Inverse Function Definition

5 min read

Reference: Stewart §6.1

Reversing the Process

When a biologist measures that a bacteria population reaches 550 at time $t = 6$, she uses a function $N = f(t)$ that takes time and outputs population. But what if she asks: “When will the population reach 600?” Now she needs the reverse: a function that takes population and outputs time.

This “reverse” function is the inverse function, written $f^{-1}$. It undoes what $f$ does.

Before You Start: Quick Self-Check

Can you answer these questions?

What does “one-to-one” mean? Different inputs give different outputs.
If $f(3) = 10$, what does this tell us? When input is 3, output is 10.
What is function composition? Applying one function after another, like $g(f(x))$.

If you struggled with these, review first

The inverse function definition builds directly on one-to-one functions. Review One-to-One Functions before continuing.

Prerequisite Map

Quick Reference

Property	Value
Concept	Inverse Functions
Chapter	6.1
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Definition

Let $f$ be a one-to-one function with domain $A$ and range $B$. The inverse function $f^{-1}$ has domain $B$ and range $A$ and is defined by:

$$\boxed{f^{-1}(y) = x \quad \Longleftrightarrow \quad f(x) = y}$$

In words: $f^{-1}(y)$ is the unique input $x$ such that $f(x) = y$.

Critical Warning

$$f^{-1}(x) \neq \frac{1}{f(x)}$$

The notation $f^{-1}$ means “the inverse function of $f$,” NOT “1 divided by $f$.”

$f^{-1}(x)$ = inverse function
$[f(x)]^{-1} = \frac{1}{f(x)}$ = reciprocal

Domain and Range Swap

Function	Domain	Range
$f$	$A$	$B$
$f^{-1}$	$B$	$A$

Key insight: The domain of $f^{-1}$ equals the range of $f$, and vice versa.

The Cancellation Equations

$$\boxed{f^{-1}(f(x)) = x \quad \text{for all } x \text{ in domain of } f}$$

$$\boxed{f(f^{-1}(x)) = x \quad \text{for all } x \text{ in domain of } f^{-1}}$$

These say that $f$ and $f^{-1}$ “undo” each other:

Apply $f$, then apply $f^{-1}$: you get back where you started
Apply $f^{-1}$, then apply $f$: you get back where you started

Visualizing the Reversal

    f                           f⁻¹
  ─────►                      ◄─────
x ═══════► f(x) = y     y ═══════► f⁻¹(y) = x
  ◄─────                      ─────►
   f⁻¹                          f

Example: $f(x) = x^3$

$f(2) = 8$, so $f^{-1}(8) = 2$
$f^{-1}(x) = \sqrt[3]{x} = x^{1/3}$

Check cancellation:

$f^{-1}(f(x)) = (x^3)^{1/3} = x$ ✓
$f(f^{-1}(x)) = (x^{1/3})^3 = x$ ✓

Reading Inverse Values from a Table

If you have a table for $f$:

$x$	1	2	3	4
$f(x)$	5	7	2	9

Then for $f^{-1}$, swap the rows:

$x$	5	7	2	9
$f^{-1}(x)$	1	2	3	4

For example: $f^{-1}(7) = 2$ because $f(2) = 7$.

Practice Problems

Level 1 Reading Inverse Values

If $f$ is one-to-one and $f(4) = 10$, $f(7) = 3$, and $f(2) = -5$, find:

(a) $f^{-1}(10)$ (b) $f^{-1}(3)$ (c) $f^{-1}(-5)$

Thought Process

Goal: Find $f^{-1}(10)$, $f^{-1}(3)$, and $f^{-1}(-5)$.

Key definition: $f^{-1}(y) = x$ means $f(x) = y$.

In words: To find $f^{-1}(y)$, ask “what input gives output $y$?”

For each part:

$f^{-1}(10) = ?$ → What input gives output 10? We know $f(4) = 10$, so answer is 4.
$f^{-1}(3) = ?$ → What input gives output 3? We know $f(7) = 3$, so answer is 7.
$f^{-1}(-5) = ?$ → What input gives output $-5$? We know $f(2) = -5$, so answer is 2.

Show Answer

(a) $f^{-1}(10) = 4$ because $f(4) = 10$

(b) $f^{-1}(3) = 7$ because $f(7) = 3$

Level 2 Using Cancellation Equations

Simplify each expression:

(a) $f^{-1}(f(5))$ (b) $f(f^{-1}(\pi))$ (c) $f^{-1}(f(f^{-1}(3)))$

Thought Process

Goal: Simplify expressions using cancellation equations.

The two key equations:

$f^{-1}(f(x)) = x$ (apply $f$, then undo with $f^{-1}$, get back $x$)
$f(f^{-1}(x)) = x$ (apply $f^{-1}$, then undo with $f$, get back $x$)

Strategy: Work from the inside out.

(a) $f^{-1}(f(5))$: The inner function is $f(5)$. Then $f^{-1}$ undoes it. Result: 5.

(b) $f(f^{-1}(\pi))$: The inner function is $f^{-1}(\pi)$. Then $f$ undoes it. Result: $\pi$.

(c) $f^{-1}(f(f^{-1}(3)))$: Start with innermost: $f^{-1}(3)$ gives some value. Then $f$ undoes it giving 3. Then $f^{-1}(3)$ again.

Show Answer

(a) $f^{-1}(f(5)) = 5$ by the first cancellation equation

(b) $f(f^{-1}(\pi)) = \pi$ by the second cancellation equation

First simplify the inner part: $f(f^{-1}(3)) = 3$

Then: $f^{-1}(3) = $ whatever value maps to 3 under $f$

But we can just apply cancellation: $f^{-1}(f(f^{-1}(3))) = f^{-1}(3)$

Level 3 Domain and Range

Let $f(x) = \sqrt{x - 3}$ with domain $[3, \infty)$.

(a) What is the range of $f$? (b) What is the domain of $f^{-1}$? (c) What is the range of $f^{-1}$?

Thought Process

First find the range of $f$ by considering what outputs are possible from $\sqrt{x-3}$ when $x \geq 3$. Then use the domain/range swap for inverses.

Show Answer

(a) When $x \geq 3$, we have $x - 3 \geq 0$, so $\sqrt{x-3} \geq 0$.

As $x \to \infty$, $\sqrt{x-3} \to \infty$.

Range of $f$: $[0, \infty)$

(b) Domain of $f^{-1}$ = Range of $f$ = $\boxed{[0, \infty)}$

Level 4 Verifying an Inverse

Let $f(x) = \frac{2x+1}{x-3}$ (for $x \neq 3$) and $g(x) = \frac{3x+1}{x-2}$ (for $x \neq 2$).

Verify that $g = f^{-1}$ by checking both cancellation equations.

Thought Process

Compute $f(g(x))$ and $g(f(x))$ and show both equal $x$. Be careful with the algebra: you’ll need to simplify complex fractions.

Show Answer

Check $f(g(x)) = x$:

$$f(g(x)) = f\left(\frac{3x+1}{x-2}\right) = \frac{2 \cdot \frac{3x+1}{x-2} + 1}{\frac{3x+1}{x-2} - 3}$$

Numerator: $\frac{2(3x+1) + (x-2)}{x-2} = \frac{6x+2+x-2}{x-2} = \frac{7x}{x-2}$

Denominator: $\frac{(3x+1) - 3(x-2)}{x-2} = \frac{3x+1-3x+6}{x-2} = \frac{7}{x-2}$

$$f(g(x)) = \frac{7x/(x-2)}{7/(x-2)} = \frac{7x}{7} = x \quad ✓$$

Check $g(f(x)) = x$:

$$g(f(x)) = g\left(\frac{2x+1}{x-3}\right) = \frac{3 \cdot \frac{2x+1}{x-3} + 1}{\frac{2x+1}{x-3} - 2}$$

Numerator: $\frac{3(2x+1) + (x-3)}{x-3} = \frac{7x}{x-3}$

Denominator: $\frac{(2x+1) - 2(x-3)}{x-3} = \frac{7}{x-3}$

$$g(f(x)) = \frac{7x/(x-3)}{7/(x-3)} = x \quad ✓$$

Both cancellation equations hold, so $g = f^{-1}$.

Level 5 Self-Inverse Functions

A function $f$ is called self-inverse if $f^{-1} = f$, meaning $f(f(x)) = x$ for all $x$ in the domain.

(a) Show that $f(x) = \frac{1}{x}$ (for $x \neq 0$) is self-inverse.

(b) Show that $g(x) = \frac{a-x}{1+ax}$ (for $ax \neq -1$, where $a$ is a constant) is self-inverse.

Thought Process

For (a) and (b), compute $f(f(x))$ and show it equals $x$. For (c), think about what happens when you reflect a self-inverse function about $y = x$.

Show Answer

(a) $f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{1/x} = x$ ✓

(b) $$g(g(x)) = g\left(\frac{a-x}{1+ax}\right) = \frac{a - \frac{a-x}{1+ax}}{1 + a \cdot \frac{a-x}{1+ax}}$$

Numerator: $\frac{a(1+ax) - (a-x)}{1+ax} = \frac{a + a^2x - a + x}{1+ax} = \frac{x(a^2+1)}{1+ax}$

Denominator: $\frac{(1+ax) + a(a-x)}{1+ax} = \frac{1+ax+a^2-ax}{1+ax} = \frac{1+a^2}{1+ax}$

$$g(g(x)) = \frac{x(a^2+1)/(1+ax)}{(1+a^2)/(1+ax)} = \frac{x(a^2+1)}{1+a^2} = x \quad ✓$$

(c) A self-inverse function equals its own inverse, so its graph is symmetric about the line $y = x$.

This is because reflecting about $y = x$ gives the inverse function’s graph, and for a self-inverse function, that’s the same graph.

CCI-Style Conceptual Questions

Question 1: If $f^{-1}(5) = 2$, which statement must be true?

(A) $f(5) = 2$ (B) $f(2) = 5$ (C) $f(5) = -2$ (D) $f^{-1}(2) = 5$

Answer

(B) By definition, $f^{-1}(5) = 2$ means that $f(2) = 5$.

Question 2: What is the domain of $f^{-1}$ if $f(x) = x^2$ with domain $[0, 3]$?

(A) $[0, 3]$ (B) $[0, 9]$ (C) $[-3, 3]$ (D) All real numbers

Answer

(B) The domain of $f^{-1}$ equals the range of $f$. When $x \in [0, 3]$, $f(x) = x^2$ ranges from $0^2 = 0$ to $3^2 = 9$, so the range is $[0, 9]$.

Common Misconceptions

Common misconception

$f^{-1}(x) = \frac{1}{f(x)}$ because the $-1$ exponent indicates a reciprocal.

This is the concept-image-conflicts-definition error. In the notation $f^{-1}$, the $-1$ is not an exponent applied to the output; it is a label indicating the inverse function. For $f(x) = 2x + 3$, the inverse is $f^{-1}(x) = \frac{x-3}{2}$, while the reciprocal is $[f(x)]^{-1} = \frac{1}{2x+3}$. The inverse satisfies $f(f^{-1}(x)) = x$, which the reciprocal does not.

Mastery Checklist

I can state the definition of the inverse function
I know that $f^{-1}(x) \neq 1/f(x)$
I can read inverse values from a table or given function values
I understand the domain/range swap between $f$ and $f^{-1}$
I can apply both cancellation equations
I can verify that two functions are inverses of each other

Mental Model

The Undo Button:

Think of $f^{-1}$ as the “undo” operation for $f$. If $f$ transforms input $x$ into output $y$, then $f^{-1}$ transforms $y$ back into $x$. It’s like hitting Ctrl+Z after a text edit: you return to exactly where you started.

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One-to-One Functions	Skills Index	Finding Inverse Functions

Last updated: 2026-01-22