Inverse Functions — AP® Calculus AB Notes

A function takes an input and produces an output. Very often, though, the question runs the other way: the thermometer reads 70 °F, so what time is it? The sine of an angle is 0.8, so what is the angle? Running a function backward is the job of an inverse function. Unit 3 asks how fast an inverse changes, and the answer comes straight out of one picture on this page.

What an inverse does

If $f$ sends $a$ to $b$ , its inverse $f^{-1}$ sends $b$ back to $a$ : $f^{-1}(b) = a \quad \text{exactly when} \quad f(a) = b$ Doing one and then the other gets you back where you started: $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ .

From a table, read an inverse backward: find the output in the output row and read the input above it. If $f(3) = 9$ , then $f^{-1}(9) = 3$ .

One warning, repeated because it costs so many points: $f^{-1}(x)$ is not $\frac{1}{f(x)}$ . The $-1$ here means "undo," not "reciprocal." If you mean the reciprocal, write $\frac{1}{f(x)}$ or $[f(x)]^{-1}$ .

When an inverse exists

An inverse must send each output back to exactly one input. So $f$ needs to be one-to-one: no two inputs share an output. On a graph, that is the horizontal line test: every horizontal line meets the graph at most once. Functions that only increase, or only decrease, are one-to-one.

$f(x) = x^2$ fails, since $f(2) = f(-2) = 4$ . The fix is to restrict the domain. On $x \ge 0$ , $x^2$ is one-to-one and its inverse is $\sqrt{x}$ . That is why $\sqrt{9}$ is 3 and not $-3$ .

Finding a formula

To find $f^{-1}$ , write the equation $x = f(y)$ (inputs and outputs swapped) and solve for $y$ .

Worked example 1. Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$ .

Swap: $x = \frac{2y + 1}{y - 3}$ . Clear the fraction: $x(y - 3) = 2y + 1$ , so $xy - 3x = 2y + 1$ .

Collect the $y$ terms: $xy - 2y = 3x + 1$ , so $y(x - 2) = 3x + 1$ , and $f^{-1}(x)$ $= \frac{3x + 1}{x - 2}$ .

Check with a number: $f(0) = -\frac{1}{3}$ , and $f^{-1}\left(-\frac{1}{3}\right)$ $= \frac{-1 + 1}{-\frac{1}{3} - 2}$ $= 0$ . It undoes $f$ .

The graph of an inverse

Every point $(a, b)$ on the graph of $f$ becomes the point $(b, a)$ on the graph of $f^{-1}$ . Swapping coordinates is a reflection across the line $y = x$ .

The curve y = f(x), with f(x) = 0.5x cubed + 0.5x + 2, passes through (1, 3); its inverse, the mirror image across the dashed line y = x, passes through (3, 1). Short lines show slope 2 on f at (1, 3) and slope 1/2 on the inverse at (3, 1). — **Figure 1.** Reflection across $y = x$ : points swap coordinates and slopes become reciprocals. LevlPrep original diagram.

Look at the slopes in the figure. Reflecting swaps the roles of rise and run, so a line of slope 2 through $(1, 3)$ reflects to a line of slope $\frac{1}{2}$ through $(3, 1)$ . In general, at matching points the slopes of $f$ and $f^{-1}$ are reciprocals. You can see it in a table too: if $f$ rises 3 while $x$ moves 1, then $f^{-1}$ rises 1 while its input moves 3.

Worked example 2. $f(2) = 7$ , and the graph of $f$ has slope 4 at $(2, 7)$ . Describe $f^{-1}$ there.

The graph of $f^{-1}$ passes through $(7, 2)$ , so $f^{-1}(7) = 2$ .

Its slope at $(7, 2)$ is $\frac{1}{4}$ . Topic 3.3 turns exactly this into a formula.

Inverse trig functions

Sine, cosine and tangent repeat forever, so none of them is one-to-one. Each gets an inverse by restricting to one stretch where it is one-to-one and covers all its outputs:

Inverse	Input	Output (an angle)	Example
$\arcsin x$	$-1 \le x \le 1$	$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$	$\arcsin\frac{1}{2} = \frac{\pi}{6}$
$\arccos x$	$-1 \le x \le 1$	$0 \le \theta \le \pi$	$\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$
$\arctan x$	every real number	$-\frac{\pi}{2} < \theta < \frac{\pi}{2}$	$\arctan(-1) = -\frac{\pi}{4}$

They are also written $\sin^{-1} x$ , $\cos^{-1} x$ and $\tan^{-1} x$ , with the same warning: $\sin^{-1} x$ is not $\frac{1}{\sin x}$ .

Worked example 3. Find $\arcsin\left(\sin\frac{5\pi}{6}\right)$ .

It is tempting to say the functions cancel and answer $\frac{5\pi}{6}$ . But arcsine only returns angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ .

Work inside out: $\sin\frac{5\pi}{6} = \frac{1}{2}$ , and $\arcsin\frac{1}{2} = \frac{\pi}{6}$ .