A7AlgebraHigherGrades 6–9

Functions

Q: Why swap x and y for the inverse?

The inverse is conventionally written as a function of x, so the letters are renamed at the end.

A function is a rule that turns an input into an output. Writing $f(x) = 2x + 3$ names the rule $f$ , so $f(4)$ means "put 4 into the rule", giving 11. AQA Higher papers use function notation to ask for outputs, solve equations, combine functions and reverse them.

45 min AQA GCSE Maths

The big picture

Think of a function as a machine. A $\textbf{composite function}$ feeds the output of one machine straight into another; an $\textbf{inverse function}$ runs a machine backwards to recover the input.

The notation is compact, and most errors come from reading it the wrong way round. $fg(x)$ means apply $g$ first, then $f$ — the function nearest the $x$ acts first.

What you'll learn

Interpret and use function notation
Solve equations written in function notation
Find composite functions such as $fg(x)$
Find inverse functions $f^{-1}(x)$

Function notation

$f(x) = 2x + 3$ gives $f(4) = 2(4) + 3 = 11$ and $f(-1) = 1$ .

You can substitute expressions as well: $f(a + 1) = 2(a + 1) + 3 = 2a + 5$ .

Solving $f(x) = 17$ means finding the input: $2x + 3 = 17$ , so $x = 7$ .

Example 1

g(x) = x^{2} - 1

. Find

g(-3)

and solve

g(x) = 48

g(-3) = 9 - 1 = 8

x^{2} - 1 = 48

, so

x^{2} = 49

Answer

g(-3) = 8

;

x = 7

x = -7

Your turnSame method, new numbers — try it before moving on.

$f(x) = 3x - 7$ . Solve $f(x) = 11$ .

Your answer

Show worked solution

Solution

1 $3x - 7 = 11$ , so $3x = 18$ .
2 $x = 6$ .

Answer: $x = 6$

Composite functions

$fg(x)$ means $f(g(x))$ : apply $g$ , then apply $f$ to the result.

With $f(x) = 2x + 3$ and $g(x) = x^{2} - 1$ : $fg(3) = f(8) = 19$ , but $gf(3) = g(9) = 80$ . Order matters.

Example 1Using

f(x) = 2x + 3

and

g(x) = x^{2} - 1

, find

fg(x)

and

gf(x)

fg(x) = f(x^{2} - 1) = 2(x^{2} - 1) + 3 = 2x^{2} + 1

gf(x) = g(2x + 3) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 8

Answer

fg(x) = 2x^{2} + 1

gf(x) = 4x^{2} + 12x + 8

Your turnSame method, new numbers — try it before moving on.

$f(x) = x + 4$ and $g(x) = 2x^2$ . Find $gf(1)$ .

Your answer

Show hint

$gf(1)$ means find $f(1)$ first.

Show worked solution

Solution

1 $f(1) = 5$ .
2 $g(5) = 2 \times 25 = 50$ .

Answer: $50$

Example 2Solve

gf(x) = 0

4x^{2} + 12x + 8 = 0

, so

x^{2} + 3x + 2 = 0

(x + 1)(x + 2) = 0

Answer

x = -1

x = -2

Tip — Replace every $x$ in the outer function with the whole inner expression in brackets.

Inverse functions

The inverse $f^{-1}$ reverses $f$ : if $f(4) = 11$ then $f^{-1}(11) = 4$ .

Method: write $y = f(x)$ , rearrange to make $x$ the subject, then swap $y$ for $x$ .

Example 1Find

f^{-1}(x)

when

f(x) = \dfrac{5x - 2}{3}

y = \dfrac{5x - 2}{3}

, so

3y = 5x - 2

x = \dfrac{3y + 2}{5}

Answer

f^{-1}(x) = \dfrac{3x + 2}{5}

Your turnSame method, new numbers — try it before moving on.

$f(x) = 4 - 3x$ . Find $f^{-1}(x)$ .

Your answer

Type it like (4 - x)/3.

Show worked solution

Solution

1 $y = 4 - 3x$ , so $3x = 4 - y$ and $x = \dfrac{4 - y}{3}$ .
2Swap: $f^{-1}(x) = \dfrac{4 - x}{3}$ .

Answer: $f^{-1}(x) = \dfrac{4 - x}{3}$

Why this works

Composing a function with its inverse gets you back where you started: $ff^{-1}(x) = x$ . That makes a quick check — substitute one value into $f$ , then into $f^{-1}$ .

Think like an examiner

Show the substitution step for composite functions.
Give the inverse in terms of $x$ , written as $f^{-1}(x) = \dots$ .
Expand and simplify composite functions unless told otherwise.

Remember these

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

Composite: g first, then f.

y = f(x) \Rightarrow x = f^{-1}(y)

Inverse: rearrange for x.

ff^{-1}(x) = x

Inverse undoes the function.

Watch out for these

Working out

fg(x)

by applying

f

first.

The function next to

x

acts first.

Thinking

f^{-1}(x) = \dfrac{1}{f(x)}

The

-1

means inverse, not reciprocal.

Writing

fg(x)

f(x) \times g(x)

It is a composition, not a product.

Forgetting brackets:

(2x + 3)^{2} - 1

written as

2x^{2} + 9 - 1

Square the whole expression.

Stretch yourself

$f(x) = 3x - 1$ . Solve $ff(x) = f^{-1}(x)$ .

Hint — Find $ff(x)$ and $f^{-1}(x)$ separately, then form an equation.

Reveal solution

Solution

ff(x) = 3(3x - 1) - 1 = 9x - 4

. For the inverse,

y = 3x - 1

gives

x = \dfrac{y + 1}{3}

, so

f^{-1}(x) = \dfrac{x + 1}{3}

. Solve

9x - 4 = \dfrac{x + 1}{3}

27x - 12 = x + 1

26x = 13

x = \tfrac12

. Check:

ff(\tfrac12) = 0.5

and

f^{-1}(\tfrac12) = 0.5

✓.

\blacksquare

Questions students ask

fg(x)

the same as

gf(x)

Usually not. Always work out which function acts first.

Why swap

x

and

y

for the inverse?

The inverse is conventionally written as a function of

x

, so the letters are renamed at the end.

Do all functions have inverses?

Only one-to-one functions. At GCSE you will be given functions that have inverses.

Key takeaways

$f(a)$ means substitute $a$ into the rule.
$fg(x) = f(g(x))$ : inside function first.
Composite functions are not products.
Find $f^{-1}$ by rearranging $y = f(x)$ for $x$ .
$f^{-1}$ undoes $f$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 12 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 65 marks

$f(x) = 2x - 5$ and $g(x) = x^2$ .

(a)Work out $fg(3)$ .[2]
Show mark scheme (a)
- M1 $g(3) = 9$
- A1 $13$
Answer: $13$
(b)Solve $gf(x) = 49$ .[3]
Show mark scheme (b)
- M1 $(2x - 5)^2 = 49$
- M1 $2x - 5 = 7$ or $2x - 5 = -7$
- A1 $x = 6$ and $x = -1$
Answer: $x = 6$ or $x = -1$

Question 2Grade 84 marks

$f(x) = \dfrac{x + 3}{2}$ and $g(x) = 4x - 1$ .

Find $(gf)^{-1}(x)$ .[4]
Show mark scheme
- M1 $gf(x) = 4 \times \frac{x + 3}{2} - 1$
- A1 $gf(x) = 2x + 5$
- M1Rearranges $y = 2x + 5$
- A1 $\dfrac{x - 5}{2}$
Answer: $\dfrac{x - 5}{2}$

Question 3Grade 93 marks

$f(x) = x^2 - 6x + 2$ . Show that $f(x) \ge -7$ for all values of $x$ .[3]
Show mark scheme
- M1Completes the square: $(x - 3)^2 - 7$
- A1Correct form
- B1Explains $(x - 3)^2 \ge 0$ , so $f(x) \ge -7$

Unlimited practice

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