Inverse Functions and One-to-One Functions

By the end of this lesson, you will be able to prove a function is one-to-one algebraically and graphically, find and verify inverse functions, understand the domain/range switch, and apply the composition identity $f(f^{-1}(x)) = x$.

The Big Idea

Inverse functions answer: "if I know the output, what was the input?" Every bijective function (one-to-one and onto) has an inverse. This lesson deepens the Algebra 2 treatment by emphasizing the algebraic verification, the graph symmetry, and the crucial role inverses play in solving equations involving exponentials, trig functions, and composition.

One-to-One Functions

Definition: $f$ is one-to-one (injective) if different inputs always produce different outputs:

$$x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$$

Equivalent: $f(x_1) = f(x_2) \implies x_1 = x_2$

Horizontal Line Test: $f$ is one-to-one iff every horizontal line intersects the graph at most once.

Proving One-to-One Algebraically

Assume $f(x_1) = f(x_2)$ and show it forces $x_1 = x_2$.

$f(x) = 3x - 5$: $3x_1 - 5 = 3x_2 - 5 \implies x_1 = x_2$ ✓ — one-to-one.

$f(x) = x^2$: $f(1) = f(-1) = 1$ — NOT one-to-one on all reals.

Finding the Inverse

Given $f$ is one-to-one:

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

Example 1 — Radical Inverse

$f(x) = \dfrac{x^3 - 1}{2}$. Find $f^{-1}(x)$.

$y = \dfrac{x^3-1}{2} \implies 2y = x^3 - 1 \implies x^3 = 2y + 1 \implies x = \sqrt[3]{2y+1}$

$f^{-1}(x) = \sqrt[3]{2x+1}$

Verify: $f(f^{-1}(x)) = \dfrac{(\sqrt[3]{2x+1})^3 - 1}{2} = \dfrac{(2x+1) - 1}{2} = x$ ✓

Example 2 — Rational Function Inverse

$f(x) = \dfrac{2x+3}{x-1}$. Find $f^{-1}(x)$ and its domain.

$y = \dfrac{2x+3}{x-1} \implies y(x-1) = 2x+3 \implies xy - y = 2x + 3 \implies x(y-2) = y+3$

$x = \dfrac{y+3}{y-2} \implies f^{-1}(x) = \dfrac{x+3}{x-2}$

Domain of $f^{-1}$: $x \neq 2$ (range of $f$ excludes 2, since $f(x) = 2$ would require $2x+3 = 2x-2$, impossible).

The Graph Relationship