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:$(f(x))^{1}=\frac{1}{f(x)}$
$(f \circ f^{1})(x)=(f^{1} \circ f)(x)=x$
StepByStep 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 $y$.
$\begin{aligned}x&=8y+5\\x5&=8y+55\\x5&=8y\\\frac{x5}{8}&=y\end{aligned}$
Step 4
Replace $y$ with $f^{1}(x)$.
$f^{1}(x)=\frac{x5}{8}$
Solution
The inverse of the function is:
Check that the composition of the function and its inverse is the identity by using the function:
Calculate:
$f^{1}(x)=\frac{x5}{8}$
$f^{1}(f(x))=x$
$\begin{aligned}f^{1}(f(x))&=f^{1}(8x+5)\\&=\frac{(8x+5)5}{8}\\&=\frac{8x}{8}\\&=x\end{aligned}$
StepByStep 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 $y$.
$\begin{aligned}x&=\sqrt{4y+6}\\(x)^2&=(\sqrt{4y+6})^2\\x^2&=4y+6\\x^26&=4y+66\\x^26&=4y\\\frac{x^26}{4}&=y\end{aligned}$
Step 4
Replace $y$ with $f^{1}(x)$.
$f^{1}(x)=\frac{x^26}{4}$
Solution
The inverse of the radical function is:
Check that composition of the function and its inverse is the identity by checking that:
Calculate:
$f^{1}(x)=\frac{x^26}{4}$
$f^{1}(f(x))=x$
$\begin{aligned} f^{1}(f(x))&=f^{1}(\sqrt{4x+6})\\&=\frac{(\sqrt{4x+6})^26}{4}\\&=\frac{4x+66}{4}\\&=\frac{4x}{4}\\&=x\end{aligned}$
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.
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 $y=x$.
$f(x) = \left \{ (4,3),(2,6),(8,4),(4,7),(8,1)\right \}$
$f^{1}(x) = \left \{(3,4),(6,2),(4,8),(7,4),(1,8) \right \}$
StepByStep Example
Graphing the Inverse of a Function
Identify the inverse of the given function:
Then graph both the given function and its inverse.
$f(x)=62x$
Step 1
Replace $f(x)$ with $y$.
$y=62x$
Step 2
Switch the positions of $x$ and $y$.
$x=62y$
Step 3
Solve for $y$.
$\begin{aligned}x&=62y\\x+2yx&=62y+2yx\\2y&=6x\\y&=3\frac{x}{2}\end{aligned}$
Step 4
Replace $y$ with $f^{1}(x)$.
$f^{1}(x)=3\frac{x}{2}$
Solution
The inverse is:
Graph both functions. The graph of $f^{1}(x)$ is the reflection of the graph of $f(x)$ across the line $y=x$.
$f^{1}(x)=3\frac{x}{2}$
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 onetoone. One way to test whether a function is onetoone 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 onetoone.
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 onetoone. 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 $y$. $y=x^2\ \rightarrow\ x=y^2$ 

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 onetoone.  
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 $y$. $\begin{aligned}x&=y^2\\\pm\sqrt{x}&=\sqrt{y^2}\\\pm\sqrt{x}&=y\end{aligned}$ $f^{1}(x)=\sqrt{x}$ 
StepByStep Example
Restricting the Domain of a Function
Identify a restriction on the domain of the function so that it has an inverse:
Then determine that inverse.
$f(x)=\left ( x3 \right )^{2}+2$
Step 1
Graph the function.
Step 2
Restrict the domain so that the new function is onetoone.
Limit the $x$values used so that each $y$value appears only once.So $f(x)$ is onetoone for $x \geq3$.
Step 3
Replace $f(x)$ with $y$.
$y=\left ( x3 \right )^{2}+2$
Step 4
Switch the positions of $x$ and $y$.
$x=\left ( y3 \right )^{2}+2$
Step 5
Solve for $y$.
$\begin{aligned}x&=\left ( y3 \right)^{2}+2\\x2&=\left ( y3 \right )^{2}+22 \\x2&=\left ( y3 \right)^{2}\\\sqrt{x2}&=\sqrt{\left ( y3 \right)^{2}}\\\sqrt{x2}&=y3\\\sqrt{x2}+3&=y3+3\\\sqrt{x2}+3&=y\end{aligned}$
Solution
Replace $y$ with $f^{1}(x)$, and include the domain.
For $x \geq3$, the inverse is:$f^{1}(x)=\sqrt{x2}+3$