Functions

• 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.

• 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 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.

• “\(f^{-1}(x)\) 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.

• Every candidate takes both written papers, and either paper can assess any part of the subject content — so Functions can appear in either, and no part of this topic belongs to one paper rather than the other. All questions are compulsory, and necessary working must be shown. Both papers are externally assessed, both carry approximately 45–55% AO1 and 45–55% AO2, and both contribute to the same A*–E grade range. The examination is available in the June and November series, and in the March series in India.
• 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.