# Inverse Functions

### Relating a Function with Its Inverse Two functions $f$ and $g$ are inverse functions if $f(g(x)) = g(f(x)) = x$. To determine the rule for an inverse function, write as $y=f(x)$, reverse $x$ and $y$, and then solve for $y$. The domain of $f$ is the range of $g$, and vice versa.

If the result of switching the inputs and outputs of a function is also a function, it is called the inverse function. The composition of a function and its inverse is the identity function.

The inverse of the function $f(x)$ is written as $f^{-1}(x)$. This is read "$f$ inverse of $x$." This notation does not mean the same thing as:
$(f(x))^{-1}=\frac{1}{f(x)}$
The inverse of a function is defined only if the function is one-to-one.
$(f \circ f^{-1})(x)=(f^{-1} \circ f)(x)=x$ If the output of a function is used as the input of its inverse function, then the result is the original input. The inverse function can be said to "undo" the operations performed by the function.
To identify the inverse of a function that is given as a rule, mapping, or graph, switch the inputs and outputs, or swap $x$ and $y$.
Step-By-Step Example
Identifying the Inverse of a Linear Function
Determine the inverse of:
$f(x)=8x+5$
Step 1
Write $f(x)$ as $y$.
$y=8x+5$
Step 2
Switch the positions of $x$ and $y$.
$x=8y+5$
Step 3
Solve for $y$.
\begin{aligned}x&=8y+5\\x-5&=8y+5-5\\x-5&=8y\\\frac{x-5}{8}&=y\end{aligned}
Step 4
Replace $y$ with $f^{-1}(x)$.
$f^{-1}(x)=\frac{x-5}{8}$
Solution
The inverse of the function is:
$f^{-1}(x)=\frac{x-5}{8}$
Check that the composition of the function and its inverse is the identity by using the function:
$f^{-1}(f(x))=x$
Calculate:
\begin{aligned}f^{-1}(f(x))&=f^{-1}(8x+5)\\&=\frac{(8x+5)-5}{8}\\&=\frac{8x}{8}\\&=x\end{aligned}

Step-By-Step Example
Identifying the Inverse of a Radical Function
Determine the inverse of the radical function:
$f(x)=\sqrt {4x+6}$
Step 1
Write $f(x)$ as $y$.
$y=\sqrt {4x+6}$
Step 2
Switch the positions of $x$ and $y$.
$x=\sqrt {4y+6}$
Step 3
Solve for $y$.
\begin{aligned}x&=\sqrt{4y+6}\\(x)^2&=(\sqrt{4y+6})^2\\x^2&=4y+6\\x^2-6&=4y+6-6\\x^2-6&=4y\\\frac{x^2-6}{4}&=y\end{aligned}
Step 4
Replace $y$ with $f^{-1}(x)$.
$f^{-1}(x)=\frac{x^2-6}{4}$
Solution
The inverse of the radical function is:
$f^{-1}(x)=\frac{x^2-6}{4}$
Check that composition of the function and its inverse is the identity by checking that:
$f^{-1}(f(x))=x$
Calculate:
\begin{aligned} f^{-1}(f(x))&=f^{-1}(\sqrt{4x+6})\\&=\frac{(\sqrt{4x+6})^2-6}{4}\\&=\frac{4x+6-6}{4}\\&=\frac{4x}{4}\\&=x\end{aligned}

### Graphs of Inverse Functions The graphs of inverse functions are reflections of each other across the line $y = x$.
The inverse of a function given as a set of ordered pairs is obtained by switching the $x$- and $y$-coordinates.
$f(x) = \left \{ (-4,3),(2,6),(8,-4),(4,7),(-8,1)\right \}$
$f^{-1}(x) = \left \{(3,-4),(6,2),(-4,8),(7,4),(1,-8) \right \}$
When the $x$- and $y$-coordinates of a function are switched, the graph is reflected (flipped) across the line $y=x$. So the graph of an inverse function is the reflection (mirror image) of the function across the line $y=x$.
Step-By-Step Example
Graphing the Inverse of a Function
Identify the inverse of the given function:
$f(x)=6-2x$
Then graph both the given function and its inverse.
Step 1
Replace $f(x)$ with $y$.
$y=6-2x$
Step 2
Switch the positions of $x$ and $y$.
$x=6-2y$
Step 3
Solve for $y$.
\begin{aligned}x&=6-2y\\x+2y-x&=6-2y+2y-x\\2y&=6-x\\y&=3-\frac{x}{2}\end{aligned}
Step 4
Replace $y$ with $f^{-1}(x)$.
$f^{-1}(x)=3-\frac{x}{2}$
Solution
The inverse is:
$f^{-1}(x)=3-\frac{x}{2}$
Graph both functions. The graph of $f^{-1}(x)$ is the reflection of the graph of $f(x)$ across the line $y=x$.

### Restricting the Domain If the inverse of a function is not a function, the domain of the function can be restricted to values that produce a function for the inverse.

The inverse of a function is defined only if the function is one-to-one. One way to test whether a function is one-to-one is the horizontal line test. If the function does not pass the horizontal line test, then the inverse of the function is not a function.

Analyzing the graph of a function can help determine restrictions on the domain that will make the function one-to-one.

### Horizontal and Vertical Line Test

Descriptions Examples
The function $f(x)=x^2$ is a function, but it does not pass the horizontal line test, so it is not one-to-one. This means the inverse of the function $f(x)=x^2$ is not a function.
The graph of the inverse of $f(x)=x^2$ is a reflection of $f(x)$ over the line $y=x$.

To identify the equation of the inverse, replace $f(x)$ with $y$, and then switch the positions of $x$ and $y$.
$y=x^2\ \rightarrow\ x=y^2$
The graph of the inverse does not pass the vertical line test, so it is not a function.
If the domain of the function $f(x)=x^2$ is restricted to $x \geq 0$, the function passes the horizontal line test, so it is one-to-one.
When the domain of $f(x)=x^2$ is restricted to $x \geq 0$, the graph of the inverse passes the vertical line test, so it is a function.

To identify the rule for the function, solve the inverse $x=y^2$ for $y$.
\begin{aligned}x&=y^2\\\pm\sqrt{x}&=\sqrt{y^2}\\\pm\sqrt{x}&=y\end{aligned}
Since the domain of the original function was restricted to positive values of $x$, the inverse function is only the positive square root, $\sqrt{x}=y$. Replace $y$ with $f^{-1}(x)$.
$f^{-1}(x)=\sqrt{x}$

A vertical line test can help determine whether or not a graph is a graph of a function. A horizontal line test can help identify whether or not the graph of a function is one-to-one. Restricting the domain of a graph of a function ensures that the inverse will be unique.

Step-By-Step Example
Restricting the Domain of a Function
Identify a restriction on the domain of the function so that it has an inverse:
$f(x)=\left ( x-3 \right )^{2}+2$
Then determine that inverse.
Step 1
Graph the function.
Step 2

Restrict the domain so that the new function is one-to-one.

Limit the $x$-values used so that each $y$-value appears only once.
So $f(x)$ is one-to-one for $x \geq3$.
Step 3
Replace $f(x)$ with $y$.
$y=\left ( x-3 \right )^{2}+2$
Step 4
Switch the positions of $x$ and $y$.
$x=\left ( y-3 \right )^{2}+2$
Step 5
Solve for $y$.
\begin{aligned}x&=\left ( y-3 \right)^{2}+2\\x-2&=\left ( y-3 \right )^{2}+2-2 \\x-2&=\left ( y-3 \right)^{2}\\\sqrt{x-2}&=\sqrt{\left ( y-3 \right)^{2}}\\\sqrt{x-2}&=y-3\\\sqrt{x-2}+3&=y-3+3\\\sqrt{x-2}+3&=y\end{aligned}
Solution

Replace $y$ with $f^{-1}(x)$, and include the domain.

For $x \geq3$, the inverse is:
$f^{-1}(x)=\sqrt{x-2}+3$