The inverse of a function undoes what the function does: it maps each output back to the original input. Inverse functions exist precisely when a function is one-to-one, and their graphs are reflections of each other across the line $y = x$.
Quick Reference
| Concept | Formula / Rule |
|---|---|
| Definition | $x = f^{-1}(y) \Leftrightarrow y = f(x)$ |
| Domain/range swap | $\operatorname{Dom}(f) = \operatorname{Rng}(f^{-1})$; $\operatorname{Rng}(f) = \operatorname{Dom}(f^{-1})$ |
| Cancellation equations | $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ |
| When an inverse exists | If and only if $f$ is one-to-one |
| Graph reflection | Graph of $f^{-1}$ is the reflection of graph of $f$ across $y = x$ |
| Warning | $f^{-1}(y) \neq \frac{1}{f(y)}$ |
The Idea of Inversion
Consider $f(x) = x^3$. Given $x = 2$, we get $f(2) = 8$. Now suppose we are told that $f(x) = 27$ and asked to find $x$. We solve $x^3 = 27$, giving $x = \sqrt[3]{27} = 3$.
In general, for any $y$ in the range of $f$, the value of $x$ satisfying $y = f(x)$ is given by $x = \sqrt[3]{y}$. This defines $x$ as a function of $y$, which we call $g(y) = \sqrt[3]{y}$. The function $g$ is the inverse of $f$, since it undoes the effect of $f$:
$g(f(x)) = \sqrt[3]{x^3} = x \qquad \text{and} \qquad f(g(y)) = \left(\sqrt[3]{y}\right)^3 = y.$
Definition
Suppose $f$ is a one-to-one function with domain $A$ and range $B$. The inverse function $f^{-1}$ is defined by
$x = f^{-1}(y) \quad \text{means} \quad y = f(x)$for every $y$ in $B$.
The domain and range of $f$ and $f^{-1}$ simply swap:
$\operatorname{Dom}(f) = \operatorname{Rng}(f^{-1}), \qquad \operatorname{Rng}(f) = \operatorname{Dom}(f^{-1}).$Important notation warning: The $-1$ in $f^{-1}$ denotes an inverse function, not an exponent. In particular, $f^{-1}(y) \neq \dfrac{1}{f(y)}$.
Given that $f$ has an inverse and $f(1) = 3$, $f(2) = -4$, and $f(5) = -1$, find $f^{-1}(3)$, $f^{-1}(-4)$, and $f^{-1}(-1)$.
Solution
By the definition of inverse function: $f^{-1}(3) = 1 \quad \text{because } f(1) = 3,$ $f^{-1}(-4) = 2 \quad \text{because } f(2) = -4,$ $f^{-1}(-1) = 5 \quad \text{because } f(5) = -1.$
Verifying an Inverse Function
Let $f$ be a one-to-one function with domain $A$ and range $B$, and let $g$ be a function with domain $B$ and range $A$. Then $g = f^{-1}$ if and only if both of the following hold:
$g(f(x)) = x \quad \text{for every } x \in A,$$f(g(y)) = y \quad \text{for every } y \in B.$Proof
Suppose $g = f^{-1}$. By definition, $x = g(y) \Leftrightarrow y = f(x)$. Substituting $f(x)$ for $y$ gives $x = g(f(x))$. Substituting $g(y)$ for $x$ gives $y = f(g(y))$. Both conditions follow. Conversely, suppose (1) and (2) hold. If $x = g(y)$, then by (2), $f(g(y)) = y$, i.e., $f(x) = y$. If $y = f(x)$, then by (1), $g(f(x)) = x$, i.e., $g(y) = x$. So $x = g(y) \Leftrightarrow y = f(x)$, which is the definition of $g = f^{-1}$.Writing the independent variable of $f^{-1}$ as $x$ (rather than $y$), the cancellation equations become:
$f^{-1}(f(x)) = x \quad \text{for every } x \text{ in the domain of } f,$$f(f^{-1}(x)) = x \quad \text{for every } x \text{ in the domain of } f^{-1}.$A function has an inverse function if and only if it is one-to-one. Equivalently, every increasing or decreasing (monotonic) function has an inverse function.
Verify that $f(x) = 5x^3 + 2$ and $g(x) = \sqrt[3]{\dfrac{x-2}{5}}$ are inverses of each other.
Solution
Both functions have domain $\mathbb{R}$. We verify both cancellation equations: $f(g(x)) = f\!\left(\sqrt[3]{\frac{x-2}{5}}\right) = 5\left(\sqrt[3]{\frac{x-2}{5}}\right)^3 + 2 = 5 \cdot \frac{x-2}{5} + 2 = x. \checkmark$ $g(f(x)) = g(5x^3 + 2) = \sqrt[3]{\frac{(5x^3 + 2) - 2}{5}} = \sqrt[3]{\frac{5x^3}{5}} = \sqrt[3]{x^3} = x. \checkmark$Determine if each function has an inverse. (a) $f(x) = x^3 + 1$. (b) $h(x) = x^3 - x$.
Solution
(a) We show $f$ is one-to-one. If $f(x_1) = f(x_2)$: $x_1^3 + 1 = x_2^3 + 1 \implies x_1^3 = x_2^3 \implies x_1 = x_2.$ So $f$ is one-to-one and has an inverse. The graph of $f$ is an upward shift of $y = x^3$ and passes the horizontal line test.
How to Find the Inverse Function
Steps to find $f^{-1}$:
- Write down the equation $y = f(x)$.
- Solve for $x$ in terms of $y$: $x = g(y)$.
- The formula $f^{-1}(y) = g(y)$ defines the inverse, with $y$ restricted to the range of $f$.
- If you prefer, replace every $y$ with $x$ to write $f^{-1}(x) = g(x)$.
Given $f(x) = 2x^5 + 3$, find its inverse function.
Solution
Set $y = f(x)$ and solve for $x$: $2x^5 + 3 = y \implies 2x^5 = y - 3 \implies x^5 = \frac{y-3}{2} \implies x = \left(\frac{y-3}{2}\right)^{1/5}.$ Therefore $f^{-1}(x) = \left(\dfrac{x-3}{2}\right)^{1/5}$.Given $g(x) = 2x - 1$, find $g^{-1}$.
Solution
Set $y = 2x - 1$ and solve: $x = \dfrac{y+1}{2}$. Replacing $y$ with $x$: $g^{-1}(x) = \frac{x+1}{2}.$ Verification: $g(g^{-1}(x)) = 2 \cdot \dfrac{x+1}{2} - 1 = x$. ✓Given $h(x) = \sqrt{2x+3}$, find $h^{-1}$.
Solution
Set $y = \sqrt{2x+3}$ and solve: $y^2 = 2x+3$, so $x = \dfrac{y^2-3}{2}$. The range of $h$ is $[0, \infty)$, so: $h^{-1}(x) = \frac{x^2 - 3}{2}, \qquad x \geq 0.$Given $u(x) = x^2 + 1$ with domain $x \leq 0$, find $u^{-1}$.
Solution
Set $y = x^2 + 1$ and solve: $x = \pm\sqrt{y-1}$. Since $x \leq 0$, we choose $x = -\sqrt{y-1}$. The range is $[1, \infty)$: $u^{-1}(x) = -\sqrt{x-1}, \qquad x \geq 1.$
Graph of the Inverse Function
If $(a, b)$ is a point on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$. The point $(b, a)$ is the reflection of $(a, b)$ across the line $y = x$.
The graphs of a function and its inverse are reflections of one another across the line $y = x$.
Given the graph of $f(x)$ below, sketch the graph of $f^{-1}(x)$.
Solution
Reflect across $y = x$ by swapping coordinates: $(2,4) \to (4,2)$, $(3,2) \to (2,3)$, $(4,1) \to (1,4)$, $(5, 0.5) \to (0.5, 5)$.
