Composite Functions

Video lesson · from 25:183 worked examples

One lesson video covers all of Functions & Graphs, so it opens at 25:18 for this topic — not from the beginning.

Theory

A composite function is when one function is 'inside' another function.

It is formed by applying one function to the result of another.

e.g., $g(f(x))$ means you compute $f(x)$ first, and plug the result into $g$ .

⚠️ Common Examiner Traps

Order matters: $f(g(x))$ means do $g$ first, then $f$ . Reading it left to right and applying $f$ first is the standard error, and gives a completely different function.
Substitute the whole function: every $x$ in the outer function is replaced by the entire inner expression, in brackets.
Use brackets, then expand: $f(x) = x^2$ with $g(x) = x+3$ gives $(x+3)^2$ , which is not $x^2+9$ .
$f(g(x))$ rarely equals $g(f(x))$ : if a question asks you to show they are equal, that is a special result worth checking carefully.

Worked examples

Example 1

Functions $f$ and $g$ are defined by $f(x) = 2x$ and $g(x) = x - 3$ .

Both are defined on suitable domains.

a) Find $f(g(x))$ .

b) Find $g(f(x))$ .

a) To find $f(g(x))$ , substitute $g(x)$ into $f$ :

f(g(x)) = f(x - 3)

f(g(x)) = 2(x - 3) = 2x - 6

b) To find $g(f(x))$ , substitute $f(x)$ into $g$ :

g(f(x)) = g(2x)

g(f(x)) = (2x) - 3 = 2x - 3

Example 2

Functions $f(x) = x^3 + 3$ and $g(x) = \frac{1}{x}$ are defined on suitable domains.

Find formulae for $h(x) = f(g(x))$ and $k(x) = (g \circ f)(x)$ .

For $h(x) = f(g(x))$ :

f(g(x)) = f\left(\frac{1}{x}\right)

h(x) = \left(\frac{1}{x}\right)^3 + 3 = \frac{1}{x^3} + 3

For $k(x) = g(f(x))$ :

g(f(x)) = g(x^3 + 3)

k(x) = \frac{1}{x^3 + 3}

Example 3

Example 3 – Composite Functions with Fractions

Functions $f$ , $g$ and $h$ are defined by $f(x) = \frac{x}{1-x}$ , $g(x) = \frac{1}{x}$ and $h(x) = \frac{1}{1-x}$ . All are defined on suitable domains.

a) Find $f(g(x))$ .

b) Find $f(h(x))$ .

a) Substitute $g(x)$ into $f$ :

\begin{aligned} f(g(x)) &= f\left(\frac{1}{x}\right) \\ &= \frac{\frac{1}{x}}{1 - \frac{1}{x}} \end{aligned}

Multiply numerator and denominator by $x$ :

f(g(x)) = \frac{1}{x - 1}

b) Substitute $h(x)$ into $f$ :

\begin{aligned} f(h(x)) &= f\left(\frac{1}{1-x}\right) \\ &= \frac{\frac{1}{1-x}}{1 - \frac{1}{1-x}} \end{aligned}

Multiply numerator and denominator by $(1-x)$ :

\begin{aligned} f(h(x)) &= \frac{1}{(1-x) - 1} \\ &= \frac{1}{-x} \\ &= -\frac{1}{x} \end{aligned}