Composite Functions (Edexcel IGCSE Maths A (Modular))

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Written by: Mark Curtis

Reviewed by: Dan Finlay

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Composite Functions

What is a composite function?

A composite function is a function applied to the output of another function
- Composite functions may also referred to as compound functions

What notation is used for composite functions?

If $f (x)$ and $g (x)$ are two functions, then
- $g (f (x))$ is a composite function
  - Also written $gf : x$
- 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 $g : x \to {(x + 2)}^{2}$ .

(a) Find $fg (4)$ .

"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 $gf (x)$ .

"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

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

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

You do not need to expand the answer

Last updated: 18 October 2024

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