Inverse Functions

By the end of this lesson, you will be able to determine if a function has an inverse, find the inverse algebraically and graphically, verify inverse functions using composition, and understand why restricting domains is sometimes necessary.

The Big Idea

A function's inverse undoes what the function does. If $f$ maps $x$ to $y$, then $f^{-1}$ maps $y$ back to $x$. The graph of $f^{-1}$ is the reflection of the graph of $f$ over the line $y = x$. Inverse functions appear in every branch of mathematics — logarithms are inverses of exponentials, arcsin is the inverse of sin, and square roots are inverses of squaring (on the correct domain).

When Does an Inverse Function Exist?

$f$ has an inverse function iff $f$ is one-to-one (injective): each output corresponds to exactly one input.

Horizontal Line Test: if every horizontal line intersects the graph at most once, $f$ is one-to-one and has an inverse.

Finding the Inverse

Algebraic method:

1. Replace $f(x)$ with $y$
2. Swap $x$ and $y$
3. Solve for $y$
4. Write as $f^{-1}(x)$

Domain and range swap: the domain of $f^{-1}$ = range of $f$; range of $f^{-1}$ = domain of $f$.

Verifying with Composition

$$f(f^{-1}(x)) = x \quad \text{AND} \quad f^{-1}(f(x)) = x$$

Both compositions must equal $x$ (the identity function) on the appropriate domains.

Example 1 — Linear Function

$f(x) = 3x - 7$. Find $f^{-1}(x)$.

$y = 3x - 7 \implies x = 3y - 7 \implies y = \dfrac{x+7}{3}$

$f^{-1}(x) = \dfrac{x+7}{3}$

Verify: $f(f^{-1}(x)) = 3\left(\dfrac{x+7}{3}\right) - 7 = x + 7 - 7 = x$ ✓

Example 2 — Radical Function

$f(x) = \sqrt{x - 2} + 5$. Find $f^{-1}(x)$.

Domain of $f$: $[2, \infty)$; Range: $[5, \infty)$

$y = \sqrt{x-2} + 5 \implies y - 5 = \sqrt{x-2} \implies (y-5)^2 = x - 2 \implies x = (y-5)^2 + 2$

$f^{-1}(x) = (x-5)^2 + 2$

Domain of $f^{-1}$: $[5, \infty)$ (= range of $f$); Range: $[2, \infty)$ (= domain of $f$)

Example 3 — Restricting the Domain

$f(x) = x^2$ is NOT one-to-one (parabola fails horizontal line test).