IGCSEMathsCambridge (CIE)ExtendedRevision NotesAlgebra & SequencesFunctionsComposite Functions

Composite Functions (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Exam code: 0580 & 0980

Written by: Amber

Reviewed by: Dan Finlay

Updated on 20 January 2026

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Understand this topic with deeper explanations and understanding checks.

Composite functions

What is a composite function?

A composite function is a function applied to the output of another function
- The input goes through the 1st function to become an output
- This output goes through the 2nd function to become a new output

What notation is used for composite functions?

If $f (x)$ and $g (x)$ are two functions, then
- $g (f (x))$ is a composite function
- It means the input $x$ goes through function $f$ first
  - This gives the output $f (x)$
  - Then this output, $f (x)$ , becomes the input of function $g$ , giving $g (f (x))$
- $gf (x)$ is the shorthand notation used for $g (f (x))$
  - It means do $f$ first, then $g$
  - The order of applying the functions goes from right to left
  - (the letter nearest the bracket goes first)
  - This is often the opposite of what people expect!
- $fg (x)$ means do $g (x)$ first then $f (x)$ second
- $ff (x)$ means apply $f (x)$ twice!
  - This can be written $f^{2} (x)$
  - This does not mean the same as ${[f (x)]}^{2}$

Examiner Tips and Tricks

A good trick in the exam is to write brackets around $gf (x)$ to make it $g (f (x))$ , to see that it is "g" of "f(x)".

How do I substitute numbers into composite functions?

If you are putting a number into a composite function
- put the number into the function closest to (x)
- then make the output of the first function the input of the second function
For example, if $f (x) = 2 x + 1$ and $g (x) = \frac{1}{x}$
- to find $gf (2)$ :
  - Put the 2 in as the input of $f$ first
  - $f (2) = 2 (2) + 1 = 5$
  - Then put 5 in as the input of $g$
  - So $gf (2) = g (f (2)) = g (5) = \frac{1}{5}$
- to find $fg (2)$ :
  - Put the 2 in as the input of $g$ first
  - $g (2) = \frac{1}{2}$
  - Then put $\frac{1}{2}$ in as the input of $f$
  - So $fg (2) = f (g (2)) = f (\frac{1}{2}) = 2 (\frac{1}{2}) + 1 = 2$
- to find $ff (2)$ :
  - $f (2) = 2 \times 2 + 1 = 5$
  - $f (5) = 2 \times 5 + 1 = 11$
  - so $ff (2) = 11$

How do I find composite functions algebraically?

If you are using algebra, substitute the whole algebraic expression as your input
- For example, if $f (x) = 2 x + 1$ and $g (x) = \frac{1}{x}$
  - $fg (x) = f (g (x)) = f (\frac{1}{x}) = 2 \times (\frac{1}{x}) + 1 = \frac{2}{x} + 1$
  - $gf (x) = g (f (x)) = g (2 x + 1) = \frac{1}{2 x + 1}$
  - $ff (x) = f (f (x)) = f (2 x + 1) = 2 (2 x + 1) + 1$ which simplifies to $ff (x) = 4 x + 3$

Worked Example

In this question, $f (x) = 2 x - 1$ and $g (x) = {(x + 2)}^{2}$ .

(a) Find $fg (4)$ .

Answer:

"g" is on the inside of the composite function, so apply g first

$g (4) = {(4 + 2)}^{2} = 6^{2} = 36$

Now apply the function "f" to 36

$\begin{array}{rcl} f (36) & = & 2 (36) - 1 \\ = & 72 - 1 \end{array}$

$fg (4) = 71$

(b) Find $gf (x)$ .

Answer:

"f" is on the inside of the composite function so substitute the function f(x) into g(x)
It can help to write $gf (x) = g (f (x))$

$gf (x) = g (f (x)) = g (2 x - 1) = {((2 x - 1) + 2)}^{2}$

Simplify inside the bracket

$\begin{array}{rcl} gf (x) & = & {(2 x - 1 + 2)}^{2} \end{array}$

$gf (x) = {(2 x + 1)}^{2}$

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Questions students ask about Composite Functions

Real questions from students revising this topic on Save My Exams

Does fg(x) mean multiplying f(x) by g(x)?

No — fg(x) is not a multiplication; it means put x into g first, then put that answer into f.

fg(x) is shorthand for f(g(x))
Work right to left: the letter nearest the bracket goes first
The output of the first function becomes the input of the second
Multiplying would be written f(x) × g(x), which is a different thing
Writing the extra brackets, f(g(x)), helps you see it is “f of g(x)”

Example from this page: f(x) = 2x − 1 and g(x) = (x + 2)². Find fg(4).

g is nearest the bracket, so do g first: g(4) = (4 + 2)² = 6² = 36
Put 36 into f: f(36) = 2(36) − 1 = 71

So fg(4) = 71.

Multiplying instead would give f(4) × g(4) = 7 × 36 = 252 — nowhere near 71.

The same idea works in algebra: for gf(x) you replace the x in g with the whole of f(x).

Do you need to find the domain of a composite function?

Not for the questions on this page — they only ask you to work out a composite function's value or its expression — but it is worth knowing which inputs a composite function cannot take.

The domain is the set of x-values you are allowed to put in. A composite function cannot take any x that breaks either step.

Example using this page's functions: f(x) = 2x + 1 and g(x) = 1/x

g(x) = 1/x cannot take x = 0
gf(x) = 1/(2x + 1): the denominator is zero when x = −½, so that value is excluded
fg(x) = 2/x + 1: x = 0 is excluded, because g(0) does not exist

You would write gf(x) = 1/(2x + 1), x ≠ −½.

If a question does ask for a domain, find the values that would break either step and leave them out. Otherwise, just answer what is asked.

Do you always need to expand brackets when finding a composite function?

No — gf(x) = (2x + 1)² is already a correct, fully simplified answer; expanding just rewrites the same function.

The page's working for part (b): f(x) = 2x − 1 and g(x) = (x + 2)²

gf(x) = g(2x − 1) = ((2x − 1) + 2)²
Simplify inside the bracket: (2x + 1)²

The question only says “Find gf(x)”, so stopping there is fine. Do always tidy up inside the bracket, though.

If you did expand:

(2x + 1)² = (2x + 1)(2x + 1) = 4x² + 4x + 1
Same function, different form — check with x = 1: 3² = 9 and 4 + 4 + 1 = 9

Expand only when the question asks for it, for example “in the form ax² + bx + c”. Expanding for no reason just gives you another chance to slip up.

On this page ff(x) = 2(2x + 1) + 1 is tidied to 4x + 3, because collecting like terms genuinely simplifies it.