Intermediate Algebra (11th edition)

Thus, is a linear function. In the function defined by

the value of yis found by starting with a value of x, multiplying by 2, and adding 5.

The equation

for the inverse has us subtract5, and then divideby 2. This shows how an inverse is

used to “undo” what a function does to the variable x.

(b)

This equation has a vertical parabola as its graph, so some horizontal lines will

intersect the graph at two points. For example, both and correspond to

Because of the -term, there are many pairs of x-values that correspond to

the same y-value. This means that the function defined by is not one-

to-one and does not have an inverse.

Alternatively, applying the steps for finding the equation of an inverse leads to

the following.

Interchange xand y.

Solve for y.

Square root property

The last step shows that there are two y-values for each choice of so we again

see that the given function is not one-to-one. It does not have an inverse.

(c)

Refer to Section 5.3to see that the graph of a cubing function like this is one-to-

one.

Replace with y.

Interchange xand y.

Take the cube root on each side.

Solve for y.

Replace ywith NOW TRY

OBJECTIVE 4 Graph , given the graph of ƒ.One way to graph the inverse

of a function ƒ whose equation is given is as follows.

1. Find several ordered pairs that belong to ƒ.

2. Interchange xand yto obtain ordered pairs that belong to

3. Plot those points, and sketch the graph of through them.

A simpler way is to select points on the graph of ƒ and use symmetry to find cor-

responding points on the graph of ƒ-^1.

ƒ-^1

ƒ-^1.

ƒ^1

ƒ-^11 x 2 = 23 x+ 2 ƒ-^11 x 2.

y= 23 x+ 2

23 x=y- 2 23 a^3 =a

23 x= 231 y- 223

x= 1 y- 223

y= 1 x- 223 ƒ 1 x 2

ƒ(x)=(x-2)^3

ƒ 1 x 2 = 1 x- 223

x 7 2,

y=  2 x- 2

y^2 = x- 2

x= y^2 + 2

y= x^2 + 2

y= x^2 + 2

y=11. x^2

x= 3 x=- 3

y=x^2 + 2

ƒ-^11 x 2 =

x- 5

2

y= 2 x+ 5,

ƒ-^1

SECTION 10.1 Inverse Functions 575

NOW TRY
EXERCISE 3
Decide whether each equation
defines a one-to-one function.
If so, find the equation that
defines the inverse.

(a)

(b)

(c) ƒ 1 x 2 =x^3 - 4

ƒ 1 x 2 = 1 x+ 122

ƒ 1 x 2 = 5 x- 7

NOW TRY ANSWERS

3. (a)one-to-one function;
   or

(b)not a one-to-one function

(c)one-to-one function;

ƒ-^11 x 2 = 23 x+ 4

ƒ-^11 x 2 =^15 x+^75

ƒ-^11 x 2 =x+ 57 ,