Home   >  Worksheets   >  GCSE questions by topic   >  Functions (harder)

MathsLinks Logo Questions by topic

Last update: 05/11/2026

Give feedback : Buy me a coffee

Functions (harder)

Free Printable Functions (harder) PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free functions (harder) worksheets (PDF), GCSE maths past paper questions and worked solutions for Functions (harder) (7J). This is a Grade 7 algebra topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, Naiker Maths, 1st Class Maths, Metatutor, Dr Austin, GCSE Maths Rocket, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our functions (harder) revision notes and worked examples with diagrams, try real GCSE past paper functions (harder) questions with step-by-step solutions, or download a PDF of all functions (harder) worksheet links.

Free Functions (harder) PDF Worksheets & Answers

Grade 7 • Topic 7J
Resource Questions Answers
GCSE past paper questions & solutions
 Qs Ans
Graded exam questions & solutions
 Qs Ans
GCSE 9–1 practice booklet & solutions
 Qs Ans
Practice Questions & answers
 Qs Ans
Textbook Exercise & answers
 Qs Ans
MME worksheet & model answers
 Qs Ans
GCSE worksheet & solutions
 Qs Ans
Exam question booklet & solutions
 Qs Ans
Edexcel exam questions & solutions
 Qs Ans
Edexcel exam questions & solutions
 Qs Ans
Edexcel exam questions & solutions
 Qs Ans
Topic worksheet & worked solutions
 Qs Ans
Practice grid / worksheet & answers
 Qs Ans
Practice questions & answers
 Qs Ans
Practice questions & answers
 Qs Ans
Edexcel exam questions & solutions
 Qs Ans
AQA exam questions & solutions
 Qs Ans
GCSE topic booklet & answers
 Qs Ans
PMT (Edxl)
Edexcel GCSE topic questions & solutions
 Qs Ans
PMT (AQA)
AQA GCSE topic questions & solutions
 Qs Ans
PMT (IGCSE)
Edexcel IGCSE topic questions & solutions
 Qs Ans
Online topic notes & examples
 Page
Interactive online lesson & practice
 Page
Self-marking online worksheet
 Page
Grade 7 topics
7A - Surds7B - Bounds
7C - Proportion (harder)7D - Algebraic fractions
7E - The quadratic formula7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs7H - Rearranging (harder)
7J - Functions (harder)7K - Iteration
7L - Area of any triangle7M - The sine rule
7N - The cosine rule7P - Congruent triangles
7Q - 3D pythagoras / trig.7R - Histograms
7S - Conditional probability7T - Exact trig. values (harder)
7U - Area of a segment
GCSE Maths Grade 7 Revision Notes & Method

Guide to Functions (harder): Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

A composite function fg(x) means apply function g first, then substitute that output into f, i.e. fg(x) = f(g(x)). Always work from the inside out, and remember that fg(x) is rarely equal to gf(x).

To find the inverse function f−1(x), write y = f(x), swap x and y (or rearrange directly to make the input variable the subject), and write your final answer in terms of x. If the original input appears twice, use the factorising rearrangement method.

Example 1: f(x) = 2x − 3 and g(x) = x2 + 1. Solve fg(x) = 15.

Solution:

Substitute g(x) into f(x):
fg(x) = 2(x2 + 1) − 3 = 2x2 − 1
Set equal to 15: 2x2 − 1 = 15 ⇒ 2x2 = 16
x2 = 8 ⇒ x = ±√8 = ±2√2
Answer: x = ±2√2

Example 2: Given h(x) = 4xx + 5, find h−1(x).

Solution:

Let y = 4xx + 5 and swap x and y: x = 4yy + 5
Multiply by (y + 5): xy + 5x = 4y
Collect y terms: 5x = 4y − xy = y(4 − x)
Divide by (4 − x): y = 5x4 − x
Answer: h−1(x) = 5x4 − x
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Functions (harder) Questions & Answers

Past Paper Example 1 AQA Jun 2017 2H Q27 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2
27 (b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360°
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2 [2 marks]
–2–112–8–6–4–22468xyO
(b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360° [2 marks]
0°90°180°270°360°–101xy
Worked Solution & Mark Scheme (4 marks)
Part (a) Solution:
The inverse function is h-1(x) = x3 M1
Values: (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
Plot these points and draw a smooth cubic curve. A1
–2–112–8–6–4–22468xyO
Answer (a): Cubic graph y = x³ drawn through (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
Part (b) Solution:
fg(x) = f(g(x)) = \sin(x + 90°) = \cos x M1
Plot the cosine wave: 0° → 1, 90° → 0, 180° → -1, 270° → 0, 360° → 1. A1
0°90°180°270°360°–101xy
Answer (b): Cosine curve drawn from (0°, 1) to (90°, 0), (180°, -1), (270°, 0), (360°, 1)
Past Paper Example 2 Eduqas Nov 2021 1H Q20 8 marks
Full Paper (PDF) • Mark Scheme (PDF)
(a) The functions f(x) and g(x) are defined by
f(x) = \sin x°,   g(x) = x - 90.
Find the value of fg(120).
(b) The diagram shows a sketch of the graph of y = h(x).
(c) The function k(x) is defined by
k(x) = x3 - 23.
Solve k-1(x) = 5.
(a) The functions f(x) and g(x) are defined by
f(x) = \sin x°,   g(x) = x - 90.
Find the value of fg(120). [2 marks]
(b) The diagram shows a sketch of the graph of y = h(x). [2 marks]
xyO(−6, 0.5)(−2, −1)
On the same diagram, sketch the graph of y = -h(x).
Write down the coordinates of the maximum and minimum points of y = -h(x).
(c) The function k(x) is defined by
k(x) = x3 - 23.
Solve k-1(x) = 5. [4 marks]
Worked Solution & Mark Scheme (8 marks)
Part (a) Solution:
g(120) = 120 - 90 = 30 M1
fg(120) = f(30) = \sin 30° = 0.5 A1
Answer (a): 0.5 (or 12)
Part (b) Solution:
The transformation y = -h(x) reflects the graph in the x-axis. M1
• The minimum point (-2, -1) becomes a maximum at (-2, 1).
• The maximum point (-6, 0.5) becomes a minimum at (-6, -0.5). A1
Answer (b): Maximum: (-2, 1); Minimum: (-6, -0.5)
Part (c) Solution:
If k-1(x) = 5, then x = k(5). M1
k(5) = 53 - 23 = 125 - 23 = 102 M1
(Alternatively: y = x3 - 23 ⇒ x = 3√(y + 23) ⇒ k^{-1}(x) = 3√(x + 23) M1
3√(x + 23) = 5 ⇒ x + 23 = 53 = 125 ⇒ x = 102) A1
Answer (c): x = x = 102
Past Paper Example 3 Eduqas Jun 2018 1H Q23 6 marks
Full Paper (PDF) • Mark Scheme (PDF)
The functions f(x) and g(x) are defined, for x ≥ 0, by
f(x) = x2,   g(x) = x + 65
(a) Find the value of gf(8). [2 marks]
Find an expression for g-1(x). [2 marks]
(b) The diagram shows a sketch of the graph of y = h(x).
The point A has coordinates (-2, -3) and the point B has coordinates (2, 1). [2 marks]
xyOA(–2, –3)B(2, 1)
On the same diagram, sketch the graph of y = h(x) + 4.
You must indicate the coordinates of the new positions of the points A and B.
Worked Solution & Mark Scheme (6 marks)
Part (a) Solution:
f(8) = 82 = 64. M1
gf(8) = g(64) = 64 + 65 = 705 = 14. A1
Answer (a): 14
Part Solution:
Let y = x + 65 ⇒ 5y = x + 6 ⇒ x = 5y - 6 M1
So g-1(x) = 5x - 6. A1
Answer : 5x - 6
Part (b) Solution:
The transformation y = h(x) + 4 is a vertical translation by vector (04). M1
A' = (-2, -3 + 4) = (-2, 1).
B' = (2, 1 + 4) = (2, 5). A1
xyOA(–2, –3)B(2, 1)A'(–2, 1)B'(2, 5)
Answer (b): Graph translated 4 units vertically up. New A: (-2, 1), new B: (2, 5)

← 7H Rearranging (harder)

Back to GCSE questions by topic

7K Iteration →