions. The best way to graph a function is to input the
function using the Y= key. Set your XMIN and XMAX using the WINDOW key. Once you do that, you can
see the graph’s total behavior by pressing ZOOM 0. To evaluate a function at a specific value of x, the easiest
way to do so is to press these keys: VARS → 1:Function 1 1:Y1 ( and input your x-value.
Other than basic calculations, and taking trig functions of angles, there are three calculator features you need to
know: evaluating functions at values of x and finding zeros of functions, which we know is finding where the
function crosses the x-axis. The other is finding the point of intersection of two graphs. Both of these features
are found on the TI-84+ calculator in the CALC menu 2ND TRACE. They are 1:value, 2: zero, and 5: intersect.
Solving equations using the calculator is accomplished by setting the equation equal to zero, graphing the
function, and using the ZERO feature. To use it, press 2ND TRACE ZERO . You will be prompted to type in a
number to the left of the approximate zero, to the right of the approximate zero, and a guess (for which you can
press ENTER). You will then see the zero (the solution) on the screen.
• Solve these equations graphically.
1. 2x 2 − 9x + 3 = 0
2. 2 cos2x − x = 0 on [ 0, 2π ) and find 2 cos ( 2e) − e .
This equation can be solved with the quadratic formula.
9 ± 81 − 24 9 ± 57
x=
=
4
4
If this were the inequality 2 cos2x − x > 0,
the answer would be [ 0, 0.626 ).
3. Find the x-coordinate of the intersection of y = x 3 and y = 2x − 3
You can use the intersection feature.
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Or set them equal to each other: x 3 = 2x − 3 or x 3 − 2x + 3 = 0
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U. Graphical Solutions to Equations and Inequalities – Assignment
• Solve these equations or inequalities graphically.
1. 3x 3 − x − 5 = 0
2. x 3 − 5x 2 + 4x −1 = 0
3. 2x 2 −1 = 2 x
4. 2 ln ( x +1) = 5cos x
5. x − 9x − 3x −15 < 0
x 2 − 4x − 4
> 0 on [ 0,8]
6.
x 2 +1
7. x sin x 2 > 0 on [ 0, 3]
8. cos−1 x > x 2 on [ −1,1]
4
2
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on [ 0, 2π )
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