Composite and inverse functions

IB Mathematics Analysis and Approaches SL· 2.2 Composite and inverse functions· 45 min read

1. Composite Functions★★☆☆☆⏱ 15 min

📘 Definition

Composite Function

f \circ g(x) = f(g(x))

A function where the output of $g (x)$ is used as the input for $f (x)$ . Order matters, so $f \circ g \neq = g \circ f$ in most cases.

Example:

If $f (x) = x + 2$ and $g (x) = x^{2}$ , then $f (g (x)) = x^{2} + 2$ , while $g (f (x)) = (x + 2)^{2}$

When finding the domain of a composite function, the domain is restricted by any restrictions on the inner function $g (x)$ , and any restrictions on the outer function $f (x)$ that come from the output of $g (x)$ .

📐 Worked Example

Given $f (x) = x + 3$ with domain $x \geq - 3$ , and $g (x) = 2 x - 4$ , find $f \circ g (x)$ and state its domain.

1
Substitute $g (x)$ into the input of $f$ to get the algebraic form:
2
$f(g(x)) = f(2x - 4) = \sqrt{(2x - 4) + 3} = \sqrt{2x - 1}$
3
Find the domain: $g (x)$ is linear so it has no domain restrictions. The expression inside the square root must be non-negative:
4
$2x - 1 \geq 0 \implies x \geq \frac{1}{2}$
5
Final result: $f \circ g (x) = 2 x - 1$ with domain $x \geq \frac{1}{2}$

Exam tip:

Always check domain restrictions for composite functions: they are almost always worth one mark in exams, even if the question only asks for the function.

2. Inverse Functions and One-to-One Condition★★★☆☆⏱ 20 min

📘 Definition

One-to-One Function

A function where every output corresponds to exactly one input. Graphically, a function is one-to-one if it passes the horizontal line test. Only one-to-one functions have inverse functions.

Example:

$f (x) = 3 x + 1$ is one-to-one; $f (x) = x^{2}$ is not one-to-one over all real numbers

If a function is not one-to-one over its natural domain, you can restrict the domain to create a one-to-one function that has an inverse. The range of the original function becomes the domain of the inverse function.

📐 Worked Example

Given $f (x) = x^{2} - 6 x$ , $x \geq 3$ , find $f^{- 1} (x)$ and state its domain.

1
Start by writing $y = f (x)$ , then swap $x$ and $y$ to prepare for the inverse:
2
$x = y^2 - 6y$
3
Complete the square to solve for $y$ :
4
$x = (y - 3)^2 - 9 \implies (y - 3)^2 = x + 9 \implies y - 3 = \pm \sqrt{x + 9}$
5
The original domain is $x \geq 3$ , so the range of $f (x)$ is $y \geq (3)^{2} - 6 (3) = - 9$ . This means the domain of $f^{- 1}$ is $x \geq - 9$ , and since $y \geq 3$ for the inverse, we take the positive root:
6
$y = 3 + \sqrt{x + 9}$
7
Final result: $f^{- 1} (x) = 3 + x + 9$ , domain $x \geq - 9$

Exam tip:

Picking the correct sign after taking the square root for quadratic inverses is a common marking point, always reference the original domain to justify your choice.

3. Graphical Relationship Between f and f⁻¹★★☆☆☆⏱ 10 min

Because finding an inverse swaps the $x$ and $y$ values of every point on the original function, the graph of $f^{- 1} (x)$ is the reflection of the graph of $f (x)$ over the line $y = x$ .

📐 Worked Example

The graph of $f (x) = 2^{x + 1}$ passes through $(0, 2)$ , has domain all real numbers, and has a horizontal asymptote at $y = 0$ . Sketch the key features of $f^{- 1} (x)$ .

1
Reflect all key features over $y = x$ : the point $(0, 2)$ on $f$ becomes $(2, 0)$ on $f^{- 1}$ .
2
The horizontal asymptote $y = 0$ on $f$ becomes a vertical asymptote $x = 0$ (the y-axis) on $f^{- 1}$ .
3
The domain of $f$ becomes the range of $f^{- 1}$ , and the range of $f$ ( $y > 0$ ) becomes the domain of $f^{- 1}$ ( $x > 0$ ).
4
Since $f$ is increasing, $f^{- 1}$ is also increasing.

Exam tip:

If you are asked to sketch an inverse from a given graph of $f$ , reflecting over $y = x$ is much faster than calculating the inverse algebraically.

4. Common Pitfalls

Wrong move:

Reversing the order of composition, calculating $g (f (x))$ instead of $f (g (x))$ for $f \circ g$ .

Why:

The notation is read left to right, leading many students to mix up which function is applied first.

Correct move:

Remember the inner function is closest to $x$ : $f \circ g (x) = f (g (x))$ , work from the inside out.

Wrong move:

Confusing inverse function notation with reciprocal: assuming $f^{- 1} (x) = \frac{1}{f ( x )}$ .

Why:

The $- 1$ exponent looks like a reciprocal exponent, leading to this common mix-up.

Correct move:

Remember $f^{- 1}$ is the inverse function that undoes $f$ , while $(f (x))^{- 1}$ is the reciprocal of $f (x)$ .

Wrong move:

Forgetting to state the domain of composite or inverse functions.

Why:

Students often focus only on finding the algebraic expression and overlook domain requirements that are marked separately.

Correct move:

Always add the domain to your final answer, even if the question does not explicitly ask for it.

Wrong move:

Claiming a non-one-to-one function has an inverse without restricting the domain.

Why:

Many students forget the inverse only exists for one-to-one functions.

Correct move:

First check if the function passes the horizontal line test, and restrict the domain if necessary before finding the inverse.

5. Quick Reference Cheatsheet

Concept	Key Rule	Exam Note
Composite $f \circ g$	$f (g (x))$ , inner function first	Order matters, always state domain
Domain of composite	Inner restrictions + outer restrictions	Worth 1 mark almost every time
Inverse exists if	Function is one-to-one (passes horizontal line test)	Restrict domain if needed
Find inverse algebraically	Swap $x$ and $y$ , solve for $y$	Pick correct sign for quadratics
Graph of inverse	Reflection of $f$ over $y = x$	$(a, b) \to (b, a)$ on the inverse

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.

2022 · 1
Find inverse of quadratic function
2023 · 1
Calculate domain of composite function
2021 · 2
Sketch inverse from given graph

Going deeper

IB SyllabusIB AA SL Syllabus Topic 2Official syllabus reference

What's Next

Composite and inverse functions are foundational for nearly all后续 function topics in IB AA SL. The most direct application is with exponential and logarithmic functions, which are inverses of each other, so the skills you learned here will help you solve exponential and logarithmic equations later. You will also use inverse functions when working with cumulative distribution functions in the statistics unit, and domain restriction skills are critical for solving all types of functional equations that appear regularly in both paper 1 and paper 2.