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.

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

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 $y$.

Graphs of Inverse Functions

The graphs of inverse functions are reflections of each other 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.

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.