Lesson 3.7: Inverse Functions

New Concept

An inverse function, denoted $f^{-1}(x)$, reverses the action of a function by swapping its inputs and outputs. For an inverse to exist, the original function must be one-to-one. We will explore how to find, verify, and graph these special functions.

What’s next

You'll start by verifying inverse functions in practice problems, then find inverses from formulas, tables, and graphs using our interactive examples and video guides.

Property

For any one-to-one function $f(x) = y$, a function $f^{-1}(x)$ is an inverse function of $f$ if $f^{-1}(y) = x$. This can also be written as $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$. It also follows that $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$ if $f^{-1}$ is the inverse of $f$. Keep in mind that $f^{-1}(x) \neq \frac{1}{f(x)}$ and not all functions have inverses.

Examples

• If a one-to-one function has $f(3)=7$ and $f(8)=15$, the inverse function has $f^{-1}(7)=3$ and $f^{-1}(15)=8$.

• Given that $g^{-1}(10) = 4$, the corresponding input and output for the original function $g$ is $g(4) = 10$.

Property

Given two functions $f(x)$ and $g(x)$, test whether the functions are inverses of each other.

1. Determine whether $f(g(x)) = x$ or $g(f(x)) = x$.
2. If either statement is true, then both are true, and $g = f^{-1}$ and $f = g^{-1}$. If either statement is false, then both are false, and $g \neq f^{-1}$ and $f \neq g^{-1}$.

Examples

• Are $f(x) = 2x+1$ and $g(x) = \frac{x-1}{2}$ inverses? Let's check: $f(g(x)) = 2(\frac{x-1}{2}) + 1 = (x-1) + 1 = x$. Yes, they are inverses.

• Are $f(x) = x^2+5$ and $g(x)=\sqrt{x-5}$ inverses on the correct domain? Let's check: $f(g(x))=(\sqrt{x-5})^2+5 = (x-5)+5=x$. Yes, for $x \ge 5$.

Property

The range of a function $f(x)$ is the domain of the inverse function $f^{-1}(x)$.

The domain of $f(x)$ is the range of $f^{-1}(x)$.

Given a function, find the domain and range of its inverse.

1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.

Property

Given a function represented by a formula, find the inverse.

1. Make sure $f$ is a one-to-one function.
2. Solve for $x$.
3. Interchange $x$ and $y$.

Examples

• To find the inverse of $f(x) = 4x-7$, write $y = 4x-7$. Solve for $x$: $y+7 = 4x$, so $x = \frac{y+7}{4}$. Swap variables to get $f^{-1}(x) = \frac{x+7}{4}$.

• To find the inverse of $f(x) = \frac{x}{x+3}$, write $y = \frac{x}{x+3}$. Then $y(x+3)=x$, so $x=\frac{3y}{1-y}$. Swap variables to get $f^{-1}(x) = \frac{3x}{1-x}$.

Property

The graph of $f^{-1}(x)$ is the graph of $f(x)$ reflected about the diagonal line $y = x$, which we will call the identity line. This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

Examples

• If the point $(2, 5)$ is on the graph of a one-to-one function $f(x)$, then the point $(5, 2)$ must be on the graph of its inverse, $f^{-1}(x)$.

• To find $f^{-1}(3)$ from the graph of $f(x)$, find the value 3 on the $y$-axis, then find the corresponding $x$-value on the graph. If the point $(1, 3)$ is on the graph, then $f^{-1}(3) = 1$.