Inverse functions: algebraic solution

IB Mathematics: Applications and Interpretation HL· 2.11 Inverse functions· 5 min read

1. Key Definitions and Verification of Inverses★★☆☆☆⏱ 15 min

🚫 No Calculator

📘 Definition

Inverse Function

$f^{- 1} (x)$

For a one-to-one function $f$ with domain $D$ and range $R$ , the inverse function $f^{- 1}$ has domain $R$ and range $D$ , and satisfies $f (f^{- 1} (x)) = x$ for all $x \in R$ and $f^{- 1} (f (x)) = x$ for all $x \in D$ .

Example:

If $f (x) = 2 x + 3$ , then $f^{- 1} (x) = \frac{x - 3}{2}$

Only one-to-one functions (functions that pass the horizontal line test) have inverses. If a function is not one-to-one over its entire domain, we restrict the domain to make it one-to-one before calculating the inverse.

📐 Worked Example

Verify that $f (x) = 4 x - 5$ and $g (x) = \frac{x + 5}{4}$ are inverses of each other.

1
First check the composition $f (g (x))$ by substituting $g (x)$ into $f$ :
$f(g(x)) = f\left(\frac{x+5}{4}\right) = 4\left(\frac{x+5}{4}\right) - 5 = (x+5) - 5 = x$
2
Next check the reverse composition $g (f (x))$ by substituting $f (x)$ into $g$ :
$g(f(x)) = g(4x - 5) = \frac{(4x - 5) + 5}{4} = \frac{4x}{4} = x$
3
Since both compositions simplify to $x$ , the functions meet the definition of inverses.

2. Step-by-Step Algebraic Method for Finding Inverses★★★☆☆⏱ 20 min

🚫 No Calculator

Start with $y = f (x)$ , note the original domain and range
Swap the positions of $x$ and $y$ (this reflects over $y = x$ )
Rearrange the equation to solve for $y$ in terms of $x$
Replace $y$ with $f^{- 1} (x)$
State the domain of $f^{- 1} (x)$ , which equals the range of the original $f (x)$

📐 Worked Example

Find the inverse of $f (x) = 2 x + 3$ for $x \geq - 3$ .

1
Set $y$ equal to $f (x)$ :
$y = 2\sqrt{x + 3}, \quad x \geq -3$
2
Swap $x$ and $y$ to get the inverse relationship:
$x = 2\sqrt{y + 3}$
3
Solve for $y$ : divide by 2 then square both sides:
$\frac{x}{2} = \sqrt{y + 3} \implies \left(\frac{x}{2}\right)^2 = y + 3$
4
Isolate $y$ :
$y = \frac{x^2}{4} - 3$
5
The original function $f (x)$ has range $y \geq 0$ , so the inverse has domain $x \geq 0$ :
$f^{-1}(x) = \frac{x^2}{4} - 3, \quad x \geq 0$

Exam tip:

Examiners almost always award a separate mark for correctly stating the domain of the inverse function — never skip this step.

3. Inverses of Restricted Quadratic Functions★★★★☆HL only⏱ 20 min

🚫 No Calculator

Quadratic functions are not one-to-one over their full domain, so we always restrict the original domain to one side of the vertex to make it invertible. The original domain restriction tells us which root to pick when solving for the inverse.

📐 Worked Example

Find the inverse of $f (x) = (x - 2)^{2} - 4$ for $x \geq 2$ .

1
Set $y$ equal to $f (x)$ :
$y = (x - 2)^2 - 4, \quad x \geq 2$
2
Swap $x$ and $y$ :
$x = (y - 2)^2 - 4$
3
Rearrange to isolate the squared term:
$x + 4 = (y - 2)^2$
4
Take the square root of both sides, keeping the absolute value:
$\sqrt{x + 4} = |y - 2|$
5
Original domain $x \geq 2$ means inverse range is $y \geq 2$ , so $y - 2$ is non-negative:
$\sqrt{x + 4} = y - 2$
6
Original range of $f (x)$ is $y \geq - 4$ , so inverse domain is $x \geq - 4$ :
$f^{-1}(x) = \sqrt{x + 4} + 2, \quad x \geq -4$

4. Common Pitfalls

Wrong move:

Forgetting to state the domain of the inverse function

Why:

Examiners require the domain to confirm you understand that inverse domain equals original range, skipping this loses easy marks

Correct move:

Always write the domain of $f^{- 1} (x)$ after finding the algebraic expression

Wrong move:

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

Why:

The $- 1$ exponent in inverse notation does not mean reciprocal — this is a common beginner mistake

Correct move:

Remember $f^{- 1} (x)$ undoes the action of $f (x)$ , it is not the reciprocal

Wrong move:

Keeping both positive and negative roots when solving for inverse quadratics

Why:

A function can only have one output per input, so two roots would not give a valid function

Correct move:

Use the original domain restriction to select only one valid root for the inverse

Wrong move:

Swapping variables after solving for $y$ instead of before

Why:

This leaves you with the inverse expressed in terms of the wrong variables, giving an incorrect final result

Correct move:

Always swap $x$ and $y$ immediately after writing $y = f (x)$

5. Quick Reference Cheatsheet

Step	Action	Exam Note
1	Start with $y = f (x)$	Write original domain/range
2	Swap $x$ and $y$	Reflects over $y = x$
3	Solve for $y$	Pick one root for quadratics
4	Write $f^{- 1} (x) = y$	Set domain = original range
5	Verify (optional)	Check $f (f^{- 1} (x)) = x$ to confirm

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find inverse of linear function
2023 · 1
Inverse of restricted quadratic

Going deeper

IB SyllabusIB AI HL Syllabus Topic 2Official syllabus reference

What's Next

Now that you can find inverses algebraically, you can extend this understanding to graphical representations of inverse functions and core applications like exponential and logarithmic inverse pairs, which are heavily tested in IB AI HL. Inverse functions are foundational for many topics in Unit 2 and beyond, including solving equations involving composite functions and modelling inverse relationships in real-world contexts. Mastering algebraic inversion will also help you quickly recognize inverse pairs on exam day, saving you valuable time on both Paper 1 and Paper 2.