# 4.1 Graphs of inverse functions

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## Graphs of inverse functions

In earlier grades, you studied various types of functions and understood the effect of various parameters in the general equation. In this section, we will consider inverse functions .

An inverse function is a function which "does the reverse" of a given function. More formally, if $f$ is a function with domain $X$ , then ${f}^{-1}$ is its inverse function if and only if for every $x\in X$ we have:

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

A simple way to think about this is that a function, say $y=f\left(x\right)$ , gives you a $y$ -value if you substitute an $x$ -value into $f\left(x\right)$ . The inverse function tells you tells you which $x$ -value was used to get a particular $y$ -value when you substitue the $y$ -value into ${f}^{-1}\left(x\right)$ . There are some things which can complicate this for example, think about a $sin$ function, there are many $x$ -values that give you a peak as the function oscillates. This means that the inverse of a $sin$ function would be tricky to define because if you substitute the peak $y$ -value into it you won't know which of the $x$ -values was used to get the peak.

$\begin{array}{ccc}\hfill y& =& f\left(x\right)\phantom{\rule{1.em}{0ex}}\mathrm{we have a function}\hfill \\ \hfill {y}_{1}& =& f\left({x}_{1}\right)\hfill \\ & & \mathrm{we substitute a specific x-value into the function to get a specific y-value}\hfill \\ & & \mathrm{consider the inverse function}\hfill \\ \hfill x& =& {f}^{-1}\left(y\right)\hfill \\ \hfill x& =& {f}^{-1}\left(y\right)\hfill \\ & & \mathrm{substituting the specific y-value into the inverse should return the specific x-value}\\ & =& {f}^{-1}\left({y}_{1}\right)\hfill \\ & =& {x}_{1}\hfill \end{array}$

This works both ways, if we don't have any complications like in the case of the $sin$ function, so we can write:

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

For example, if the function $x\to 3x+2$ is given, then its inverse function is $x\to \frac{\left(x-2\right)}{3}$ . This is usually written as:

$\begin{array}{ccc}\hfill f& :& x\to 3x+2\hfill \\ \hfill {f}^{-1}& :& x\to \frac{\left(x-2\right)}{3}\hfill \end{array}$

The superscript "-1" is not an exponent.

If a function $f$ has an inverse then $f$ is said to be invertible.

If $f$ is a real-valued function, then for $f$ to have a valid inverse, it must pass the horizontal line test , that is a horizontal line $y=k$ placed anywhere on the graph of $f$ must pass through $f$ exactly once for all real $k$ .

It is possible to work around this condition, by defining a “multi-valued“ function as an inverse.

If one represents the function $f$ graphically in a $xy$ -coordinate system, the inverse function of the equation of a straight line, ${f}^{-1}$ , is the reflection of the graph of $f$ across the line $y=x$ .

Algebraically, one computes the inverse function of $f$ by solving the equation

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

for $x$ , and then exchanging $y$ and $x$ to get

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

## Inverse function of $y=ax+q$

The inverse function of $y=ax+q$ is determined by solving for $x$ as:

$\begin{array}{ccc}\hfill y& =& ax+q\hfill \\ \hfill ax& =& y-q\hfill \\ \hfill x& =& \frac{y-q}{a}\hfill \\ & =& \frac{1}{a}y-\frac{q}{a}\hfill \end{array}$

Therefore the inverse of $y=ax+q$ is $y=\frac{1}{a}x-\frac{q}{a}$ .

The inverse function of a straight line is also a straight line, except for the case where the straight line is a perfectly horizontal line, in which case the inverse is undefined.

For example, the straight line equation given by $y=2x-3$ has as inverse the function, $y=\frac{1}{2}x+\frac{3}{2}$ . The graphs of these functions are shown in [link] . It can be seen that the two graphs are reflections of each other across the line $y=x$ .

## Domain and range

We have seen that the domain of a function of the form $y=ax+q$ is $\left\{x:x\in \mathbb{R}\right\}$ and the range is $\left\{y:y\in \mathbb{R}\right\}$ . Since the inverse function of a straight line is also a straight line, the inverse function will have the same domain and range as the original function.

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