#### Read an introductory example

#### Hide the 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\qquad\text{and}\qquad f(g(y))=f(\sqrt[3]{y})=(\sqrt[3]{y})^{3}=y.

\]

## 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)\qquad\Leftrightarrow\qquad 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)}

\]

**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)\qquad\text{means}\qquad y=f(x)

\] for every $y$ in $B$.

### 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!

\[\bbox[#F2F2F2,5px,border:2px solid black]{Dom(f) =Rng(f^{-1})}\]
and

\[\bbox[#F2F2F2,5px,border:2px solid black]{Rng(f) =Dom(f^{-1})}\]

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* $A = Dom(f)$

and

(2) $f(g(y))=y$ *for every* $y$ *in* $B=Dom(g)$.

#### Click to Read the Proof

#### Hide 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)\quad\Leftrightarrow\quad 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))\qquad\text{for every }x\text{ in }A.

\]
Similarly, if we substitute $g(y)$ for $x$ in $y=f(x)$, we obtain

\[

y=f(x)=f(g(y))\qquad\text{for every }y\text{ 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, $f\big(\underbrace{g(y)}_{=x}\big)=y$;

that is $f(x)=y$. This shows that

\[

x=g(y)\implies y=f(x)

\]
Next suppose $y=f(x)$. Because (1) is true, $g\big(\underbrace{f(x)}_{=y}\big)=x$; that is $g(y)=x$. This shows that

\[

y=f(x)\implies 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.

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\large \text{if } \ g=f^{-1}\ \text{ then }\ g^{-1}=f}\]

### 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 $f^{-1}$.

### 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,

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.

## 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.

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

## 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)=2x^{5}+3 $ and$ f^{-1}(x)=\left(\dfrac{x-3}{2}\right)^{1/5} $ | (b) $g(x)=2x-1$ and $ g^{-1}(x)=\frac{x+1}{2} $ | (c) $h(x)=\sqrt{2x+3} $ and $ h^{-1}(x)=\frac{1}{2}(x^{2}-3),x\geq0$ |

**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 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 $y=f(x)$.