Finding Composite and Inverse Functions

In this chapter, we will introduce two new types of functions, exponential functions and logarithmic functions. These functions are used extensively in business and the sciences as we will see.

Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called composition. In composition, the output of one function is the input of a second function. For functions

f

becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

We have actually used composition without using the notation many times before. When we graphed quadratic functions using translations, we were composing functions. For example, if we first graphed

g (x) = x^{2}

as a parabola and then shifted it down vertically four units, we were using the composition defined by

(f \circ g) (x) = f (g (x))

For functions $f (x) = 4 x - 5$

and $g (x) = 2 x + 3,$

find: ⓐ $(f \circ g) (x),$

ⓑ $(g \circ f) (x),$

and ⓒ $(f \cdot g) (x) .$

ⓐ* * *

Use the definition of

(f \circ g) (x) .

| | {: valign=”top”}| | | {: valign=”top”}| | | {: valign=”top”}| Distribute. | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”In the first step, we use the definition of f of g of x to obtain that f of g of x equals f of g of x. In the second step, we substitute 2 x plus 3 for g of x. This means that f of g of x equals f of 2 x plus 3. In the third step, we find f of 2 x plus 3 where f of x equals 4 x minus 5. This means that f of g of x equals 4 times the quantity 2 x plus 3, minus 5. In the fourth step, we distribute. This means that f of g of x equals 8 x plus 12 minus 5. In the fifth step, we simplify. This means that f of g of x equals 8 x plus 7.” data-label=””}

ⓑ* * *

Use the definition of

(f \circ g) (x) .

| | {: valign=”top”}| | | {: valign=”top”}| | | {: valign=”top”}| Distribute. | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”In the first step, we use the definition of g of f of x to obtain that g of f of x equals g of f of x. In the second step, we substitute 4 x minus 5 for f of x. This means that g of f of x equals g of 4 x minus 5. In the third step, we find g of 4 x minus 5 where g of x equals 2 x plus 3. This means that g of f of x equals 2 times the quantity 4 x minus 5, plus 3. In the fourth step, we distribute. This means that g of f of x equals 8 x minus 10 plus 3. In the fifth step, we simplify. This means that g of f of x equals 8 x minus 7.” data-label=””}

Notice the difference in the result in part ⓐ and part ⓑ.

is different than $(f \circ g) (x) .$

In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.* * *

\begin{matrix} Use the definition of (f \cdot g) (x) . & (f \cdot g) (x) = f (x) \cdot g (x) \\ Substitute f (x) = 4 x - 5 and g (x) = 2 x + 3 . & (f \cdot g) (x) = (4 x - 5) \cdot (2 x + 3) \\ Multiply. & (f \cdot g) (x) = 8 x^{2} + 2 x - 15 \end{matrix}

For functions $f (x) = x^{2} - 4,$

and $g (x) = 3 x + 2,$

find: ⓐ $(f \circ g) (−3),$

ⓑ $(g \circ f) (−1),$

and ⓒ $(f \circ f) (2) .$

ⓐ* * *

Use the definition of

(f \circ g) (−3) .

| | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”In the first step, we use the definition of f of g of negative 3 to obtain that f of g of negative 3 equals f evaluated at g of negative 3. In the second step, we find g of negative 3 where g of x equals 3 x plus 2. This means that f of g of negative 3 equals f evaluated at 3 times negative 3 plus 2. In the third step, we simplify to obtain that f of g of negative 3 equals f of negative 7. In the fourth step, we find f of negative 7 where f of x equals x squared minus 4. This means that f of g of negative 3 equals negative 7 squared minus 4. In the fifth step, we simplify to obtain that f of g of negative 3 equals 45.” data-label=””}

ⓑ* * *

Use the definition of

(g \circ f) (−1) .

| | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”In the first step, we use the definition of g of f of negative 1 to obtain that g of f of negative 1 equals g evaluated at f of negative 1. In the second step, we find f of negative 1 where f of x equals x squared minus 4. This means that g of f of negative 1 equals g of negative 1 squared minus 4. In the third step, we simplify. This means that g of f of negative 1 equals g of negative 3. In the fourth step, we find g of negative 3 where g of x equals 3 x plus 2. This means that g of f of negative 1 equals 3 times negative 3 plus 2. In the fifth step, we simplify. This means that g of f of negative 1 equals negative 7.” data-label=””}

ⓒ* * *

Use the definition of

(f \circ f) (2) .

| | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}| | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”In the first step, we use the definition of f of f of 2 to obtain that f of f of 2 equals f evaluated at f of 2. In the second step, we find f of x where f of x equals x squared minus 4. This means that f of f of 2 equals f of 2 squared minus 4. In the third step, we simplify to obtain that f of f of 2 equals f of 0. In the fourth step, we find f of 0 where f of x equals x squared minus 4. This means that f of f of 2 equals 0 squared minus 4. In the fifth step, we simplify to obtain that f of f of 2 equals negative 4.” data-label=””}

Determine Whether a Function is One-to-One

When we first introduced functions, we said a function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x-value is matched with only one y-value.

We used the birthday example to help us understand the definition. Every person has a birthday, but no one has two birthdays and it is okay for two people to share a birthday. Since each person has exactly one birthday, that relation is a function.

This figure shows two tables. To the left is the table labeled Name, which from top to bottom reads Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, and Danny. The table on the right is labeled Birthday, which from top to bottom reads January 12, February 3, April 25, May 10, May 23, July 24, August 2, and September 15. There are arrows going from Alison to April 25, Penelope to May 23, June to August 2, Gregory to September 15, Geoffrey to January 12, Lauren to May 10, Stephen to July 24, Alice to February 3, Liz to July 24, and Danny to no birthday.

A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value.

Our example of the birthday relation is not a one-to-one function. Two people can share the same birthday. The range value August 2 is the birthday of Liz and June, and so one range value has two domain values. Therefore, the function is not one-to-one.

To help us determine whether a relation is a function, we use the vertical line test. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. Also, if any vertical line intersects the graph in more than one point, the graph does not represent a function.

The vertical line is representing an x-value and we check that it intersects the graph in only one y-value. Then it is a function.

To check if a function is one-to-one, we use a similar process. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a y-value and we check that it intersects the graph in only one x-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the horizontal line test.

We can test whether a graph of a relation is a function by using the vertical line test. We can then tell if the function is one-to-one by applying the horizontal line test.

Find the Inverse of a Function

In the next example we will find the inverse of a function defined by ordered pairs.

We will use this concept to graph the inverse of a function in the next example.

When we began our discussion of an inverse function, we talked about how the inverse function ‘undoes’ what the original function did to a value in its domain in order to get back to the original x-value.

We can use this property to verify that two functions are inverses of each other.

We have found inverses of function defined by ordered pairs and from a graph. We will now look at how to find an inverse using an algebraic equation. The method uses the idea that if

f (x)

If we reverse the x and y in the function and then solve for y, we get our inverse function.

Key Concepts

Practice Makes Perfect

Find and Evaluate Composite Functions

In the following exercises, find ⓐ (f ∘ g)(x), ⓑ (g ∘ f)(x), and ⓒ (f · g)(x).

f (x) = 4 x + 3

and $g (x) = 2 x + 5$

ⓐ $8 x + 23$

ⓑ $8 x + 11$

ⓒ* * *

8 x^{2} + 26 x + 15

f (x) = 3 x - 1

and $g (x) = 5 x - 3$

f (x) = 6 x - 5

and $g (x) = 4 x + 1$

ⓐ $24 x + 1$

ⓑ $24 x - 19$

f (x) = 2 x + 7

and $g (x) = 3 x - 4$

f (x) = 3 x

and $g (x) = 2 x^{2} - 3 x$

ⓐ $6 x^{2} - 9 x$

ⓑ $18 x^{2} - 9 x$

f (x) = 2 x

and $g (x) = 3 x^{2} - 1$

f (x) = 2 x - 1

and $g (x) = x^{2} + 2$

ⓐ $2 x^{2} + 3$

ⓑ $4 x^{2} - 4 x + 3$

f (x) = 4 x + 3

and $g (x) = x^{2} - 4$

In the following exercises, find the values described.

For functions $f (x) = 2 x^{2} + 3$

and $g (x) = 5 x - 1,$

find* * *

ⓐ $(f \circ g) (−2)$

ⓑ $(g \circ f) (−3)$

ⓐ 245 ⓑ 104 ⓒ 53

For functions $f (x) = 5 x^{2} - 1$

and $g (x) = 4 x - 1,$

find* * *

ⓐ $(f \circ g) (1)$

ⓑ $(g \circ f) (−1)$

For functions $f (x) = 2 x^{3}$

and $g (x) = 3 x^{2} + 2,$

find* * *

ⓐ $(f \circ g) (−1)$

ⓑ $(g \circ f) (1)$

ⓐ 250 ⓑ 14 ⓒ 77

For functions $f (x) = 3 x^{3} + 1$

and $g (x) = 2 x^{2} - 3,$

find* * *

ⓐ $(f \circ g) (−2)$

ⓑ $(g \circ f) (−1)$

Determine Whether a Function is One-to-One

In the following exercises, determine if the set of ordered pairs represents a function and if so, is the function one-to-one.

{(−3, 9), (−2, 4), (−1, 1), (0, 0)

,* * *

(1, 1), (2, 4), (3, 9)}

Function; not one-to-one

{(9, −3), (4, −2), (1, −1), (0, 0)

,* * *

(1, 1), (4, 2), (9, 3)}

{(−3, −5), (−2, −3), (−1, −1)

,* * *

(0, 1), (1, 3), (2, 5), (3, 7)}

One-to-one function

{(5, 3), (4, 2), (3, 1), (2, 0)

,* * *

(1, −1), (0, −2), (−1, −3)}

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ* * *

This figure shows a graph of a circle with center at the origin and radius 3.

ⓑ* * *

This figure shows a graph of a parabola opening upward with vertex at (0k, 2).

ⓐ Not a function ⓑ Function; not one-to-one

ⓐ* * *

This figure shows a parabola opening to the right with vertex at (negative 2, 0).

ⓑ* * *

This figure shows a graph of a polynomial with odd order, so that it starts in the third quadrant, increases to the origin and then continues increasing through the first quadrant.

ⓐ* * *

This figure shows a graph of a curve that starts at (negative 6 negative 2) increases to the origin and then continues increasing slowly to (6, 2).

ⓑ* * *

This figure shows a parabola opening upward with vertex at (0, negative 4).

ⓐ One-to-one function* * *

ⓑ Function; not one-to-one

ⓐ* * *

This figure shows a straight line segment decreasing from (negative 4, 6) to (2, 0), after which it increases from (2, 0) to (6, 4).

ⓑ* * *

This figure shows a circle with radius 4 and center at the origin.

In the following exercises, find the inverse of each function. Determine the domain and range of the inverse function.

{(2, 1), (4, 2), (6, 3), (8, 4)}

Inverse function: ${(1, 2), (2, 4), (3, 6), (4, 8)} .$

Domain: ${1, 2, 3, 4} .$

Range: ${2, 4, 6, 8} .$

{(6, 2), (9, 5), (12, 8), (15, 11)}

{(0, −2), (1, 3), (2, 7), (3, 12)}

Inverse function: ${(−2, 0), (3, 1), (7, 2), (12, 3)} .$

Domain: ${−2, 3, 7, 12} .$

Range: ${0, 1, 2, 3} .$

{(0, 0), (1, 1), (2, 4), (3, 9)}

{(−2, −3), (−1, −1), (0, 1), (1, 3)}

Inverse function: ${(−3, − 2), (−1, −1), (1, 0), (3, 1)} .$

Domain: ${−3, − 1, 1, 3} .$

Range: ${−2, −1, 0, 1} .$

{(5, 3), (4, 2), (3, 1), (2, 0)}

In the following exercises, graph, on the same coordinate system, the inverse of the one-to-one function shown.

This figure shows a series of line segments from (negative 4, negative 3) to (negative 3, 0) then to (negative 1, 2) and then to (3, 4).

![This figure shows a series of line segments from (negative 3, negative 4) to (0, negative 3) then to (2, negative 1), and then to (4, 3).](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_10_01_303_img.jpg)

This figure shows a series of line segments from (negative 4, negative 4) to (negative 3, 1) then to (0, 2) and then to (2, 4).

This figure shows a series of line segments from (negative 4, 4) to (0, 3) then to (3, 2) and then to (4, negative 1).

![This figure shows a series of line segments from (negative 1, 4) to (2, 3) then to (3, 0), and then to (4, negative 4).](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_10_01_305_img.jpg)

This figure shows a series of line segments from (negative 4, negative 4) to (negative 1, negative 3) then to (0, 1), then to (1, 3), and then to (4, 4).

In the following exercises, determine whether or not the given functions are inverses.

f (x) = x + 8

and $g (x) = x - 8$

g (f (x)) = x,

and $f (g (x)) = x,$

so they are inverses.

f (x) = x - 9

and $g (x) = x + 9$

f (x) = 7 x

and $g (x) = \frac{x}{7}$

g (f (x)) = x,

and $f (g (x)) = x,$

so they are inverses.

f (x) = \frac{x}{11}

and $g (x) = 11 x$

f (x) = 7 x + 3

and $g (x) = \frac{x - 3}{7}$

g (f (x)) = x,

and $f (g (x)) = x,$

so they are inverses.

f (x) = 5 x - 4

and $g (x) = \frac{x - 4}{5}$

f (x) = \sqrt{x + 2}

and $g (x) = x^{2} - 2$

g (f (x)) = x,

and $f (g (x)) = x,$

so they are inverses (for nonnegative $x) .$

f (x) = \sqrt[3]{x - 4}

and $g (x) = x^{3} + 4$

In the following exercises, find the inverse of each function.

f (x) = x - 12

f^{−1} (x) = x + 12

f (x) = x + 17

f (x) = 9 x

f^{−1} (x) = \frac{x}{9}

f (x) = 8 x

f (x) = \frac{x}{6}

f^{−1} (x) = 6 x

f (x) = \frac{x}{4}

f (x) = 6 x - 7

f^{−1} (x) = \frac{x + 7}{6}

f (x) = 7 x - 1

f (x) = −2 x + 5

f^{−1} (x) = \frac{x - 5}{−2}

f (x) = −5 x - 4

f (x) = x^{2} + 6,

x \geq 0

f^{−1} (x) = \sqrt{x - 6}

f (x) = x^{2} - 9,

x \geq 0

f (x) = x^{3} - 4

f^{−1} (x) = \sqrt[3]{x + 4}

f (x) = x^{3} + 6

f (x) = \frac{1}{x + 2}

f^{−1} (x) = \frac{1}{x} - 2

f (x) = \frac{1}{x - 6}

f (x) = \sqrt{x - 2},

x \geq 2

f^{−1} (x) = x^{2} + 2

, $x \geq 0$

f (x) = \sqrt{x + 8},

x \geq −8

f (x) = \sqrt[3]{x - 3}

f^{−1} (x) = x^{3} + 3

f (x) = \sqrt[3]{x + 5}

f (x) = \sqrt[4]{9 x - 5},

x \geq \frac{5}{9}

f^{−1} (x) = \frac{x^{4} + 5}{9}

, $x \geq 0$

f (x) = \sqrt[4]{8 x - 3},

x \geq \frac{3}{8}

f (x) = \sqrt[5]{−3 x + 5}

f^{−1} (x) = \frac{x^{5} - 5}{−3}

f (x) = \sqrt[5]{−4 x - 3}

Writing Exercises

Explain how the graph of the inverse of a function is related to the graph of the function.

Answers will vary.

Explain how to find the inverse of a function from its equation. Use an example to demonstrate the steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Find and evaluate composite functions, determine whether a function is one-to-one, and find the inverse of a function. The rest of the cells are blank. ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.


{: valign=”top”}
{: valign=”top”}	Simplify.
{: valign=”top”}	Simplify.
{: valign=”top”}
{: valign=”top”}
{: valign=”top”}	Substitute $\frac{x + 1}{5}$

Finding Composite and Inverse Functions

Find and Evaluate Composite Functions

Determine Whether a Function is One-to-One

Find the Inverse of a Function

Key Concepts

Practice Makes Perfect

Writing Exercises

Self Check

Glossary