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

We read $f (g (x))$ as $“ f$ of $g$ of $x .”$

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, $g (x),$ 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 $(f \circ g) (x) = f (g (x))$ where $f (x) = x - 4 .$

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 $(f \circ g) (x),$ $(g \circ f) (x)$ and $(f \cdot g) (x)$ usually result in different outputs.

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

Solution

ⓐ

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


Distribute.
Simplify.

ⓑ

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


Distribute.
Simplify.

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

ⓒ Notice that $(f \cdot g) (x)$ 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.

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

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

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

Solution

ⓐ

Use the definition of $(f \circ g) (−3) .$

Simplify.

Simplify.

ⓑ

Use the definition of $(g \circ f) (−1) .$

Simplify.

Simplify.

ⓒ

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

Simplify.

Simplify.

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.

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

ⓐ ${(−3, 27), (−2, 8), (−1, 1), (0, 0), (1, 1), (2, 8), (3, 27)}$ and ⓑ ${(0, 0), (1, 1), (4, 2), (9, 3), (16, 4)} .$

Solution

ⓐ
$\begin{matrix} {(−3, 27), (−2, 8), (−1, 1), (0, 0), (1, 1), (2, 8), (3, 27)} \end{matrix}$

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, $(−3, 27)$ and $(3, 27),$ for example. So this function is not one-to-one.

ⓑ
$\begin{matrix} {(0, 0), (1, 1), (4, 2), (9, 3), (16, 4)} \end{matrix}$

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.

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.

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).

Solution

ⓐ
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.

Find the Inverse of a Function

Let’s look at a one-to one function, $f$ , represented by the ordered pairs ${(0, 5), (1, 6), (2, 7), (3, 8)} .$ 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 $f^{−1} .$

This figure shows the set (0, 5), (1, 6), (2, 7) and (3, 8) on the left side of an oval. The oval contains the numbers 0, 1, 2, and 3. There are black arrows from these numbers that point to the numbers 5, 6, 7, and 8, respectively in a second oval to the right of the first. Above this, there is a black arrow labeled “f add 5” coming from the left oval to the right oval. There are red arrows from the numbers 5, 6, 7, and 8 in the right oval to the numbers 0, 1, 2, and 3, respectively, in the left oval. Below this, we have a red arrow labeled “f with a superscript negative 1” and “subtract 5”. To the right of this, we have the set (5, 0), (6, 1), (7, 2) and (8, 3).

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 $f .$

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

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

Solution

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.

A function, its inverse, and the domain and range of the inverse function represented as sets.
Function	${(0, 3), (1, 5), (2, 7), (3, 9)}$
Inverse Function	${(3, 0), (5, 1), (7, 2), (9, 3)}$
Domain of Inverse Function	${3, 5, 7, 9}$
Range of Inverse Function	${0, 1, 2, 3}$

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

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

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 $(a, b)$ and $(b, a)$ 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 $f (x)$ is a mirror image of a point on the graph of $f^{−1} (x),$ 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).

Solution

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

So, the inverse function will contain the points: $(−3, −5), (−1, −3), (0, −1), (2, 0), (4, 3)$ .
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 $y = x .$

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.

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

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

Solution

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


Substitute $5 x - 1$ for $f (x) .$

Simplify.
Simplify.


Substitute $\frac{x + 1}{5}$ for $g (x) .$
$A mathematical problem asks to find the value of the function f((x+1)/5) given that f(x) = 5x - 1. The input to the function f is shown as a fraction with 'x+1' in red in the numerator and '5' in red in the denominator.$
Simplify.
Simplify.

Since both $g (f (x)) = x$ and $f (g (x)) = x$ are true, the functions $f (x) = 5 x - 1$ and $g (x) = \frac{x + 1}{5}$ are inverse functions. That is, they 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 $f (x)$ is a one-to-one function with ordered pairs $(x, y),$ then its inverse function $f^{−1} (x)$ is the set of ordered pairs $(y, x) .$

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 $f (x) = 4 x + 7 .$

Solution

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.

We summarize the steps below.

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

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

Solution

Step-by-step process for finding the inverse of the function f(x) = (2x-3)^(1/5) and verifying the inverse relationship.
	$f (x) = \sqrt[5]{2 x - 3}$
Substitute $y$ for $f (x)$ .	$y = \sqrt[5]{2 x - 3}$
Interchange the variables $x$ and $y$ .	$x = \sqrt[5]{2 y - 3}$
Solve for $y$ .	${(x)}^{5} = {(\sqrt[5]{2 y - 3})}^{5}$
	$x^{5} = 2 y - 3$
	$x^{5} + 3 = 2 y$
	$\frac{x^{5} + 3}{2} = y$
Substitute $f^{- 1} (x)$ for $y$ .	$f^{- 1} (x) = \frac{x^{5} + 3}{2}$
Verify that the functions are inverses.
$f^{- 1} (f (x)) \overset{?}{=} x$	$f (f^{- 1} (x)) \overset{?}{=} x$
$f^{- 1} (\sqrt[5]{2 x - 3}) \overset{?}{=} x$	$f (\frac{x^{5} + 3}{2}) \overset{?}{=} x$
$\frac{{(\sqrt[5]{2 x - 3})}^{5} + 3}{2} \overset{?}{=} x$	$\sqrt[5]{2 (\frac{x^{5} + 3}{2}) - 3} \overset{?}{=} x$
$\frac{2 x - 3 + 3}{2} \overset{?}{=} x$	$\sqrt[5]{x^{5} + 3 - 3} \overset{?}{=} x$
$\frac{2 x}{2} \overset{?}{=} x$	$\sqrt[5]{x^{5}} \overset{?}{=} x$
$x = x ✓$	$x = x ✓$

Key Concepts

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

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

We read $f (g (x))$ 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 $f (x)$ is a one-to-one function whose ordered pairs are of the form $(x, y),$ then its inverse function $f^{−1} (x)$ is the set of ordered pairs $(y, x) .$
Inverse Functions: For every $x$ in the domain of one-to-one function $f$ and $f^{−1},$

$\begin{matrix} f^{−1} (f (x)) & = & x \\ f (f^{−1} (x)) & = & x \end{matrix}$
How to Find the Inverse of a One-to-One Function:
1. Substitute y for $f (x) .$
2. Interchange the variables x and y.
3. Solve for y.
4. Substitute $f^{−1} (x)$ 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).

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

Solution

ⓐ $8 x + 23$ ⓑ $8 x + 11$ ⓒ
$8 x^{2} + 26 x + 15$

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

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

Solution

ⓐ $24 x + 1$ ⓑ $24 x - 19$
ⓒ $24 x^{2} - 14 x - 5$

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

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

Solution

ⓐ $6 x^{2} - 9 x$ ⓑ $18 x^{2} - 9 x$
ⓒ $6 x^{3} - 9 x^{2}$

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

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

Solution

ⓐ $2 x^{2} + 3$ ⓑ $4 x^{2} - 4 x + 3$
ⓒ $2 x^{3} - x^{2} + 4 x - 2$

$f (x) = 4 x + 3$ 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 ⓐ $(f \circ g) (−2)$ ⓑ $(g \circ f) (−3)$ ⓒ $(f \circ f) (−1)$

Solution

ⓐ 245 ⓑ 104 ⓒ 53

For functions $f (x) = 5 x^{2} - 1$ and $g (x) = 4 x - 1,$ find ⓐ $(f \circ g) (1)$ ⓑ $(g \circ f) (−1)$ ⓒ $(f \circ f) (2)$

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

Solution

ⓐ 250 ⓑ 14 ⓒ 77

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

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.

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

Solution

Function; not one-to-one

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

${(−3, −5), (−2, −3), (−1, −1)$ ,
$(0, 1), (1, 3), (2, 5), (3, 7)}$

Solution

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.
ⓑ

Solution

ⓐ 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).

Solution

ⓐ 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.

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

Solution

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

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

${(0, −2), (1, 3), (2, 7), (3, 12)}$

Solution

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

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

${(−2, −3), (−1, −1), (0, 1), (1, 3)}$

Solution

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).

Solution

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).

Solution

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.

$f (x) = x + 8$ and $g (x) = x - 8$

Solution

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

$f (x) = x - 9$ and $g (x) = x + 9$

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

Solution

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

Solution

$g (f (x)) = x,$ 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 $g (x) = x^{2} - 2 (x > 0)$

Solution

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

$f (x) = x - 12$

Solution

$f^{−1} (x) = x + 12$

$f (x) = x + 17$

$f (x) = 9 x$

Solution

$f^{−1} (x) = \frac{x}{9}$

$f (x) = 8 x$

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

Solution

$f^{−1} (x) = 6 x$

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

$f (x) = 6 x - 7$

Solution

$f^{−1} (x) = \frac{x + 7}{6}$

$f (x) = 7 x - 1$

$f (x) = −2 x + 5$

Solution

$f^{−1} (x) = \frac{x - 5}{−2}$

$f (x) = −5 x - 4$

$f (x) = x^{2} + 6,$ $x \geq 0$

Solution

$f^{−1} (x) = \sqrt{x - 6}$

$f (x) = x^{2} - 9,$ $x \geq 0$

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

Solution

$f^{−1} (x) = \sqrt[3]{x + 4}$

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

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

Solution

$f^{−1} (x) = \frac{1}{x} - 2$

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

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

Solution

$f^{−1} (x) = x^{2} + 2$ , $x \geq 0$

$f (x) = \sqrt{x + 8},$ $x \geq −8$

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

Solution

$f^{−1} (x) = x^{3} + 3$

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

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

Solution

$f^{−1} (x) = \frac{x^{4} + 5}{9}$ , $x \geq 0$

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

$f (x) = \sqrt[5]{−3 x + 5}$

Solution

$f^{−1} (x) = \frac{x^{5} - 5}{−3}$

$f (x) = \sqrt[5]{−4 x - 3}$

Writing Exercises

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

Solution

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. Whom 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.

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.