Composite Functions (CIE IGCSE Additional Maths)

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Composite Functions

What is a composite function?

A composite function is where one function is applied after another function

Composite function fg(x) applies g first then f

The ‘output’ of one function will be the ‘input’ of the next one
Sometimes called function-of-a-function
A composite function can be denoted
- $fg (x)$
- $f (g (x))$
- All of these mean “ $f$ of $g (x)$ ”

Composite functions as function machines

How do I work with composite functions?

Notation for composite functions

Recognise the notation
- $fg (x)$ means “f of g of x”
The order matters
- First apply $g$ to $x$ to get $g (x)$
- Then apply $f$ to $g (x)$ to get $f (g (x))$
- Always start with the function closest to the variable
- $fg (x)$ is not usually equal to $gf (x)$

Special cases

$fg (x)$ and $gf (x)$ are generally different but can sometimes be the same
$ff (x)$ is written as $f^{2} (x)$
- Note that trig functions are exceptions to this rule
  - e.g. $\sin^{2} (x)$ means ${(\sin (x))}^{2}$ not $\sin (\sin (x))$
For inverse functions, ${ff}^{- 1} (x) = f^{- 1} f (x) = x$

Worked example

Two functions, $f (x)$ and $g (x)$ are

$f (x) = x^{2} + 3 x - 2 g (x) = x + 3$

Find

f (3)

and

g (3)

$\begin{array}{rcl} f (3) & = & {(3)}^{2} + 3 (3) - 2 \\ = & 9 + 6 - 2 \end{array}$

$g (3) = 3 + 3$

$f (3) = 13, g (3) = 6$

Find, in terms of

x

fg (x)

$g$ is the first function to be applied ...

$\begin{array}{rcl} ∴ fg (x) & = & f [x + 3] \\ = & {(x + 3)}^{2} + 3 (x + 3) - 2 \\ = & x^{2} + 6 x + 9 + 3 x + 9 - 2 \end{array}$

$fg (x) = x^{2} + 9 x + 16$

Domain & Range of Composite Functions

How do I find the domain and range of composite functions?

Use logic to determine the domain and range of a composite function
For $fg (x)$ the first function to be applied will be $g$
- So, at best, the domain of $fg (x)$ will be the same as the domain of $g (x)$
However, for this to be the case, the range of $g (x)$ must be contained within the domain of $f (x)$
- If this is not the case, then restrictions on the domain of $fg (x)$ will be required
Similarly, at best, the range of $fg (x)$ will be the same as the range of $f (x)$
- But if the domain of $f (x)$ has been affected, the range of $fg (x)$ will also be affected

Exam Tip

Domain and range are important in composite funcitons like $fg (x)$
- the ‘output’ (range) of g must be in the domain of f(x), so $fg (x)$ could exist,
  but $gf (x)$ may not (or not for some values of $x$ )

Worked example

Two functions, $f (x)$ and $g (x)$ are defined as follows

$f (x) = \frac{1}{x}, 0 < x < 1$

$g (x) = x^{2}, x > 1$

Write down the range of

f (x)

and the range of

g (x)

As the domain of $f (x)$ is $0 < x < 1$ , $\frac{1}{x}$ will always be greater than 1,

The range of $f (x)$ is $f > 1$

The square of any value will be positive or zero, but here, $x = 0$ is not included in the domain for $g (x)$ .

The range of $g (x)$ is $g > 1$

Use your answers to (a) to help explain why

fg (x)

does not exist.

$g$ is the first function to be applied
The range of $g$ would need to be contained within the domain of $f$
But the range of $g$ is $g > 1$ which is outside the domain of $f$ which is $0 < x < 1$
$∴ fg (x)$ does not exist

Find the range of

gf (x)

$f$ is the first function. The range of $f$ is $f > 1$ . This is the same as the domain of $g (x > 1)$ .

The range of $gf$ is $gf > 1$

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