Inverse Functions

Verifying That Two Functions Are Inverse Functions

Betty is traveling to Milan for a fashion show and wants to know what the temperature will be. She is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, Betty wants to convert 75 degrees Fahrenheit to degrees Celsius, using the formula

C = \frac{5}{9} (F - 32)

and substitutes 75 for $F$ to calculate

\frac{5}{9} (75 - 32) \approx 24 °C .

Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, Betty gets the week’s weather forecast from Figure 2 for Milan, and wants to convert all of the temperatures to degrees Fahrenheit.

A forecast of Monday’s through Thursday’s weather. — Figure 2

At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for $F$ after substituting a value for $C .$ For example, to convert 26 degrees Celsius, she could write

\begin{array}{l} \begin{array}{l} 26 = \frac{5}{9} (F - 32) \\ 26 \cdot \frac{9}{5} = F - 32 \\ F = 26 \cdot \frac{9}{5} + 32 \approx 79 \end{array} \end{array}

After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.

The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.

Given a function $f (x),$ we represent its inverse as $f^{- 1} (x),$ read as $“ f$ inverse of $x . ”$ The raised $−1$ is part of the notation. It is not an exponent; it does not imply a power of $−1$ . In other words, $f^{- 1} (x)$ does not mean $\frac{1}{f (x)}$ because $\frac{1}{f (x)}$ is the reciprocal of $f$ and not the inverse.

The “exponent-like” notation comes from an analogy between function composition and multiplication: just as $a^{- 1} a = 1$ (1 is the identity element for multiplication) for any nonzero number $a,$ so $f^{- 1} \circ f$ equals the identity function, that is,

(f^{- 1} \circ f) (x) = f^{- 1} (f (x)) = f^{- 1} (y) = x

This holds for all $x$ in the domain of $f .$ Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses.

Given a function $f (x),$ we can verify whether some other function $g (x)$ is the inverse of $f (x)$ by checking if both $g (f (x)) = x$ and $f (g (x)) = x$ are true.

For example, $y = 4 x$ and $y = \frac{1}{4} x$ are inverse functions.

(f^{- 1} \circ f) (x) = f^{- 1} (4 x) = \frac{1}{4} (4 x) = x

and

(f^{} \circ f^{- 1}) (x) = f (\frac{1}{4} x) = 4 (\frac{1}{4} x) = x

A few coordinate pairs from the graph of the function $y = 4 x$ are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function $y = \frac{1}{4} x$ are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.

Example 1

Identifying an Inverse Function for a Given Input-Output Pair

If for a particular one-to-one function $f (2) = 4$ and $f (5) = 12,$ what are the corresponding input and output values for the inverse function?

Solution

The inverse function reverses the input and output quantities, so if

\begin{matrix} f (2) = 4, then f^{- 1} (4) = 2; \\ f (5) = 12, {then f}^{- 1} (12) = 5. \end{matrix}

Alternatively, if we want to name the inverse function $g,$ then $g (4) = 2$ and $g (12) = 5.$

Analysis

Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1.

Table 1 For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).
$(x, f (x))$	$(x, g (x))$
$(2, 4)$	$(4, 2)$
$(5, 12)$	$(12, 5)$

Example 2

Testing Inverse Relationships Algebraically

If $f (x) = \frac{1}{x + 2}$ and $g (x) = \frac{1}{x} - 2,$ is $g = f^{- 1} ?$

Solution

\begin{array}{l} \begin{array}{l} g (f (x)) = \frac{1}{(\frac{1}{x + 2})} - 2 \end{array} \\ = x + 2 - 2 \\ = x \end{array}

We must also verify the other formula.

\begin{array}{l} \begin{array}{l} f (g (x)) = \frac{1}{\frac{1}{x} - 2 + 2} \end{array} \\ = \frac{1}{\frac{1}{x}} \\ = x \end{array}

so

g = f^{- 1} and f = g^{- 1}

Analysis

Notice the inverse operations are in reverse order of the operations from the original function.

Example 3

Determining Inverse Relationships for Power Functions

If $f (x) = x^{3}$ (the cube function) and $g (x) = \frac{1}{3} x,$ is $g = f^{- 1} ?$

Solution

f (g (x)) = \frac{x^{3}}{27} \neq x

No, the functions are not inverses.

Analysis

The correct inverse to the cube is, of course, the cube root $\sqrt[3]{x} = x^{\frac{1}{3}},$ that is, the one-third is an exponent, not a multiplier.

Finding Domain and Range of Inverse Functions

The outputs of the function $f$ are the inputs to $f^{- 1},$ so the range of $f$ is also the domain of $f^{- 1} .$ Likewise, because the inputs to $f$ are the outputs of $f^{- 1},$ the domain of $f$ is the range of $f^{- 1} .$ We can visualize the situation as in Figure 3.

Domain and range of a function and its inverse. — Figure 3 Domain and range of a function and its inverse

When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of $f (x) = \sqrt{x}$ is $f^{- 1} (x) = x^{2},$ because a square “undoes” a square root; but the square is only the inverse of the square root on the domain $[0, \infty),$ since that is the range of $f (x) = \sqrt{x} .$

We can look at this problem from the other side, starting with the square (toolkit quadratic) function $f (x) = x^{2} .$ If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function $f (x) = x^{2}$ with its domain limited to $[0, \infty),$ which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If $f (x) = {(x - 1)}^{2}$ on $[1, \infty),$ then the inverse function is $f^{- 1} (x) = \sqrt{x} + 1.$

The domain of $f$ = range of $f^{- 1}$ = $[1, \infty) .$
The domain of $f^{- 1}$ = range of $f$ = $[0, \infty) .$

Example 4

Finding the Inverses of Toolkit Functions

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 2. We restrict the domain in such a fashion that the function assumes all y-values exactly once.

Table 2 A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1/x; the reciprocal squared function is f(x)=1/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1/3).
Constant	Identity	Quadratic	Cubic	Reciprocal
$f (x) = c$	$f (x) = x$	$f (x) = x^{2}$	$f (x) = x^{3}$	$f (x) = \frac{1}{x}$
Reciprocal squared	Cube root	Square root	Absolute value
$f (x) = \frac{1}{x^{2}}$	$f (x) = \sqrt[3]{x}$	$f (x) = \sqrt{x}$	$f (x) = \| x \|$

Solution

The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.

The absolute value function can be restricted to the domain $[0, \infty),$ where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain $(0, \infty) .$

Analysis

We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.

Graph of an absolute function. — Figure 4 (a) Absolute value (b) Reciprocal squared

Finding and Evaluating Inverse Functions

Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.

Inverting Tabular Functions

Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.

Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.

Example 5

Interpreting the Inverse of a Tabular Function

A function $f (t)$ is given in Table 3, showing distance in miles that a car has traveled in $t$ minutes. Find and interpret $f^{- 1} (70) .$

Table 3 Two rows and five columns. The first row is labeled “t (minutes)”, and the second row is labeled “f(x) (miles)”. Reading the columns as ordered pairs, we have the following values (30, 20), (50, 40), (70, 60), and (90, 70).
$t (minutes)$	30	50	70	90
$f (t) (miles)$	20	40	60	70

Solution

The inverse function takes an output of $f$ and returns an input for $f .$ So in the expression $f^{- 1} (70),$ 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function $f,$ 90 minutes, so $f^{- 1} (70) = 90.$ The interpretation of this is that, to drive 70 miles, it took 90 minutes.

Alternatively, recall that the definition of the inverse was that if $f (a) = b,$ then $f^{- 1} (b) = a .$ By this definition, if we are given $f^{- 1} (70) = a,$ then we are looking for a value $a$ so that $f (a) = 70.$ In this case, we are looking for a $t$ so that $f (t) = 70,$ which is when $t = 90.$

Try It #5

Using Table 4, find and interpret (a) $f (60),$ and (b) $f^{- 1} (60) .$

Table 4 Two rows and five columns. The first row is labeled “t (minutes)”, and the second row is labeled “f(t)”. Reading the columns as ordered pairs, we have the following values (30, 20), (50, 40), (70, 60), and (90, 70).
$t (minutes)$	30	50	60	70	90
$f (t) (miles)$	20	40	50	60	70

Solution

$f (60) = 50.$ In 60 minutes, 50 miles are traveled.
$f^{- 1} (60) = 70.$ To travel 60 miles, it will take 70 minutes.

Evaluating the Inverse of a Function, Given a Graph of the Original Function

We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph.

Example 6

Evaluating a Function and Its Inverse from a Graph at Specific Points

A function $g (x)$ is given in Figure 5. Find $g (3)$ and $g^{- 1} (3) .$

A graph displays an exponential growth function g(x) on a coordinate plane. The blue curve rises continuously, starting near the x-axis for negative x values and increasing sharply for positive x values. — Figure 5

Solution

To evaluate $g (3),$ we find 3 on the x-axis and find the corresponding output value on the y-axis. The point $(3, 1)$ tells us that $g (3) = 1.$

To evaluate $g^{- 1} (3),$ recall that by definition $g^{- 1} (3)$ means the value of x for which $g (x) = 3.$ By looking for the output value 3 on the vertical axis, we find the point $(5, 3)$ on the graph, which means $g (5) = 3,$ so by definition, $g^{- 1} (3) = 5.$ See Figure 6.

Finding Inverses of Functions Represented by Formulas

Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula— for example, $y$ as a function of $x —$ we can often find the inverse function by solving to obtain $x$ as a function of $y .$

Example 7

Inverting the Fahrenheit-to-Celsius Function

Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.

C = \frac{5}{9} (F - 32)

Solution

\begin{array}{l} \begin{array}{l} C = \frac{5}{9} (F - 32) \\ C \cdot \frac{9}{5} = F - 32 \\ F = \frac{9}{5} C + 32 \end{array} \end{array}

By solving in general, we have uncovered the inverse function. If

C = h (F) = \frac{5}{9} (F - 32),

then

F = h^{- 1} (C) = \frac{9}{5} C + 32.

In this case, we introduced a function $h$ to represent the conversion because the input and output variables are descriptive, and writing $C^{- 1}$ could get confusing.

Example 8

Solving to Find an Inverse Function

Find the inverse of the function $f (x) = \frac{2}{x - 3} + 4.$

Solution

\begin{array}{l} y = \frac{2}{x - 3} + 4 \begin{matrix} \begin{matrix}  \end{matrix} \end{matrix} & Set up an equation . \\ y - 4 = \frac{2}{x - 3} & Subtract 4 from both sides . \\ x - 3 = \frac{2}{y - 4} & Multiply both sides by x - 3 and divide by y - 4. \\ x = \frac{2}{y - 4} + 3 & Add 3 to both sides . \end{array}

So $f^{- 1} (y) = \frac{2}{y - 4} + 3$ or $f^{- 1} (x) = \frac{2}{x - 4} + 3.$

Analysis

The domain and range of $f$ exclude the values 3 and 4, respectively. $f$ and $f^{- 1}$ are equal at two points but are not the same function, as we can see by creating Table 5.

Table 5 The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.
$x$	1	2	5	$f^{- 1} (y)$
$f (x)$	3	2	5	$y$

Example 9

Solving to Find an Inverse with Radicals

Find the inverse of the function $f (x) = 2 + \sqrt{x - 4} .$

Solution

\begin{array}{l} y = 2 + \sqrt{x - 4} \\ {(y - 2)}^{2} = x - 4 \\ x = {(y - 2)}^{2} + 4 \end{array}

So $f^{- 1} (x) = {(x - 2)}^{2} + 4.$

The domain of $f$ is $[4, \infty) .$ Notice that the range of $f$ is $[2, \infty),$ so this means that the domain of the inverse function $f^{- 1}$ is also $[2, \infty) .$

Analysis

The formula we found for $f^{- 1} (x)$ looks like it would be valid for all real $x .$ However, $f^{- 1}$ itself must have an inverse (namely, $f$ ) so we have to restrict the domain of $f^{- 1}$ to $[2, \infty)$ in order to make $f^{- 1}$ a one-to-one function. This domain of $f^{- 1}$ is exactly the range of $f .$

Finding Inverse Functions and Their Graphs

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function $f (x) = x^{2}$ restricted to the domain $[0, \infty),$ on which this function is one-to-one, and graph it as in Figure 7.

Graph of f(x). — Figure 7 Quadratic function with domain restricted to [0, ∞).

Restricting the domain to $[0, \infty)$ makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.

We already know that the inverse of the toolkit quadratic function is the square root function, that is, $f^{- 1} (x) = \sqrt{x} .$ What happens if we graph both $f$ and $f^{- 1}$ on the same set of axes, using the $x -$ axis for the input to both $f and f^{- 1} ?$

We notice a distinct relationship: The graph of $f^{- 1} (x)$ is the graph of $f (x)$ reflected about the diagonal line $y = x,$ which we will call the identity line, shown in Figure 8.

Graph of f(x) and f^(-1)(x). — Figure 8 Square and square-root functions on the non-negative domain

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

Example 10

Finding the Inverse of a Function Using Reflection about the Identity Line

Given the graph of $f (x)$ in Figure 9, sketch a graph of $f^{- 1} (x) .$

A graph showing a logarithmic function. The curve starts near the bottom of the y-axis (as x approaches 0 from the right), passes through (1,0), and curves upwards to the right. — Figure 9

Solution

This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of $(0, \infty)$ and range of $(- \infty, \infty),$ so the inverse will have a domain of $(- \infty, \infty)$ and range of $(0, \infty) .$

If we reflect this graph over the line $y = x,$ the point $(1, 0)$ reflects to $(0, 1)$ and the point $(4, 2)$ reflects to $(2, 4) .$ Sketching the inverse on the same axes as the original graph gives Figure 10.

Key Concepts

If $g (x)$ is the inverse of $f (x),$ then $g (f (x)) = f (g (x)) = x .$ See Example 1, Example 2, and Example 3.
Each of the toolkit functions has an inverse. See Example 4.
For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5.
The inverse of a function can be determined at specific points on its graph. See Example 6.
To find the inverse of a formula, solve the equation $y = f (x)$ for $x$ as a function of $y .$ Then exchange the labels $x$ and $y .$ See Example 7, Example 8, and Example 9.
The graph of an inverse function is the reflection of the graph of the original function across the line $y = x .$ See Example 10.

Section Exercises

Verbal

Exercise 1

Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?

Solution

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that $y$ -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no $y$ -values repeat and the function is one-to-one.

Exercise 2

Why do we restrict the domain of the function $f (x) = x^{2}$ to find the function’s inverse?

Exercise 3

Can a function be its own inverse? Explain.

Solution

Yes. For example, $f (x) = \frac{1}{x}$ is its own inverse.

Exercise 4

Are one-to-one functions either always increasing or always decreasing? Why or why not?

Exercise 5

How do you find the inverse of a function algebraically?

Solution

Given a function $y = f (x),$ solve for $x$ in terms of $y .$ Interchange the $x$ and $y .$ Solve the new equation for $y .$ The expression for $y$ is the inverse, $y = f^{- 1} (x) .$

Algebraic

Exercise 6

Show that the function $f (x) = a - x$ is its own inverse for all real numbers $a .$

For the following exercises, find $f^{- 1} (x)$ for each function.

Exercise 7

$f (x) = x + 3$

Solution

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

Exercise 8

$f (x) = x + 5$

Exercise 9

$f (x) = 2 - x$

Solution

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

Exercise 10

$f (x) = 3 - x$

Exercise 11

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

Solution

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

Exercise 12

$f (x) = \frac{2 x + 3}{5 x + 4}$

For the following exercises, find a domain on which each function $f$ is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of $f$ restricted to that domain.

Exercise 13

$f (x) = {(x + 7)}^{2}$

Solution

domain of $f (x) : [- 7, \infty); f^{- 1} (x) = \sqrt{x} - 7$

Exercise 14

$f (x) = {(x - 6)}^{2}$

Exercise 15

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

Solution

domain of $f (x) : [0, \infty); f^{- 1} (x) = \sqrt{x + 5}$

Exercise 16

Given $f (x) = \frac{x}{2 + x}$ and $g (x) = \frac{2 x}{1 - x} :$

ⓐ Find $f (g (x))$ and $g (f (x)) .$
ⓑ What does the answer tell us about the relationship between $f (x)$ and $g (x) ?$

Solution

ⓐ $f (g (x)) = x$ and $g (f (x)) = x .$
ⓑ This tells us that $f$ and $g$ are inverse functions

For the following exercises, use function composition to verify that $f (x)$ and $g (x)$ are inverse functions.

Exercise 17

$f (x) = \sqrt[3]{x - 1}$ and $g (x) = x^{3} + 1$

Solution

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

Exercise 18

$f (x) = - 3 x + 5$ and $g (x) = \frac{x - 5}{- 3}$

Graphical

For the following exercises, use a graphing utility to determine whether each function is one-to-one.

Exercise 19

$f (x) = \sqrt{x}$

Solution

one-to-one

Exercise 20

$f (x) = \sqrt[3]{3 x + 1}$

Exercise 21

$f (x) = −5 x + 1$

Solution

one-to-one

Exercise 22

$f (x) = x^{3} - 27$

For the following exercises, determine whether the graph represents a one-to-one function.

Exercise 23

Solution

not one-to-one

Exercise 24

For the following exercises, use the graph of $f$ shown in Figure 11.

A graph displays a downward-sloping line 'f' on a Cartesian coordinate system. The line intersects the y-axis at (0, 3) and the x-axis at (2, 0). — Figure 11

Exercise 25

Find $f (0) .$

Solution

$3$

Exercise 26

Solve $f (x) = 0.$

Exercise 27

Find $f^{- 1} (0) .$

Solution

$2$

Exercise 28

Solve $f^{- 1} (x) = 0.$

For the following exercises, use the graph of the one-to-one function shown in Figure 12.

Graph of a square root function. — Figure 12

Exercise 29

Sketch the graph of $f^{- 1} .$

Solution

Graph of a square root function and its inverse.

Exercise 30

Find $f (6) and f^{- 1} (2) .$

Exercise 31

If the complete graph of $f$ is shown, find the domain of $f .$

Solution

$[2, 10]$

Exercise 32

If the complete graph of $f$ is shown, find the range of $f .$

Numeric

For the following exercises, evaluate or solve, assuming that the function $f$ is one-to-one.

Exercise 33

If $f (6) = 7,$ find $f^{- 1} (7) .$

Solution

$6$

Exercise 34

If $f (3) = 2,$ find $f^{- 1} (2) .$

Exercise 35

If $f^{- 1} (- 4) = - 8,$ find $f (- 8) .$

Solution

$- 4$

Exercise 36

If $f^{- 1} (- 2) = - 1,$ find $f (- 1) .$

For the following exercises, use the values listed in Table 6 to evaluate or solve.

Table 6 Two rows and ten columns. The first row is labeled, “x”, and the second is labeled, “f(x)”. The values of x are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. So for f(0)=8, f(1)=0, f(2)=7, f(3)=4, f(4)=2, f(5)=6, f(6)=5, f(7)=8, f(8)=9, and f(9)=1.
$x$	0	1	2	3	4	5	6	7	8	9
$f (x)$	8	0	7	4	2	6	5	3	9	1

Exercise 37

Find $f (1) .$

Solution

$0$

Exercise 38

Solve $f (x) = 3.$

Exercise 39

Find $f^{- 1} (0) .$

Solution

$1$

Exercise 40

Solve $f^{- 1} (x) = 7.$

Exercise 41

Use the tabular representation of $f$ in Table 7 to create a table for $f^{- 1} (x) .$

Table 7 Two rows and six columns. The first row is labeled, “x”, and the second is labeled, “f(x)”. The values of x are 3, 6, 9, 13, and 14. So for f(3)=1, f(6)=4, f(9)=7, f(13)=12, and f(14)=16.
$x$	3	6	9	13	14
$f (x)$	1	4	7	12	16

Solution

Two rows and six columns. The first row is labeled, “x”, and the second is labeled, “f^(-1)(x)”. The values of x are 1, 4, 7, 12, and 16. So for f^(-1) (1)=1, f^(-1) (4)=6, f^(-1) (7)=9, f^(-1) (12)=13, and f^(-1)f(16)=14.
$x$	1	4	7	12	16
$f^{- 1} (x)$	3	6	9	13	14

Technology

For the following exercises, find the inverse function. Then, graph the function and its inverse.

Exercise 42

$f (x) = \frac{3}{x - 2}$

Exercise 43

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

Solution

$f^{- 1} (x) = {(1 + x)}^{1 / 3}$

Graph of a cubic function and its inverse.

Exercise 44

Find the inverse function of $f (x) = \frac{1}{x - 1} .$ Use a graphing utility to find its domain and range. Write the domain and range in interval notation.

Real-World Applications

Exercise 45

To convert from $x$ degrees Celsius to $y$ degrees Fahrenheit, we use the formula $f (x) = \frac{9}{5} x + 32.$ Find the inverse function, if it exists, and explain its meaning.

Solution

$f^{- 1} (x) = \frac{5}{9} (x - 32) .$ Given the Fahrenheit temperature, $x,$ this formula allows you to calculate the Celsius temperature.

Exercise 46

The circumference $C$ of a circle is a function of its radius given by $C (r) = 2 π r .$ Express the radius of a circle as a function of its circumference. Call this function $r (C) .$ Find $r (36 π)$ and interpret its meaning.

Exercise 47

A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time, $t,$ in hours given by $d (t) = 50 t .$ Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function $t (d) .$ Find $t (180)$ and interpret its meaning.

Solution

$t (d) = \frac{d}{50},$ $t (180) = \frac{180}{50} .$ The time for the car to travel 180 miles is 3.6 hours.

Chapter Review Exercises

Functions and Function Notation

For the following exercises, determine whether the relation is a function.

${(a, b), (c, d), (e, d)}$

Solution

function

${(5, 2), (6, 1), (6, 2), (4, 8)}$

$y^{2} + 4 = x,$ for $x$ the independent variable and $y$ the dependent variable

Solution

not a function

Is the graph in Figure 13 a function?

For the following exercises, evaluate the function at the indicated values: $f (- 3); f (2); f (- a); - f (a); f (a + h) .$

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

Solution

$f (- 3) = - 27;$ $f (2) = - 2;$ $f (- a) = - 2 a^{2} - 3 a;$
$- f (a) = 2 a^{2} - 3 a;$ $f (a + h) = - 2 a^{2} + 3 a - 4 a h + 3 h - 2 h^{2}$

$f (x) = 2 | 3 x - 1 |$

For the following exercises, determine whether the functions are one-to-one.

$f (x) = - 3 x + 5$

Solution

one-to-one

$f (x) = | x - 3 |$

For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function.

Solution

function

Solution

function

For the following exercises, graph the functions.

$f (x) = | x + 1 |$

$f (x) = x^{2} - 2$

Solution

A graph of an upward-opening parabola with its vertex at (0, -2) and passing through approximately (-1.4, 0) and (1.4, 0). The curve extends upwards on both sides.

For the following exercises, use Figure 14 to approximate the values.

$f (2)$

$f (−2)$

Solution

$2$

If $f (x) = −2,$ then solve for $x .$

If $f (x) = 1,$ then solve for $x .$

Solution

$x = - 1.8$ or $or x = 1.8$

For the following exercises, use the function $h (t) = - 16 t^{2} + 80 t$ to find the values.

$\frac{h (2) - h (1)}{2 - 1}$

$\frac{h (a) - h (1)}{a - 1}$

Solution

$\frac{- 64 + 80 a - 16 a^{2}}{- 1 + a} = - 16 a + 64$

Domain and Range

For the following exercises, find the domain of each function, expressing answers using interval notation.

$f (x) = \frac{2}{3 x + 2}$

$f (x) = \frac{x - 3}{x^{2} - 4 x - 12}$

Solution

$(- \infty, - 2) \cup (- 2, 6) \cup (6, \infty)$

$f (x) = \frac{\sqrt{x - 6}}{\sqrt{x - 4}}$

Graph this piecewise function: $f (x) = {\begin{cases} x + 1 x < - 2 \\ - 2 x - 3 x \geq - 2 \end{cases}$

Solution

A piecewise linear graph featuring a jump discontinuity at x = -2. The function approaches -1 from the left with an open circle, and has a value of 1 from the right, continuing downwards.

Rates of Change and Behavior of Graphs

For the following exercises, find the average rate of change of the functions from $x = 1 to x = 2.$

$f (x) = 4 x - 3$

$f (x) = 10 x^{2} + x$

Solution

$31$

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

For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant.

Solution

increasing $(2, \infty);$ decreasing $(- \infty, 2)$

Solution

increasing $(- 3, 1);$ constant $(- \infty, - 3) \cup (1, \infty)$

Find the local minimum of the function graphed in Exercise.

Find the local extrema for the function graphed in Exercise.

Solution

local minimum $(- 2, - 3);$ local maximum $(1, 3)$

For the graph in Figure 15, the domain of the function is $[- 3, 3] .$ The range is $[- 10, 10] .$ Find the absolute minimum of the function on this interval.

Find the absolute maximum of the function graphed in Figure 15.

Solution

Absolute Maximum: 10

Composition of Functions

For the following exercises, find $(f \circ g) (x)$ and $(g \circ f) (x)$ for each pair of functions.

$f (x) = 4 - x, g (x) = - 4 x$

$f (x) = 3 x + 2, g (x) = 5 - 6 x$

Solution

$(f \circ g) (x) = 17 - 18 x; (g \circ f) (x) = - 7 - 18 x$

$f (x) = x^{2} + 2 x, g (x) = 5 x + 1$

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

Solution

$(f \circ g) (x) = \sqrt{\frac{1}{x} + 2};$ $(g \circ f) (x) = \frac{1}{\sqrt{x + 2}}$

$f (x) = \frac{x + 3}{2}, g (x) = \sqrt{1 - x}$

For the following exercises, find $(f \circ g)$ and the domain for $(f \circ g) (x)$ for each pair of functions.

$f (x) = \frac{x + 1}{x + 4}, g (x) = \frac{1}{x}$

Solution

$(f \circ g) (x) = \frac{1 + x}{1 + 4 x}, x \neq 0, x \neq - \frac{1}{4}$

$f (x) = \frac{1}{x + 3}, g (x) = \frac{1}{x - 9}$

$f (x) = \frac{1}{x}, g (x) = \sqrt{x}$

Solution

$(f \circ g) (x) = \frac{1}{\sqrt{x}}, x > 0$

$f (x) = \frac{1}{x^{2} - 1}, g (x) = \sqrt{x + 1}$

For the following exercises, express each function $H$ as a composition of two functions $f$ and $g$ where $H (x) = (f \circ g) (x) .$

$H (x) = \sqrt{\frac{2 x - 1}{3 x + 4}}$

Solution

sample: $g (x) = \frac{2 x - 1}{3 x + 4}; f (x) = \sqrt{x}$

$H (x) = \frac{1}{{(3 x^{2} - 4)}^{- 3}}$

Transformation of Functions

For the following exercises, sketch a graph of the given function.

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

Solution

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

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

Solution

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

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

Solution

$f (x) = 5 \sqrt{- x} - 4$

$f (x) = 4 [| x - 2 | - 6]$

Solution

A V-shaped graph resembling an absolute value function is shown on a Cartesian plane. Its vertex is at (2, -24), and it crosses the x-axis at (-4, 0) and (8, 0).

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

For the following exercises, sketch the graph of the function $g$ if the graph of the function $f$ is shown in Figure 16.

This graph shows a semi-circle centered at the origin with a radius of 2. The curve originates at (-2, 0), reaches its apex at (0, 2), and terminates at (2, 0) on the Cartesian plane. — Figure 16

$g (x) = f (x - 1)$

Solution

$g (x) = 3 f (x)$

For the following exercises, write the equation for the standard function represented by each of the graphs below.

Solution

$f (x) = | x - 3 |$

For the following exercises, determine whether each function below is even, odd, or neither.

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

Solution

even

$g (x) = \sqrt{x}$

$h (x) = \frac{1}{x} + 3 x$

Solution

odd

For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither.

Solution

even

Absolute Value Functions

For the following exercises, write an equation for the transformation of $f (x) = | x | .$

Solution

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

A blue V-shaped graph on a Cartesian coordinate plane, representing an absolute value function. The graph opens upwards, with its vertex located at the point (1.5, -3).

Solution

$f (x) = - 3 | x - 3 | + 3$

For the following exercises, graph the absolute value function.

$f (x) = | x - 5 |$

$f (x) = - | x - 3 |$

Solution

A graph displays a blue V-shaped function, symmetrical around x=3, with its vertex at (3, 0). The two linear segments extend downwards, passing through points (0, -3) and (6, -3).

$f (x) = | 2 x - 4 |$

For the following exercises, solve the absolute value equation.

$| x + 4 | = 18$

Solution

$x = - 22, x = 14$

$| \frac{1}{3} x + 5 | = | \frac{3}{4} x - 2 |$

For the following exercises, solve the inequality and express the solution using interval notation.

$| 3 x - 2 | < 7$

Solution

$(- \frac{5}{3}, 3)$

$| \frac{1}{3} x - 2 | \leq 7$

For the following exercises, find $f^{- 1} (x)$ for each function.

$f (x) = 9 + 10 x$

Solution

$f^{- 1} (x) = \frac{x - 9}{10}$

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

For the following exercise, find a domain on which the function $f$ is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of $f$ restricted to that domain.

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

Given $f (x) = x^{3} - 5$ and $g (x) = \sqrt[3]{x + 5} :$

Find $f (g (x))$ and $g (f (x)) .$
What does the answer tell us about the relationship between $f (x)$ and $g (x) ?$

For the following exercises, use a graphing utility to determine whether each function is one-to-one.

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

Solution

The function is one-to-one.

A graph of the reciprocal function y = 1/x, showing a curve in the first and third quadrants of the Cartesian coordinate system. The x-axis and y-axis are labeled with tick marks from -4 to 4. The curve has a vertical asymptote at x=0 (the y-axis) and a horizontal asymptote at y=0 (the x-axis). In the first quadrant, the curve starts high near the positive y-axis, passes through (1,1) and extends towards the positive x-axis. In the third quadrant, the curve starts low near the negative y-axis, passes through (-1,-1) and extends towards the negative x-axis.

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

Solution

The function is not one-to-one.

A graph shows a parabola opening downwards, with its vertex at the origin (0,0). The curve passes through points such as (-1, -1) and (1, -1), and extends downwards as x moves away from 0. The x-axis is labeled from -4 to 4, and the y-axis is labeled from -4 to 4.

If $f (5) = 2,$ find $f^{- 1} (2) .$

Solution

$5$

If $f (1) = 4,$ find $f^{- 1} (4) .$

Practice Test

For the following exercises, determine whether each of the following relations is a function.

$y = 2 x + 8$

Solution

The relation is a function.

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

For the following exercises, evaluate the function $f (x) = - 3 x^{2} + 2 x$ at the given input.

$f (−2)$

Solution

−16

$f (a)$

Show that the function $f (x) = - 2 {(x - 1)}^{2} + 3$ is not one-to-one.

Solution

The graph is a parabola and the graph fails the horizontal line test.

Write the domain of the function $f (x) = \sqrt{3 - x}$ in interval notation.

Given $f (x) = 2 x^{2} - 5 x,$ find $f (a + 1) - f (1) .$

Solution

$2 a^{2} - a$

Graph the function $f (x) = {\begin{matrix} x + 1 if & - 2 < x < 3 \\ - x if & x \geq 3 \end{matrix}$

Find the average rate of change of the function $f (x) = 3 - 2 x^{2} + x$ by finding $\frac{f (b) - f (a)}{b - a} .$

Solution

$- 2 (a + b) + 1$

For the following exercises, use the functions $f (x) = 3 - 2 x^{2} + x and g (x) = \sqrt{x}$ to find the composite functions.

$(g \circ f) (x)$

$(g \circ f) (1)$

Solution

$\sqrt{2}$

Express $H (x) = \sqrt[3]{5 x^{2} - 3 x}$ as a composition of two functions, $f$ and $g,$ where $(f \circ g) (x) = H (x) .$

For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function.

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

Solution

A graph displays a curve resembling a square root function, starting at approximately (-6, -1) and extending upward and to the right, passing through (-5, 0).

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

For the following exercises, determine whether the functions are even, odd, or neither.

$f (x) = - \frac{5}{x^{2}} + 9 x^{6}$

Solution

$even$

$f (x) = - \frac{5}{x^{3}} + 9 x^{5}$

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

Solution

$odd$

Graph the absolute value function $f (x) = - 2 | x - 1 | + 3.$

Solve $| 2 x - 3 | = 17.$

Solution

$x = - 7$ and $x = 10$

Solve $- | \frac{1}{3} x - 3 | \geq 17.$ Express the solution in interval notation.

For the following exercises, find the inverse of the function.

$f (x) = 3 x - 5$

Solution

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

$f (x) = \frac{4}{x + 7}$

For the following exercises, use the graph of $g$ shown in Figure 17.

On what intervals is the function increasing?

Solution

$(- \infty, - 1.1) and (1.1, \infty)$

On what intervals is the function decreasing?

Approximate the local minimum of the function. Express the answer as an ordered pair. The ordered pair should include both the $x$ value as well as the $g (x)$ value.

Solution

$(1.1, - 0.9)$

Approximate the local maximum of the function. Express the answer as an ordered pair. The ordered pair should include both the $x$ value as well as the $g (x)$ value.

For the following exercises, use the graph of the piecewise function shown in Figure 18.

Graph of absolute function and step function. — Figure 18

Find $f (2) .$

Solution

$f (2) = 2$

Find $f (−2) .$

Write an equation for the piecewise function.

Solution

$f (x) = {\begin{matrix} | x | if x \leq 2 \\ 3 if x > 2 \end{matrix}$

For the following exercises, use the values listed in Table 8.

Table 8 ..
$x$	0	1	2	3	4	5	6	7	8
$f (x)$	1	3	5	7	9	11	13	15	17

Find $f (6) .$

Solve the equation $f (x) = 5.$

Solution

$x = 2$

Is the graph increasing or decreasing on its domain?

Is the function represented by the graph one-to-one?

Solution

yes

Find $f^{- 1} (15) .$

Given $f (x) = - 2 x + 11,$ find $f^{- 1} (x) .$

Solution

$f^{- 1} (x) = - \frac{x - 11}{2}$