2.16 Inverse Functions

Read an introductory example

Consider a function $y = f (x)$ . If a value of $x$ is given, we just need to substitute it in the formula of $f$ to obtain the corresponding value of $y$ . For example if $f (x) = x^{3}$ , and $x = 2$ is given, the corresponding value of $y$ is 8. Now suppose we want to know at which $x$ , the value of $f$ is $27$ . To answer this question, we need to solve $x^{3} = 27$ for $x$ . If we take the cube root of both sides, we have $x = \sqrt[3]{27} = 3$ . In general, for every $y$ , there exists a number $x$ for which $y = f (x)$ , and that $x$ is given by
$x = \sqrt[3]{y} .$ This equation defines $x$ as a function of $y$ . If we denote this function by $g$ , then
$g (y) = \sqrt[3]{y} .$ The function $g$ is called the inverse of $f$ because it undoes the effect of $f$ :
$g (f (x)) = g (x^{3}) = \sqrt[3]{x^{3}} = x and f (g (y)) = f (\sqrt[3]{y}) = (\sqrt[3]{y})^{3} = y .$

Table of Contents

The Inverse of a Function

Consider a general function $f$ with domain $A$ and range $B$ . For every $y$ in $B$ , there is at least one number $x$ in $A$ such that $y = f (x)$ . If $f$ is one-to-one, there is exactly one $x$ in $A$ such that $y = f (x)$ . Because $x$ is unique, we can define a function $g$ from $B$ to $A$ as follows:
$x = g (y) \Leftrightarrow y = f (x) .$ As in Figure 1, the function $g$ reverses the correspondence given by $f$ and is called the inverse of $f$ and the process of obtaining $g$ from $f$ is called inversion.

The process of inversion can be applied to any one-to-one function. If $f$ is one-to-one, we often denote its inverse function by $f^{- 1}$ . The symbol $f^{- 1}$ is read “ $f$ inverse.”

Note that the “ $- 1$ ” in the symbol $f^{- 1}$ denotes an inverse and not an exponent. In other words $f^{- 1} (y)$ is NOT the same as $1 / [f (y)]$ . The reciprocal $1 / [f (y)]$ is denoted by $[f (y)]^{- 1}$ .
$f^{- 1} (y) \neq \frac{1}{f (y)}$

Figure 1:The function and its inverse undo the effects of each other.

Definition 1: Suppose

f

is a one-to-one function on a domain

A

with range

B

. The inverse function

f^{- 1}

is
defined as follows:

x = f^{- 1} (y) means y = f (x)

for every

y

B

Domains and Ranges of Inverse Functions

From the above definition, it is clear that the range and domain of $f$ and $f^{- 1}$ simply switch!

$D o m (f) = R n g (f^{- 1})$ and
$R n g (f) = D o m (f^{- 1})$

Example 1

Given the function

f

has an inverse and

f (1) = 3

f (2) = - 4

, and

f (5) = - 1

, find

f^{- 1} (3), f^{- 1} (- 4)

, and

f^{- 1} (- 1)

Solution

From Definition 1 we have

\begin{aligned} f^{- 1} (3) & = 1 because f (1) = 3 \\ f^{- 1} (- 4) & = 2 because f (2) = - 4 \\ f^{- 1} (- 1) & = 5 because f (5) = - 1 \end{aligned}

As we can see in the following arrow diagram, the effect of

f^{- 1}

simply undoes the effect of

f


(a)	(b)

Figure 2

The following theorem can be used to verify that a function $g$ is the inverse of $f$ .

Theorem 1: Let $f$ be a one-to-one function with domain $A$ and range $B$ , and $g$ be a function with domain $B$ and range $A$ . The function $g$ is the inverse function of $f$ (i.e. $g = f^{- 1}$ ) if and only if the following conditions hold:

(1) $g (f (x)) = x$ for every $x$ in $A = D o m (f)$

and

(2) $f (g (y)) = y$ for every $y$ in $B = D o m (g)$ .

Click to Read the Proof

First, suppose $g = f^{- 1}$ ; we are going to show that (1) and (2) are true. By the definition of an inverse function, we have
$x = g (y) \Leftrightarrow y = f (x)$ for every $x$ in $A$ and every $y$ in $B$ . If we substitute $f (x)$ for $y$ in $x = g (y)$ we obtain
$x = g (y) = g (f (x)) for every x in A .$ Similarly, if we substitute $g (y)$ for $x$ in $y = f (x)$ , we obtain
$y = f (x) = f (g (y)) for every y in B .$ Thereby we proved that if $g = f^{- 1}$ both (1) and (2) are true.

Conversely, suppose $g$ is a function for which both (1) and (2) are true, then we have to show $x = g (y)$ implies $y = f (x)$ and $y = f (x)$ implies $x = g (y)$ .

If $x = g (y)$ , because (2) is true, $f (\underset{= x}{\underset{⏟}{g (y)}}) = y$ ;
that is $f (x) = y$ . This shows that
$x = g (y) ⟹ y = f (x)$ Next suppose $y = f (x)$ . Because (1) is true, $g (\underset{= y}{\underset{⏟}{f (x)}}) = x$ ; that is $g (y) = x$ . This shows that
$y = f (x) ⟹ x = g (y) .$ Thereby we proved that if (1) and (2) hold, then $x = g (y) \Leftrightarrow y = f (x)$ ,and the proof is complete.

Inverse of Inverse

The above theorem indicates that if $g$ is the inverse of $f$ then the inverse of $g$ is $f$ . In other words the inverse of inverse of $f$ is $f$ , or $f$ and $f^{- 1}$ are inverses of each other.

$if g = f^{- 1} then g^{- 1} = f$

Changing the Independent Variable

Note that $y$ in $f^{- 1} (y)$ simply represents an arbitrary number in the domain of the function $f^{- 1}$ . There is nothing special about $y$ . Therefore, $y$ can be replaced by any letter that you like such as $u, z$ , or even $x$ ! So, instead of $f^{- 1} (y)$ , we may write $f^{- 1} (x)$ . In this case the two conditions of the above theorem can be written as

$f^{- 1} (f (x)) = x$ for every $x$ in the domain of $f$

and

$f (f^{- 1} (x)) = x$ for every $x$ in the domain of $f^{- 1}$ .

Example 2

Verify that

f (x) = 5 x^{3} + 2 and g (x) = \sqrt[3]{\frac{x - 2}{5}}

are inverses of each other.

Solution

The domain of $f$ and the domain of $g$ are both the entire set of real numbers $R$ . Thus to prove that they are inverse of each other, we have to show that for all $x$ , we have $f (g (x)) = x$ and $g (f (x)) = x$ :
$\begin{aligned} f (g (x)) & = f (\sqrt[3]{\frac{x - 2}{5}}) \\ = 5 {(\sqrt[3]{\frac{x - 2}{5}})}^{3} + 2 \\ = 5 \frac{x - 2}{5} + 2 \\ = x \end{aligned}$
and
$\begin{aligned} g (f (x)) & = g (5 x^{3} + 2) \\ = {(\sqrt[3]{\frac{(5 x^{3} + 2) - 2}{5}})}^{3} \\ = {(\sqrt[3]{\frac{5 x^{3}}{5}})}^{3} \\ = {(\sqrt[3]{x})}^{3} \\ = x . \end{aligned}$

When Does a Function Have an Inverse?

As explained at the beginning of this section, if a function $f$ is one-to-one, then it has an inverse function; otherwise, it does not. That is,

A function has an inverse function if an only if it is one-to-one

In Section 2.15, we learned that every monotonic (= increasing or decreasing) function is one-to-one. Thus we can say:

Every increasing or decreasing function has an inverse function.

Example 3

Determine if each of the following functions has an inverse function.
(a)

f (x) = x^{3} + 1

(b)

h (x) = x^{3} - x

Solution

(a) We know how

y = x^{3}

looks like. So we can easily sketch the graph of

f (x) = x^{3} + 1

by shifting the graph of

y = x^{3}

upward one unit. From Figure 3(a), it is clear that

f

is a one-to-one function, so it has an inverse function. Alternatively, we can algebraically show that

f

is one-to-one. To do so, we have to show that

f (x_{1}) = f (x_{2})

implies

x_{1} = x_{2}

\begin{aligned} f (x_{1}) & = f (x_{2}) \\ x_{1}^{3} + 1 & = x_{2}^{3} + 1 \\ x_{1}^{3} & = x_{2}^{3} \\ \sqrt[3]{x_{1}^{3}} & = \sqrt[3]{x_{2}^{3}} & (take cube roots of both sides) \\ x_{1} & = x_{2} \end{aligned}

(b) We note that $h (x) = x^{3} - x = x (x^{2} - 1)$ , and thus $h$ has three real roots $- 1, 0,$ and 1; that is,
$h (- 1) = h (1) = h (0) = 0.$ Because the value of $h$ at three different values of $x$ is the same, $h$ is not a one-to-one function We may confirm the fact that $h$ is not one-to-one by sketching its graph using a computer or a graphing calculator. As it is clear from the Figure 3(b),the horizontal test is not passed, and hence $h$ is not one-to-one.Therefore, $h$ does not have an inverse function.


(a) Graph of $y = x^{3} + 1$ . The horizontal line test is passed.	(b) Graph of $y = x^{3} - x$ . The horizontal line test is not passed.

Figure 3

How to Find the Inverse Function

To find the inverse of a one-to-one function $y = f (x)$ , we have to solve the equation $y = f (x)$ for $x$ . Solving this equation for $x$ may not be easy if not impossible, but the fact that $f$ is one-to-one assures that there is a unique solution for $x$ in terms of $y$ provided that $y$ lies in the range of $f$ . The process of inversion is more explained through the following examples.

Example 4

Given

f (x) = 2 x^{5} + 3

, find its inverse function.

Solution

Let

y = f (x)

and then solve for

x

in terms of

y

to find

x = f^{- 1} (y)

\begin{aligned} 2 x^{5} + 3 & = y & (let y = f (x)) \\ 2 x^{5} & = y - 3 \\ x^{5} & = \frac{y - 3}{2} \\ x & = {(\frac{y - 3}{2})}^{1 / 5} & (bring both sides to the power 1 / 5) \end{aligned}

Therefore,

f^{- 1} (y) = {(\frac{y - 3}{2})}^{1 / 5}

. Note that

y

must lie within the range of

f

. Because

f

is a polynomial of an odd degree, its range is

(- \infty, \infty)

. Therefore,

y

can be any number, and the domain of

f^{- 1}

(- \infty, \infty)

R

Recall that $y$ in $f^{- 1} (y)$ represents an arbitrary number in the domain of $f^{- 1}$ . That is, $f^{- 1} (y) = {(\frac{y - 3}{2})}^{1 / 5}$ simply means that $f^{- 1}$ transforms the input $y$ to the output ${(\frac{y - 3}{2})}^{1 / 5}$ . So if you wish to denote the input variable by $x$ instead of $y$ then just replace every “ $y$ ” in $f^{- 1} (y)$ with an “ $x$ .” That is,
$f^{- 1} (x) = {(\frac{x - 3}{2})}^{1 / 5} .$

The steps of finding the inverse of a function can be summarized as

Write down the equation $y = f (x)$ .
Solve the equation $y = f (x)$ for $x$ and express $x$ as a function of $y$ : $x = g (y)$
The equation $x = g (y)$ provides a formula for the inverse function $f^{- 1}$ ; that is, $f^{- 1} (y) = g (y)$ with $y$ as the independent variable.
If $y$ is acceptable as the independent variable for $f^{- 1}$ , you are finished. However, if you wish to denote the independent variable, as usual, by $x$ , simply replace every $y$ in the formula $f^{- 1} (y)$ by $x$ to obtain a formula for $f^{- 1} (x)$ . Now as usual, we can show the output of $f^{- 1} (x)$ by $y$ and write $y = f^{- 1} (x)$ .

Example 5

Given

g (x) = 2 x - 1

, find the inverse of

g

Solution

First let

y = g (x)

and solve for

x

in terms of

y

to find

x = g^{- 1} (y)

\begin{aligned} 2 x - 1 & = y \\ x & = \frac{y + 1}{2} . \end{aligned}

Therefore,

x = g^{- 1} (y) = (y + 1) / 2.

Again we note that

y

must lie within the range of

g

. The graph of

g

is a line which shows that its range is

(- \infty, \infty)

and hence

y

can be any real number.

We can show the input of $g^{- 1}$ by “ $x$ ” instead of “ $y$ ”. That is,
$g^{- 1} (x) = \frac{x + 1}{2} .$ We can easily verify that $g (g^{- 1} (x)) = x$ and $g^{- 1} (g (x)) = x$ .

Example 6

Given

h (x) = \sqrt{2 x + 3}

, find the inverse of

h

Solution

Let

y = h (x)

and solve for

x

in terms of

y

to find

x = h^{- 1} (y)

\begin{aligned} \sqrt{2 x + 3} & = y \\ 2 x + 3 & = y^{2} \\ x & = \frac{y^{2} - 3}{2} \end{aligned}

Therefore,

h^{- 1} (y) = (y^{2} - 3) / 2

. We note that

y

must lie within the range of

h

. We know that the output of the square root is always nonnegative, and because

h (- 3 / 2) = 0

, the range of

h

[0, \infty)

.
Therefore

h^{- 1} (y) = \frac{1}{2} (y^{2} - 3), y \geq 0.

If we wish, we can replace “ $y$ ” in $h^{- 1} (y)$ with “ $x$ ” and write:
$h^{- 1} (x) = \frac{1}{2} (x^{2} - 3), x \geq 0.$

Example 7

Given

u (x) = x^{2} + 1

(

x \leq 0

), find

u^{- 1}

Solution

Let

y = u (x)

, solve this equation for

x

in terms of

y

to find

u^{- 1}

\begin{aligned} x^{2} + 1 & = y \\ x^{2} & = y - 1 \end{aligned}

Now there are two solutions for

x

x = \pm \sqrt{y - 1}

. Because the domain of

u

is restricted to

x \leq 0

, we choose

x = - \sqrt{y - 1} .

Therefore,

u^{- 1} (y) = - \sqrt{y - 1}

. As we can see the range of

u

[1, \infty)

[simply plot the graph of

u

to see that], and the domain of

u^{- 1}

[1, \infty)

. Again we can denote the independent variable of

u^{- 1}

by “

x

” instead of “

y

”. Thus

u^{- 1} (x) = - \sqrt{x - 1} .

Note that if we do not restrict the domain of

u (x) = x^{2} + 1

to negative numbers (or positive numbers), its graph does not pass the horizontal test, which shows, in this case, the function would not be one-to-one and hence would not have an inverse function.

Graph of the Inverse Function

Let’s graph a number of functions we have seen in the examples of this section and their inverses.


(a) $f (x) = 2 x^{5} + 3$ and $f^{- 1} (x) = {(\frac{x - 3}{2})}^{1 / 5}$	(b) $g (x) = 2 x - 1$ and $g^{- 1} (x) = \frac{x + 1}{2}$	(c) $h (x) = \sqrt{2 x + 3}$ and $h^{- 1} (x) = \frac{1}{2} (x^{2} - 3), x \geq 0$

Figure 4

In these figures, the graphs of each function and its inverse appear to be mirror images of each other with respect to the bisector of the first and third quadrants $y = x$ . Those are not coincidences. If $(a, b)$ is a point on the graph of $f$ , then $b = f (a)$ . It follows from $b = f (a)$ that $a = f^{- 1} (b)$ , which means $(b, a)$ is on the graph of $f^{- 1}$ . A similar argument will verify that if $(b, a)$ is on the graph of $f^{- 1}$ , then $(a, b)$ will be on the graph of $f$ . We get the point $(b, a)$ from $(a, b)$ by reflecting through the line $y = x$ .

The graphs of a function and its inverse are the reflections of one another through the line

y = x

The line $y = x$ is the bisector of the first and third quadrants.
Here we have plotted the graph of $y = f^{- 1} (x)$ ; that is, if the independent variable is laid off along the horizontal axis (= $x$ -axis) and the dependent variable along the $y$ -axis. But if we plot $x = f^{- 1} (y)$ ;
that is, if the independent variable is marked off along the vertical axis (= $y$ -axis) and the dependent along the $x$ -axis, then the graph of the inverse function $x = f^{- 1} (y)$ coincides with the graph of $y = f (x)$ .

Example 8

Given

f (x)

is graphed in Figure 5, sketch the graph of

f^{- 1} (x)

Figure 5

Solution

The graph of

f^{- 1}

will be obtained by reflecting the graph of

f

across the line

y = x

. The reflection of the point

(a, b)

across the line

y = x

is the point

(b, a)

. Because

(2, 4), (3, 2), (4, 1)

and

(5, 0.5)

are on the graph of

f

, the points

(4, 2), (2, 3), (1, 4)

, and

(0.5, 5)

are on graph of

f^{- 1}

. The graph of

f^{- 1}

is depicted in Figure 6.

Figure 6 : The graph of

f

(orange) and the graph of its inverse

f^{- 1}

are reflections of one another across

y = x

Read an introductory example

Hide the introductory example

The Inverse of a Function

Domains and Ranges of Inverse Functions

Click to Read the Proof

Hide the Proof

Inverse of Inverse

Changing the Independent Variable

When Does a Function Have an Inverse?

How to Find the Inverse Function

Graph of the Inverse Function