1MA1 · A7AlgebraHigherGrades 7–9

Functions

A function is a rule that takes an input and produces exactly one output — like a machine. Writing $\mathrm{f}(x) = 3x - 2$ gives the machine a name, so you can ask for $\mathrm{f}(4)$ , feed one machine’s output into another, or reverse the machine to find which input produced a given output.

45 min Sequences & Functions

The big picture

Function notation is compact and precise. $\mathrm{f}(4)$ means "put 4 into the rule f"; $\mathrm{fg}(x)$ means "put $x$ into g, then put the result into f". The order matters, and reading it correctly is where most marks are won or lost.

The inverse function $\mathrm{f}^{-1}$ undoes f. Finding it is rearranging a formula, and Edexcel often combines all three ideas in one question — evaluate, compose, then solve an equation involving an inverse.

What you'll learn

Use function notation to evaluate functions and solve equations
Find composite functions such as $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$
Find inverse functions $\mathrm{f}^{-1}(x)$
Solve problems combining composite and inverse functions

Function notation

$\mathrm{f}(x) = 3x - 2$ means "the function f multiplies by 3 and subtracts 2".

To find $\mathrm{f}(4)$ , substitute: $3(4) - 2 = 10$ . Negative inputs go in brackets: $\mathrm{f}(-1) = 3(-1) - 2 = -5$ .

$\mathrm{f}(x) = 13$ is an equation: $3x - 2 = 13$ , so $x = 5$ .

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

$f(x) = 5 - 2x$ . Find $f(-3)$ .

Your answer

Show worked solution

Solution

1 $f(-3) = 5 - 2(-3) = 5 + 6 = 11$ .

Answer: $11$

Tip — Read $\mathrm{f}(a) = b$ as "input $a$ gives output $b$ ". Deciding whether a question gives you the input or the output tells you whether to substitute or solve.

Composite functions

$\mathrm{fg}(x)$ means $\mathrm{f}(\mathrm{g}(x))$ : apply g first, then f. The function nearest to $x$ acts first.

Numerically, work from the inside out. Algebraically, substitute the whole expression for $\mathrm{g}(x)$ wherever $x$ appears in f.

In general $\mathrm{fg}(x) \neq \mathrm{gf}(x)$ .

\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))

g first, then f.

Example 1

\mathrm{f}(x) = 3x - 2

and

\mathrm{g}(x) = x^{2} + 1

. Find (a)

\mathrm{fg}(2)

(b)

\mathrm{gf}(2)

(c)

\mathrm{fg}(x)

1(a)

\mathrm{g}(2) = 5

, then

\mathrm{f}(5) = 13

2(b)

\mathrm{f}(2) = 4

, then

\mathrm{g}(4) = 17

3(c)

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

Answer(a)

13

(b)

17

(c)

3x^{2} + 1

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

$f(x) = 2x + 1$ and $g(x) = x^2$ . Find $gf(3)$ .

Your answer

Show worked solution

Solution

1 $f(3) = 7$ .
2 $g(7) = 49$ .

Answer: $49$

Why this works

$\mathrm{fg}(2) = 13$ but $\mathrm{gf}(2) = 17$ : the same two machines in a different order give different results, just as "add 1 then square" differs from "square then add 1".

Inverse functions

The inverse $\mathrm{f}^{-1}$ reverses f: if $\mathrm{f}(5) = 13$ then $\mathrm{f}^{-1}(13) = 5$ .

To find it, write $y = \mathrm{f}(x)$ , rearrange to make $x$ the subject, then replace $y$ with $x$ .

$\mathrm{f}^{-1}(x)$ does not mean $\dfrac{1}{\mathrm{f}(x)}$ .

Example 1Find

\mathrm{f}^{-1}(x)

when

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

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

, so

3y = 2x + 5

2x = 3y - 5

, so

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

3Replace

y

with

x

Answer

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

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

$f(x) = \dfrac{x - 4}{5}$ . Find $f^{-1}(x)$ .

Your answer

Show worked solution

Solution

1 $y = \frac{x - 4}{5}$ , so $5y = x - 4$ and $x = 5y + 4$ .
2 $f^{-1}(x) = 5x + 4$ .

Answer: $f^{-1}(x) = 5x + 4$

Tip — Check an inverse by composing: $\mathrm{f}^{-1}(\mathrm{f}(1)) = \mathrm{f}^{-1}\left(\tfrac73\right) = \dfrac{7 - 5}{2} = 1$ ✓.

Solving with functions

Equations such as $\mathrm{fg}(x) = 28$ or $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ are solved by writing each side as an expression first, then solving.

Example 1With

\mathrm{f}(x) = 3x - 2

and

\mathrm{g}(x) = x^{2} + 1

, solve

\mathrm{fg}(x) = 28

\mathrm{fg}(x) = 3x^{2} + 1

3x^{2} + 1 = 28

, so

x^{2} = 9

Answer

x = 3

x = -3

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

With $f(x) = 2x + 1$ and $g(x) = x^2$ , solve $gf(x) = 25$ .

Your answer

Separate with a comma.

Show worked solution

Solution

1 $(2x + 1)^2 = 25$ , so $2x + 1 = \pm 5$ .
2 $x = 2$ or $x = -3$ .

Answer: $x = 2$ or $x = -3$

Think like an examiner

Show the inside function evaluated before the outside one.
Use brackets when substituting an expression into a function.
Write inverses in terms of $x$ , using the notation $\mathrm{f}^{-1}(x)$ .

Functions

\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))

Composite: g first.

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

Inverse undoes f.

y = \mathrm{f}(x) \to x = \dots \to \mathrm{f}^{-1}(x)

Finding an inverse.

Watch out for these

Applying f first in

\mathrm{fg}(x)

The function next to

x

, g, acts first.

Writing

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

The inverse reverses the operations:

\dfrac{x + 2}{3}

Substituting without brackets:

3x^{2} + 1 - 2

for

\mathrm{f}(x^{2} + 1)

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

Giving only

x = 3

for

x^{2} = 9

Include

x = -3

Stretch yourself

$\mathrm{f}(x) = 2x + 5$ . Solve $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ .

Hint — Find $\mathrm{f}^{-1}(x)$ first, then set the two expressions equal.

Reveal solution

SolutionFrom

y = 2x + 5

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

, so

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

. Setting

2x + 5 = \dfrac{x - 5}{2}

4x + 10 = x - 5

, so

3x = -15

and

x = -5

. Check:

\mathrm{f}(-5) = -5

and

\mathrm{f}^{-1}(-5) = \dfrac{-10}{2} = -5

✓. (Graphs of a function and its inverse meet on the line

y = x

, which is why the output equals the input here.)

\blacksquare

Questions students ask

What does

\mathrm{ff}(x)

mean?

Apply f twice:

\mathrm{f}(\mathrm{f}(x))

How do I check a composite function?

Pick a number, evaluate step by step, and compare with substituting into your algebraic answer.

Does every function have an inverse?

At GCSE the functions you invert are one-to-one. Functions like

x^{2}

need a restricted domain, which is A-Level content.

Key takeaways

$\mathrm{f}(a)$ : substitute $a$ . $\mathrm{f}(x) = b$ : solve for $x$ .
$\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$ : g first, then f.
Usually $\mathrm{fg}(x) \neq \mathrm{gf}(x)$ .
Find $\mathrm{f}^{-1}$ by rearranging $y = \mathrm{f}(x)$ for $x$ .
Check inverses by composing: $\mathrm{f}^{-1}(\mathrm{f}(x)) = x$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 8 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 72 marks

$f(x) = x^2 - 3$ and $g(x) = 2x + 1$ . Find $fg(x)$ in its simplest form.[2]
Show mark scheme
- M1 $(2x + 1)^2 - 3$
- A1 $4x^2 + 4x - 2$
Answer: $4x^2 + 4x - 2$

Question 2Grade 83 marks

$f(x) = 3x + 2$ . Solve $f(x) = f^{-1}(x)$ .[3]
Show mark scheme
- M1 $f^{-1}(x) = \frac{x - 2}{3}$
- M1 $3x + 2 = \frac{x - 2}{3}$
- A1 $x = -1$
Answer: $x = -1$

Question 3Grade 93 marks

$f(x) = \dfrac{2x + 3}{x - 1}$ . Find $f^{-1}(x)$ .[3]
Show mark scheme
- M1 $y(x - 1) = 2x + 3$
- M1 $x(y - 2) = y + 3$
- A1 $f^{-1}(x) = \dfrac{x + 3}{x - 2}$
Answer: $\dfrac{x + 3}{x - 2}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

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Test yourself

Answer a few questions, then review anything you missed.

Next step

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