Slide 2.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Inverse Functions

Learn the definition of an inverse function and a relation.

Learn to identify one-to-one functions.

Learn a procedure for finding an inverse function.

Learn to use inverse functions to find the range of a function.

Learn to apply inverse functions in the real world.

SECTION 2.8

1

2

3

4

5


EXAMPLE 1Determining whether an Inverse Relation is a Function

Ray’s Music Mart has six employees. The first table lists the first names and the Social Security numbers of the employees, and the second table lists the first names and the ages of the employees.

a. Find the inverse of the function defined by the first table, and determine whether the inverse relation is a function.

b. Find the inverse of the function defined by the second table, and determine whether the inverse relation is a function.



Solution

Every y–value corresponds to exactly one x–value. Thus the inverse of the function defined in this table is a function.

Dwayne 590-56-4932

Sophia 599-23-1746

Desmonde 264-31-4958

Carl 432-77-6602

Anna 195-37-4165

Sal 543-71-8026



Solution continued

There is more than one x–value that corresponds to a y–value. For example, the age of 24 yields the names Dwayne and Anna. Thus the inverse of the function defined in this table is not a function.

Dwayne 24

Sophia 26

Desmonde 42

Carl 51

Anna 24

Sal 26


DEFINITION OF A ONE-TO-ONE FUNCTION

A function is called a one-to-one function if each y-value in its range corresponds to only one x-value in its domain.


A ONE-TO-ONE FUNCTION

Each y-value in the range corresponds to only one x-value in the domain.


NOT A ONE-TO-ONE FUNCTION

The y-value y2 in the range corresponds to two x-values, x2 and x3, in the domain.


NOT A FUNCTION

The x-value x2 in the domain corresponds to the two y-values, y2 and y3, in the range.


HORIZONTAL- LINE TEST

A function f is one-to-one if no horizontal line intersects the graph of f in more than one point.


EXAMPLE 2 Using the Horizontal-Line Test.

Use the horizontal-line test to determine which of the following functions are one-to-one.

a. f x 2x 5 b. g x x2 1 c. h x 2 x

Solution

No horizontal line intersects the graph of f in more than one point, therefore the function f is one-to-one.



Solution continued

There are many horizontal lines that intersect the graph of f in more than one point, therefore the function f is not one-to-one.

b. g x x2 1



Solution continued

No horizontal line intersects the graph of f in more than one point, therefore the function f is one-to-one.

c. h x 2 x


DEFINITION OF f –1 FOR AONE-TO-ONE FUNCTION f

Let f represent a one-to-one function. The inverse of f is also a function, called the inverse function of f, and is denoted by f –1. If (x, y) is an ordered pair of f, then (y, x) is an ordered pair of f –1, and we write x = f –1(y). We have y = f (x) if and only if f –1(y) = x.


EXAMPLE 3Relating the Values of a Function and Its Inverse

Assume that f is a one-to-one function.• If f (3) = 5, find f –1(5).• If f –1(–1) = 7, find f (7).

a. Let x = 3 and y = 5. Now 5 = f (3) if and only if f –1(5) = 3. Thus, f –1(5) = 3.

b. Let y = –1 and x = 7. Now, f –1(–1) = 7 if and only if f (7) = –1. Thus, f (7) = –1.

Solution

We know that y = f (x) if and only if f –1(y) = x.


INVERSE FUNCTION PROPERTY

Let f denote a one-to-one function. Then

f o f 1 x f f 1 x x

for every x in the domain of f –1.

1.

f 1 o f x f 1 f x x

for every x in the domain of f .

2.


UNIQUE INVERSE FUNCTION PROPERTY

Let f denote a one-to-one function. Then if g is any function such that

g = f –1. That is, g is the inverse function of f.

f g x x for every x in the domain of g and

g f x x for every x in the domain of f, then


EXAMPLE 4 Verifying Inverse Functions

Verify that the following pairs of functions are inverses of each other:

f x 2x 3 and g x x 3

2.

Solution

Form the composition of f and g.

f og x f g x fx 3

2

2x 3

2

3 x 3 3

x


EXAMPLE 4 Verifying Inverse Functions

Solution continued

Now find

go f x g f x g 2x 3

2x 3 3

2x

g f x .

Since f g x g f x x, we conclude that

f and g are inverses of each other.


SYMMETRY PROPERTY OFTHE GRAPHS OF f AND f –1

The graph of the function f and the graph of f –1 are symmetric with respect to the line y = x.


EXAMPLE 5 Finding the Graph of f –1 from the Graph of f

The graph of the function f is shown. Sketch the graph of the f –1.

Solution


PROCEDURE FOR FINDING f –1

Step 1 Replace f (x) by y in the equation for f (x).

Step 2 Interchange x and y.Step 3 Solve the equation in Step 2 for y.Step 4 Replace y with f –1(x).


EXAMPLE 6 Finding the Inverse Function

Find the inverse of the one-to-one function

f x x 1

x 2, x 2.

Solution

y x 1

x 2Step 1

x y 1

y 2Step 2

x y 2 y 1

xy 2y y 1Step 3


EXAMPLE 6 Finding the Inverse Function

Solution continued

f 1 x 2x 1

x 1, x 1Step 4

Step 3(cont.)

xy 2y 2x y y 1 2x y

xy y 2x 1

y x 1 2x 1

y 2x 1

x 1


EXAMPLE 7 Finding the Domain and Range

Find the domain and the range of the function

f x x 1

x 2, x 2.

Solution

Domain of f, all real numbers x such that x ≠ 2, in interval notation (–∞, 2) U (2, –∞).

Range of f is the domain of f –1.

f 1 x 2x 1

x 1, x 1

Range of f is (–∞, 1) U (1, –∞).


EXAMPLE 8 Finding an Inverse Function

Find the inverse of g(x) = x2 – 1, x ≥ 0.Solution

Step 1 y = x2 – 1, x ≥ 0

Step 2 x = y2 – 1, y ≥ 0

Step 4 g 1 x x 1

Step 3

y x 1, y 0

Since y ≥ 0, reject y x 1.

x 1 y2 , y 0


EXAMPLE 8 Finding an Inverse Function

Solution continued

Here are the graphs of g and g –1.


EXAMPLE 9 Water Pressure on Underwater Devices

The formula for finding the water pressure p (in pounds per square inch), at a depth d (in feet)

p 15d

33.

pressure gauge on a diving bell breaks and shows a reading of 1800 psi. Determine how far below the surface the bell was when the gauge failed.

below the surface is Suppose the

p 15d

33.

Solution

The depth is given by the inverse of


EXAMPLE 9 Water Pressure on Underwater Devices

Solution continued

p 15d

3333p 15d

d 33p

15

Solve the equation for d.

Let p = 1800.

d 33 1800

15d 3960

The device was 3960 feet below the surface when the gauge failed.

Slide 2.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Documents

Transcript of Slide 2.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.