Inverse Functions

Key Fact	Statement
Domain/Range swap	$\mathrm{Dom}(f)=\mathrm{Rng}(f^{-1})$ and $\mathrm{Rng}(f)=\mathrm{Dom}(f^{-1})$
Cancellation	$f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$
Graph	Reflect the graph of $f$ in the line $y=x$

Definition

Consider a function $f$ with domain $A$ and range $B$. For every $y$ in $B$, there is at least one $x$ in $A$ such that $y=f(x)$. If $f$ is one-to-one, there is exactly one such $x$. This uniqueness allows us to define a new function $g$ from $B$ to $A$:

$$x=g(y)\text{means}y=f(x).$$

This function $g$ is called the inverse of $f$ and is denoted $f^{-1}$.

The function and its inverse undo the effects of each other. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

• The process of obtaining $f^{-1}$ from $f$ is called inversion.
• The "$-1$" in $f^{-1}$ is not an exponent. $f^{-1}(x)$ is the inverse function, not the reciprocal $1/f(x)$. The reciprocal is written $[f(x){]}^{-1}$: $$f^{-1}(x)\ne [f(x){]}^{-1}=\frac{1}{f(x)}.$$
• Every one-to-one function has an inverse.

Domain, Range, and Cancellation

The domain and range of $f$ and $f^{-1}$ simply swap:

Notice also that $(f^{-1}{)}^{-1}=f$.

Graphs of Inverse Functions

Suppose $f$ has an inverse. If $(a,b)$ is a point on the graph of $f$, then $b=f(a)$, which means $f^{-1}(b)=a$, so $(b,a)$ is on the graph of $f^{-1}$.

We get $(b,a)$ from $(a,b)$ by reflecting through the line $y=x$. Therefore:

The graphs of a function and its inverse are the mirror images of each other with respect to the line $y=x$.

For example, compare the graphs of $h(x)=\sqrt{2x+3}$ and its inverse $h^{-1}(x)=\frac{1}{2}(x^2-3)$ (defined for $x\ge 0$). You can verify the inverse relationship: $h(h^{-1}(x))=x$ and $h^{-1}(h(x))=x$.

Frequently Asked Questions

How do I find the formula for $f^{-1}$ from the formula for $f$?

Write $y=f(x)$, then solve for $x$ in terms of $y$ to get $x=f^{-1}(y)$. If you prefer the input variable to be called $x$, swap $x$ and $y$ at the end. For example, if $f(x)=2x+3$: set $y=2x+3$, solve to get $x=\frac{y-3}{2}$, so $f^{-1}(x)=\frac{x-3}{2}$.

What happens to the derivative when you invert a function?

The inverse function theorem states: if $f$ is differentiable and one-to-one near $a$, and f'(a) \ne 0, then $f^{-1}$ is differentiable at $b=f(a)$ and (f^{-1})'(b) = \frac{1}{f'(a)}. Geometrically, reflecting the graph in $y=x$ turns a slope of $m$ into a slope of $1/m$.

Is ${\sin }^{-1}(x)$ the same as $1/\sin (x)$?

No. ${\sin }^{-1}(x)$ (also written $\arcsin (x)$) is the inverse function of the restricted sine function. It gives you the angle whose sine is $x$, for example ${\sin }^{-1}(1/2)=\pi /6$. The reciprocal of $\sin (x)$ is $\csc (x)=1/\sin (x)$. This notational collision is regrettable but unavoidable: always read "$-1$" as "inverse" when it is a superscript on a function name.