Composite functions and inverses · Functions

In General Maths you met function notation and the inverse of a function (see algebraic expressions): $f(3)$ means “put 3 in place of every $x$ ”, and $f^{-1}$ undoes $f$ . Further Maths joins functions together. This lesson shows how to put one function inside another, then how inverses fit in.

WAEC often writes a function with an arrow: $f: x \to 2x + 1$ means the same as $f(x) = 2x + 1$ .

Values and unknowns

Two facts carry over from General Maths:

Put the input in everywhere. $f(x + 2)$ means replace every $x$ by $(x + 2)$ , in brackets.
Given values give equations. If $f(1) = 3$ , substitute $x = 1$ and set the result equal to 3. Two unknowns need two values, solved simultaneously (see polynomials).

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Composite functions: one inside another

$f \circ g$ , also written $fg$ , means do $g$ first, then $f$ :

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

Read it from the inside out: the function nearest to $x$ acts first. The order matters, because doing the same two steps the other way round usually gives a different answer:

Order mattersf∘g: g first. g∘f: f first

To find $f \circ g(x)$ as an expression, write out $g(x)$ , then put the whole of it, in brackets, in place of every $x$ in $f$ .

For example, with ${f(x) = 2x + 1}$ and ${g(x) = x^2}$ :

$f \circ g(x) = f(x^2)$ . Replace $x$ in $f$ by $x^2$ : ${2x^2 + 1}$ .
$g \circ f(x) = g(2x + 1)$ . Replace $x$ in $g$ by $2x + 1$ : ${(2x + 1)^2 = 4x^2 + 4x + 1}$ .

Run both orders on the same input:

Which machine first?Pick a pair, slide x

f∘g: g first, with f(x) = 2x + 1 and g(x) = x²

2→g→4→f→9

g∘f: f first

2→f→5→g→25

9f∘g(x) = 2x² + 125g∘f(x) = (2x + 1)² = 4x² + 4x + 1

x = 2

f∘g means do g first, then f. Doing them the other way round gives 25, not 9: the order matters.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q9 (b)

Given that $f: x \to x^2 + 1$ and $g: x \to \dfrac{3x + 1}{x - 1}$ , $x \ne 1$ , find: (i) $g\circ f$ ; (ii) $g\circ f(2)$ .

Which function first?
- $g \circ f$ means $f$ first, then $g$ .
- So ${g \circ f(x) = g(f(x)) = g(x^2 + 1)}$ .
Think first. In g∘f, which acts first?
Put f(x) into g
- The top: ${3(x^2 + 1) + 1 = 3x^2 + 4}$ .
- The bottom: ${(x^2 + 1) - 1 = x^2}$ .
- So ${g \circ f(x) = \frac{3x^2 + 4}{x^2}}$ .
- Divide each term by $x^2$ : ${g \circ f(x) = 3 + \frac{4}{x^2}}$ .
Think first. Replace every x in (3x + 1)/(x − 1) by x² + 1.
The value at 2
- ${g \circ f(2) = 3 + \frac44 = 4}$ .
- Check the long way: ${f(2) = 5}$ , then ${g(5) = \frac{16}{4} = 4}$ ✓.
Think first. Put x = 2 into 3 + 4/x².

More: composite functions

Inverse functions

The inverse $f^{-1}$ undoes $f$ : if $f$ takes 3 to 4, then $f^{-1}$ takes 4 back to 3. To find it:

write $y = f(x)$ ;
make $x$ the subject;
swap the letter: write the result with $x$ in place of $y$ .

On a graph, swapping $x$ and $y$ reflects the curve in the line $y = x$ :

f and f⁻¹Reflections in y = x: (3, 4) on f, (4, 3) on f⁻¹

For example, for $f(x) = 2x - 2$ :

Write ${y = 2x - 2}$ .
Add 2 to both sides: ${y + 2 = 2x}$ .
Divide by 2: ${x = \frac{y + 2}{2}}$ .
So ${f^{-1}(x) = \frac{x + 2}{2}}$ . Check: $f(3) = 4$ and $f^{-1}(4) = 3$ ✓.

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Inverses inside composites

Questions such as “find $g^{-1} \circ f$ ” combine both ideas. Find the inverse first, on its own line. Then compose as usual: the function nearest to $x$ acts first.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q10 (b)

Two functions $f$ and $g$ are defined by $f: x \to \frac12x + 1$ and $g: x \to \frac{5x - 1}{3}$ . Find: (i) $g^{-1}$ ; (ii) $g^{-1}\circ f$ .

The inverse of g
- Write ${y = \frac{5x - 1}{3}}$ .
- Multiply by 3: ${3y = 5x - 1}$ .
- Add 1: ${3y + 1 = 5x}$ .
- Divide by 5: ${x = \frac{3y + 1}{5}}$ , so ${g^{-1}(x) = \frac{3x + 1}{5}}$ .
Think first. Write y = (5x − 1)/3 and make x the subject.
Compose
- $f$ acts first: ${g^{-1} \circ f(x) = g^{-1}\left(\frac12x + 1\right)}$ .
- Replace $x$ in $g^{-1}$ : ${\frac{3\left(\frac12x + 1\right) + 1}{5}}$ .
- Multiply out the top: ${\frac{\frac32x + 4}{5}}$ .
- Multiply top and bottom by 2: ${g^{-1} \circ f(x) = \frac{3x + 8}{10}}$ .
Think first. g⁻¹∘f: which acts first?

More: inverses inside composites

WAEC 2008 · Paper 2 · Q1Two functions $f$ and $g$ are defined on the set $\mathbb{R}$ of real numbers by $f : x \to \dfrac{x + 1}{x - 2}$ , $x \neq 2$ , …

Your turn

WAEC 2017 · Paper 2 · Q1 (a)

(a)
If $f(x) = \dfrac{4 - 5x}{2}$ and $g(x) = x + 6$ , $x \in \mathbb R$ , find $f \circ g^{-1}$ .

Worked solution (try it first)

(a)

Find $g^{-1}$ first: $y = x + 6$ gives $x = y - 6$ , so $g^{-1}(x) = x - 6$ .
$g^{-1}$ acts first: $f \circ g^{-1}(x) = f(x - 6)$
$= \dfrac{4 - 5(x - 6)}{2}$ .
Multiply out the top: $4 - 5x + 30 = 34 - 5x$ .
Divide each term by 2: $f \circ g^{-1}(x) = 17 - \frac52x$ .

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