Inverse functions
A function takes an input and gives an output. An inverse function goes backward: it takes the output and gives back the input. For example, the function "multiply by ${3}$" is undone by "divide by ${3}$". This page shows when a function has an inverse, and how to find it.
One-to-one functions
To go backward, each output must come from only one input. Look at ${g(x)=x^2}$. Both ${g(2)}$ and ${g(-2)}$ are ${4}$. So from the output ${4}$, we cannot tell if the input was ${2}$ or ${-2}$.
A function ${f}$ is one-to-one if different inputs always give different outputs: if ${x_1\ne x_2}$, then ${f(x_1)\ne f(x_2)}$.
On a graph, two inputs with the same output are two points at the same height. A horizontal line through them meets the graph twice. This gives a quick test:
Horizontal line test: A function is one-to-one if and only if no horizontal line meets its graph more than once.
Example 1: Is each function one-to-one?
(a) ${f(x)=2x+3}$
(b) ${g(x)=x^2}$
(c) ${h(x)=x^3}$
Solution:
(a) Yes. Suppose ${f(x_1)=f(x_2)}$. Then ${2x_1+3=2x_2+3}$, so ${x_1=x_2}$. Different inputs cannot give the same output. (The graph is a rising line, so it passes the horizontal line test.)
(b) No. ${g(2)=4}$ and ${g(-2)=4}$.
(c) Yes. The graph of ${y=x^3}$ always rises, so every horizontal line meets it exactly once.
The inverse of a function
Let ${f}$ be one-to-one, with domain ${A}$ and range ${B}$. Its inverse function ${f^{-1}}$ has domain ${B}$ and range ${A}$, and
${f^{-1}(y)=x\quad}$ exactly when ${\quad f(x)=y}$.
So ${f^{-1}}$ sends each output of ${f}$ back to its input. The domain and the range trade places.
Watch out: The ${-1}$ in ${f^{-1}}$ is not an exponent. ${f^{-1}(x)}$ does not mean ${\dfrac{1}{f(x)}}$.
If we apply ${f}$ and then ${f^{-1}}$, we get back where we started. The same is true in the other order:
${f^{-1}\big(f(x)\big)=x}$ for every ${x}$ in the domain of ${f}$
${f\big(f^{-1}(x)\big)=x}$ for every ${x}$ in the domain of ${f^{-1}}$
These two equations are called the inverse function property. We can use them to check that two functions are inverses.
Example 2: Show that ${f(x)=2x+3}$ and ${g(x)=\dfrac{x-3}{2}}$ are inverses of each other.
Solution:
Compose them in both orders (see combining functions):
$\begin{align*}&g\big(f(x)\big)\\&=\dfrac{(2x+3)-3}{2}\\&=\dfrac{2x}{2}=x\end{align*}$
$\begin{align*}&f\big(g(x)\big)\\&=2\cdot\dfrac{x-3}{2}+3\\&=x-3+3=x\end{align*}$
Both give ${x}$, so ${g=f^{-1}}$. In words: ${f}$ doubles and then adds ${3}$. The inverse undoes these steps in reverse order: it subtracts ${3}$ and then halves.
Finding the inverse of a function
To find the inverse of a one-to-one function ${f}$:
- Write ${y=f(x)}$.
- Solve this equation for ${x}$ in terms of ${y}$.
- Swap ${x}$ and ${y}$. The result is ${y=f^{-1}(x)}$.
We swap in step 3 because we usually call the input of any function ${x}$.
Example 3: Find the inverse of ${f(x)=3x-5}$.
Solution:
Write ${y=3x-5}$, and solve for ${x}$:
$\begin{align*}y+5&=3x\\x&=\dfrac{y+5}{3}\end{align*}$
Swap ${x}$ and ${y}$:
${f^{-1}(x)=\dfrac{x+5}{3}}$
Check: ${f(2)=1}$, and ${f^{-1}(1)=\dfrac{1+5}{3}=2}$.
Example 4: Find the inverse of ${f(x)=\dfrac{2x+1}{x-3}}$.
Solution:
Write ${y=\dfrac{2x+1}{x-3}}$. Multiply both sides by ${x-3}$:
${y(x-3)=2x+1}$
Collect the terms with ${x}$ on the left. Then factor out ${x}$:
$\begin{align*}xy-3y&=2x+1\\xy-2x&=3y+1\\x(y-2)&=3y+1\\x&=\dfrac{3y+1}{y-2}\end{align*}$
Swap ${x}$ and ${y}$:
${f^{-1}(x)=\dfrac{3x+1}{x-2}}$
The domain of ${f}$ is ${x\ne 3}$. The domain of ${f^{-1}}$ is ${x\ne 2}$, so the range of ${f}$ is all numbers except ${2}$. (This matches the horizontal asymptote ${y=2}$ of ${f}$.)
Check: ${f(0)=\dfrac{1}{-3}=-\dfrac{1}{3}}$. Then
$\begin{align*}&f^{-1}\left(-\dfrac{1}{3}\right)\\&=\dfrac{-1+1}{-\frac{1}{3}-2}=0\end{align*}$
The graph of an inverse function
If ${(a,b)}$ is a point on the graph of ${f}$, then ${f(a)=b}$. So ${f^{-1}(b)=a}$, and ${(b,a)}$ is on the graph of ${f^{-1}}$. Swapping the coordinates reflects a point across the line ${y=x}$. So:
The graph of ${f^{-1}}$ is the reflection of the graph of ${f}$ across the line ${y=x}$.
A function that is not one-to-one has no inverse. But we can often restrict its domain: use only a part of the graph that passes the horizontal line test.
Example 5: The function ${f(x)=x^2+1}$ is not one-to-one. Restrict it to ${x\ge 0}$. Then find ${f^{-1}}$, and sketch both graphs.
Solution:
For ${x\ge 0}$, the graph is the right half of a parabola. It always rises, so it is one-to-one. Its domain is ${[0,\infty)}$ and its range is ${[1,\infty)}$.
Write ${y=x^2+1}$, and solve for ${x}$:
$\begin{align*}x^2&=y-1\\x&=\sqrt{y-1}\end{align*}$
We take the positive root, because ${x\ge 0}$. Swap ${x}$ and ${y}$:
${f^{-1}(x)=\sqrt{x-1}}$, with domain ${[1,\infty)}$.
For example, ${(2,5)}$ is on the graph of ${f}$, so ${(5,2)}$ is on the graph of ${f^{-1}}$.
An application
Example 6: The formula ${F=\dfrac{9}{5}C+32}$ changes a temperature from degrees Celsius to degrees Fahrenheit. Find the inverse formula, and change ${68}$°F to Celsius.
Solution:
Solve for ${C}$. Subtract ${32}$, then multiply by ${\dfrac{5}{9}}$:
$\begin{align*}F-32&=\dfrac{9}{5}C\\C&=\dfrac{5}{9}(F-32)\end{align*}$
Here we do not swap the letters, because ${C}$ and ${F}$ have a meaning. The inverse formula changes Fahrenheit back to Celsius. For ${68}$°F:
$C=\dfrac{5}{9}(68-32)=\dfrac{5}{9}\cdot 36=20$
So ${68}$°F is ${20}$°C.
Summary
- A function is one-to-one if different inputs give different outputs. Test: no horizontal line meets the graph twice.
- Only one-to-one functions have inverses. ${f^{-1}(y)=x}$ exactly when ${f(x)=y}$.
- ${f^{-1}(f(x))=x}$ and ${f(f^{-1}(x))=x}$. The domain and range trade places.
- To find ${f^{-1}}$: write ${y=f(x)}$, solve for ${x}$, then swap ${x}$ and ${y}$.
- The graph of ${f^{-1}}$ is the graph of ${f}$ reflected across ${y=x}$.