Composite Functions

IB Mathematics Analysis and Approaches SL· 2.3 Composite functions· 40 min read

1. What is a Composite Function?★★☆☆☆⏱ 10 min

📘 Definition

Composite Function

$f \circ g$ , read as "f after g"

A function formed by using the output of one function $g$ as the input to a second function $f$ , so $(f \circ g) (x) = f (g (x))$ .

Example:

If $f (x) = x^{2}$ and $g (x) = x + 1$ , then $(f \circ g) (x) = (x + 1)^{2}$

A core property of composition is that order matters: $(f \circ g) (x)$ is almost never equal to $(g \circ f) (x)$ .

📐 Worked Example

Given $f (x) = 2 x + 3$ and $g (x) = x^{2} - 1$ , find $(f \circ g) (2)$

1
First evaluate the innermost function $g$ at $x = 2$ :
$g(2) = (2)^2 - 1 = 4 - 1 = 3$
2
Substitute this result into the outer function $f$ :
$f(g(2)) = f(3) = 2(3) + 3 = 9$

Exam tip:

Always work from the inside out: evaluate the function closest to $x$ first, then work outward to the outermost function.

2. Finding Algebraic Expressions★★★☆☆⏱ 15 min

To find the general algebraic form of $(f \circ g) (x)$ , replace every occurrence of $x$ in the outer function $f$ with the full expression of the inner function $g (x)$ , then simplify if required.

📐 Worked Example

Given $f (x) = x + 4$ and $g (x) = 3 x - 2$ , find $(f \circ g) (x)$ in simplest form

1
Replace $x$ in $f (x)$ with the full expression for $g (x)$ :
$f(g(x)) = f(3x - 2) = \sqrt{(3x - 2) + 4}$
2
Simplify the expression inside the square root:
$\sqrt{3x - 2 + 4} = \sqrt{3x + 2}$

3. Domain of a Composite Function★★★★☆⏱ 15 min

The domain of $(f \circ g) (x)$ follows a simple two-part rule:

All $x$ must be in the domain of the inner function $g (x)$
The output $g (x)$ must be in the domain of the outer function $f (x)$

📐 Worked Example

Find the domain of $(f \circ g) (x)$ where $f (x) = x + 4$ and $g (x) = 3 x - 2$

1
Find the domain of the inner function $g (x)$ : $g (x)$ is linear, so its domain is all real numbers:
$\text{dom}(g) = \mathbb{R}$
2
Find the domain of the outer function $f (x)$ : the input to the square root must be non-negative:
$x + 4 \geq 0 \implies \text{dom}(f) = [-4, \infty)$
3
Require $g (x) \in dom (f)$ and solve for $x$ :
$3x - 2 \geq -4 \implies 3x \geq -2 \implies x \geq -\frac{2}{3}$

Exam tip:

Never just use the domain of the simplified composite expression: always check restrictions from the original inner function first.

4. Decomposing Functions★★★☆☆⏱ 10 min

Exam questions often ask you to reverse the composition process: write a given complex function as a composite of two simpler non-trivial functions. This is an important skill for later topics like the chain rule.

📐 Worked Example

Write $h (x) = (2 x + 5)^{3}$ as $h (x) = (f \circ g) (x)$

1
Identify the inner function: the expression that is being substituted into a basic outer function. Here $2 x + 5$ is the inner function:
$g(x) = 2x + 5$
2
Identify the outer function that acts on the inner expression:
$f(x) = x^3$
3
Check that the composition matches the original function:
$f(g(x)) = f(2x + 5) = (2x + 5)^3 = h(x)$

5. Common Pitfalls

Wrong move:

Swapping order of composition, calculating $g (f (x))$ when asked for $f (g (x))$

Why:

Many students misread $f \circ g$ as "f first then g", instead of "f after g"

Correct move:

Remember: $f \circ g = f (g (x))$ , the function closest to $x$ is always applied first

Wrong move:

Taking the domain of the simplified composite expression instead of applying the two-step rule

Why:

Simplification can remove restrictions that originated from the inner function

Correct move:

Always check that $x$ is in the domain of $g$ , and $g (x)$ is in the domain of $f$

Wrong move:

Confusing composition with multiplication, calculating $f (a) \times g (a)$ for $f (g (a))$

Why:

Similar notation leads to this common error, especially in exam pressure

Correct move:

Recall $f (g (x))$ is function composition, not multiplication: substitute inner output into outer function

Wrong move:

Giving a trivial decomposition $f (x) = x, g (x) = h (x)$ for a decomposition question

Why:

While technically correct, this does not answer what the question asks for

Correct move:

Decompose into two non-trivial simpler functions, with the inner function as the substituted expression and outer as the basic function acting on it

6. Quick Reference Cheatsheet

Concept	Notation	Key Rule
Composite (f after g)	$f \circ g$	$(f \circ g) (x) = f (g (x))$
Evaluate at $x = a$	$f (g (a))$	Calculate $g (a)$ first, substitute into $f$
Domain of $f \circ g$	$dom (f \circ g)$	$x \in dom (g)$ , $g (x) \in dom (f)$
Decompose $h (x)$	$h = f \circ g$	Inner = substituted expression, Outer = basic function

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
Evaluate composite function at a point
2022 · 1
Find domain of composite function
2023 · 2
Find algebraic form of composite

What's Next

Composite functions are a foundational building block for almost all advanced function topics in IB AI SL. You will use composition to describe transformations of functions, and later apply it to differentiation (via the chain rule) and integration (via $u$ -substitution). Errors in composition order or domain calculation are common sources of lost marks in these higher-weight topics, so mastering this sub-topic pays off across the whole syllabus. Practice the skills here to build a solid base for what comes next.