Cambridge A Level Mathematics 9709 · Pure 1

Functions

A function is a rule that assigns exactly one output to each input in its domain; the set of outputs it actually produces is its range. Cambridge 9709 Pure 1 builds four question types on that: finding a range (usually by completing the square), forming a composite $\mathrm{gf}(x) = \mathrm{g}(\mathrm{f}(x))$ , finding an inverse $\mathrm{f}^{-1}(x)$ and stating its domain, and describing or applying transformations of a graph.

Section 1.2 carries 427 marks across the 40 Pure 1 papers Quanta has mapped — 14.2% of every mark in the corpus — and it appears on all 40. The recurring trap is the same one every year: an inverse exists only for a one-to-one function, so a quadratic has to be restricted before it has one.

Updated 15 September 2026

Domain and range

The domain is the set of inputs the function is defined for; the range is the set of outputs it produces over that domain. P1 states a domain in the question and asks for the range, and the reliable method for a quadratic is to complete the square:

\mathrm{f}(x) = x^2 - 6x + 11 = (x-3)^2 + 2 \quad\Rightarrow\quad \mathrm{f}(x) \ge 2

The square is never negative, so the least value is the constant — 2, reached at $x = 3$ . If the domain excludes the vertex the answer changes: on $x \ge 4$ the function is increasing throughout, so the range starts at $\mathrm{f}(4) = 3$ and the answer is $\mathrm{f}(x) \ge 3$ . Always check whether the given domain contains the turning point before quoting the vertex.

Notation matters for the mark: the range is written in terms of $\mathrm{f}(x)$ or $y$ , not $x$ . Writing $x \ge 2$ for a range is marked wrong even when 2 is right.

Composite functions

$\mathrm{gf}(x)$ means “do $\mathrm{f}$ first, then $\mathrm{g}$ ” — the function nearer $x$ acts first. With $\mathrm{f}(x) = 2x + 1$ and $\mathrm{g}(x) = x^2$ :

\mathrm{gf}(x) = \mathrm{g}(2x+1) = (2x+1)^2 \qquad \mathrm{fg}(x) = \mathrm{f}(x^2) = 2x^2 + 1

They are different, and questions exploit that. A composite exists only if the range of the inner function lies inside the domain of the outer one — so if $\mathrm{g}$ is defined only for $x \ge 0$ , then $\mathrm{gf}$ needs $\mathrm{f}(x) \ge 0$ , which restricts the composite’s own domain. “Solve $\mathrm{gf}(x) = 25$ ” is the standard follow-up: form the composite, set it equal, solve, then reject any root outside the domain.

Inverse functions

The inverse undoes the function: $\mathrm{f}^{-1}\mathrm{f}(x) = x$ . It exists only when the function is one-to-one — each output coming from exactly one input. A quadratic on its natural domain is two-to-one, which is why P1 questions restrict it (to $x \ge 3$ , say, one side of the vertex) before asking for an inverse.

To find it: write $y = \mathrm{f}(x)$ , make $x$ the subject, then swap the letters.

y = (x-3)^2 + 2 \;\Rightarrow\; (x-3)^2 = y - 2 \;\Rightarrow\; x = 3 + \sqrt{y-2}

\mathrm{f}^{-1}(x) = 3 + \sqrt{x-2}

The $+$ is a decision, not a default: the domain $x \ge 3$ means the outputs of the inverse must be $\ge 3$ , so the positive root is the right one. Had the restriction been $x \le 3$ , the answer would be $3 - \sqrt{x-2}$ .

Two facts that questions test directly: the domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ (here $x \ge 2$ ), and the graph of $y = \mathrm{f}^{-1}(x)$ is the reflection of $y = \mathrm{f}(x)$ in the line $y = x$ . That reflection is also why solving $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ can be done by solving $\mathrm{f}(x) = x$ instead.

Worked example

f(x) = x² − 4x + 7 for x ≥ 2. (i) Express f(x) in completed-square form and state the range. (ii) Find f⁻¹(x) and state its domain. (iii) g(x) = 2x − 1 for all x. Solve fg(x) = 7.

(i) $x^2 - 4x + 7 = (x-2)^2 + 3$ . The domain $x \ge 2$ starts exactly at the vertex and the function increases from there, so the range is $\mathrm{f}(x) \ge 3$ .

(ii) $y = (x-2)^2 + 3 \Rightarrow (x-2)^2 = y-3 \Rightarrow x = 2 + \sqrt{y-3}$ — positive root because $x \ge 2$ . So

\mathrm{f}^{-1}(x) = 2 + \sqrt{x-3}, \qquad x \ge 3

The domain of the inverse is the range of $\mathrm{f}$ from part (i) — a mark of its own, and one most candidates forget to write down.

(iii) $\mathrm{fg}(x) = \mathrm{f}(2x-1) = (2x-1)^2 - 4(2x-1) + 7$ . Expanding: $4x^2 - 4x + 1 - 8x + 4 + 7 = 4x^2 - 12x + 12$ . Set equal to 7:

4x^2 - 12x + 5 = 0 \;\Rightarrow\; (2x-1)(2x-5) = 0 \;\Rightarrow\; x = \tfrac{1}{2} \ \text{or}\ \tfrac{5}{2}

Now the check almost everyone skips. $\mathrm{f}$ is defined only for inputs $\ge 2$ , and $\mathrm{g}(\tfrac{1}{2}) = 0$ , which is not in that domain — so $x = \tfrac{1}{2}$ is rejected. $\mathrm{g}(\tfrac{5}{2}) = 4 \ge 2$ , so the only solution is $x = \tfrac{5}{2}$ .

Graph transformations

The largest single sub-skill in this section. Two families, and the rule is that changes inside the bracket affect $x$ and do the opposite of what they look like:

From $y = \mathrm{f}(x)$	Transformation
y = f(x) + a	Translation a units up — vector (0, a).
y = f(x + a)	Translation a units LEFT — vector (−a, 0). Inside the bracket, opposite direction.
y = k f(x)	Stretch parallel to the y-axis, factor k.
y = f(kx)	Stretch parallel to the x-axis, factor 1/k. Again the reciprocal.
y = −f(x)	Reflection in the x-axis.
y = f(−x)	Reflection in the y-axis.

Combined transformations are where the marks are, and order matters. To describe the move from $y = x^2$ to $y = 3(x-2)^2 + 1$ : translate 2 right, stretch by factor 3 parallel to the $y$ -axis, then translate 1 up. Describing a translation needs the word “translation” and a vector; describing a stretch needs the factor and the direction. “Moved up 1” does not score.

Common mistakes

1.Range written in terms of x
A range describes outputs: $\mathrm{f}(x) \ge 3$ or $y \ge 3$ . $x \ge 3$ is a domain, and loses the mark.
2.fg and gf the wrong way round
The function next to the $x$ goes first. $\mathrm{gf}(x) = \mathrm{g}(\mathrm{f}(x))$ — substitute $\mathrm{f}$ into $\mathrm{g}$ .
3.Taking ± on an inverse
An inverse is a function, so it takes one sign. Choose it from the restricted domain: outputs of $\mathrm{f}^{-1}$ must lie in the domain of $\mathrm{f}$ .
4.Not stating the inverse's domain
It is the range of the original function, and it carries its own mark. Part (i) usually hands it to you.
5.f(x + 2) translated right
Inside the bracket, the effect is reversed: $\mathrm{f}(x+2)$ moves the graph 2 to the left. Same for stretches: $\mathrm{f}(2x)$ is a stretch of factor $\tfrac{1}{2}$ .
6.Keeping a root outside the domain
After solving a composite equation, test each root against the domain of the inner and outer functions and reject what does not fit.

Common questions

How do you find the range of a quadratic function?

Complete the square. $a(x+p)^2 + q$ has minimum $q$ when $a > 0$ , so the range is $\mathrm{f}(x) \ge q$ — provided the given domain includes the vertex. If it does not, evaluate the function at the end of the domain instead.

What does gf(x) mean?

Apply $\mathrm{f}$ first, then $\mathrm{g}$ to the result: $\mathrm{gf}(x) = \mathrm{g}(\mathrm{f}(x))$ . It is generally not the same as $\mathrm{fg}(x)$ .

Why does a function need to be one-to-one to have an inverse?

Because the inverse must return a single input for each output. If two inputs give the same output — as $x = 1$ and $x = 5$ do for $(x-3)^2$ — the reverse rule would have to return both, which no function can do. Restricting the domain to one side of the vertex fixes it.

How are graph transformations described for full marks?

A translation needs the word “translation” and a column vector; a stretch needs the scale factor and the direction (parallel to which axis); a reflection needs the line. Vague directional language scores nothing even when the numbers are right.

Practise functions against real mark schemes

Quanta has real Cambridge A Level Maths 9709 past-paper questions, broken into skill checkpoints, marked criterion by criterion the way examiners mark — and it tracks which skills you’re missing. Free for individual students.

Start practising free