Recipe

Find the inverse of a function

Use when: you have a formula for a one-to-one function f (or one already restricted to a piece where it's one-to-one) and need a formula for f⁻¹

Goal: a verified formula for f⁻¹(x), confirmed to undo f in both directions, with its domain stated whenever it isn't all real numbers

Taught in Composition and inverse functions

Quick recipe

Write the function as y = f(x)
Swap x and y
Solve the new equation for y
Verify by composing both directions
State the domain restriction, if the original function needed one

1
Replace the function notation $f(x)$ with the single letter $y$ , so the rule becomes an equation in two variables, $x$ and $y$ .

Example

For $f(x) = 4x + 5$ , write $y = 4x + 5$ .

Why: the next step swaps two variables in an equation, but $f(x)$ names an output, not a variable you can trade places with $x$ . Writing the rule as $y = \ldots$ first gives you two symmetric letters to actually swap.
2
Exchange every $x$ for a $y$ and every $y$ for an $x$ in the equation.

Example

$y = 4x + 5$ becomes $x = 4y + 5$ .

Why: a point $(a, b)$ on $f$ 's graph becomes the point $(b, a)$ on $f^{-1}$ 's graph, the reflection across $y = x$ from the definition above. Swapping the letters is that same coordinate swap done as algebra instead of geometry.
3
Treat $x$ as a known number and solve the equation for $y$ : isolate $y$ on one side using ordinary equation-solving steps.

Example

$x = 4y + 5$ . Subtract 5, then divide by 4: $y = \dfrac{x-5}{4}$ .
- $y$ appears once, in a simple expression — isolate it directly, the way the example below does.
- $y$ appears more than once, such as in a fraction with $y$ in the denominator — clear the denominator first. Then collect every $y$ -term on one side and factor $y$ out before dividing (see the rational-function worked examples following this recipe).
Why: $x = 4y + 5$ describes the same relationship as $y = 4x + 5$ with the two letters' roles swapped. Solving for $y$ turns that swapped relationship back into a usable formula; rename $y$ to $f^{-1}(x)$ and the algebra is done.
4
Check the result by confirming both $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ . One direction checking out doesn't guarantee the other does.

Example

$f^{-1}(f(x)) = \dfrac{(4x+5)-5}{4} = \dfrac{4x}{4} = x$ ✓, and $f(f^{-1}(x)) = 4 \cdot \dfrac{x-5}{4} + 5 = (x-5)+5 = x$ ✓. Both round trips return $x$ , so $f^{-1}(x) = \dfrac{x-5}{4}$ is confirmed.

Why: $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ are two separate equalities in the definition of an inverse. An algebra slip while solving for $y$ can produce a formula that undoes $f$ in one direction only; composing both ways catches what a single spot-check would miss.
5
If $f$ needed a restricted domain to pass the horizontal line test, that restriction carries forward into $f^{-1}$ 's range, and any input $f$ 's formula can't produce becomes an input $f^{-1}$ can't accept.

Example

For $f(x) = 4x + 5$ , every real number is a legal input and output, so $f^{-1}(x) = \dfrac{x-5}{4}$ has domain all real numbers, nothing to restrict.
- $f$ was already one-to-one on its full natural domain — the domain of $f^{-1}$ is simply the range of $f$ ; state it only if that range isn't all real numbers.
- $f$ needed a restricted domain to pass the horizontal line test — report that restriction as part of $f$ 's own domain. Remember $f^{-1}$ 's domain is $f$ 's range on that restricted piece, not on all real numbers.
Why: the domain of $f^{-1}$ is defined to be the range of $f$ and vice versa. Skipping this check is how a formula that looks fine on paper ends up handed an input $f$ never actually produced.

Common mistakes

Solving for $x$ instead of $y$ after swapping: the target is a new formula for $y$ , which becomes $f^{-1}(x)$ , not a trip back to the original equation.
Writing $f^{-1}(x) = \dfrac{1}{f(x)}$ . The $-1$ is functional notation for "the inverse function," not an exponent; $f^{-1}$ is not "one over $f$ ."
Checking only $f^{-1}(f(x)) = x$ and stopping there. That equality can hold for a wrong formula by algebraic coincidence; $f(f^{-1}(x)) = x$ has to hold too.
Forgetting that a restricted domain on $f$ carries forward into a restriction on $f^{-1}$ , or skipping the domain step entirely just because the formula for $f^{-1}$ happens to simplify cleanly.