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256 CHAPTER 4 Exponential and Logarithmic Functions
4.2 One-to-One Functions; Inverse Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
• Functions (Section 2.1,pp.56–67) • Increasing/Decreasing Functions (Section 2.3,pp.82–85)
Now work the ‘Are You Prepared?’ problems on page 267.
OBJECTIVES
1
Determine Whether a Function Is One-to-One
2
Determine the Inverse of a Function Defined by a Map or an Ordered Pair
3
Obtain the Graph of the Inverse Function from the Graph of the Function
4
Find the Inverse of a Function Defined by an Equation
✓
1 Determine Whether a Function Is One-to-One
In Section 2.1, we presented four different ways to represent a function: as (1) a
map,(2) a set of ordered pairs,(3) a graph,and (4) an equation.For example,Figures
6 and 7 illustrate two different functions represented as mappings.The function in
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 257
Indiana
Washington
South Dakota
North Carolina
Tennessee
State
6,159,068
6,068,996
761,063
8,320,146
5,797,289
Population
Figure 6
Dog
Cat
Duck
Lion
Pig
Rabbit
Animal
11
10
7
Life Expectancy
Figure 7
x
3
x
1
y
1
x
2
y
2
One-to-one function:
Each
x
in the domain has
one and only one image
in the range
(a)
y
3
Domain Range
x
3
x
1
y
1
x
2
Not a one-to-one function:
y
1
is the image of both
x
1
and
x
2
.
(b)
y
3
Domain Range
x
3
x
1
y
1
y
2
Not a function:
x
1
has two images,
y
1
and
y
2
.
(c)
y
3
Figure 8
In Words
A function is not one-to-one if
two different inputs correspond
to the same output.
Figure 6 shows the correspondence between states and their population. The func-
tion in Figure 7 shows a correspondence between animals and life expectancy.
Suppose we asked a group of people to name the state that has a population of
761,063 based on the function in Figure 6. Everyone in the group would respond
South Dakota. Now, if we asked the same group of people to name the animal
whose life expectancy is 11 years based on the function in Figure 7, some would re-
spond dog,while others would respond cat.What is the difference between the func-
tions in Figures 6 and 7? In Figure 6, we can see that each element in the domain
corresponds to exactly one element in the range.In Figure 7,this is not the case:two
different elements in the domain correspond to the same element in the range.We
give functions such as the one in Figure 6 a special name.
A function is one-to-one if any two different inputs in the domain correspond
to two different outputs in the range.That is, if and are two different in-
puts of a function then
Put another way, a function is one-to-one if no y in the range is the image of
more than one x in the domain. A function is not one-to-one if two different
elements in the domain correspond to the same element in the range. So, the func-
tion in Figure 7 is not one-to-one because two different elements in the domain,dog
and cat, both correspond to 11. Figure 8 illustrates the distinction of one-to-one
functions,non–one-to-one functions,and nonfunctions.
f
f1x
1
2 Z f1x
2
2.f,
x
2
x
1
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
258 CHAPTER 4 Exponential and Logarithmic Functions
x
h
y
(
x
1
,
h
)(
x
2
,
h
)
y
f
(
x
)
y
h
x
1
x
2
Figure 9
and
is not a
one-to-one function.
fx
1
Z x
2
;
f(x
1
) = f(x
2
) = h
Determining Whether a Function Is One-to-One
Determine whether the following functions are one-to-one.
(a) For the following function, the domain represents the age of five males and the
range represents their HDL (good) cholesterol
(b)
Solution (a) The function is not one-to-one because there are two different inputs,55 and 61,
that correspond to the same output,38.
(b) The function is one-to-one because there are no two distinct inputs that
correspond to the same output.
NOW WORK PROBLEMS 9 AND 13.
If the graph of a function is known,there is a simple test,called the horizontal-
line test,to determine whether is one-to-one.
Theorem
Horizontal-line Test
If every horizontal line intersects the graph of a function in at most one
point,then is one-to-one.
The reason that this test works can be seen in Figure 9,where the horizontal line
intersects the graph at two distinct points, and Since h is the
image of both and is not one-to-one. Based on Figure 9, we can
state the horizontal-line test in another way: If the graph of any horizontal line in-
tersects the graph of a function at more than one point, then is not one-to-one.
Using the Horizontal-line Test
For each function, use the graph to determine whether the function is one-to-one.
(a) (b)
Solution (a) Figure 10(a) illustrates the horizontal-line test for The horizontal
line intersects the graph of twice, at and at so is not
one-to-one.
(b) Figure 10(b) illustrates the horizontal-line test for Because every
horizontal line will intersect the graph of exactly once, it follows that is
one-to-one.
gg
g1x2 = x
3
.
f1-1, 12,11, 12fy = 1
f1x2 = x
2
.
g1x2 = x
3
f1x2 = x
2
EXAMPLE 2
ff
fx
2
, x
1
Z x
2
,x
1
1x
2
, h2.1x
1
, h2y = h
f
f
f
f
51-2, 62, 1-1, 32, 10, 22, 11, 52, 12, 826
38
42
46
55
61
Age
57
54
34
38
HDL Cholesterol
(mg>dL).
EXAMPLE 1
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 259
A horizontal line intersects the graph
twice;
f
is not one-to-one
x
y
33
(1, 1)
y
1
y
x
2
(1, 1)
3
3
(a) Every horizontal line intersects the graph
exactly once;
g
is one-to-one
(b)
x
y
3
3
y
x
3
3
3
Figure 10
NOW WORK PROBLEM
17.
Let’s look more closely at the one-to-one function This function is
an increasing function. Because an increasing (or decreasing) function will always
have different y-values for unequal x-values, it follows that a function that is
increasing (or decreasing) over its domain is also a one-to-one function.
Theorem
A function that is increasing on an interval I is a one-to-one function on I.
A function that is decreasing on an interval I is a one-to-one function on I.
✓
2 Determine the Inverse of a Function Defined by a Map or an
Ordered Pair
In Section 2.1,we said that a function can be thought of as a machine that receives
an input, say x, from the domain, manipulates it, and outputs the value The
inverse function of receives as input manipulates it, and outputs x.For
example,if then Here,the input is 3 and the output is 11.
The inverse of would “undo” what does and receive as input 11 and provide as
output 3. So, if a function takes inputs and multiplies them by 2, the inverse would
take inputs and divide them by 2. For the inverse of a function to itself be a func-
tion, must be one-to-one.
Recall that we have a variety of ways of representing functions.We will discuss
how to find inverses for all four representations of functions: (1) maps, (2) sets of
ordered pairs (3) graphs, and (4) equations. We begin with finding inverses of
functions represented by maps or sets of ordered pairs.
Finding the Inverse of a Function Defined by a Map
Find the inverse of the following function. Let the domain of the function represent
certain states, and let the range represent the state’s population. State the domain
and the range of the inverse function.
Indiana
Washington
South Dakota
North Carolina
Tennessee
State
6,159,068
6,068,996
761,063
8,320,146
5,797,289
Population
EXAMPLE 3
f
f
ff
f132 = 11.f1x2 = 2x + 5,
f1x2,f
f1x2.
f
g1x2 = x
3
.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
260 CHAPTER 4 Exponential and Logarithmic Functions
f
f
1
Domain of
f
Range of
f
Domain of
f
1
Range of
f
1
Figure 11
WARNING
Be careful! is a symbol for the
inverse function of The used
in is not an exponent. That is,
does not mean the reciprocal
of ; is not equal to .
■
1
f1x2
f
-1
1x2f
f
-1
f
-1
-1f.
f
-1
Solution The function is one-to-one,so the inverse will be a function.To find the inverse func-
tion,we interchange the elements in the domain with the elements in the range. For
example, the function receives as input Indiana and outputs 6,159,068. So, the in-
verse receives as input 6,159,068 and outputs Indiana.The inverse function is shown
next.
The domain of the inverse function is
The range of the inverse function is Washington, South
Dakota, North Carolina,
If the function is a set of ordered pairs then the inverse of is the set of
ordered pairs
Finding the Inverse of a Function Defined
By a Set of Ordered Pairs
Find the inverse of the following one-to-one function:
Solution The inverse of the given function is found by interchanging the entries in each
ordered pair and so is given by
NOW WORK PROBLEMS 23 AND 27.
Remember, if is a one-to-one function, its inverse is a function.Then, to each
x in the domain of there is exactly one y in the range (because is a function);and
to each y in the range of there is exactly one x in the domain (because is one-to-
one).The correspondence from the range of back to the domain of is called the
inverse function of and is denoted by the symbol Figure 11 illustrates this
definition.
Based on Figure 11, two facts are now apparent about a function and its
inverse
Look again at Figure 11 to visualize the relationship. If we start with x, apply
and then apply we get x back again. If we start with x, apply and then
apply we get the number x back again.To put it simply,what does, undoes,
and vice versa. See the illustration that follows.
f
-1
ff,
f
-1
,f
-1
,f,
Domain of f = Range of f
-1
Range of f = Domain of f
-1
f
-1
.
f
f
-1
.f
ff
ff,
ff,
f
51-27, -32, 1-8, -22, 1-1, -12, 10, 02, 11, 12, 18, 22, 127, 326
51-3, -272, 1-2, -82, 1-1, -12, 10, 02, 11, 12, 12, 82, 13, 2726
EXAMPLE 4
1y, x2.
f1x, y2,f
Tennessee6.
5Indiana,5 797 2896.
56 159 068, 6 068 996, 761 063, 8 320 146,
Indiana
Washington
South Dakota
North Carolina
Tennessee
State
6,159,068
6,068,996
761,063
8,320,146
5,797,289
Population
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 261
x
f
f
1
(2
x
) =
xf
1
(2
x
) =
f
(
x
) = 2
x
1
–
2
Figure 12
Exploration
Simultaneously graph
and on a square screen with
What do you observe
about the graphs of its inverse
and the line
Repeat this experiment by simulta-
neously graphing
and on a square screen
with Do you see the
symmetry of the graph of and its
inverse with respect to the line
Y
1
= x?
Y
3
Y
2
-6 … x … 3.
Y
3
=
1
2
(x - 3)
Y
1
= x, Y
2
= 2x + 3,
Y
1
= x?Y
3
= 13 x,
Y
2
= x
3
,
-3 … x … 3.
Y
3
= 13 x
Y
1
= x, Y
2
= x
3
,
In other words,
where x is in the domain of
where x is in the domain of
Consider the function which multiplies the argument x by 2. The
inverse function undoes whatever does. So the inverse function of is
which divides the argument by 2. For example, and
so undoes what did.We can verify this by showing that
See Figure 12.
Verifying Inverse Functions
(a) We verify that the inverse of is by showing that
for all x in the domain of g
for all x in the domain of
(b) We verify that the inverse of is by showing that
f1f
-1
1x22 = f a
1
2
1x - 32b = 2c
1
2
1x - 32d + 3 = 1x - 32 + 3 = x
for all x in the
domain of f
f
-1
1f1x22 = f
-1
12x + 32 =
1
2
312x + 32 - 34 =
1
2
12x2 = x
f
-1
1x2 =
1
2
1x - 32f1x2 = 2x + 3
g
-1
.g1g
-1
1x22 = g113 x2 = 113 x2
3
= x
g
-1
1g1x22 = g
-1
1x
3
2 = 33 x
3
= x
g
-1
1x2 = 13 xg1x2 = x
3
EXAMPLE 5
f
-1
1f1x22 = f
-1
12x2 =
1
2
12x2 = x
and
f1f
-1
1x22 = fa
1
2
xb = 2a
1
2
xb = x
ff
-1
f
-1
162 =
1
2
162 = 3,
f132 = 2132 = 6f
-1
1x2 =
1
2
x,
fff
-1
f1x2 = 2x,
f
-1
f1f
-1
1x22 = x
ff
-1
1f1x22 = x
Input x
Apply f
"
f1x2
Apply f
-1
"
f
-1
1f1x22 = x
Input x
Apply f
-1
"
f
-1
1x2
Apply f
"
f1f
-1
1x22 = x
for all x in the
domain of
f
-1
.
Verifying Inverse Functions
Verify that the inverse of is For what values of x is
For what values of x is
Solution The domain of is and the domain of is Now,
NOW WORK PROBLEM 37.
f1f
-1
1x22 = fa
1
x
+ 1b =
1
1
x
+ 1 - 1
=
1
1
x
= x
provided x Z 0
f
-1
1f1x22 = f
-1
a
1
x - 1
b =
1
1
x - 1
+ 1 = x - 1 + 1 = x
provided x Z 1.
5x
ƒ
x Z 06f
-1
5x
ƒ
x Z 16f
f1f
-1
1x22 = x?f
-1
1f1x22 = x?
f
-1
1x2 =
1
x
+ 1.f1x2 =
1
x - 1
EXAMPLE 6
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
262 CHAPTER 4 Exponential and Logarithmic Functions
✓
3 Obtain the Graph of the Inverse Function from the Graph of
the Function
For the functions in Example 5(b), we list points on the graph of and on the
graph of in Table 1.
We notice that whenever is on the graph of then is on the graph
of Figure 13 shows these points plotted. Also shown is the graph of
which you should observe is a line of symmetry of the points.
y = x,f
-1
.
1b, a2f1a, b2
f
-1
= Y
2
f = Y
1
Graphing the Inverse Function
The graph in Figure 16(a) is that of a one-to-one function Draw the graph
of its inverse.
y = f1x2.
EXAMPLE 7
x
y
x
y
ba
b
a
(
b
,
a
)
(
a
,
b
)
Figure 14
Table 1
x
y
x
y
f
(
x
)
y
f
1
(
x
)
y
(
a
3
,
b
3
)
(
a
2
,
b
2
)
(
a
1
,
b
1
)
(
b
3
,
a
3
)
(
b
2
,
a
2
)
(
b
1
,
a
1
)
Figure 15
8
y
x
8
1113
Figure 13
Suppose that is a point on the graph of a one-to-one function defined by
Then This means that so is a point on the
graph of the inverse function The relationship between the point on
and the point on is shown in Figure 14.The line segment containing
and is perpendicular to the line and is bisected by the line (Do
you see why?) It follows that the point on is the reflection about the line
of the point on
Theorem
The graph of a function and the graph of its inverse are symmetric with
respect to the line
Figure 15 illustrates this result. Notice that, once the graph of is known, the
graph of may be obtained by reflecting the graph of about the line y = x.ff
-1
f
y = x.
f
-1
f
f.1a, b2y = x
f
-1
1b, a2
y = x.y = x1b, a2
1a, b2f
-1
1b, a2
f1a, b2f
-1
.
1b, a2a = f
-1
1b2,b = f1a2.y = f1x2.
f1a, b2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 263
Solution We begin by adding the graph of to Figure 16(a). Since the points
and are on the graph of we know that the points
and must be on the graph of Keeping in mind that the
graph of is the reflection about the line of the graph of we can draw
See Figure 16(b).f
-1
.
f,y = xf
-1
f
-1
.11, 221-1, -22, 10, -12,
f,12, 121-2, -12, 1-1, 02,
y = x
x
y
33
(2, 1)
(1, 0)
(2, 1)
3
3
y
f
(
x
)
(a)
x
y
33
(2, 1)
(1, 2)
(1, 0)
(2, 1)
(0, 1)
(1, 2)
3
3
y
f
(
x
)
y
f
1
(
x
)
(b)
y
x
Figure 16
NOW WORK PROBLEM 31.
✓
4 Find the Inverse of a Function Defined by an Equation
The fact that the graphs of a one-to-one function and its inverse function are
symmetric with respect to the line tells us more.It says that we can obtain
by interchanging the roles of x and y in Look again at Figure 15.If is defined by
the equation
then is defined by the equation
The equation defines implicitly.If we can solve this equation for y,we
will have the explicit form of that is,
Let’s use this procedure to find the inverse of (Since is a
linear function and is increasing,we know that is one-to-one and so has an inverse
function.)
Finding the Inverse Function
Find the inverse of Also find the domain and range of and
Graph and on the same coordinate axes.
Solution In the equation interchange the variables x and y.The result,
is an equation that defines the inverse implicitly. To find the explicit form, we
solve for y.
y =
1
2
1x - 32
2 y = x - 3
2 y + 3 = x
f
-1
x = 2y + 3
y = 2x + 3,
f
-1
f
f
-1
.ff1x2 = 2x + 3.
EXAMPLE 8
f
ff1x2 = 2x + 3.
y = f
-1
1x2
f
-1
,
f
-1
x = f1y2
x = f1y2
f
-1
y = f1x2
ff.
f
-1
y = x
f
-1
f
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
264 CHAPTER 4 Exponential and Logarithmic Functions
The explicit form of the inverse is therefore
which we verified in Example 5(c).
Next we find
The graphs of and its inverse
are shown in Figure 17. Note the symmetry of the graphs with respect to the line
Y
3
= x.
Y
2
= f
-1
1x2 =
1
2
1x - 32Y
1
= f1x2 = 2x + 3
Range of f = Domain of f
-1
= 1-
q
,
q
2
Domain of f = Range of f
-1
= 1-
q
,
q
2
f
-1
1x2 =
1
2
1x - 32
f
-1
6
Y
3
x
Y
1
Y
2
6
88
y
x
f
(
x
) 2
x
3
x
y
55
5
5
f
1
(
x
) (
x
3)
1
–
2
Figure 17
We now outline the steps to follow for finding the inverse of a one-to-one
function.
Procedure for Finding the Inverse of a One-to-One Function
STEP 1: In interchange the variables x and y to obtain
This equation defines the inverse function implicitly.
STEP 2: If possible,solve the implicit equation for y in terms of x to obtain the
explicit form of
STEP 3: Check the result by showing that
Finding the Inverse Function
The function
is one-to-one. Find its inverse and check the result.
f1x2 =
2x + 1
x - 1
,
x Z 1
EXAMPLE 9
f
-1
1f1x22 = x
and
f1f
-1
1x22 = x
y = f
-1
1x2
f
-1
f
-1
x = f1y2
y = f1x2,
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 265
Solution
STEP 1: Interchange the variables x and y in
to obtain
STEP 2: Solve for y.
Multiply both sides by
Apply the Distributive Property.
Subtract 2x from both sides; add x to both sides.
Factor.
Divide by
The inverse is
Replace y by
S
TEP 3: ✔ CHECK:
f1f
-1
1x22 = fa
x + 1
x - 2
b =
2a
x + 1
x - 2
b + 1
x + 1
x - 2
- 1
=
21x + 12 + x - 2
x + 1 - 1x - 22
=
3x
3
= x,
x Z 2
f
-1
1f1x22 = f
-1
a
2x + 1
x - 1
b =
2x + 1
x - 1
+ 1
2x + 1
x - 1
- 2
=
2x + 1 + x - 1
2x + 1 - 21x - 12
=
3x
3
= x,
x Z 1
f
-1
(x).f
-1
1x2 =
x + 1
x - 2
,
x Z 2
x - 2. y =
x + 1
x - 2
1x - 22y = x + 1
xy - 2y = x + 1
xy - x = 2y + 1
x - 1. x1y - 12 = 2y + 1
x =
2y + 1
y - 1
x =
2y + 1
y - 1
y =
2x + 1
x - 1
Exploration
In Example 9, we found that, if then Compare the vertical and
horizontal asymptotes of and What did you find? Are you surprised?
Result You should have determined that the vertical asymptote of is and the horizontal
asymptote is The vertical asymptote of is and the horizontal asymptote is
NOW WORK PROBLEM
49.
We said in Chapter 2 that finding the range of a function is not easy.However,
if is one-to-one, we can find its range by finding the domain of the inverse
function f
-1
.
f
f
y = 1.x = 2,f
-1
y = 2.
x = 1f
f
-1
.f
f
-1
(x) =
x + 1
x - 2
.f(x) =
2x + 1
x - 1
,
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
266 CHAPTER 4 Exponential and Logarithmic Functions
Finding the Range of a Function
Find the domain and range of
Solution The domain of is To find the range of we first find the inverse
Based on Example 9,we have
The domain of is so the range of is
NOW WORK PROBLEM 63.
If a function is not one-to-one, then its inverse is not a function. Sometimes,
though,an appropriate restriction on the domain of such a function will yield a new
function that is one-to-one. Let’s look at an example of this common practice.
Finding the Inverse of a Domain-restricted Function
Find the inverse of if
Solution The function is not one-to-one. [Refer to Example 2(a).] However, if we
restrict the domain of this function to as indicated, we have a new function
that is increasing and therefore is one-to-one. As a result, the function defined by
has an inverse function,
We follow the steps given previously to find
STEP 1: In the equation interchange the variables x and y.The result is
This equation defines (implicitly) the inverse function.
S
TEP 2: We solve for y to get the explicit form of the inverse.Since only one
solution for y is obtained,namely, So
S
TEP 3: ✔ CHECK: since
Figure 18 illustrates the graphs of and
Y
2
= f
-1
1x2 = 1x.
Y
1
= f1x2 = x
2
, x Ú 0,
f1f
-1
1x22 = f11x2 = 11x2
2
= x.
x Ú 0f
-1
1f1x22 = f
-1
1x
2
2 = 3x
2
=
ƒ
x
ƒ
= x,
f
-1
1x2 = 1x.
y = 1x.
y Ú 0,
x = y
2
,
y Ú 0
y = x
2
, x Ú 0,
f
-1
.
f
-1
.y = f1x2 = x
2
, x Ú 0,
x Ú 0,
y = x
2
x Ú 0.y = f1x2 = x
2
EXAMPLE 11
5y
ƒ
y Z 26.f5x
ƒ
x Z 26,f
-1
f
-1
1x2 =
x + 1
x - 2
f
-1
.f,5x
ƒ
x Z 16.f
f1x2 =
2x + 1
x - 1
EXAMPLE 10
0
4
Y
1
x
2
Y
2
x
06
Y
3
x
x
y
2
2
f
1
(
x
)
x
y
x
f
(
x
)
x
2
,
x
≥
0
Figure 18
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.2 One-to-One Functions; Inverse Functions 267
Summary
1. If a function is one-to-one,then it has an inverse function
2. Domain of of Range of of
3. To verify that is the inverse of show that for every x in the domain of and
for every x in the domain of
4. The graphs of and are symmetric with respect to the line
5. To find the range of a one-to-one function find the domain of the inverse function f
-1
.f,
y = x.f
-1
f
f
-1
.
f1f
-1
1x22 = xff
-1
1f1x22 = xf,f
-1
f
-1
.f = Domainf
-1
;f = Range
f
-1
.f
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
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