Functions · Study Notes · Edexcel IGCSE 4MA1 Higher

Function notation, evaluation, composite functions, and inverses. A* topic — guaranteed on every Higher Tier paper.

At a glance

$f(x)$ : rule mapping input to output.
Evaluate: substitute number for $x$ .
$fg(x) = f(g(x))$ : apply $g$ FIRST, then $f$ .
Inverse $f^{-1}(x)$ : undoes $f$ .
Method for inverse: swap $x, y$ and solve.
$f^{-1}$ is NOT the reciprocal $1/f$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.2 — Use function notation $f(x)$ .
3.2 — Evaluate functions for given values.
3.2 — Find composite functions $fg(x)$ .
3.2 — Find inverse functions $f^{-1}(x)$ .

Function notation and evaluation

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$f(x)$ is a recipe. $f(5)$ means 'apply the recipe to $5$ '.

Function notation:

$f(x)$ reads 'f of x'.
$x$ is the INPUT.
$f(x)$ is the OUTPUT.

Examples:

A function takes an input, applies its rule, and produces one output.

If $f(x) = 2x + 3$ :

$f(0) = 3$ .
$f(5) = 13$ .
$f(-1) = 1$ .
$f(a) = 2a + 3$ .
$f(x + 1) = 2(x + 1) + 3 = 2x + 5$ .

Substitution rule.

Wherever you see $x$ in the function definition, replace with the input. Use BRACKETS around the input.

For $f(x) = x^{2} + 3x$ :

$f(2) = (2)^{2} + 3(2) = 4 + 6 = 10$ .
$f(-2) = (-2)^{2} + 3(-2) = 4 - 6 = -2$ . (Note: brackets matter!)

Edexcel tip. Always show the substitution explicitly: ' $f(5) = 3(5) - 7 = 8$ '. Mark schemes credit each step.

$f(x)$ = function notation.
Input goes in; output comes out.
Use brackets when substituting.
Show every step.

Composite functions

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$fg(x) = f(g(x))$ . Apply $g$ first, $f$ second.

Composite function: $fg(x) = f(g(x))$ .

Read right-to-left like onion layers — the inner one is applied first.

Method (algebraic):

Find $g(x)$ as an expression.
Substitute that into $f$ wherever $x$ appears.
Simplify.

Example. $f(x) = 2x + 3$ , $g(x) = x^{2} - 1$ .

(a) $fg(x) = f(g(x)) = f(x^{2} - 1) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .

(b) $gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .

Numerical composite. Easier — work inside-out.

$fg(4)$ with same $f, g$ :

$g(4) = 16 - 1 = 15$ .
$f(15) = 2(15) + 3 = 33$ .

In fg(x) the function nearest the bracket, g, is applied first; then f acts on its output.

Worked qualitative. Why does $fg(x) \ne gf(x)$ in general?

Order matters. 'Take square then add 1' is NOT the same as 'add 1 then square'.
Example: $f(x) = x^{2}, g(x) = x + 1$ .
$fg(x) = (x + 1)^{2} = x^{2} + 2x + 1$ .
$gf(x) = x^{2} + 1$ .
Different.

Edexcel tip. Mark schemes give M1 for correct order, M1 for substitution, A1 for simplified answer.

$fg(x) = f(g(x))$ .
$g$ first, then $f$ .
$fg \ne gf$ in general.
Inside-out for numerical inputs.

Inverse functions

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$f^{-1}$ undoes $f$ . Method: swap $x, y$ , solve.

Inverse function $f^{-1}(x)$ undoes $f$ .

If $f(2) = 7$ , then $f^{-1}(7) = 2$ .

Property: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

The inverse f⁻¹ undoes f — it maps every output straight back to its input.

Method to find $f^{-1}$ :

Write $y = f(x)$ .
SWAP $x$ and $y$ .
SOLVE for $y$ .
The result is $f^{-1}(x)$ .

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

$y = \dfrac{2x - 5}{3}$ .
Swap: $x = \dfrac{2y - 5}{3}$ .
Solve for $y$ $y$ :
- $3x = 2y - 5$ .
- $2y = 3x + 5$ .
- $y = \dfrac{3x + 5}{2}$ .
$f^{-1}(x) = \dfrac{3x + 5}{2}$ .

Verify. $f(f^{-1}(x)) = f\left(\dfrac{3x + 5}{2}\right) = \dfrac{2 \cdot (3x + 5)/2 - 5}{3} = \dfrac{3x + 5 - 5}{3} = \dfrac{3x}{3} = x$ . ✓

Worked qualitative. Why is $f^{-1}(x) \ne 1/f(x)$ ?

$f^{-1}$ means INVERSE FUNCTION (undo).
$1/f(x)$ means RECIPROCAL.
The notation is overloaded but they're DIFFERENT.
$f(x) = 2x$ has inverse $f^{-1}(x) = x/2$ , but reciprocal $1/(2x) = 1/(2x)$ . Different.

Edexcel tip. When asked for $f^{-1}$ , ALWAYS use the swap method. Showing the swap step earns the M1.

$f^{-1}$ undoes $f$ .
Method: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ — different.
Verify by composition.

Quick recap

$f(x)$ : function notation.
Evaluate by substitution.
$fg(x) = f(g(x))$ — $g$ first.
Inverse: swap $x, y$ , solve.
$f^{-1} \ne 1/f$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$fg(x) = f(g(x))$ — composite.
Inverse method: swap, solve.
$f^{-1}$ undoes $f$ .
$f^{-1}$ is NOT $1/f$ .

How it’s examined

Functions on every Higher Tier paper (4-7 marks). Composite + inverse standard. Examiner reports flag (1) wrong order in composite, (2) interpreting $f^{-1}$ as reciprocal, (3) skipping swap step.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on functions, written exactly the way a tutor would explain them at the board.

1Evaluate a function
Foundation• function, evaluate
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Question
$f(x) = 3x - 7$ . Find $f(5)$ and $f(-2)$ .
Step-by-step solution
1. Step 1
  $f(5) = 3(5) - 7 = 8$ .
2. Step 2
  $f(-2) = 3(-2) - 7 = -13$ .
Answer
$f(5) = 8$ , $f(-2) = -13$
2Composite functions $fg(x)$
Higher• Adapted from 4MA1/1H May/Jun 2024 Q16• composite
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Question
$f(x) = 2x + 3$ and $g(x) = x^{2} - 1$ . Find (a) $fg(x)$ , (b) $gf(x)$ .
Step-by-step solution
1. Step 1
  (a) $fg(x) = f(g(x))$ . Apply $g$ FIRST, then $f$ .
2. Step 2
  $g(x) = x^{2} - 1$ . Substitute into $f$ : $f(g(x)) = 2(x^{2} - 1) + 3 = 2x^{2} + 1$ .
3. Step 3
  (b) $gf(x) = g(f(x))$ . Apply $f$ first.
4. Step 4
  $f(x) = 2x + 3$ . Substitute: $g(f(x)) = (2x + 3)^{2} - 1 = 4x^{2} + 12x + 9 - 1 = 4x^{2} + 12x + 8$ .
Answer
(a) $fg(x) = 2x^{2} + 1$ (b) $gf(x) = 4x^{2} + 12x + 8$
Examiner tip
$fg(x)$ means 'do $g$ first, then $f$ '. Read RIGHT to LEFT — like onion layers.
3Find an inverse function
Higher• Adapted from 4MA1/2H Jan 2024 Q13• inverse
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Question
$f(x) = \dfrac{2x - 5}{3}$ . Find $f^{-1}(x)$ .
Step-by-step solution
1. Step 1
  Set $y = f(x) = \dfrac{2x - 5}{3}$ .
2. Step 2
  Swap $x$ and $y$ : $x = \dfrac{2y - 5}{3}$ .
3. Step 3
  Solve for $y$ : $3x = 2y - 5 \implies 2y = 3x + 5 \implies y = \dfrac{3x + 5}{2}$ .
4. Step 4
  So $f^{-1}(x) = \dfrac{3x + 5}{2}$ .
Answer
$f^{-1}(x) = \dfrac{3x + 5}{2}$
Examiner tip
Method: write $y = f(x)$ , swap $x$ and $y$ , solve for $y$ . The result is $f^{-1}(x)$ .
4Evaluate a composite
Higher• composite, evaluate
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Question
$f(x) = 2x + 1$ and $g(x) = 3x - 2$ . Find $fg(4)$ .
Step-by-step solution
1. Step 1
  Compute $g(4) = 3(4) - 2 = 10$ .
2. Step 2
  Compute $f(10) = 2(10) + 1 = 21$ .
Answer
$fg(4) = 21$
Examiner tip
For numerical composite: evaluate inside-out. $g(4)$ first, then $f$ of that result.
5Evaluating a function at a negative fraction [1 mark]
Getting started• Paper 1H/2H style — 1 mark• functions, evaluate
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Question
$f(x) = \dfrac{12}{x}$

Find $f\left(-\dfrac{1}{3}\right)$ [1]
Step-by-step solution
1. Step 1
  Substitute $-\dfrac{1}{3}$ for $x$ .
  $f\left(-\dfrac{1}{3}\right) = 12 \div \left(-\dfrac{1}{3}\right)$
2. Step 2
  B1 Dividing by $\dfrac{1}{3}$ is the same as multiplying by $3$ , and the sign is negative.
  $12 \times (-3) = -36$
Answer
$-36$
Examiner tip
Examiner tip: Dividing by a number between $0$ and $1$ makes the answer bigger in size. If your answer is smaller than $12$ , you have multiplied instead of divided.

Common mistake: Writing $12 \times \left(-\dfrac{1}{3}\right) = -4$ . The function divides $12$ by $x$ ; it does not multiply.
6Two values excluded from a domain [2 marks]
Getting started• Paper 1H/2H style — 2 marks• functions, domain
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Question
$h(x) = \dfrac{5}{x^{2} - 16}$

Find the values of $x$ that must be excluded from any domain of $h$ . [2]
Step-by-step solution
1. Step 1
  M1 $h(x)$ cannot be worked out when the denominator is $0$ .
  $x^{2} - 16 = 0$
2. Step 2
  A1 Solve. $16$ has two square roots.
  $x^{2} = 16 \quad \Rightarrow \quad x = 4 \quad \text{or} \quad x = -4$
Answer
$x = 4$ and $x = -4$
Examiner tip
Examiner tip: Only the denominator matters here. The numerator $5$ can never make the function undefined.

Common mistake: Giving only $x = 4$ . Substituting $x = -4$ also gives $16 - 16 = 0$ , so both values must be excluded.
7A composite written as a single fraction [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• functions, composite, algebraic fractions
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Question
$f(x) = \dfrac{1}{x - 2}$ , $x \neq 2$ , and $g(x) = 3x + 1$

Express $gf(x)$ as a single fraction in its simplest form. [3]
Step-by-step solution
1. Step 1
  M1 $gf(x)$ means do $f$ first. Put $f(x)$ in place of $x$ in $g$ .
  $gf(x) = 3\left(\dfrac{1}{x - 2}\right) + 1 = \dfrac{3}{x - 2} + 1$
2. Step 2
  M1 Write $1$ as $\dfrac{x - 2}{x - 2}$ so that both terms have the same denominator.
  $\dfrac{3}{x - 2} + \dfrac{x - 2}{x - 2} = \dfrac{3 + x - 2}{x - 2}$
3. Step 3
  A1 Simplify the numerator.
  $gf(x) = \dfrac{x + 1}{x - 2}$
Answer
$gf(x) = \dfrac{x + 1}{x - 2}$
Examiner tip
Examiner tip: Check with $x = 3$ : $f(3) = 1$ and $g(1) = 4$ . Your fraction gives $\dfrac{4}{1} = 4$ .

Common mistake: Working out $fg(x) = \dfrac{1}{3x - 1}$ instead. In $gf(x)$ , $f$ is next to $x$ , so $f$ is applied first. Another slip is writing $\dfrac{3}{x - 2} + 1 = \dfrac{4}{x - 2}$ , adding the $1$ to the numerator only.
8The inverse is not the reciprocal [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• functions, inverse
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Question
$f(x) = 2x + 6$

(a) Find $f^{-1}(x)$ [2]

(b) Show that $f^{-1}(8)$ is not equal to $\dfrac{1}{f(8)}$ [1]
Step-by-step solution
1. Step 1
  M1 (a) Write $y = 2x + 6$ and rearrange for $x$ .
  $2x = y - 6$
2. Step 2
  A1 Divide by $2$ and write the answer in terms of $x$ .
  $f^{-1}(x) = \dfrac{x - 6}{2}$
3. Step 3
  B1 (b) Work out both values and compare them.
  $f^{-1}(8) = \dfrac{8 - 6}{2} = 1, \qquad \dfrac{1}{f(8)} = \dfrac{1}{22}, \qquad 1 \neq \dfrac{1}{22}$
Answer
(a) $f^{-1}(x) = \dfrac{x - 6}{2}$ (b) $f^{-1}(8) = 1$ but $\dfrac{1}{f(8)} = \dfrac{1}{22}$
Examiner tip
Examiner tip: The $-1$ in $f^{-1}$ is notation for the inverse, not a power. $f^{-1}$ undoes $f$ : $f(1) = 8$ , so $f^{-1}(8) = 1$ .

Common mistake: Writing $f^{-1}(x) = \dfrac{1}{2x + 6}$ . That is the reciprocal of $f(x)$ , and it scores nothing in (a).
9Excluded values of a composite, then solve [4 marks]
Stretch• Paper 1H/2H style — 4 marks• functions, composite, domain
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Question
$f(x) = \dfrac{6}{x}$ , $x \neq 0$ , and $g(x) = x^{2} - 3x - 10$

(a) Find the values of $x$ that must be excluded from the domain of $fg$ . [2]

(b) Solve $fg(x) = -\dfrac{1}{2}$ [2]
Step-by-step solution
1. Step 1
  M1 (a) Apply $g$ first, then $f$ . $fg(x)$ cannot be worked out when its denominator, $g(x)$ , is $0$ .
  $fg(x) = \dfrac{6}{x^{2} - 3x - 10}, \qquad x^{2} - 3x - 10 = (x - 5)(x + 2) = 0$
2. Step 2
  A1 Both roots of the denominator are excluded.
  $x = 5 \quad \text{and} \quad x = -2$
3. Step 3
  M1 (b) Set $fg(x) = -\dfrac{1}{2}$ . Cross-multiplying gives $g(x) = -12$ ; rearrange so that one side is $0$ .
  $x^{2} - 3x - 10 = -12 \quad \Rightarrow \quad x^{2} - 3x + 2 = 0$
4. Step 4
  A1 Factorise and solve. Neither answer is an excluded value.
  $(x - 1)(x - 2) = 0 \quad \Rightarrow \quad x = 1 \quad \text{or} \quad x = 2$
Answer
(a) $x = 5$ and $x = -2$ (b) $x = 1$ or $x = 2$
Examiner tip
Examiner tip: Check (b): $g(1) = 1 - 3 - 10 = -12$ and $f(-12) = -\dfrac{1}{2}$ . Also $g(2) = 4 - 6 - 10 = -12$ , so both answers work.

Common mistake: Working out $gf(x) = \dfrac{36}{x^{2}} - \dfrac{18}{x} - 10$ instead of $fg(x)$ , and then excluding only $x = 0$ . In $fg$ , $g$ is applied first, so the whole of $g(x)$ becomes the denominator.
10Solving ff(x) = k with four solutions [4 marks]
Stretch• Paper 1H/2H style — 4 marks• functions, composite, quadratic
▼
Question
$f(x) = x^{2} - 6$

Solve $ff(x) = 3$

Give your answers in exact form. Show clear algebraic working. [4]
Step-by-step solution
1. Step 1
  M1 Write $ff(x) = 3$ in terms of $f(x)$ : apply $f$ to $f(x)$ .
  $(f(x))^{2} - 6 = 3 \quad \Rightarrow \quad (f(x))^{2} = 9$
2. Step 2
  M1 Take both square roots, so $f(x) = 3$ or $f(x) = -3$ .
  $x^{2} - 6 = 3 \quad \text{or} \quad x^{2} - 6 = -3$
3. Step 3
  M1 Solve each equation for $x^{2}$ .
  $x^{2} = 9 \quad \text{or} \quad x^{2} = 3$
4. Step 4
  A1 Each gives two square roots, so there are four solutions.
  $x = 3, \; -3, \; \sqrt{3}, \; -\sqrt{3}$
Answer
$x = \pm 3$ or $x = \pm \sqrt{3}$
Examiner tip
Examiner tip: Check $x = \sqrt{3}$ : $f(\sqrt{3}) = 3 - 6 = -3$ and $f(-3) = 9 - 6 = 3$ . It works even though $f(\sqrt{3})$ is negative.

Common mistake: Taking only the positive root at each stage. That gives $f(x) = 3$ only and then $x = 3$ only, finding one solution out of four and losing the A mark.

Question 1
Paper 1H/2H style — 1 mark1 mark
$f(x) = 10 - x$

Show that $ff(x) = x$ [1]
Model answer
$ff(x) = f(10 - x) = 10 - (10 - x) = 10 - 10 + x = x$
Why this scores
Mark scheme:
- B1 for $10 - (10 - x)$ leading to $x$
Why this scores full marks: The candidate substitutes the whole of $10 - x$ in brackets, so the minus sign changes both terms. Writing $10 - 10 - x = -x$ is the error weaker scripts make.
Question 2
Paper 1H/2H style — 2 marks2 marks
$f(x) = 7 - 2x$ for $-1 \leqslant x \leqslant 3$

Find the range of $f$ . [2]
Model answer
$f(-1) = 7 + 2 = 9$

$f(3) = 7 - 6 = 1$

$f$ is linear, so its outputs run from $1$ to $9$ .

Answer: $1 \leqslant f(x) \leqslant 9$
Why this scores
Mark scheme:
- M1 for $f(-1) = 9$ or $f(3) = 1$
- A1 for $1 \leqslant f(x) \leqslant 9$ oe
Why this scores full marks: The candidate works out the outputs at both ends of the domain and writes the range in terms of $f(x)$ , smaller value first. Because $f$ decreases, the left end of the domain gives the larger output; writing $9 \leqslant f(x) \leqslant 1$ or $-1 \leqslant f(x) \leqslant 3$ (the domain again) loses the A mark.
Question 3
Paper 1H/2H style — 2 marks2 marks
$h(x) = \sqrt{3x + 1}$ , $x \geqslant -\dfrac{1}{3}$

(a) Work out $h(5)$ [1]

(b) Solve $h(x) = 7$ [1]
Model answer
(a) $h(5) = \sqrt{15 + 1} = \sqrt{16} = 4$

(b) $\sqrt{3x + 1} = 7$

$3x + 1 = 49$

$3x = 48$

Answer: $x = 16$
Why this scores
Mark scheme:
- B1 for $4$ (a)
- B1 for $16$ (b)
Why this scores full marks: In (b) the candidate squares both sides first to remove the root, then solves the linear equation. Check: $\sqrt{49} = 7$ . A common wrong answer is $x = 2$ , from solving $3x + 1 = 7$ without squaring.
Question 4
Paper 1H/2H style — 3 marks3 marks
$f(x) = 3 - x^{2}$ and $g(x) = x + 2$

Show that $fg(x) = -x^{2} - 4x - 1$ [3]
Model answer
$fg(x) = f(x + 2) = 3 - (x + 2)^{2}$

$= 3 - (x^{2} + 4x + 4)$

$= 3 - x^{2} - 4x - 4$

$= -x^{2} - 4x - 1$
Why this scores
Mark scheme:
- M1 for $3 - (x + 2)^{2}$
- M1 for $x^{2} + 4x + 4$ seen as the expansion of $(x + 2)^{2}$
- A1 for a complete, correct argument leading to $-x^{2} - 4x - 1$
Why this scores full marks: The candidate expands $(x + 2)^{2}$ inside a bracket before subtracting, so the minus sign changes all three terms. Scripts that write $3 - x^{2} + 4x + 4$ lose the bracket, and the result cannot reach the printed answer honestly.
Question 5
Paper 1H/2H style — 3 marks3 marks
$f(x) = \dfrac{2x + 1}{x - 4}$

(a) State the value of $x$ that must be excluded from the domain of $f$ . [1]

(b) Solve $f(x) = 3$ [2]
Model answer
(a) $x = 4$

(b) $\dfrac{2x + 1}{x - 4} = 3$

$2x + 1 = 3(x - 4) = 3x - 12$

Answer: $x = 13$
Why this scores
Mark scheme:
- B1 for $4$ (a)
- M1 for $2x + 1 = 3(x - 4)$ oe (b)
- A1 for $13$ cao (b)
Why this scores full marks: The candidate multiplies by the whole denominator, $(x - 4)$ , in a bracket, so the $3$ multiplies both terms. Check: $f(13) = \dfrac{27}{9} = 3$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
The function $f$ is defined as $f : x \mapsto \dfrac{x - 5}{2x + 1}$ , $x \neq -\dfrac{1}{2}$

Express the inverse function $f^{-1}$ in the form $f^{-1} : x \mapsto \ldots$ [3]
Model answer
$y = \dfrac{x - 5}{2x + 1}$

$2xy + y = x - 5$

$2xy - x = -5 - y$

$x(2y - 1) = -5 - y$

$x = \dfrac{-5 - y}{2y - 1} = \dfrac{y + 5}{1 - 2y}$

Answer: $f^{-1} : x \mapsto \dfrac{x + 5}{1 - 2x}$
Why this scores
Mark scheme:
- M1 for $y(2x + 1) = x - 5$ or $x(2y + 1) = y - 5$
- M1 for collecting the terms in the new subject and factorising, e.g. $x(2y - 1) = -5 - y$
- A1 for $\dfrac{x + 5}{1 - 2x}$ oe
Why this scores full marks: Once $x$ appears in two terms, the candidate gathers them on one side and takes $x$ out as a factor. The answer is written in the mapping notation the question asks for. Check: $f(0) = -5$ and $\dfrac{-5 + 5}{1 + 10} = 0$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
$f(x) = \sqrt{x + 5}$ and $g(x) = x^{2} - 2x$

Solve $fg(x) = 3$

Give your answers correct to 3 significant figures. [4]
Model answer
$fg(x) = f(x^{2} - 2x) = \sqrt{x^{2} - 2x + 5}$

$\sqrt{x^{2} - 2x + 5} = 3$ , so $x^{2} - 2x + 5 = 9$

$x^{2} - 2x - 4 = 0$

$x = \dfrac{2 \pm \sqrt{(-2)^{2} - 4 \times 1 \times (-4)}}{2 \times 1} = \dfrac{2 \pm \sqrt{20}}{2}$

Answer: $x = 3.24$ or $x = -1.24$
Why this scores
Mark scheme:
- M1 for $\sqrt{x^{2} - 2x + 5}$
- M1 for squaring to get $x^{2} - 2x + 5 = 9$ oe, leading to $x^{2} - 2x - 4 = 0$
- M1 for a correct substitution into the quadratic formula for their 3-term quadratic
- A1 for $3.24$ and $-1.24$ (awrt)
Why this scores full marks: $g$ is applied first, so the whole of $x^{2} - 2x$ goes under the square root with the $+5$ . The candidate squares both sides to remove the root, keeps $\sqrt{20}$ exact until the last line and gives both answers to 3 significant figures. Working out $gf(x) = x + 5 - 2\sqrt{x + 5}$ instead applies the functions in the wrong order. Check: $(3.236)^{2} - 2(3.236) + 5 = 9.00$ , and $\sqrt{9} = 3$ .
Question 8
Paper 1H/2H style — 5 marks5 marks
$f(x) = x + 3$ and $g(x) = \dfrac{12}{x - 1}$ , $x \neq 1$

(a) Which value of $x$ must be excluded from the domain of $gf$ ? [1]

(b) Solve $gf(x) = f(x)$ [4]
Model answer
$gf(x) = \dfrac{12}{(x + 3) - 1} = \dfrac{12}{x + 2}$

(a) $x = -2$

(b) $\dfrac{12}{x + 2} = x + 3$

$12 = (x + 3)(x + 2) = x^{2} + 5x + 6$

$x^{2} + 5x - 6 = 0$

$(x + 6)(x - 1) = 0$

Answer: $x = 1$ or $x = -6$
Why this scores
Mark scheme:
- B1 for $-2$ (a)
- M1 for $gf(x) = \dfrac{12}{x + 3 - 1}$ oe (b)
- M1 for $12 = (x + 3)(x + 2)$ oe (b)
- M1 for $(x + 6)(x - 1) = 0$ or a correct substitution into the quadratic formula (b)
- A1 for $1$ and $-6$ (b)
Why this scores full marks: Simplifying $gf(x)$ to $\dfrac{12}{x + 2}$ shows the excluded value at once: $gf$ fails when $f(x) = 1$ , which is $x = -2$ . Neither solution in (b) is $-2$ , so both stand. Check: $gf(1) = g(4) = 4 = f(1)$ , and $gf(-6) = g(-3) = -3 = f(-6)$ . Giving $x = 1$ as the excluded value (from $g$ alone) is the usual slip in (a).
Question 9
Paper 1H/2H style — 5 marks5 marks
$f(x) = \dfrac{1}{1 - x}$ , $x \neq 1$

(a) Show that $ff(x) = \dfrac{x - 1}{x}$ [3]

(b) Hence show that $fff(x) = x$ [2]
Model answer
(a) $ff(x) = \dfrac{1}{1 - \dfrac{1}{1 - x}}$

$1 - \dfrac{1}{1 - x} = \dfrac{1 - x - 1}{1 - x} = \dfrac{-x}{1 - x}$

$ff(x) = 1 \div \dfrac{-x}{1 - x} = \dfrac{1 - x}{-x} = \dfrac{x - 1}{x}$

(b) $fff(x) = f\left(\dfrac{x - 1}{x}\right) = \dfrac{1}{1 - \dfrac{x - 1}{x}}$

$1 - \dfrac{x - 1}{x} = \dfrac{x - (x - 1)}{x} = \dfrac{1}{x}$

$fff(x) = 1 \div \dfrac{1}{x} = x$
Why this scores
Mark scheme:
- M1 for $\dfrac{1}{1 - \dfrac{1}{1 - x}}$ (a)
- M1 for $1 - \dfrac{1}{1 - x} = \dfrac{-x}{1 - x}$ oe (a)
- A1 for a complete, correct argument leading to $\dfrac{x - 1}{x}$ (a)
- M1 for $\dfrac{1}{1 - \dfrac{x - 1}{x}}$ (b)
- A1 for a complete, correct argument leading to $x$ (b)
Why this scores full marks: Both answers are printed, so the marks are for the working. The candidate writes each subtraction as a single fraction, keeps $-(x - 1)$ in brackets, and turns $\dfrac{1 - x}{-x}$ into $\dfrac{x - 1}{x}$ by changing both signs. "Hence" is followed: (b) uses the result of (a).
Question 10
Paper 1H/2H style — 6 marks6 marks
$f(x) = x^{2} + 6x$ and $g(x) = x - 3$

(a) Show that $fg(x) = x^{2} - 9$ [2]

(b) Write down the range of $fg$ . [1]

(c) Solve $fg(x) = 2g(x)$ [3]
Model answer
(a) $fg(x) = (x - 3)^{2} + 6(x - 3) = x^{2} - 6x + 9 + 6x - 18 = x^{2} - 9$

(b) $x^{2} \geqslant 0$ , so $fg(x) \geqslant -9$

(c) $x^{2} - 9 = 2(x - 3)$

$x^{2} - 2x - 3 = 0$

$(x - 3)(x + 1) = 0$

Answer: $x = 3$ or $x = -1$
Why this scores
Mark scheme:
- M1 for $(x - 3)^{2} + 6(x - 3)$ (a)
- A1 for $x^{2} - 6x + 9 + 6x - 18$ leading to $x^{2} - 9$ with no errors (a)
- B1 for $fg(x) \geqslant -9$ oe (b)
- M1 for $x^{2} - 9 = 2(x - 3)$ and rearranging to $x^{2} - 2x - 3 = 0$ (c)
- M1 for $(x - 3)(x + 1) = 0$ or a correct substitution into the quadratic formula (c)
- A1 for $3$ and $-1$ (c)
Why this scores full marks: In (c) the candidate rearranges to $0$ and factorises. A tempting shortcut is to write $x^{2} - 9 = (x - 3)(x + 3)$ and divide both sides by $(x - 3)$ . That gives $x + 3 = 2$ and loses the solution $x = 3$ , and with it the final A mark. Check: $fg(3) = 0 = 2g(3)$ and $fg(-1) = -8 = 2g(-1)$ .

4MA1 Functions

At a glance

What you’ll learn

Function notation and evaluation

Composite functions

Inverse functions

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Evaluate a function

2Composite functions $fg(x)$

3Find an inverse function

4Evaluate a composite

5Evaluating a function at a negative fraction [1 mark]

6Two values excluded from a domain [2 marks]

7A composite written as a single fraction [3 marks]

8The inverse is not the reciprocal [3 marks]

9Excluded values of a composite, then solve [4 marks]

10Solving ff(x) = k with four solutions [4 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Functions

At a glance

What you’ll learn

Function notation and evaluation

Composite functions

Inverse functions

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Evaluate a function

2Composite functions fg(x)fg(x)fg(x)

3Find an inverse function

4Evaluate a composite

5Evaluating a function at a negative fraction [1 mark]

6Two values excluded from a domain [2 marks]

7A composite written as a single fraction [3 marks]

8The inverse is not the reciprocal [3 marks]

9Excluded values of a composite, then solve [4 marks]

10Solving ff(x) = k with four solutions [4 marks]

Model AnswersFull mark-scheme-style answers

2Composite functions $fg(x)$