Inverse Functions

Reversing a function is a four-move routine: confirm it is one-to-one, swap $x$ and $y$, solve for $y$, and verify by composition. The underlying idea is that an inverse undoes a function, so if $f(a) = b$, then $f^{-1}(b) = a$.

But the first move is not optional. An inverse exists only when the original function is one-to-one on the domain you are using. If two inputs share the same output, the inverse cannot decide which input to return.

When this method applies

Use this routine whenever you need to recover an original input from an output: solving equations, working with graphs, or describing operation pairs where one undoes the other. Subtraction undoes addition, division undoes multiplication, and logarithms undo exponentials. In calculus the same idea governs $\ln x$, $\arcsin x$, and $\arctan x$. The method works only on a domain where the function is one-to-one, so part of the job is checking — or arranging — that condition before you start the algebra.

The procedure, step by step

1. Check the condition. Confirm the function is one-to-one on the given domain, or restrict the domain so that it is. Each output must come from exactly one input. This is also why domain and range switch: the domain of $f^{-1}$ is the range of $f$, and vice versa.
2. Swap variables. Write $y = f(x)$, then exchange $x$ and $y$ to reverse the roles of input and output.
3. Solve for $y$. Rearrange until $y$ is alone, then rename it $f^{-1}(x)$.
4. Verify. Check by composition that the function and its inverse undo each other.

A full run-through

Start with a one-to-one function:

Step 1: a line with nonzero slope is one-to-one everywhere, so no restriction is needed. Step 2, write and swap:

Step 3, solve for $y$:

so

Step 4, verify by composition:

Checking the other direction, $f^{-1}(f(x)) = x$, confirms it. If both compositions return $x$ on the relevant domain, the inverse is correct.

Where each step trips people up

• At "check the condition," skipping it entirely. Run the algebra on $f(x) = x^2$ over all reals and you get a tidy-looking expression, but no inverse function exists there: $f(2) = 4$ and $f(-2) = 4$, so the output $4$ has two sources. Restrict to $x \ge 0$ and the function becomes one-to-one, with inverse $f^{-1}(x) = \sqrt{x}$. Without that restriction, calling the inverse $\sqrt{x}$ is incomplete.
• At "solve for $y$," confusing $f^{-1}(x)$ with $\frac{1}{f(x)}$. One undoes a function; the other takes a reciprocal. They are different objects.
• After solving, forgetting the domain–range switch. It matters when you state where the inverse is defined. Self-check: if two functions are inverses, their graphs are reflections across the line $y = x$. A quick sketch confirms a result makes sense.

FAQ

Run the four steps on $f(x) = 5x - 7$: write $y = f(x)$, swap, solve for $y$, and check by composition. Then try $f(x) = x^2$ and decide what domain restriction is required before an inverse can exist at all.