# Inverse Functions

> IB Mathematics: Applications and Interpretation SL · Unit 2: Functions
> Source: https://www.owlsprep.com/study/ib-math-ai-sl-u2-inverse-functions/

This module covers how to identify invertible one-to-one functions, find inverses algebraically and graphically, and determine the domain and range of inverse functions. You will also learn to restrict domains for non-invertible functions.

**Prerequisites:** [Function notation, domain and range](https://www.owlsprep.com/study/ib-math-ai-sl-u2-functions-basics/); [Graphing polynomial functions](https://www.owlsprep.com/study/ib-math-ai-sl-u2-polynomial-functions/)

## Learning objectives

- Identify one-to-one functions that have valid inverses
- Find the inverse of a function algebraically and graphically
- Determine the domain and range of inverse functions
- Restrict domains of non-invertible functions to create inverses

## One-to-One Functions and Invertibility

A function can only have a valid inverse if it is **one-to-one (injective)**. A one-to-one function maps every output to exactly one input, so it must pass both the vertical line test (to qualify as a function) and the horizontal line test.

**One-to-one function** — A function $f$ is one-to-one if whenever $f(a) = f(b)$, it follows that $a = b$. No two distinct inputs produce the same output.

*Example:* $f(x) = 2x + 3$ is one-to-one; $f(x) = x^2$ is not one-to-one over all real numbers.

**Worked example:** Prove that $f(x) = 3x - 2$ is one-to-one over all real $x$.

1. Assume $f(a) = f(b)$ for inputs $a$ and $b$:
2. $$3a - 2 = 3b - 2$$
3. Add 2 to both sides, then divide by 3:
4. $$3a = 3b \implies a = b$$
5. Since $f(a) = f(b)$ implies $a = b$, $f(x)$ satisfies the definition of a one-to-one function.

> **Exam tip:** Always check for one-to-one status before finding an inverse; non one-to-one functions do not have valid inverses over their full domain.

## Finding Inverses Algebraically

Follow this standard 4-step process to find the inverse of any one-to-one function algebraically:

1. Start with the equation $y = f(x)$
2. Swap the positions of $x$ and $y$
3. Rearrange the equation to solve for $y$
4. The resulting $y$ is $f^{-1}(x)$

**Worked example:** Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$.

1. Step 1: Write the function with $y$:
2. $$y = \frac{2x + 1}{x - 3}$$
3. Step 2: Swap $x$ and $y$:
4. $$x = \frac{2y + 1}{y - 3}$$
5. Step 3: Multiply both sides by $(y-3)$ and expand:
6. $$x(y - 3) = 2y + 1 \implies xy - 3x = 2y + 1$$
7. Step 4: Collect terms with $y$ on one side and factor:
8. $$xy - 2y = 3x + 1 \implies y(x - 2) = 3x + 1$$
9. Step 5: Solve for $y$ to get the inverse:
10. $$f^{-1}(x) = \frac{3x + 1}{x - 2}$$

## Graphical Properties of Inverse Functions

Swapping $x$ and $y$ for inverse functions means the graph of $f^{-1}(x)$ is the reflection of the graph of $f(x)$ over the line $y = x$. Any point $(a, b)$ on $f(x)$ maps to $(b, a)$ on $f^{-1}(x)$.

> **info**
>
> Any intersection points between $f(x)$ and $f^{-1}(x)$ always lie on the line $y = x$.

**Worked example:** Given $f(x) = 2x + 1$ crosses the axes at $(0, 1)$ and $(-0.5, 0)$, sketch the inverse by reflection.

1. Step 1: Swap the coordinates of the intercepts to get points on $f^{-1}(x)$:
2. Original $(0, 1)$ → $(1, 0)$; original $(-0.5, 0)$ → $(0, -0.5)$
3. Step 2: Reflect all points of $f(x)$ over the line $y = x$ to get the full graph of $f^{-1}(x)$.
4. Step 3: Verify: the inverse is $f^{-1}(x) = \frac{x - 1}{2}$, which passes through both new points, confirming the reflection is correct.

## Domain and Range of Inverse Functions

Because inputs and outputs are swapped between a function and its inverse, there is a fixed relationship between their domains and ranges:

**Domain-Range Relationship** — For any invertible function $f$: $\text{dom}(f^{-1}) = \text{range}(f)$ and $\text{range}(f^{-1}) = \text{dom}(f)$

*Example:* If $f(x) = x^2$ with domain $x \geq 0$, range $y \geq 0$, so $f^{-1}(x) = \sqrt{x}$ has domain $x \geq 0$, range $y \geq 0$.

**Worked example:** Given $f(x) = x^2$ with domain $x \geq 2$ and range $y \geq 4$, find the domain and range of $f^{-1}(x)$.

1. Apply the domain-range relationship:
2. 1. Domain of $f^{-1}$ = Range of original $f$
3. $$\text{dom}(f^{-1}) = x \geq 4$$
4. 2. Range of $f^{-1}$ = Domain of original $f$
5. $$\text{range}(f^{-1}) = y \geq 2$$
6. This matches the inverse $f^{-1}(x) = \sqrt{x}$, confirming the result.

## Common pitfalls

- **Wrong:** Confusing $f^{-1}(x)$ with the reciprocal $\frac{1}{f(x)}$
  - Why it fails: The notation $f^{-1}$ denotes inverse, not exponent -1.
  - Correct: Remember $f^{-1}(x)$ undoes the action of $f(x)$, it is not the reciprocal. For $f(x)=2x$, $f^{-1}(x) = \frac{x}{2}$, not $\frac{1}{2x}$.
- **Wrong:** Finding an inverse for a non one-to-one function without restricting the domain
  - Why it fails: Non one-to-one functions do not pass the horizontal line test, so their inverse would not be a valid function.
  - Correct: Always check for one-to-one status first, and restrict the domain of the original function to make it one-to-one if needed.
- **Wrong:** Swapping the domain and range of the inverse function
  - Why it fails: It is easy to mix up which set belongs to which function.
  - Correct: Memorize: domain of inverse = range of original, range of inverse = domain of original to avoid this mistake.
- **Wrong:** Reflecting the inverse over the x-axis or y-axis instead of $y=x$
  - Why it fails: Common mistake when sketching inverse graphs from memory.
  - Correct: Always use $y=x$ as the mirror line, and swap the coordinates of 1-2 key points to confirm your reflection is correct.

## Cheatsheet

| Concept | Key Rule/Property |
| --- | --- |
| Inverse existence | Only for one-to-one functions (pass horizontal line test) |
| Algebraic steps | 1. $y=f(x)$ 2. Swap $x,y$ 3. Solve for $y$ 4. $y=f^{-1}(x)$ |
| Graph property | $f^{-1}(x)$ is reflection of $f(x)$ over $y=x$ |
| Domain/Range | $\text{dom}(f^{-1}) = \text{range}(f)$, $\text{range}(f^{-1}) = \text{dom}(f)$ |
| Notation | $f^{-1}(x) \neq \frac{1}{f(x)}$ |

## What's next

Inverse functions are a foundational concept for all future work with functions in IB AI SL. Most importantly, logarithms are defined as the inverse of exponential functions, so understanding inverses will make learning exponential and logarithmic models much simpler. Inverses are also commonly used to solve application problems where you need to reverse a function to find an input for a given output, which is a frequent requirement in exam questions. Mastering inverses will set you up for success for all upcoming function topics.

- [Linear functions and their graphs](https://www.owlsprep.com/study/ib-math-ai-sl-u2-linear-functions-and-their-graphs/)
- [Quadratic functions, roots, and vertices](https://www.owlsprep.com/study/ib-math-ai-sl-u2-quadratic-functions-roots-and-vertices/)
- [Exponential and logarithmic functions](https://www.owlsprep.com/study/ib-math-ai-sl-u2-exponential-and-logarithmic-functions/)

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