Composite Functions (Cambridge International IGCSE Further Maths)

Composite functions

What is a composite function?

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

• 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

  • $\mathrm{fg}(x)$
  • $f(g(x))$
  • All of these mean “$f$ of $g(x)$”

How do I work with composite functions?

• Recognise the notation The order matters

  • $\mathrm{fg}(x)$ means “f of g of x”

  • First apply $g$* to $x$ *to get $g(x)$

  • Then apply $f$ to $\mathrm{g} ( x )$ to get $f(g(x))$

  • Always start with the function closest to the variable

  • $\mathrm{fg}(x)$ is not usually equal to $\mathrm{gf}(x)$

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, $ff^{- 1} ( x ) = \mathrm{f}^{- 1} \mathrm{f} ( x ) = x$

Worked Example Two functions, $\mathrm{f} ( x )$ and $\mathrm{g} ( x )$ are

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

a) Find $\mathrm{f} ( 3 )$ and $\mathrm{g} ( 3 )$.

$\begin{gathered} \mathrm{f} ( 3 ) & = ( 3 )^{2} + 3 ( 3 ) - 2 \\ = 9 + 9 - 2 \\ = 16 \end{gathered}$

$\mathrm{g} ( 3 ) = 3 + 3$

$f(3)=16,g(3)=6$

b) Find, in terms of $x$, $fg ( x )$.

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

$\begin{gathered} \therefore \text{fg} ( x ) & = \mathrm{f} [ x + 3 ] \\ = ( x + 3 )^{2} + 3 ( x + 3 ) - 2 \\ = x^{2} + 6 x + 9 + 3 x + 9 - 2 \end{gathered}$

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

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 $\mathrm{g} ( x )$

• However, for this to be the case, the range of $\mathrm{g} ( x )$ must be contained within the domain of $\mathrm{f} ( x )$

  • 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 $\mathrm{f} ( x )$

  • But if the domain of $\mathrm{f} ( x )$ has been affected, the range of $fg ( x )$ will also be affected

Exam Hint

• 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, $\mathrm{f} ( x )$ and $\mathrm{g} ( x )$ are defined as follows

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

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

a) Write down the range of $\mathrm{f} ( x )$ and the range of $\mathrm{g} ( x )$.

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

Final answer: 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 $\mathrm{g} ( x )$.

Final answer: The range of$g(x)$is$g>1$

b) Use your answers to (a) to help explain why $fg ( x )$ does not exist.

Final answer: $g$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$0<x<1$ $\therefore\mathrm{fg}(x)$does not exist

c) 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 $\mathrm{g} ( x > 1 )$.

Final answer: The range of$\mathrm{gf}$is$\mathrm{gf}>1$