Learn9709Paper 1 · Pure Mathematics 1FunctionsComposite functions

Functions · Subtopic 2 of 5

Composite Functions

Applying one function after another: what $fg(x)$ means, finding and solving with composite expressions, and when a composite function exists.

Cambridge 9709 Paper 1Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Understand that $fg(x)$ means $f\big(g(x)\big)$ : apply $g$ first, then $f$ .
Evaluate composites such as $fg(2)$ , and find expressions for $fg(x)$ , $gf(x)$ and $ff(x)$ .
Solve equations involving composite functions.
Decide whether a composite exists, using the range of the first function and the domain of the second.
Find the range of a composite function on a given domain.

Pause and think

Warm-up: if $f(x)=2x+3$ , what is $f(5)$ ? What is $f(\text{banana})$ ? What is $f(x^2-1)$ ?

Section 2 of 14

Core idea

Start hereTwo machines in a row

A factory paints a block (machine $g$ ), then packs it (machine $f$ ). The finished product is "pack(paint(block))". The order matters: painting a packed box is not the same thing. When the output of one function is fed straight into another, we get a composite function.

The big idea

$fg(x)$ means $f(g(x))$ : apply $g$ first, then apply $f$ to the result. Usually $fg\neq gf$ . For $fg$ to exist, every output of $g$ must lie in the domain of $f$ .

1. Reading the notation

fg(3): g first, then f

fg(3)=f(g(3))=f(9)=19

. The function written nearest the

x

acts first.

fg(x)=f\big(g(x)\big)

$g(x)$: the inner function, applied first.
$f(\ldots)$: the outer function, applied to whatever $g$ produced.

Worked example $f(x)=2x+1$ , $g(x)=x^2$

Find $fg(3)$ , $gf(3)$ , and expressions for $fg(x)$ and $gf(x)$ .

Goal: Numerical values and formulas.
Strategy: Work from the inside out.

$g(3)=9,\quad fg(3)=f(9)=19$
$g$ first.
$f(3)=7,\quad gf(3)=g(7)=49$
$f$ first.
$fg(x)=f(x^2)=2x^2+1$
Replace the $x$ in $f$ by $x^2$ .
$gf(x)=g(2x+1)=(2x+1)^2=4x^2+4x+1$
Square the whole of $f(x)$ .

Answer

19

49

;

fg(x)=2x^2+1

gf(x)=4x^2+4x+1

✓ Check

fg(3)=2(9)+1=19

✓;

gf(3)=36+12+1=49

✓.

Common misconception $fg$ is not $f\times g$

fg(x)

is "

f

g

x

", not

f(x)\cdot g(x)

. And

f^2(x)

means

ff(x)

, applying

f

twice.

2. When does a composite exist?

Worked exampleDomain condition

$f(x)=\sqrt{x-4}$ for $x\ge4$ ; $g(x)=x+k$ for $x\ge0$ . Find the least $k$ so that $fg$ exists.

Goal: Smallest $k$ .
Strategy: The range of $g$ must sit inside the domain of $f$ .

$\text{range of }g:\ g(x)\ge k$
Smallest output at $x=0$ .
$\text{need }k\ge4$
All outputs of $g$ must be $\ge4$ .

Answer

k=4

✓ CheckWith

k=4

fg(0)=\sqrt{0}=0

is defined ✓; with

k=3

fg(0)=\sqrt{-1}

is not.

3. Solving composite equations

Worked exampleSolve $gf(x)=25$

With $f(x)=2x+1$ , $g(x)=x^2$ , solve $gf(x)=25$ .

Goal: All $x$ .
Strategy: Write the composite out, then solve.

$(2x+1)^2=25$
$gf(x)=(2x+1)^2$ .
$2x+1=\pm5$
Square root, both signs.
$x=2\quad\text{or}\quad x=-3$
Solve each.

Answer

x=2

x=-3

✓ Check

gf(-3)=g(-5)=25

✓.

Pause and think

With $f(x)=2x+1$ , find $ff(x)$ .

Summary

$fg(x)=f(g(x))$ : right-hand function first.
Order matters.
Range of $g$ ⊆ domain of $f$ .

Composite function: One function applied to the output of another.

Notation

fg(x)=f\big(g(x)\big)\qquad gf(x)=g\big(f(x)\big)\qquad ff(x)=f^2(x)=f\big(f(x)\big)

With $f(x)=2x+3$ and $g(x)=x^2-1$ : $fg(2)=f(3)=9$ but $gf(2)=g(7)=48$ . Order matters, and in general $fg\ne gf$ .

Fun factSocks, then shoes. Getting dressed is a composite: socks first, then shoes. Do it in the other order and you get a very different result. Composite functions work the same way, which is why

fg

and

gf

are usually different.

In the real worldCurrency conversion is a composite. To change Pakistani rupees to pounds through dollars, you apply "rupees to dollars" and then "dollars to pounds". If

d(r)=\tfrac{r}{280}

and

p(d)=0.79d

, then rupees to pounds is

pd(r)=\tfrac{0.79r}{280}

. (These rates are only examples.)

When does fg exist?

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Try this

The third curve is $fg(x)=\sqrt{(3-x^2)-1}$ . Why does it only appear between about $x=-1.4$ and $x=1.4$ ?
$f$ only accepts inputs $\ge1$ . Find where $g(x)\ge1$ : that is exactly where $fg$ exists.

$fg(x)$ exists only when $g(x)$ lands inside the domain of $f$ . On all real numbers the range of $g$ is $g(x)\le3$ , which includes values below $1$ , so $fg$ is not defined for every $x$ .

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