Finding Composite and Inverse Functions

Lynn Marecek

74 Finding Composite and Inverse Functions

Learning Objectives

By the end of this section, you will be able to:

Find and evaluate composite functions
Determine whether a function is one-to-one
Find the inverse of a function

Before you get started, take this readiness quiz.

If $f\left(x\right)=2x-3$ and $g\left(x\right)={x}^{2}+2x-3,$ find $f\left(4\right).$

If you missed this problem, review (Figure).
Solve for $x,$ $3x+2y=12.$

If you missed this problem, review (Figure).
Simplify: $5\frac{\left(x+4\right)}{5}-4.$

If you missed this problem, review (Figure).

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 $f$ and $g,$ the composition is written $f\circ g$ and is defined by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right).$

We read $f\left(g\left(x\right)\right)$ as $\text{``}f$ of $g$ of $x\text{.''}$

This figure shows x as the input to a box denoted as function g with g of x as the output of the box. Then, g of x is the input to a box denoted as function f with f of g of x as the output of the box.

To do a composition, the output of the first function, $g\left(x\right),$ becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

Composition of Functions

The composition of functions f and g is written $f·g$ and is defined by

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

We read $f\left(g\left(x\right)\right)$ as $f$ of $g$ of x.

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 $g\left(x\right)={x}^{2}$ as a parabola and then shifted it down vertically four units, we were using the composition defined by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ where $f\left(x\right)=x-4.$

This figure shows x as the input to a box denoted as g of x equals x squared with x squared as the output of the box. Then, x squared is the input to a box denoted as f of x equals x minus 4 with f of g of x equals x squared minus 4 as the output of the box.

The next example will demonstrate that $\left(f\circ g\right)\left(x\right),$ $\left(g\circ f\right)\left(x\right)$ and $\left(f·g\right)\left(x\right)$ usually result in different outputs.

For functions $f\left(x\right)=4x-5$ and $g\left(x\right)=2x+3,$ find: ⓐ $\left(f\circ g\right)\left(x\right),$ ⓑ $\left(g\circ f\right)\left(x\right),$ and ⓒ $\left(f·g\right)\left(x\right).$

ⓐ

Use the definition of $\left(f\circ g\right)\left(x\right).$


Distribute.
Simplify.

ⓑ

Use the definition of $\left(f\circ g\right)\left(x\right).$


Distribute.
Simplify.

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

ⓒ Notice that $\left(f·g\right)\left(x\right)$ is different than $\left(f\circ g\right)\left(x\right).$ In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.

$\begin{array}{cccccc}\text{Use the definition of}\phantom{\rule{0.2em}{0ex}}\left(f·g\right)\left(x\right).\hfill & & & & & \left(f·g\right)\left(x\right)=f\left(x\right)·g\left(x\right)\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=4x-5\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}g\left(x\right)=2x+3.\hfill & & & & & \left(f·g\right)\left(x\right)=\left(4x-5\right)·\left(2x+3\right)\hfill \\ \text{Multiply.}\hfill & & & & & \left(f·g\right)\left(x\right)=8{x}^{2}+2x-15\hfill \end{array}$

For functions $f\left(x\right)=3x-2$ and $g\left(x\right)=5x+1,$ find ⓐ $\left(f\circ g\right)\left(x\right)$ ⓑ $\left(g\circ f\right)\left(x\right)$ ⓒ $\left(f·g\right)\left(x\right)$ .

ⓐ $15x+1$ ⓑ $15x-9$

For functions $f\left(x\right)=4x-3,$ and $g\left(x\right)=6x-5,$ find ⓐ $\left(f\circ g\right)\left(x\right),$ ⓑ $\left(g\circ f\right)\left(x\right),$ and ⓒ $\left(f·g\right)\left(x\right).$

ⓐ $24x-23$ ⓑ $24x-23$

In the next example we will evaluate a composition for a specific value.

For functions $f\left(x\right)={x}^{2}-4,$ and $g\left(x\right)=3x+2,$ find: ⓐ $\left(f\circ g\right)\left(-3\right),$ ⓑ $\left(g\circ f\right)\left(-1\right),$ and ⓒ $\left(f\circ f\right)\left(2\right).$

ⓐ

Use the definition of $\left(f\circ g\right)\left(-3\right).$

Simplify.

Simplify.

ⓑ

Use the definition of $\left(g\circ f\right)\left(-1\right).$

Simplify.

Simplify.

ⓒ

Use the definition of $\left(f\circ f\right)\left(2\right).$

Simplify.

Simplify.

For functions $f\left(x\right)={x}^{2}-9,$ and $g\left(x\right)=2x+5,$ find ⓐ $\left(f\circ g\right)\left(-2\right),$ ⓑ $\left(g\circ f\right)\left(-3\right),$ and ⓒ $\left(f\circ f\right)\left(4\right).$

ⓐ –8 ⓑ 5 ⓒ 40

For functions $f\left(x\right)={x}^{2}+1,$ and $g\left(x\right)=3x-5,$ find ⓐ $\left(f\circ g\right)\left(-1\right),$ ⓑ $\left(g\circ f\right)\left(2\right),$ and ⓒ $\left(f\circ f\right)\left(-1\right).$

ⓐ 65 ⓑ 10 ⓒ 5

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.

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.

One-to-One Function

A function is one-to-one if each value in the range corresponds to one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value. There are no repeated y-values.

For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.

ⓐ $\left\{\left(-3,27\right),\left(-2,8\right),\left(-1,1\right),\left(0,0\right),\left(1,1\right),\left(2,8\right),\left(3,27\right)\right\}$ and ⓑ $\left\{\left(0,0\right),\left(1,1\right),\left(4,2\right),\left(9,3\right),\left(16,4\right)\right\}.$

ⓐ

$\begin{array}{ccccc}& & & & \phantom{\rule{5em}{0ex}}\left\{\left(-3,27\right),\left(-2,8\right),\left(-1,1\right),\left(0,0\right),\left(1,1\right),\left(2,8\right),\left(3,27\right)\right\}\hfill \end{array}$

Each x-value is matched with only one y-value. So this relation is a function.

But each y-value is not paired with only one x-value, $\left(-3,27\right)$ and $\left(3,27\right),$ for example. So this function is not one-to-one.

ⓑ

$\begin{array}{ccccc}& & & & \phantom{\rule{5em}{0ex}}\left\{\left(0,0\right),\left(1,1\right),\left(4,2\right),\left(9,3\right),\left(16,4\right)\right\}\hfill \end{array}$

Each x-value is matched with only one y-value. So this relation is a function.

Since each y-value is paired with only one x-value, this function is one-to-one.

For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ $\left\{\left(-3,-6\right),\left(-2,-4\right),\left(-1,-2\right),\left(0,0\right),\left(1,2\right),\left(2,4\right),\left(3,6\right)\right\}$

ⓑ $\left\{\left(-4,8\right),\left(-2,4\right),\left(-1,2\right),\left(0,0\right),\left(1,2\right),\left(2,4\right),\left(4,8\right)\right\}$

ⓐ One-to-one function

ⓑ Function; not one-to-one

For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ $\left\{\left(27,-3\right),\left(8,-2\right),\left(1,-1\right),\left(0,0\right),\left(1,1\right),\left(8,2\right),\left(27,3\right)\right\}$

ⓑ $\left\{\left(7,-3\right),\left(-5,-4\right),\left(8,0\right),\left(0,0\right),\left(-6,4\right),\left(-2,2\right),\left(-1,3\right)\right\}$

ⓐ Not a function

ⓑ Function; 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.

Horizontal Line Test

If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function.

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.

Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

This first graph shows a straight line passing through (0, 2) and (3, 0). This second shows a parabola opening up with vertex at (0, negative 1).

ⓐ

This figure shows a straight line passing through (0, 2) and (3, 0), with a red vertical line that only passes through one point and a blue horizontal line that only passes through one point.

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. Since any horizontal line intersects the graph in at most one point, the graph is the graph of a one-to-one function.

ⓑ

This figure shows a parabola opening up with vertex at (0, negative 1), with a red vertical line that only passes through one point and a blue horizontal line that passes through two points.

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. The horizontal line shown on the graph intersects it in two points. This graph does not represent a one-to-one function.

Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

Graph a shows a parabola opening to the right with vertex at (negative 1, 0). Graph b shows an exponential function that does not cross the x axis and that passes through (0, 1) before increasing rapidly.

ⓐ Not a function ⓑ One-to-one function

Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

Graph a shows a parabola opening up with vertex at (0, 3). Graph b shows a straight line passing through (0, negative 2) and (2, 0).

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

Find the Inverse of a Function

Let’s look at a one-to one function, $f$ , represented by the ordered pairs $\left\{\left(0,5\right),\left(1,6\right),\left(2,7\right),\left(3,8\right)\right\}.$ For each $x$ -value, $f$ adds 5 to get the $y$ -value. To ‘undo’ the addition of 5, we subtract 5 from each $y$ -value and get back to the original $x$ -value. We can call this “taking the inverse of $f$ ” and name the function ${f}^{-1}.$

Notice that that the ordered pairs of $f$ and ${f}^{-1}$ have their $x$ -values and $y$ -values reversed. The domain of $f$ is the range of ${f}^{-1}$ and the domain of ${f}^{-1}$ is the range of $f.$

Inverse of a Function Defined by Ordered Pairs

If $f\left(x\right)$ is a one-to-one function whose ordered pairs are of the form $\left(x,y\right),$ then its inverse function ${f}^{-1}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$

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

Find the inverse of the function $\left\{\left(0,3\right),\left(1,5\right),\left(2,7\right),\left(3,9\right)\right\}.$ Determine the domain and range of the inverse function.

This function is one-to-one since every $x$ -value is paired with exactly one $y$ -value.

To find the inverse we reverse the $x$ -values and $y$ -values in the ordered pairs of the function.

$\begin{array}{cccccc}\text{Function}\hfill & & & & & \left\{\left(0,3\right),\left(1,5\right),\left(2,7\right),\left(3,9\right)\right\}\hfill \\ \text{Inverse Function}\hfill & & & & & \left\{\left(3,0\right),\left(5,1\right),\left(7,2\right),\left(9,3\right)\right\}\hfill \\ \text{Domain of Inverse Function}\hfill & & & & & \left\{3,5,7,9\right\}\hfill \\ \text{Range of Inverse Function}\hfill & & & & & \left\{0,1,2,3\right\}\hfill \end{array}$

Find the inverse of $\left\{\left(0,4\right),\left(1,7\right),\left(2,10\right),\left(3,13\right)\right\}.$ Determine the domain and range of the inverse function.

Inverse function: $\left\{\left(4,0\right),\left(7,1\right),\left(10,2\right),\left(13,3\right)\right\}.$ Domain: $\left\{4,7,10,13\right\}.$ Range: $\left\{0,1,2,3\right\}.$

Find the inverse of $\left\{\left(-1,4\right),\left(-2,1\right),\left(-3,0\right),\left(-4,2\right)\right\}.$ Determine the domain and range of the inverse function.

Inverse function: $\left\{\left(4,-1\right),\left(1,-2\right),\left(0,-3\right),\left(2,-4\right)\right\}.$ Domain: $\left\{0,1,2,4\right\}.$ Range: $\left\{-4,-3,-2,-1\right\}.$

We just noted that if $f\left(x\right)$ is a one-to-one function whose ordered pairs are of the form $\left(x,y\right),$ then its inverse function ${f}^{-1}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$

So if a point $\left(a,b\right)$ is on the graph of a function $f\left(x\right),$ then the ordered pair $\left(b,a\right)$ is on the graph of ${f}^{-1}\left(x\right).$ See (Figure).

This figure shows the line y equals x with points (3,1) and (1,3) on either side of the line. These two points are connected by a dashed blue line segment.

The distance between any two pairs $\left(a,b\right)$ and $\left(b,a\right)$ is cut in half by the line $y=x.$ So we say the points are mirror images of each other through the line $y=x.$

Since every point on the graph of a function $f\left(x\right)$ is a mirror image of a point on the graph of ${f}^{-1}\left(x\right),$ we say the graphs are mirror images of each other through the line $y=x.$ We will use this concept to graph the inverse of a function in the next example.

Graph, on the same coordinate system, the inverse of the one-to one function shown.

This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1,0) then to (0,2) and then to (3, 4).

We can use points on the graph to find points on the inverse graph. Some points on the graph are: $\left(-5,-3\right),\left(-3,-1\right),\left(-1,0\right),\left(0,2\right),\left(3,4\right)$ .

So, the inverse function will contain the points: $\left(-3,-5\right),\left(-1,-3\right),\left(0,-1\right),\left(2,0\right),\left(4,3\right)$ .

This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1, 0) then to (0,2) and then to (3, 4). Then there is a dashed line to denote y equals x. There is also a line from (negative 3, negative 5) to (negative 1, negative 3) then to (0, negative 1), then to (2, 0) and then to (4, 3).

Notice how the graph of the original function and the graph of the inverse functions are mirror images through the line $y=x.$

Graph, on the same coordinate system, the inverse of the one-to one function.

The graph shows a line from (negative 3, negative 4) to (negative 2, negative 2) then to (0, negative 1), then to (1, 2) and then to (4, 3). The graph shows a line from (negative 3, 4) to (0, 3) then to (1, 2) and then to (4, 1).

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

Graph, on the same coordinate system, the inverse of the one-to one function.

Graph extends from negative 4 to 4 on both axes. Points plotted are (negative 3, 4), (0, 3), (1, 2), and (4, 1). Line segments connect points.

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.

This figure shows x as the input to a box denoted as function f with f of x as the output of the box. Then, f of x is the input to a box denoted as function f superscript negative 1 with f superscript negative 1 of f of x equals x as the output of the box.

Inverse Functions

$\begin{array}{ccc}\hfill {f}^{-1}\left(f\left(x\right)\right)& =\hfill & x,\phantom{\rule{0.2em}{0ex}}\text{for all}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{in the domain of}\phantom{\rule{0.2em}{0ex}}f\hfill \\ \hfill f\left({f}^{-1}\left(x\right)\right)& =\hfill & x,\phantom{\rule{0.2em}{0ex}}\text{for all}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{in the domain of}\phantom{\rule{0.2em}{0ex}}{f}^{-1}\hfill \end{array}$

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

Verify that $f\left(x\right)=5x-1$ and $g\left(x\right)=\frac{x+1}{5}$ are inverse functions.

The functions are inverses of each other if $g\left(f\left(x\right)\right)=x$ and $f\left(g\left(x\right)\right)=x.$


Substitute $5x-1$ for $f\left(x\right).$

Simplify.
Simplify.


Substitute $\frac{x+1}{5}$ for $g\left(x\right).$

Simplify.
Simplify.

Since both $g\left(f\left(x\right)\right)=x$ and $f\left(g\left(x\right)\right)=x$ are true, the functions $f\left(x\right)=5x-1$ and $g\left(x\right)=\frac{x+1}{5}$ are inverse functions. That is, they are inverses of each other.

Verify that the functions are inverse functions.

$f\left(x\right)=4x-3$ and $g\left(x\right)=\frac{x+3}{4}.$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

Verify that the functions are inverse functions.

$f\left(x\right)=2x+6$ and $g\left(x\right)=\frac{x-6}{2}.$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

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 $f\left(x\right)$ is a one-to-one function with ordered pairs $\left(x,y\right),$ then its inverse function ${f}^{-1}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$

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

How to Find the inverse of a One-to-One Function

Find the inverse of $f\left(x\right)=4x+7.$

Step 1 is to substitute y for f of x. To do so, we replace f of x with y. Hence, f of x equals 4 x plus 7 becomes y equals 4 x plus 7.

Step 2 is to interchange the variables x and y. To do so, we replace x with y and then y with x. Hence, we obtain x equals 4y plus 7.

Step 3 is to solve for y. To do so, we subtract 7 from each side and then divide by 4. Hence, we have x minus 7 equals 4y and then the quantity x minus 7 divided by 4 equals y.

Step 4 is to substitute f superscript negative 1 of x for y. To do so, we replace y with f superscript negative 1 of x. Hence, the quantity x minus 7 divided by 4 equals f superscript negative 1 of x.

Step 5 is to verify that the functions are inverses. To do so, we show that f superscript negative 1 of f of x equals x and that f of f superscript negative 1of x equals x. Hence, we ask whether f inverse of 4x plus 7 equals x. This becomes a question of whether 4 x plus 7 minus 7 all divided by 4 equals x. This becomes a question of whether 4x divided by 4 equals x. This is true. To show the other side, we examine whether f of f inverse of x equals x. This becomes a question of whether f of the quantity x minus 7 divided by 4 equals x. This becomes a question of whether 4 times the quantity x minus 7 divided by 4 equals x. This becomes a question of whether x minus 7 plus 7 equals x. This is true.

Find the inverse of the function $f\left(x\right)=5x-3.$

${f}^{-1}\left(x\right)=\frac{x+3}{5}$

Find the inverse of the function $f\left(x\right)=8x+5.$

${f}^{-1}\left(x\right)=\frac{x-5}{8}$

We summarize the steps below.

How to Find the inverse of a One-to-One Function

Substitute y for $f\left(x\right).$
Interchange the variables x and y.
Solve for y.
Substitute ${f}^{-1}\left(x\right)$ for y.
Verify that the functions are inverses.

How to Find the Inverse of a One-to-One Function

Find the inverse of $f\left(x\right)=\sqrt[5]{2x-3}.$

$\begin{array}{cccccccc}& & & & & \hfill f\left(x\right)& =\hfill & \sqrt[5]{2x-3}\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}f\left(x\right).\hfill & & & & & \hfill y& =\hfill & \sqrt[5]{2x-3}\hfill \\ \text{Interchange the variables}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & & & \hfill x& =\hfill & \sqrt[5]{2y-3}\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & & & \hfill {\left(x\right)}^{5}& =\hfill & {\left(\sqrt[5]{2y-3}\right)}^{5}\hfill \\ & & & & & \hfill {x}^{5}& =\hfill & 2y-3\hfill \\ & & & & & \hfill {x}^{5}+3& =\hfill & 2y\hfill \\ & & & & & \hfill \frac{{x}^{5}+3}{2}& =\hfill & y\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}{f}^{-1}\left(x\right)\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & & & \hfill {f}^{-1}\left(x\right)& =\hfill & \frac{{x}^{5}+3}{2}\hfill \end{array}$

Verify that the functions are inverses.

$\begin{array}{}\\ \\ \hfill {f}^{-1}\left(f\left(x\right)\right)& \stackrel{?}{=}\hfill & x\hfill & & & & & \hfill f\left({f}^{-1}\left(x\right)\right)& \stackrel{?}{=}\hfill & x\hfill \\ \hfill {f}^{-1}\left(\sqrt[5]{2x-3}\right)& \stackrel{?}{=}\hfill & x\hfill & & & & & \hfill f\left(\frac{{x}^{5}+3}{2}\right)& \stackrel{?}{=}\hfill & x\hfill \\ \hfill \frac{{\left(\sqrt[5]{2x-3}\right)}^{5}+3}{2}& \stackrel{?}{=}\hfill & x\hfill & & & & & \hfill \sqrt[5]{2\left(\frac{{x}^{5}+3}{2}\right)-3}& \stackrel{?}{=}\hfill & x\hfill \\ \hfill \frac{2x-3+3}{2}& \stackrel{?}{=}\hfill & x\hfill & & & & & \hfill \sqrt[5]{{x}^{5}+3-3}& \stackrel{?}{=}\hfill & x\hfill \\ \hfill \frac{2x}{2}& \stackrel{?}{=}\hfill & x\hfill & & & & & \hfill \sqrt[5]{{x}^{5}}& \stackrel{?}{=}\hfill & x\hfill \\ \hfill x& =\hfill & x✓\hfill & & & & & \hfill x& =\hfill & x✓\hfill \end{array}$

Find the inverse of the function $f\left(x\right)=\sqrt[5]{3x-2}.$

${f}^{-1}\left(x\right)=\frac{{x}^{5}+2}{3}$

Find the inverse of the function $f\left(x\right)=\sqrt[4]{6x-7}.$

${f}^{-1}\left(x\right)=\frac{{x}^{4}+7}{6}$

Key Concepts

Composition of Functions: The composition of functions $f$ and $g,$ is written $f\circ g$ and is defined by

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

We read $f\left(g\left(x\right)\right)$ as $f$ of $g$ of $x.$
Horizontal Line Test: If every horizontal line, intersects the graph of a function in at most one point, it is a one-to-one function.
Inverse of a Function Defined by Ordered Pairs: If $f\left(x\right)$ is a one-to-one function whose ordered pairs are of the form $\left(x,y\right),$ then its inverse function ${f}^{-1}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$
Inverse Functions: For every $x$ in the domain of one-to-one function $f$ and ${f}^{-1},$

$\begin{array}{ccc}\hfill {f}^{-1}\left(f\left(x\right)\right)& =\hfill & x\hfill \\ \hfill f\left({f}^{-1}\left(x\right)\right)& =\hfill & x\hfill \end{array}$
How to Find the Inverse of a One-to-One Function:
1. Substitute y for $f\left(x\right).$
2. Interchange the variables x and y.
3. Solve for y.
4. Substitute ${f}^{-1}\left(x\right)$ for $y.$
5. Verify that the functions are inverses.

Practice Makes Perfect

Find and Evaluate Composite Functions

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

$f\left(x\right)=4x+3$ and $g\left(x\right)=2x+5$

ⓐ $8x+23$ ⓑ $8x+11$ ⓒ

$8{x}^{2}+26x+15$

$f\left(x\right)=3x-1$ and $g\left(x\right)=5x-3$

$f\left(x\right)=6x-5$ and $g\left(x\right)=4x+1$

ⓐ $24x+1$ ⓑ $24x-19$

$f\left(x\right)=2x+7$ and $g\left(x\right)=3x-4$

$f\left(x\right)=3x$ and $g\left(x\right)=2{x}^{2}-3x$

ⓐ $6{x}^{2}-9x$ ⓑ $18{x}^{2}-9x$

$f\left(x\right)=2x$ and $g\left(x\right)=3{x}^{2}-1$

$f\left(x\right)=2x-1$ and $g\left(x\right)={x}^{2}+2$

ⓐ $2{x}^{2}+3$ ⓑ $4{x}^{2}-4x+3$

$f\left(x\right)=4x+3$ and $g\left(x\right)={x}^{2}-4$

In the following exercises, find the values described.

For functions $f\left(x\right)=2{x}^{2}+3$ and $g\left(x\right)=5x-1,$ find

ⓐ $\left(f\circ g\right)\left(-2\right)$

ⓑ $\left(g\circ f\right)\left(-3\right)$

ⓐ 245 ⓑ 104 ⓒ 53

For functions $f\left(x\right)=5{x}^{2}-1$ and $g\left(x\right)=4x-1,$ find

ⓐ $\left(f\circ g\right)\left(1\right)$

ⓑ $\left(g\circ f\right)\left(-1\right)$

For functions $f\left(x\right)=2{x}^{3}$ and $g\left(x\right)=3{x}^{2}+2,$ find

ⓐ $\left(f\circ g\right)\left(-1\right)$

ⓑ $\left(g\circ f\right)\left(1\right)$

ⓐ 250 ⓑ 14 ⓒ 77

For functions $f\left(x\right)=3{x}^{3}+1$ and $g\left(x\right)=2{x}^{2}-3,$ find

ⓐ $\left(f\circ g\right)\left(-2\right)$

ⓑ $\left(g\circ f\right)\left(-1\right)$

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.

$\left\{\left(-3,9\right),\left(-2,4\right),\left(-1,1\right),\left(0,0\right)$ ,

$\left(1,1\right),\left(2,4\right),\left(3,9\right)\right\}$

Function; not one-to-one

$\left\{\left(9,-3\right),\left(4,-2\right),\left(1,-1\right),\left(0,0\right)$ ,

$\left(1,1\right),\left(4,2\right),\left(9,3\right)\right\}$

$\left\{\left(-3,-5\right),\left(-2,-3\right),\left(-1,-1\right)$ ,

$\left(0,1\right),\left(1,3\right),\left(2,5\right),\left(3,7\right)\right\}$

One-to-one function

$\left\{\left(5,3\right),\left(4,2\right),\left(3,1\right),\left(2,0\right)$ ,

$\left(1,-1\right),\left(0,-2\right),\left(-1,-3\right)\right\}$

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.

$\left\{\left(2,1\right),\left(4,2\right),\left(6,3\right),\left(8,4\right)\right\}$

Inverse function: $\left\{\left(1,2\right),\left(2,4\right),\left(3,6\right),\left(4,8\right)\right\}.$ Domain: $\left\{1,2,3,4\right\}.$ Range: $\left\{2,4,6,8\right\}.$

$\left\{\left(6,2\right),\left(9,5\right),\left(12,8\right),\left(15,11\right)\right\}$

$\left\{\left(0,-2\right),\left(1,3\right),\left(2,7\right),\left(3,12\right)\right\}$

Inverse function: $\left\{\left(-2,0\right),\left(3,1\right),\left(7,2\right),\left(12,3\right)\right\}.$ Domain: $\left\{-2,3,7,12\right\}.$ Range: $\left\{0,1,2,3\right\}.$

$\left\{\left(0,0\right),\left(1,1\right),\left(2,4\right),\left(3,9\right)\right\}$

$\left\{\left(-2,-3\right),\left(-1,-1\right),\left(0,1\right),\left(1,3\right)\right\}$

Inverse function: $\left\{\left(-3,\text{−}2\right),\left(-1,-1\right),\left(1,0\right),\left(3,1\right)\right\}.$ Domain: $\left\{-3,\text{−}1,1,3\right\}.$ Range: $\left\{-2,-1,0,1\right\}.$

$\left\{\left(5,3\right),\left(4,2\right),\left(3,1\right),\left(2,0\right)\right\}$

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).

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).

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.

$f\left(x\right)=x+8$ and $g\left(x\right)=x-8$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

$f\left(x\right)=x-9$ and $g\left(x\right)=x+9$

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

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

$f\left(x\right)=\frac{x}{11}$ and $g\left(x\right)=11x$

$f\left(x\right)=7x+3$ and $g\left(x\right)=\frac{x-3}{7}$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

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

$f\left(x\right)=\sqrt{x+2}$ and $g\left(x\right)={x}^{2}-2$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses (for nonnegative $x\right).$

$f\left(x\right)=\sqrt[3]{x-4}$ and $g\left(x\right)={x}^{3}+4$

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

$f\left(x\right)=x-12$

${f}^{-1}\left(x\right)=x+12$

$f\left(x\right)=x+17$

$f\left(x\right)=9x$

${f}^{-1}\left(x\right)=\frac{x}{9}$

$f\left(x\right)=8x$

$f\left(x\right)=\frac{x}{6}$

${f}^{-1}\left(x\right)=6x$

$f\left(x\right)=\frac{x}{4}$

$f\left(x\right)=6x-7$

${f}^{-1}\left(x\right)=\frac{x+7}{6}$

$f\left(x\right)=7x-1$

$f\left(x\right)=-2x+5$

${f}^{-1}\left(x\right)=\frac{x-5}{-2}$

$f\left(x\right)=-5x-4$

$f\left(x\right)={x}^{2}+6,$ $x\ge 0$

${f}^{-1}\left(x\right)=\sqrt{x-6}$

$f\left(x\right)={x}^{2}-9,$ $x\ge 0$

$f\left(x\right)={x}^{3}-4$

${f}^{-1}\left(x\right)=\sqrt[3]{x+4}$

$f\left(x\right)={x}^{3}+6$

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

${f}^{-1}\left(x\right)=\frac{1}{x}-2$

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

$f\left(x\right)=\sqrt{x-2},$ $x\ge 2$

${f}^{-1}\left(x\right)={x}^{2}+2$ , $x\ge 0$

$f\left(x\right)=\sqrt{x+8},$ $x\ge -8$

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

${f}^{-1}\left(x\right)={x}^{3}+3$

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

$f\left(x\right)=\sqrt[4]{9x-5},$ $x\ge \frac{5}{9}$

${f}^{-1}\left(x\right)=\frac{{x}^{4}+5}{9}$ , $x\ge 0$

$f\left(x\right)=\sqrt[4]{8x-3},$ $x\ge \frac{3}{8}$

$f\left(x\right)=\sqrt[5]{-3x+5}$

${f}^{-1}\left(x\right)=\frac{{x}^{5}-5}{-3}$

$f\left(x\right)=\sqrt[5]{-4x-3}$

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.

Glossary

one-to-one function: 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.

License

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Learning Objectives

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

License

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