# Module 5 Lecture Notes

```Module 5 Lecture Notes
Contents
5.1
5.1
An Introductory Example . . . . . . . . . . . . . . . . . . . . . . . . . .
1
5.2
Symbolic Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
5.3
One-to-One and Invertibility . . . . . . . . . . . . . . . . . . . . . . . . .
7
5.4
Numerical Examples
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
5.5
Graphing Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . .
9
5.6
Exponential Function Introduction . . . . . . . . . . . . . . . . . . . . .
10
5.7
Modeling with Exponential Functions . . . . . . . . . . . . . . . . . . .
12
5.8
Transformations of Exponential Functions . . . . . . . . . . . . . . . .
15
5.9
What’s “e”? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
5.10 Finding the Formula for an Exponential Function (REQUIRED) . . .
22
An Introductory Example
Example 1: Temperature in degrees Fahrenheit, F , can be written as a function of temperature
in degrees Celsius, C. This relationship is given by F = g(C) = 59 C + 32.
(a) Find and interpret g(100).
9
g(100) = (100) + 32
5
= 180 + 32
= 212
The temperature 100o C is equivalent to the temperature 212o F.
1
Math 111 Module 5 Lecture Notes
(b) Solve and interpret the solution to g(C) = 32.
g(C) = 32
9
C + 32 = 32
5
9
C=0
5
C=0
The temperature 32o F is equivalent to the temperature 0o C.
(c) Solve the equation F = 59 C + 32 for C.
9
F = C + 32
5
9
F − 32 = C
5
5
(F − 32) = C
9
5
C = (F − 32)
9
A function f is said to be one-to-one if for every y-value in the range of f there is exactly
one x-value in the domain of f .
A function must be one-to-one in order to have an inverse. The inverse function of f
reverses the process of the original function. In other words, the input and output switch
roles. The original function is given by y = f (x). The inverse function is given by x = f −1 (y).
If we want to graph both of these functions in the (x, y)-plane, then we use y = f −1 (x).
The inverse function of f is denoted by f −1 . It is important to note that this
1
notation is not denoting a reciprocal. That is, f −1 (x) 6=
.
f (x)
Instructor: A.E.Cary
Page 2 of 25
Math 111 Module 5 Lecture Notes
5.2
Symbolic Examples
Example 2: The function f defined by f (x) = 3x + 2 is one-to-one. Find its inverse. Then graph
y = f (x) and y = f −1 (x) in Figure 5.1. Include the graph of y = x also.
Original Function: y = 3x + 2
To find the inverse, we switch the variables x and y
and solve for y:
Figure 5.1
y
y=x
y = f (x)
y = f −1 (x)
4
x = 3y + 2
x − 2 = 3y
1
(x − 2) = y
3
1
y = (x − 2)
3
1
−1
f (x) = (x − 2)
3
2
x
−4
−2
2
4
−2
−4
In the example above, the variables x and y were switched. Notice how the graph of y = f −1 (x)
is a reflection of the graph of y = f (x) about the line y = x. This is because all of the (x, y) pairs
switched – the point (0, 2) is on the graph of y = f (x) while the point (2, 0) is on the graph of
y = f −1 (x). Similarly, the point (1, 5) is on the graph of y = f (x) whereas the point (5, 1) is on
the graph of y = f −1(x).
Example 3: To verify that two functions are inverses, we show that f (f −1 (x)) = x and that
f −1 (f (x)) = x. Do this for the previous example.
f f
−1
1
(x − 2)
3
1
=3
(x − 2) + 2
3
(x) = f
Instructor: A.E.Cary
f −1 (f (x)) = f −1 (3x + 2)
= (x − 2) + 2
1
((3x + 2) − 2)
3
1
= (3x)
3
= xX
= xX
=
Page 3 of 25
Math 111 Module 5 Lecture Notes
2x
is one-to-one. Find the inverse funcx−1
tion. Confirm that the inverse function you found is correct by showing f (f −1 (x)) = x and
f −1 (f (x)) = x.
Example 4: The function f defined by f (x) = −
Original Function: y = −
Finding the Inverse:
2x
x−1
x=−
2y
y−1
x(y − 1) = −2y
xy − x = −2y
−x = −2y − xy
−x = y(−2 − x)
−x
=y
−2 − x
−x
−2 − x
−x
y=
−1(x + 2)
x
y=
x+2
x
f −1 (x) =
x+2
y=
Confirming the Inverse:
Instructor: A.E.Cary
Page 4 of 25
Math 111 Module 5 Lecture Notes
f (f −1(x)) = f
=−
=−
=−
x
x+2
x
x+2
2
x
x+2
2
f
−1
x
x+2
x
−
x+2
x+2
1 x+2
•
−1
2x
−
(f (x)) = f
x−1
2x
− x−1
=
2x
− x−1
+2
2x
− x−1
x−1
=
•
2x
− x−1 + 2 x − 1
2x
x − 1(x + 2)
−1
=
−2x
−2x + 2(x − 1)
−2x
−2x + 2x − 2
−2x
=
−2
2x
x−x−2
2x
=−
−2
=−
=
=x
=x
To state the domain and range of each f and f −1 , we need to recognize that the inputs of f are
the outputs of f −1 . Likewise, the outputs of f are the inputs of f −1 .
The domain of f is the range of f −1 . Similarly, the range of f is the domain of f −1 .
Table 5.1
Function
Domain
Range
f
f −1
{x | x 6= 1}
{x | x 6= −2}
{x | x 6= −2}
{x | x 6= 1}
Instructor: A.E.Cary
Page 5 of 25
Math 111 Module 5 Lecture Notes
√
Example 5: The function g defined by g(x) = 3 x + 8 is one-to-one. Find the inverse function
and confirm that it is the inverse by showing g (g −1 (x)) = x and g −1 (g(x)) = x. In Figure 5.2, use
transformations to sketch y = g(x), y = g −1 (x) and y = x.
√
Original Function y = 3 x + 8
Finding the Inverse:
p
x= 3 y+8
x3 = y + 8
x3 − 8 = y
y = x3 − 8
g −1 (x) = x3 − 8
Confirming the Inverse:
g −1 (g(x)) = g −1
−1
g g (x) = g(x3 − 8)
p
= 3 (x3 − 8) + 8
√
3
= x3
=
√
3
√
3
x+8
x+8
3
−8
=x+8−8
= xX
= xX
Figure 5.2
y
y=x
10
y = g(x)
y = g −1 (x)
8
6
4
2
x
−10
−8
−6
−4
−2
−2
2
4
6
8
10
−4
−6
−8
−10
√
√
To graph y = 3 x + 8, shift the graph of y = 3 x left 8 units. To graph y = x3 − 8, shift the
graph of y = x3 down 8 units.
Instructor: A.E.Cary
Page 6 of 25
Math 111 Module 5 Lecture Notes
5.3
One-to-One and Invertibility
The horizontal line test is a way of determining if a function is one-to-one. It states that if
every horizontal line passes through a graph at most once, then the function is one-to-one.
In the same way that the vertical line test verifies if a graph represents a function, the horizontal
line test verifies if the graph of a function is one-to-one (and thus invertible).
Example 6: Graph y = f (x) for f (x) = x2 in Figure 5.3. Then graph y = g(x) for g(x) = x2 , x ≥
0 in Figure 5.4. Is either function invertible? Why or why not?
Figure 5.4: Graph of y = g(x)
Figure 5.3: Graph of y = f (x)
y
y
4
4
2
2
x
−4
−2
2
4
x
−4
−2
−2
y=x
y = x2
x=y
2
−4
2
4
−2
y=x
−4
y = g(x)
y=g
−1
(x)
We can see from the graph of y = x2 that this function cannot be inverted. Since this function
fails the horizontal line test, when we reflect it across the line y = x, the resulting graph is not a
function as it fails the vertical line test.
However, if we restict the domain on y = x2 to x ≥ 0, the resulting function is indeed invertible.
This is shown in Figure ??.
Instructor: A.E.Cary
Page 7 of 25
Math 111 Module 5 Lecture Notes
5.4
Numerical Examples
Example 7: Use the functions f and g given in Table 5.2 to determine the following.
Table 5.2
x
-2
-1
0
1
2
f (x)
5
4
2
-1
1
g(x)
7
2
0
-2
9
Recall that if y = g(x) and g is invertible, then g −1 (y) = x.
(a) g −1 (−2)
g −1 (−2) = 1 as g(1) = −2
(b) f −1 (2)
f −1 (2) = 0 as f (0) = 2
(c) f −1 (0)
f −1 (0) is undefined as 0 is not in the range of f
(d) f g −1 (0)
f g −1 (0) = f (0)
=2
Instructor: A.E.Cary
Page 8 of 25
Math 111 Module 5 Lecture Notes
5.5
Graphing Inverse Functions
Example 8: Graph the inverse function of f in Figure 5.5.
To graph the inverse function, we will switch the inputs and outputs of the original function. For
y = f (x), the points (−4, −4), (−3, −2), (−2, 0), (−1, 1), (0, 2), (1, 3) and (2, 4) are on the graph.
Therefore the points (−4, −4), (−2, −3), (0, −2), (1, −1), (2, 0), (3, 1) and (4, 2) are on the graph of
y = f −1 (x).
Figure 5.5
y
y=x
y = f (x)
y = f −1 (x)
4
2
x
−4
−2
2
4
−2
−4
Instructor: A.E.Cary
Page 9 of 25
Math 111 Module 5 Lecture Notes
5.6
Exponential Function Introduction
In 1988, a judge in Yonkers, New York instituted an exponential fine on the city of Yonkers. Below
is the background and scenario, published in the New York Times1 :
Dec. 1, 1980: Justice Department sues Board of Education, City of Yonkers
and Yonkers Community Development Agency, charging that the city racially
discriminated in education and public housing.
Nov. 20, 1985: Judge Leonard B. Sand of Federal District Court in
Manhattan rules that Yonkers’s housing and schools were intentionally
segregated by race. A housing remedy order directs the city to build 200
units of public housing and to plan additional subsidized housing.
Jan. 28, 1988: City Council approves consent decree that sets timetable
for building 200 units of public housing and commits city to an
July 26, 1988: Court sets Aug. 1 deadline for Council to adopt zoning
amendment needed to build the 800 units.
Aug. 1, 1988: Council rejects amendment in a 4-to-3 vote.
Aug. 2, 1988: Judge Sand finds city and the four Councilmen who voted
against the amendment in contempt of court and imposes fines. The city’s
fines start at \$100 and double every day. The Councilmen’s fines start
at \$500 a day and increase by \$500 each day.
Example 9: Let P be the amount fined (in dollars) t days after the fine begins. Complete the
first three entries of each Table 5.3 and Table 5.4.
Table 5.3: Councilmen
t
P
Formula
t
P
Formula
0
500
500
0
100
100
1
1000
500 + 500
1
200
100 · 2
2
1500
500 + 500(2)
2
400
(100 · 2) · 2
3
2000
500 + 500(3)
3
800
(100 · 2 · 2) · 2
4
2500
500 + 500(4)
4
1600
(100 · 2 · 2 · 2) · 2
5
..
.
3000
500 + 500(5)
5
..
.
3200
(100 · 2 · 2 · 2 · 2) · 2
..
.
500 + 500t
t
t
1
Table 5.4: City of Yonkers
100 · 2t
http://www.nytimes.com/1988/09/10/nyregion/yonkers-legal-battle-how-it-unfolded.html
Instructor: A.E.Cary
Page 10 of 25
Math 111 Module 5 Lecture Notes
Example 10:
Figure 5.6: Councilmen: P = 500 + 500t
Figure 5.7: City of Yonkers
3,600
3,000
P , fine in dollars
P , fine in dollars
3,500
2,500
2,000
1,500
1,000
500
1
2
3
4
5
t, time in days
• Linear function
3,200
2,800
2,400
2,000
1,600
1,200
800
400
6
1
2
3
4
5
6
t, time in days
• Exponential function
• Increasing by a constant amount
• Increasing by a constant factor
• The initial value is 500
• The initial value is 100
Example 11: On what day will will the city of Yonkers’ fine reach over \$1,000,000?
On day 13, the fine will be \$819,200. On day 14, the fine will be \$1,638,400.
Example 12: How much will the city of Yonkers be fined on day 30? What will each of the
Councilmen’s fines be on that day?
P = 500 + 500(30)
P = 100(2)30
= 15500
= 107,374,182,400
The Councilmen’s fine would be \$15,500. The city’s fine would be \$107,374,182,400.
Instructor: A.E.Cary
Page 11 of 25
Math 111 Module 5 Lecture Notes
An exponential function is of the form
f (x) = C ax
where
• C is the initial value
• a is the growth factor and a > 0
Consequently, an exponential function is a function that increases or decreases at a constant percent rate. Let’s review percent increase and decrease as we work through these examples.
5.7
Modeling with Exponential Functions
Example 13: You start a new job with an initial salary of \$36,000 per year. Each year thereafter,
you receive a 3% raise. Let S(t) be your salary t years after you start your new job.
(a) Write the formula for S(t).
t
S(t)
0
36000
1
3
36000 + 0.03(36000) = 36000(1.03)
36000(1.03)
(1.03) = 36000(1.03)2
36000(1.03) (1.03) = 36000(1.03)3
4
36000(1.03)4
5
..
.
36000(1.03)5
..
.
t
36000(1.03)t
2
The formula is S(t) = 36000(1.03)t
(b) What will your salary be after 10 years?
S(10) = 36000(1.03)10
≈ 48380.99
Instructor: A.E.Cary
Page 12 of 25
Math 111 Module 5 Lecture Notes
(c) When will your salary reach \$72,000? (Use your graphing calculator to solve this).
We need to solve 72000 = 36000(1.03)t . To do so with a graphing calculator, graph:
• Graph y1(x) = 36000(1.03)x
• Graph y2(x) = 72000
• Find the point of intersection
The point of intersection is about (23.45, 72000).
\$72000 in 24 years.
Instructor: A.E.Cary
Page 13 of 25
Math 111 Module 5 Lecture Notes
Example 14: A compost pile has 27 cubic feet of waste and decays at a rate of 10% per month.
Let Q(t) be the volume of compost (in cubic feet) t months since decay began. Write the formula
for this decreasing exponential function.
Decreasing by 10% means that 90% remains. Thus to find the volume after 1 month, we multiply
by the factor 0.9.
Q, volume of compost in ft3
Figure 5.8: Exponential Compost Decay
30
25
27
20
15
15.94
10
9.41
5
5.56
5
10
15
20
25
3.28
1.94
1.14
30
t, time in months
• Exponential function
• Decaying by a constant factor
• Decreasing by a smaller and smaller amount
Table 5.5: Exponential Compost Decay
t
Q
Formula
0
27
27(0.9)0
1
27(0.9) = 24.3
27(0.9)1
2
(27(0.9)) (0.9) = 21.87
27(0.9)2
3
..
.
(27(0.9)(0.9)) (0.9) = 19.683
..
.
27(0.9)3
10
27(0.9)10 ≈ 9.414
27(0.9)10
t
Instructor: A.E.Cary
27(0.9)t
Page 14 of 25
Math 111 Module 5 Lecture Notes
5.8
Transformations of Exponential Functions
Example 15: Graph of y = 2x in Figure 5.9. Use this to graph the various transformations listed.
Figure 5.10: y = 2x + 1
Figure 5.9: y = 2x
y
y
y=0
x
y=2
6
6
4
4
2
2
x
x
−6
−4
−2
2
4
−6
6
−4
−2
2
4
6
−2
−2
−4
−4
y=0
y=1
y = 2x
−6
−6
y = 2x + 1
Shift the graph of y = 2x up 1 unit.
Figure 5.11: y = 3 · 2x
Figure 5.12: y = −2x
y
y
6
6
4
4
2
2
x
−6
−4
−2
2
4
6
x
−6
−4
−2
y=0
−4
y=1
y = 2x
2
4
6
−2
y=0
−4
y=1
−6
y = 3 · 2x
Stretch the graph of y = 2x vertically
by a factor of 3.
Instructor: A.E.Cary
−2
y = 2x
−6
y = −2x
Reflect the graph of y = 2x across the
x-axis.
Page 15 of 25
Math 111 Module 5 Lecture Notes
Figure 5.13: y = 2−x
Figure 5.14: y = 1 − 2x
y
y
y=0
y=1
6
6
y = 2x
y = 1 − 2x
4
2
4
2
x
−6
−4
−2
y=0
2
4
6
x
−6
−4
−2
2
−2
−2
−4
−4
−6
−6
4
6
y=1
y = 2x
y = 2−x
Reflect the graph of y = 2x across the
y-axis.
Instructor: A.E.Cary
Reflect the graph of y = 2x across the
x-axis and then shift up 1 unit.
Page 16 of 25
Math 111 Module 5 Lecture Notes
If ax = ay , then x = y.
Pertinent Exponent Rules:
• a−n =
1
an
• (am )n = amn
• am an = am+n
Solving Exponential Equations
Example 16: Solve the following equations for x. List the solution set.
(a) 5x = 5−6
5x = 5−6
x = −6
Solution Set: {−6}
(b) 22x−1 = 4
22x−1 = 4
22x−1 = 22
2x − 1 = 2
2x = 3
x=
Solution Set:
(c) 42x−5 =
3
2
3
2
1
16
Instructor: A.E.Cary
Page 17 of 25
Math 111 Module 5 Lecture Notes
1
16
1
= 2
4
42x−5 =
42x−5
42x−5 = 4−2
2x − 5 = −2
2x = 3
x=
Solution Set:
3
2
3
2
(d) 53x−7 = 125
53x−7 = 125
53x−7 = 53
3x − 7 = 3
3x = 10
x=
Solution Set:
(e) 4x
2 −7
10
3
10 3
= 642x
4x
2 −7
4x
2 −7
4x
2 −7
= 642x
2x
= 43
= 43∗2x
x2 − 7 = 6x
x2 − 6x − 7 = 0
(x − 7)(x + 1) = 0
x − 7 = 0 or x + 1 = 0
x = 7 or x = −1
Solution Set: {−1, 7}
Instructor: A.E.Cary
Page 18 of 25
Math 111 Module 5 Lecture Notes
2
(f) 92x · 27x = 3−1
2
3 · 27x = 92x
x2
2x
3 · 33
= 32
2
31 · 33x = 34x
2
31+3x = 34x
1 + 3x2 = 4x
3x2 − 4x + 1 = 0
(3x − 1)(x − 1) = 03x − 1
x=
Solution Set:
Instructor: A.E.Cary
1
3
1
3
= 0 or x − 1 = 0
or x = 1
,1
Page 19 of 25
Math 111 Module 5 Lecture Notes
5.9
What’s “e”?
The number e is a number that occurs in nature, and is a frequent base for exponential and
logarithmic expressions. It is defined by:
n
1
e = lim 1 +
n→∞
n
It can also be expressed by the following:
e=
1
1
1
1
1
1
+ + + + + + ···
0! 1! 2! 3! 4! 5!
This number is irrational and is approximated by 2.718281828. The graph of the function
given by y = ex looks a lot like the graphs of the functions given by y = 2x and y = 3x , as shown
in Figure 5.15. In calculus, you will study that the special property of e is that the slope of the
tangent line at zero is exactly 1, as shown in Figure 5.16.
Figure 5.15
Figure 5.16
y
y=0
y
y=0
5
y = 2x
5
y = ex
y = ex
y =x+1
y = 3x
4
4
3
3
2
2
1
1
x
−3
−2
−1
1
2
x
−3
3
−1
−2
−1
1
2
3
−1
Example 17: Solve the following equation.
e3x = e2−x
e5x = e3+x
5x = 3 + x
4x = 3
x=
Solution Set:
3
4
3
Instructor: A.E.Cary
4
Page 20 of 25
Math 111 Module 5 Lecture Notes
Example 18: In 1990, the population of Oregon was 2.84 million people. In 2010, the population
of Oregon was 3.83 million people. Let P (t) be the population of Oregon in millions, where t is
the number of years after 2000. This can be modeled by P (t) = 3.298e0.015t .
(a) According to this model, what will the population be in 2020?
P (20) = 3.298e0.015∗20
≈ 4.452
In 2020, there will be approximately 4.452 million people in Oregon.
(b) According to this model, when will the population reach 4 million people? Use your graphing
calculator to solve this.
Graphing y1(x) = 4 and y2(x) = 3.298e0.015x , we find the point of intersection to be
approximately (12.87,4). The population is expected to reach 4 million in about 12.87 years.
Instructor: A.E.Cary
Page 21 of 25
Math 111 Module 5 Lecture Notes
5.10
Finding the Formula for an Exponential Function (REQUIRED)
Example 19: Find an algebraic rule (or formula) for an exponential function f that passes
through the points (−1, 8) and (2, 1). Also find the algebraic rule (or formula) for a linear function
g that passes through the points (−1, 8) and (2, 1).
The general equation for an exponential function is f (x) = Cax . We need to find the appropriate
values for C and a. As the function passes through the points (−1, 8) and (2, 1), we have f (−1) = 8
and f (2) = 1. Since f (x) = Cax , we then have Ca−1 = 8 and Ca2 = 1. Thus we can write and
solve the following ratio based on these two equations:
1
Ca2
=
−1
Ca
8
12
3
4
=
Ca2
Ca−2
1
Ca2
=
8
Ca−1
1
a3 =
8
1
a=
2
Now that we have the value of a, we can solve for C using one of the original function values:
Ca2 = 1
2
1
C
=1
2
C
1
=1
4
C=4
Therefore f (x) = 4
Note that 4
1 x
2
1 x
.
2
6= 2x .
We can (and should) check that this function satisfies f (−1) = 8 and f (2) = 1:
Check:
Instructor: A.E.Cary
Page 22 of 25
Math 111 Module 5 Lecture Notes
−1
1
f (−1) = 4
2
=4·2
=8X
2
1
f (2) = 4
2
=1X
To find the linear function passing through the points (−1, 8) and (2, 1), we need to find the slope
and the y-intercept:
1−8
m=
2 − (−1)
−7
3
7
=−
3
7
Thus y = − 3 x + b. We can use one of the points to find b:
=
7
1 = − (2) + b
3
14
1=− +b
3
17
=b
3
The formula of the linear function passing through (−1, 8) and (2, 1) is g(x) = − 73 x +
17
.
3
Figure 5.17
y
10
y=
x
1
2
7
−3x
y=4
+ 17
3
8
6
4
2
x
−2
Instructor: A.E.Cary
2
4
6
8
Page 23 of 25
Math 111 Module 5 Lecture Notes
Example 20: Find an algebraic
rule (or formula) for an exponential function f that passes
3
through the points −2, 4 and (2, 12).
The general equation for an exponential function is f (x) = Cax . We needto find the appropriate
values for C and a. As the function passes through the points −2, 34 and (2, 12), we have
f (−2) = 43 and f (2) = 12. Since f (x) = Cax , we then have Ca−2 = 34 and Ca2 = 12. Thus we can
write the following two ratios:
Ca2
12
= 3
−2
Ca
4
Ca2
= 16
Ca−2
a2
= 16
a−2
a4 = 16
a = ±(16)1/4
a = ±2
Note that a = 2 is the only relevant value.
Now that we have the value of a, we can solve for C using one of the original function values:
12 = Ca2
12 = C(2)2
12 = 4C
3=C
Therefore f (x) = 3(2)x .
We can (and should) check that this function satisfies f (−2) =
Check:
3
4
and f (2) = 12:
f (−2) = 3(2)−2
1
=3
4
=
3
X
4
f (2) = 3(2)2
= 12 X
Instructor: A.E.Cary
Page 24 of 25
Math 111 Module 5 Lecture Notes
Example 21: Find an algebraic rule (or formula) for an exponential function f that passes
through the points (1, 8) and (3, 128).
The general equation for an exponential function is f (x) = Cax . We need to find the appropriate
values for C and a. As the function passes through the points (1, 8) and (3, 128), we have f (1) = 8
and f (3) = 128. Since f (x) = Cax , we then have Ca1 = 8 and Ca3 = 128. Thus we can write and
solve the following ratios:
Ca3
128
=
1
Ca
8
Ca3
= 16
Ca1
a3
= 16
a1
a2 = 16
√
a = ± 16
a = ±4
Note that a = 4 is the only relevant value.
Now that we have the value of a, we can solve for C using one of the original function values:
8 = Ca1
8 = C(4)1
8 = 4C
2=C
Therefore f (x) = 2(4)x .
We can (and should) check that this function satisfies f (1) = 8 and f (3) = 128:
Check:
f (1) = 2(4)1
=8X
f (3) = 2(4)3
= 2 · 64
= 128 X
Instructor: A.E.Cary
Page 25 of 25
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