Functions
Cambridge O Level Additional Mathematics 4037 Topic 1 revision chapter covering the complete Functions syllabus for the 2025-2027 examination cycle. The chapter builds the whole topic along a single spine: input, rule, output, domain and range, composition, inverse. It defines a function as a rule giving exactly one output for every permitted input, and separates one-one mappings from many-one mappings using mapping diagrams and the horizontal-line test. Domain and range are treated as objects to be recorded before any algebra happens, with the four standard exclusions set out explicitly: a denominator may not be zero, a square-root argument must be non-negative over the reals, a logarithm argument must be strictly positive, and a many-one function must have its domain restricted before it can be inverted. The full notation set required by the syllabus is taught and used consistently: f(x), the mapping form f: x maps to f(x), the inverse f inverse of x, the composite fg(x) meaning f(g(x)), and the iterate f squared of x meaning f(f(x)) rather than the square of f(x), a distinction the syllabus tests directly and which is never applied to trigonometric functions in this course. The relationship between y = f(x) and the modulus graph y = |f(x)| is derived rather than asserted: sections on or above the x-axis are kept unchanged, sections below are reflected in the x-axis, roots are preserved and cusps appear where the reflected arc meets the axis. All four required forms are worked through - linear, quadratic, cubic and trigonometric - and the trigonometric case is given the syllabus families a sin(bx) + c, a cos(bx) + c and a tan(bx) + c together with the stated limits on a, b and c, showing why the constant term decides which sections of the wave reflect and why the roots must be found by solving f(x) = 0 rather than reused from the parent curve. Inverse functions are found by the interchange method, with the domain of the inverse always taken as the range of the original, illustrated on the exponential f(x) = e to the power 2x and its inverse one half of the natural logarithm of x. Composite functions are formed in both orders to show that fg and gf are generally different, and that a valid input must lie in the domain of the inner function while its output must lie in the domain of the outer function. Finally, sketch graphs establish that y = f(x) and y = f inverse of x are reflections in the line y = x, that coordinates exchange as (a, b) to (b, a), and that domain and range swap. The chapter includes eight original inline diagrams, nineteen fully worked examples with justified steps, a mistake clinic drawn from the errors that cost marks in this topic, a retrieval check with accessible answer reveals, an exam-style mixed challenge, a mastery checklist and a spaced-review schedule. All content is original and independent; the current Cambridge syllabus remains the authority for scope and assessment.Show moreShow less
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What is Functions about?
A function is a rule that gives exactly one output for every input it is allowed to take. Everything else in this topic is a consequence of that one sentence: the domain is the set of inputs you are allowed to use, the range is the set of outputs actually produced, a composite feeds one function's output into another, and an inverse undoes the rule — which is only possible when no two inputs share an output.
The set-up. \(f(x)=2x+5\) for \(x\in\mathbb{R}\), and \(g(x)=x^{2}-4x\) for \(x\ge 2\).
Key ideas to remember
- Anchor: a function is a machine with a doorway. The domain is who is allowed through the doorway; the range is what comes out the other side. Composition bolts two machines together; an inverse runs one machine backwards — and you can only run it backwards if no two people came out looking identical.
- Anchor for the superscripts: read \(f^{-1}\) as “\(f\) backwards” and \(f^{2}\) as “\(f\) twice”. Neither is ever “\(f\) to the power of something”. If you find yourself writing \(\dfrac{1}{f(x)}\) or \([f(x)]^{2}\), you have translated the symbol as an index, and you are now answering a different question from the one in front of you.
- Anchor: an inverse is a receipt. \(f\) takes the input and hands you the output; \(f^{-1}\) takes the output and hands the input back. A receipt only works if no two purchases produced identical slips — which is exactly the one‑one condition.
- The single sentence that prevents six of these eight: write down the domain before you touch the algebra, and carry it to the end of the answer.
What you need to be able to do
- 1.1 — I can define function, domain, range (image set), one‑one function, many‑one function, inverse function and composition of functions, and say which of them a given mapping is.
- 1.2 — I can find domains and ranges, including the restrictions that a composite or an inverse forces on them.
- 1.3 — I can read and write \(f(x)\), \(f:x\mapsto f(x)\), \(f^{-1}(x)\), \(fg(x)=f(g(x))\) and \(f^{2}(x)=f(f(x))\), and I know that \(f^{2}(x)\) is not \([f(x)]^{2}\).
- 1.4 — I can explain and sketch the relationship between \(y=f(x)\) and \(y=\lvert f(x)\rvert\) for linear, quadratic, cubic and trigonometric forms.
- 1.5 — I can explain clearly, with evidence, why a given function has no inverse over its stated domain.
- 1.6 — I can find the inverse of a one‑one function and state its domain correctly.
- 1.7 — I can form composite functions in either order, show that \(fg\) and \(gf\) generally differ, and keep every restriction.
- 1.8 — I can use sketch graphs to show that \(y=f(x)\) and \(y=f^{-1}(x)\) are reflections in the line \(y=x\).
Why Functions matters
Why this topic carries so much weight. Functions is not a self-contained topic you can finish and forget. The domain habits you build here are the same habits that stop you taking the logarithm of a negative number in Topic 6, dividing by an expression that could be zero wherever a rational function appears, or losing a solution in Topic 10. Later chapters will assume you already write restrictions down without being asked.
Key terms in Functions
- Domain and Range
- The domain of a function is the set of input values it is permitted to take; the range, also called the image set, is the set of output values that domain actually produces. Changing the domain changes the range, so a domain must be stated or deduced before any range can be found.
- Inverse Function
- The inverse of a one-one function f, written f inverse, is the function that reverses the mapping: if f sends a to b then f inverse sends b back to a. It exists only when f is one-one over its stated domain, its domain is the range of f, and its range is the domain of f.
- Reflection in the Line y = x
- The geometric relationship between the graph of a one-one function and the graph of its inverse. Because reflecting a point in the line y equals x exchanges its coordinates, the point (a, b) on y = f(x) corresponds to the point (b, a) on y = f inverse of x; consequently the domain and range of the two functions are interchanged, and any point where the two graphs meet a fixed point lies on the mirror line itself.
- Function
- A rule that assigns exactly one output to every input in a stated set called the domain. Two different inputs are allowed to share an output, but a single input may never produce two different outputs; a rule that does so is a relation rather than a function.
- Modulus of a Function
- The graph of y equals the modulus of f of x is obtained from the graph of y equals f of x by keeping every part that lies on or above the x-axis unchanged and reflecting every part that lies below the x-axis in the x-axis. Roots are unaltered, no output is ever negative, and a cusp appears wherever a reflected section meets the axis.
- Horizontal Line Test
- A graphical test for invertibility: a function has an inverse function over a stated domain exactly when every horizontal line meets its graph at most once. A horizontal line meeting the graph twice exhibits two different inputs sharing one output, which makes the function many-one and therefore not invertible on that domain.
- Function Notation
- The set of symbols used to describe functions in Additional Mathematics: f(x) for the output at x, f: x maps to f(x) for the mapping form, f inverse for the reverse mapping, fg(x) for the composite f(g(x)) in which g acts first, and f squared of x for the iterate f(f(x)) rather than the square of f(x).
- Composite Function
- A function formed by applying one function to the output of another. In fg(x), which means f(g(x)), the inner function g acts first and f acts on its result. A value x is a legal input only if x lies in the domain of g and g(x) lies in the domain of f, and fg is generally a different function from gf.
Common mistakes to avoid
- “\(f^{-1}(x)\) means \(\dfrac{1}{f(x)}\).” Correct The superscript \(-1\) in \(f^{-1}\) is not an index. It names the reverse mapping. If \(f(2)=7\) then \(f^{-1}(7)=2\), whereas \(\dfrac{1}{f(2)}=\dfrac{1}{7}\). The two are unrelated.
- “\(f^{2}(x)\) means \([f(x)]^{2}\).” Correct \(f^{2}(x)=f(f(x))\) — apply \(f\) twice. For \(f(x)=2x+1\): \(f^{2}(3)=f(7)=15\), while \([f(3)]^{2}=7^{2}=49\). The syllabus does not use this iteration notation with trigonometric functions.
- “Composites can be read left to right, like English.” Correct In \(fg(x)\) the function nearest \(x\) acts first, so \(g\) goes first and \(f\) second. \(fg\) and \(gf\) are different functions in general.
- “Once I have simplified, I can drop the restriction.” Correct Simplifying never creates permission. If \(g(x)=\dfrac{1}{x-1}\) and \(f(x)=x^{2}\), then \(gf(x)=\dfrac{1}{x^{2}-1}\) still carries \(x\ne\pm 1\), because those inputs were illegal before you simplified.
- “Every function has an inverse; I just have to find it.” Correct Only a one‑one function has an inverse function. \(f(x)=x^{2}\) on \(\mathbb{R}\) has none, because \(f(2)=f(-2)=4\). Restricting the domain to \(x\ge 0\) is what makes an inverse possible.
- “\(y=\lvert f(x)\rvert\) means reflect the whole curve.” Correct Reflect only the parts that lie below the \(x\)-axis. Everything on or above the axis is unchanged, and the roots stay exactly where they were.
- “I found the rule for \(f^{-1}(x)\), so I know everything about it.” Correct The rule is half of the function; the domain is the other half, and it is not free to choose — the domain of \(f^{-1}\) is the range of \(f\). For \(f(x)=e^{2x}\) the rule \(f^{-1}(x)=\tfrac{1}{2}\ln x\) carries the domain \(x>0\). Find it every time, so that you can state it the moment a question asks.
How Functions is examined
- Both written papers can assess any part of the subject content, so Functions can appear in either. All questions are compulsory, and necessary working must be shown.
- Nothing in Topic 1 needs a calculator. Inverses, composites, domains and modulus sketches are all exact-form work, so this topic is fully live on the non-calculator paper. If you are reaching for a calculator here, you are probably converting an exact answer into a decimal the question never asked for.
- Answers stay exact. Keep \(\sqrt{\ }\), \(\ln\), \(e\), \(\pi\) and fractions. The syllabus asks for answers in their simplest form, and where a question asks for exact values the answer may need to stay in terms of \(\pi\), \(e\), natural logarithms or surds. Do not round something that is already exact.
- A domain is part of a function, not decoration. Outcome 1.2 makes the domain and range of inverse and composite functions examinable in their own right, so a question can ask for the rule, for the domain, or for both. State the domain whenever you are asked for it — and work it out even when you are not, because the next part of the question usually needs it.
- A sketch is an answer, not an illustration. A sketch must show relevant intercepts, symmetry, asymptotes, the correct quadrants and long-term behaviour — and for a modulus of a non-linear graph, the cusps where the reflected parts meet the axis.
- “Explain” means give a reason. The syllabus wording for outcome 1.5 is “explain in words why a given function does not have an inverse”, so the argument is the answer — the conclusion on its own is not.
Frequently asked questions
What is the difference between the domain and the range of a function?
The domain is the set of inputs a function is allowed to take; the range (or image set) is the set of outputs that domain actually produces. The domain is stated or chosen first, and the range follows from it — change the domain and the range changes with it. For \(f(x)=x^{2}\) with domain \(x\ge 0\), the range is \(f(x)\ge 0\).
Does \(f^{-1}(x)\) mean \(\dfrac{1}{f(x)}\)?
No. The superscript \(-1\) in \(f^{-1}\) is not an index; it names the inverse function, the mapping that runs \(f\) backwards. If \(f(2)=7\) then \(f^{-1}(7)=2\), whereas \(\dfrac{1}{f(2)}=\dfrac{1}{7}\). In the same way \(f^{2}(x)\) means \(f(f(x))\), apply \(f\) twice, and not \([f(x)]^{2}\).
Why does a many-one function have no inverse?
An inverse must send each output back to exactly one input. If two different inputs share an output, the inverse would have to return two values for that one input, which a function cannot do. Use the horizontal line test: a horizontal line meeting the graph more than once shows the function is many-one, so \(f(x)=x^{2}\) on \(\mathbb{R}\) has no inverse because \(f(2)=f(-2)=4\). Restricting the domain to \(x\ge 0\) makes it one-one and an inverse exists.
In \(fg(x)\), which function is applied first?
The function nearest \(x\) acts first, so in \(fg(x)=f(g(x))\) you apply \(g\) first and then \(f\) to its output. Composites are read from the inside out, not left to right like English, and \(fg\) and \(gf\) are in general different functions. A value \(x\) is a legal input only if it is in the domain of \(g\) and \(g(x)\) is in the domain of \(f\).
How do you find the domain of an inverse function?
The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\). So after rearranging to find the rule for \(f^{-1}(x)\), go back to the original function, find its range, and state that as the domain of the inverse. The rule is only half the answer; an inverse without its domain is incomplete, and the domain is not free to choose.
How does the graph of \(y=\lvert f(x)\rvert\) relate to the graph of \(y=f(x)\)?
Keep every part of \(y=f(x)\) that lies on or above the \(x\)-axis unchanged, and reflect only the parts below the \(x\)-axis in the \(x\)-axis. The roots stay exactly where they were, no output is ever negative, and a cusp appears wherever a reflected section meets the axis. A common error is to reflect the whole curve; only the negative sections move.
How do you show that a function and its inverse are reflections in \(y=x\)?
Reflecting a point in the line \(y=x\) swaps its coordinates, so the point \((a,b)\) on \(y=f(x)\) corresponds to \((b,a)\) on \(y=f^{-1}(x)\). In an exam, sketch \(y=f(x)\), draw the dashed line \(y=x\), and reflect the curve in it, making sure the domain and range have swapped over. Any point where the two graphs meet lies on the line \(y=x\) itself.
Syllabus reference and sources
Written against: Cambridge O Level Additional Mathematics (4037) 2025–2027 Syllabus (Subject Content, Topic 1: Functions).
Written by: Academiq Edu Instructor Panel
Source documents
- Cambridge O Level Additional Mathematics 4037 syllabus for 2025, 2026 and 2027
- Syllabus update notice, Cambridge O Level Additional Mathematics 4037, 2025–2027
- Cambridge O Level Additional Mathematics 4037 syllabus for 2028, 2029 and 2030 (version 1), consulted only to confirm that no significant change affects this topic
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