Study Guides
Functions
Function notation, domain and range, inverse functions and composite functions, for Cambridge O Level Mathematics (Syllabus D) 4024.
- Subject
- Mathematics
- Level
- O LEVELS
- Topic
- Algebra and graphs
- Author
- Muhammad Ghazali Siddiqui
- Updated
- Reviewed by
- Sajawal Zahid (what this means)
Aligned to Cambridge O Level Mathematics (4024), 2025-2027. Official specification .
Syllabus page (what it covers and how it is assessed): Cambridge O Level Mathematics.
Syllabus points this page covers
4024
- 2.12 Functions
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This guide covers subtopic 2.12 Functions, from Topic 2, Algebra and graphs, for Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series — the final subtopic of Topic 2.
Where this fits in 4024
Functions formalise the input–output relationships used informally throughout Topic 2 — every equation y = … graphed in Graphs of Functions and Sketching Curves is a function — and introduces notation and operations (inverse, composite) that are new to this subtopic specifically.
Syllabus coverage
CAMBRIDGE O LEVEL MATHEMATICS (SYLLABUS D) 4024
- Understand functions, domain and range, and use function notation (2.12)
- Understand and find inverse functions, f⁻¹(x) (2.12)
- Form composite functions as defined by gf(x) = g(f(x)) (2.12)
4024 is not tiered — every candidate covers all of the above. Candidates are not expected to find the domain and range of composite functions. This topic may include mapping diagrams.
Function notation, domain and range
A function takes an input and produces exactly one output, written in function notation as f(x) — read “f of x” — where x is the input.
Worked example. If f(x) = 3x − 5, find f(4).
f(4) = 3(4) − 5 = 7
The domain of a function is the set of allowed inputs; the range is the resulting set of outputs. A mapping diagram shows this input-to-output relationship visually, with arrows connecting each domain value to its corresponding range value. A one-to-one function has exactly one arrow arriving at each range value; a many-to-one function can have two or more arrows arriving at the same range value.
Restrictions on the domain
Some values must be excluded from the domain because they would make the function undefined:
- If the function has a denominator, exclude any x that makes it zero.
- If the function involves a square root, the expression inside must be greater than or equal to zero.
Worked example. g(x) = 1/(x − 2). State the value excluded from the domain.
x − 2 = 0
x = 2 must be excluded
The denominator is zero at x = 2, so g(2) is undefined — every other real number is a valid input.
Worked example. h(x) = √(x − 3). State the restriction on the domain.
x − 3 >= 0
x >= 3
The expression under the square root cannot be negative, so only inputs of 3 or more are valid — h(2) would require the square root of a negative number, which is not defined.
Inverse functions
The inverse function, f⁻¹(x), reverses what f does — if f(x) = y, then f⁻¹(y) = x. To find an inverse function algebraically: write y = f(x), rearrange to make x the subject, then swap x and y (or simply replace the now-isolated x with f⁻¹(x)).
Worked example. Find the inverse of f(x) = 3x − 5.
y = 3x − 5
y + 5 = 3x
x = (y + 5) / 3
so f⁻¹(x) = (x + 5) / 3
Check: f⁻¹(f(4)) should return 4. f(4) = 7, and f⁻¹(7) = (7+5)/3 = 4. ✓
Composite functions
A composite function applies one function to the result of another. gf(x) means “apply f first, then apply g to the result” — formally, gf(x) = g(f(x)).
Worked example. f(x) = 3/(x + 2) and g(x) = (3x + 5)². Find fg(x) as a fraction in its simplest form.
fg(x) = f(g(x)) = f((3x + 5)²)
= 3 / ((3x + 5)² + 2)
Since (3x + 5)² + 2 does not factorise or cancel with the numerator, this is already in its simplest form.
Worked example. h(x) = 2x² + 3 and f(x) = 3x − 5. Find fh(x).
fh(x) = f(h(x)) = f(2x² + 3) = 3(2x² + 3) − 5 = 6x² + 9 − 5 = 6x² + 4
Note the order: gf(x) means f acts first (innermost), then g — reading the notation from right to left in terms of the order of operation, even though it’s written left to right.
Candidates are not expected to find the domain and range of composite functions — only to form and evaluate them.
Common mistakes
- Applying functions in the wrong order. gf(x) means f first, then g — not the reverse. Reversing the order gives fg(x), a generally different function.
- Forgetting to swap x and y (or equivalent) when finding an inverse. Rearranging y = f(x) for x gives an expression in terms of y — this becomes f⁻¹(x) only once relabelled.
- Not simplifying a composite function fraction fully, when the question specifically asks for the simplest form.
- Confusing f⁻¹(x) with 1/f(x). The inverse function and the reciprocal of a function are entirely different things, despite the similar-looking notation.
- Attempting to find the domain/range of a composite function — not required by this syllabus.
- Forgetting to exclude the value that makes a denominator zero, or allowing a negative value under a square root, when stating a domain restriction.
Quick revision checklist
- Function notation f(x), and evaluating a function for a given input
- Domain and range, and mapping diagrams, including one-to-one versus many-to-one
- Excluding a zero denominator or a negative value under a square root from the domain
- Finding an inverse function: rearrange for x, then relabel
- Composite functions: gf(x) = g(f(x)), applied innermost-function-first
- Domain/range of composite functions is not required
Related resources
- Graphs of Functions and Sketching Curves — the previous subtopics
- Algebraic Manipulation — subtopics 2.1–2.3, the foundation for algebraic notation used throughout this topic
- Cambridge O Level Mathematics subject hub
Written against Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series. Always check the current syllabus for your examination year.
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Condensed recall notes on function notation, domain and range, composite and inverse functions for Cambridge O Level Mathematics 4024.
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