Inverse Functions: Undoing What a Function Does

Reversing the question

A function turns an input into an output. Its inverse turns that output back into the input it came from.

$$f(x) = x + 7 \qquad f^{-1}(x) = x - 7$$

Adding $7$ is undone by subtracting $7$. That is the whole idea, and the machinery below is for cases where the undoing takes more than one step.

Written formally, applying one and then the other gets you back where you started:

$$f^{-1}(f(x)) = x \qquad f(f^{-1}(x)) = x$$

The $-1$ is notation for reversing, not an exponent. $f^{-1}(x)$ is not $\tfrac{1}{f(x)}$, and this is one of the more expensive pieces of notation in algebra to misread.

Why swapping x and y finds it

A function’s equation says how to get $y$ from $x$. The inverse asks the reverse question: how do you get $x$ from $y$?

So swap the letters and solve again.

Find the inverse of $f(x) = 3x - 5$.

1. Write it with $y$.

$$y = 3x - 5$$

2. Swap $x$ and $y$.

$$x = 3y - 5$$

3. Solve for $y$.

$$x + 5 = 3y \quad\Rightarrow\quad y = \frac{x + 5}{3}$$ $$f^{-1}(x) = \frac{x + 5}{3}$$

Step 3 is solving a literal equation — the same skill, applied to a letter you are less used to isolating.

Checking by composing

Substituting one function into the other has to return the input untouched.

$$f\left(f^{-1}(x)\right) = 3\left(\frac{x+5}{3}\right) - 5 = (x + 5) - 5 = x \quad\checkmark$$

This check is complete. If the composition simplifies to $x$, the inverse is right; if it does not, something went wrong. A single number works too: $f(4) = 7$ and $f^{-1}(7) = 4$.

Why the graph reflects across y = x

Swapping the letters swaps the coordinates. Every point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$.

Reflecting across the line $y = x$ is exactly the move that swaps a point’s two coordinates, so the two graphs are mirror images in that line.

• f(x) = 3x - 5
• f inverse
• y = x

Two consequences follow directly:

On $f$	On $f^{-1}$
the domain	the range
the range	the domain
the point $(a, b)$	the point $(b, a)$
$y$-intercept $(0, c)$	$x$-intercept $(c, 0)$

The domain and range trade places, because the inputs of one are the outputs of the other.

When the inverse is not a function

Reversing always produces a relation. It produces a function only if no output of $f$ was ever produced twice.

Take $f(x) = x^2$. Both $3$ and $-3$ give $9$. Reversing asks “which input gave $9$?” and there are two answers, so the reverse fails the one-output rule.

The test is visual:

Test	Checks
vertical line test	is the graph a function?
horizontal line test	does it have an inverse function?

A horizontal line crossing the graph twice means two inputs shared an output.

$y = x^2$ fails it, which is why $\sqrt{x}$ is defined as the positive root: restricting the domain to $x \ge 0$ throws away the duplicate and leaves something invertible.

Restricting the domain is the standard repair, and it is why calculators report one square root rather than two.

Worked examples

Common mistakes

Practice problems