Functions Revision Notes — 9709 Paper 1

The language, used precisely

Syllabus requirement · §1.2

“

understand the terms function, domain, range, one-one function, inverse function and composition of functions.

”

A function has three parts, and a question can ask about any of them.

The rule — what happens to the input, e.g. $\mathrm{f}(x) = 2 - \dfrac{5}{x+2}$ .
The domain — the set of inputs you are allowed to use, e.g. $x > -2$ . It is given to you; it is not something you work out.
The range — the set of outputs the rule actually produces from that domain. This one you do work out.

Domain in, range out

Domain lives on the $x$ -axis, range lives on the $y$ -axis. Almost every notation error in this topic comes from mixing those up.

The domain x ⩾ 1 is marked along the x-axis and the range f(x) ⩾ 2 up the y-axis. Read each off the axis it belongs to.

Reading a function definition

Every functions question opens by defining the functions, and the definition is dense. Learn to unpack it slowly, because everything later depends on it.

\mathrm{f}(x) = 2 - \frac{5}{x+2} \quad \text{ for } x > -2

$\mathrm{f}$ is the name of the function. It is a label, like a name on a machine.
$\mathrm{f}(x)$ is the output when you feed in $x$ . It is not "f times x".
$2 - \frac{5}{x+2}$ is the rule.
$x > -2$ is the domain — the fence.

Two notations mean exactly the same thing, and papers use both:

\mathrm{f}(x) = 3x + 1 \qquad\text{and}\qquad \mathrm{f} : x \mapsto 3x + 1

Read the second as " $\mathrm{f}$ maps $x$ to $3x+1$ ". Older papers prefer the arrow form; recent ones prefer the bracket form. Nothing about the mathematics changes.

You will also meet $x \in \mathbb{R}$ , which reads " $x$ is a real number" and means the domain is every number there is — no fence at all. That is a signal worth noticing: if the domain is all of $\mathbb{R}$ and the rule is a quadratic, the function is certainly not one-one, so a later part will restrict it before asking for an inverse.

The notation that costs marks

When you state a range, name the output: write $\mathrm{f}(x) \geqslant 2$ or $y \geqslant 2$ . Writing $x \geqslant 2$ is marked wrong, because $x$ is the input. When you state a domain, name the input: $x > -2$ .

Mark schemes say this explicitly — one 2023 scheme reads "Do not accept $x < 2$ " for a range, and another "Not $x \geqslant -2$ " where $\mathrm{g}^{-1}(x) \geqslant -2$ was wanted.

One-one and many-one

A function is one-one if different inputs always give different outputs. If two inputs share an output it is many-one, and it has no inverse — given the output you could not say which input it came from.

A many-one function is not broken — it just cannot be reversed. Restricting the domain until only one arrow arrives at each output is what makes an inverse possible.

The horizontal line test

On a graph: if any horizontal line meets the curve more than once, the function is many-one. $y = x^2$ fails it; $y = x^2$ for $x \geqslant 0$ passes, because you have thrown away the left half.

The quickest one-one argument: always going the same way

There is a second test, and in the exam it is usually the faster one.

A function is increasing on its domain if the graph only ever goes up as you move right, and decreasing if it only ever goes down. Either way the curve never doubles back, so a horizontal line can never catch it twice.

Increasing on its domain $\Rightarrow$ one-one. Decreasing on its domain $\Rightarrow$ one-one.

A function fails to be one-one only when it turns round — and a curve turns round at a turning point: the top or bottom of a bend, like the vertex of a parabola. That gives you a rule of thumb that decides almost every exam case in one glance:

No turning point inside the domain $\Rightarrow$ one-one.
A turning point strictly inside the domain $\Rightarrow$ many-one.

So $\mathrm{f}(x) = x^2$ for $x \in \mathbb{R}$ is many-one: it has a minimum at $x = 0$ , which is inside the domain, and the curve comes down and then goes back up. But $\mathrm{f}(x) = x^2$ for $x \geqslant 0$ is one-one: the turning point is now at the edge of the domain, not inside it, so the curve only ever rises.

Answering “does f⁻¹ exist? Give a reason”

This is usually a 1-mark question, and it wants one clean sentence. Any of these earns the mark:

"Yes — $\mathrm{f}$ is one-one."
"Yes — $\mathrm{f}$ is a decreasing function" (or increasing).
"Yes — it passes the horizontal line test."
"No — $\mathrm{f}$ is not one-one" / "it is many-one".

What earns nothing: "yes, because you can rearrange it", or "no, because it's a quadratic". The reason must be about one-one-ness, not about the algebra.

Explaining, in words, why a function is one-one

9709/11 O/N 2024 Q11(c)2 marks

The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = 3 + 6x - 2x^2$ for $x \leqslant 0$ .

Sketch the graph of $y = \mathrm{g}(x)$ and explain why $\mathrm{g}$ is a one-one function. You are not required to find the coordinates of any intersections with the axes.

Show full working

1
First work out what the unrestricted curve looks like. The coefficient of $x^2$ is $-2$ , which is negative, so the parabola opens downwards — it has a maximum, not a minimum.
Always settle the direction of opening before anything else. It decides where the turning point is relative to the domain.
2
Find where the turning point is. Part (a) of this question completed the square (the working is the first exercise of the next section): $\mathrm{g}(x) = \tfrac{15}{2} - 2\left(x - \tfrac32\right)^2$ The bracket is zero when $x = \tfrac32$ , so the maximum is at $x = \tfrac32$ .
3
Now compare that with the domain. The domain is $x \leqslant 0$ , and the turning point sits at $x = \tfrac32$ , which is outside it — to the right of everything we are allowed to use.
This single comparison is the whole question. Turning point outside the domain ⇒ the curve never turns round ⇒ one-one.
4
So on $x \leqslant 0$ we are only ever on the left-hand arm of the downward parabola, which is rising all the way from bottom-left up to the point $\left(0,\,3\right)$ .
5
The sketch. Draw only that rising arm: it comes up from the bottom left, stops at $x = 0$ , and does not turn over. At $x = 0$ , $\mathrm{g}(0) = 3$ , so the curve ends at $(0, 3)$ on the positive $y$ -axis. The curve therefore lives in the second and third quadrants only.
The mark scheme awards its first mark for exactly that: “graph appearing in second and third quadrants only”. Drawing the whole parabola loses it.
6
The explanation. Say it in terms of inputs and outputs: each $y$ -value on the curve comes from exactly one $x$ -value, or equivalently any horizontal line meets the curve at most once.
The mark scheme accepts the horizontal line test but explicitly says to ignore the vertical line test — that tests whether it is a function at all, which was never in doubt.

Answer

A curve rising from the bottom left and stopping at $(0, 3)$ , drawn in the second and third quadrants only. Reason: the maximum is at $x = \tfrac32$ , which is outside the domain $x \leqslant 0$ , so the curve is increasing throughout and each output comes from exactly one input.

Whenever a question restricts a quadratic and then asks "explain why it is one-one", the answer is always the same shape: locate the vertex, show it is not inside the domain, conclude the curve never turns back.

Common mistakes

" $\mathrm{f}$ is a function, so it has an inverse"
" $\mathrm{f}$ is one-one, so it has an inverse"
Every rule with a fence is a function. Only the one-one ones can be reversed.
Using the vertical line test to justify one-one
Using the horizontal line test
The vertical test only checks that the rule gives one output per input — i.e. that it is a function at all. One-one is about the other direction.
Writing the domain of $\mathrm{f}(x) = \dfrac{1}{x-3}$ as " $x \neq 3$ " when the question already stated $x > 3$
The domain is whatever the question printed: $x > 3$
The domain is given, never deduced. Working out 'where the formula is defined' answers a question CAIE did not ask.

Your turn

Three short ones. In each case the marks are for naming the right idea in the right words, not for algebra — so write full sentences, and always say which of domain and range you are talking about.

14 marks
The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = 4 - 3x$ for $-1 \leqslant x \leqslant 5$ .

(a) Write down the domain of $\mathrm{f}$ .
(b) Find the range of $\mathrm{f}$ , using correct notation.
(c) Is $\mathrm{f}$ one-one? Give a reason.
Stuck? Show hint
For (b), the rule is linear, so the outputs run in a straight line between the outputs at the two ends of the domain. Watch what the minus sign in $-3x$ does to the order.
Show solution
1. 1
  (a) The domain is read straight off the definition, with no work at all: $-1 \leqslant x \leqslant 5$
  Written with x, because the domain is a set of inputs.
2. 2
  (b) Substitute the left-hand end of the domain, $x = -1$ : $\mathrm{f}(-1) = 4 - 3(-1) = 4 + 3 = 7$
3. 3
  Substitute the right-hand end, $x = 5$ : $\mathrm{f}(5) = 4 - 3(5) = 4 - 15 = -11$
4. 4
  So the two extreme outputs are $7$ and $-11$ . Because the rule is linear, every value in between is also produced, and nothing outside.
  A straight line has no turning points, so it sweeps steadily from one end value to the other — no output can be skipped and none can be exceeded.
5. 5
  Write the range in increasing order, in terms of the output: $-11 \leqslant \mathrm{f}(x) \leqslant 7$
  The gradient is negative, so the larger input gave the smaller output. Writing “7 ⩽ f(x) ⩽ −11” is a real and common slip.
6. 6
  (c) The graph of $\mathrm{f}$ is a straight line with gradient $-3$ . A straight line has no turning point, so it is decreasing throughout its domain and can never take the same value twice.
7. 7
  So yes, $\mathrm{f}$ is one-one, because it is a decreasing function.
Answer
(a) $-1 \leqslant x \leqslant 5$ (b) $-11 \leqslant \mathrm{f}(x) \leqslant 7$ (c) Yes — it is a decreasing (linear) function, so no output is repeated.
29709/12 O/N 2022 Q9(c)1 mark
The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = x + \dfrac{1}{x}$ for $x > 0$ .

Given that the graph of $y = \mathrm{f}(x)$ has a minimum point when $x = 1$ , explain whether or not $\mathrm{f}$ has an inverse.
Stuck? Show hint
You are handed the position of the turning point. The only thing left to decide is whether it lies inside the domain.
Show solution
1. 1
  The domain is $x > 0$ . The minimum point is at $x = 1$ , and $1 > 0$ , so the turning point lies inside the domain.
2. 2
  That means the curve falls until $x = 1$ and then rises again. Values on the way down are repeated on the way up — for instance $\mathrm{f}(2) = 2 + \tfrac12 = 2\tfrac12$ and $\mathrm{f}\!\left(\tfrac12\right) = \tfrac12 + 2 = 2\tfrac12$ , the same output from two different inputs.
  Producing an actual repeated pair is not required for the mark, but it is the clearest possible evidence and takes ten seconds.
3. 3
  A repeated output means $\mathrm{f}$ is many-one, so it cannot be reversed: given the output $2\tfrac12$ you could not say which input it came from.
Answer
No. The minimum at $x = 1$ lies inside the domain $x > 0$ , so the curve turns round and $\mathrm{f}$ is many-one, not one-one. A many-one function has no inverse.
39709/12 O/N 2025 Q9(b)1 mark
The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = \dfrac{4}{(3x-6)^2} + \dfrac{1}{(3x-6)^3}$ for $x > 2$ .

(In part (a) of this question you showed, by differentiating, that $\mathrm{f}$ is a decreasing function.)

State whether $\mathrm{f}^{-1}$ exists. Give a reason for your answer.
Stuck? Show hint
Do not try to analyse that formula. The one fact carried over from part (a) is all the question needs.
Show solution
1. 1
  From part (a), $\mathrm{f}$ is decreasing: its graph falls all the way across the domain.
  Part (a) used differentiation, which comes later in Paper 1. For this part you only need its conclusion.
2. 2
  A decreasing function never turns back upwards, so it can never return to an output it has already produced. Every output therefore comes from exactly one input, which is precisely what one-one means.
3. 3
  One-one functions can be reversed, so $\mathrm{f}^{-1}$ exists.
Answer
Yes, $\mathrm{f}^{-1}$ exists, because $\mathrm{f}$ is a decreasing function on its domain and is therefore one-one.

In the exam

162 parts · 429 marks · 2.6 marks per part, 2021–2025

Functions usually comes as one long structured question near the end of the paper, in several short parts that move through range, composite, inverse and a graph. The algebra in each part is short; the marks that get lost are mostly lost on notation — the wrong letter in a range, or a missing reason.

Practise the function vocabularyReal past-paper questions · Function terminology (domain, range, one-one, inverse, composition)

Checking your access…

Can you do all of these?

State the range of $\mathrm{f}(x) = (x-3)^2 + 1$ for $x \geqslant 3$ , using the right letter
State the range of $\mathrm{f}(x) = (x+3)^2 - 12$ for $x \geqslant 0$ — and explain why it is not $\geqslant -12$
Find the range of $\mathrm{f}(x) = 2 - 3\cos x$ for $0 \leqslant x \leqslant 2\pi$ , getting the inequality signs the right way round
State the range of $\mathrm{f}(x) = \dfrac{1}{x}$ for $x \geqslant 1$ , with $<$ at one end and $\leqslant$ at the other
Explain why $\mathrm{f}(x) = x^2$ for $x \in \mathbb{R}$ has no inverse, and how to fix it
Answer "does $\mathrm{f}^{-1}$ exist? Give a reason" in one sentence that mentions one-one-ness
Form $\mathrm{g}\mathrm{f}(x)$ and $\mathrm{f}\mathrm{g}(x)$ for two given functions and see that they differ
Form $\mathrm{f}\mathrm{f}(x)$ , and evaluate $\mathrm{g}\mathrm{f}(3)$ without building the algebraic composite
Solve $\mathrm{g}\mathrm{f}(x) = k$ by both routes, and reject the root that breaks the domain
Say when a composite $\mathrm{g}\mathrm{f}$ cannot be formed, in one sentence naming both sets
Find the range of a composite such as $\mathrm{g}\mathrm{f}$ , and use a given range to find a constant
Recover $\mathrm{h}$ from $\mathrm{g}\mathrm{h}(x)$ , and recover $\mathrm{g}$ from $\mathrm{g}^{-1}(x)$
State the least $k$ for which $\mathrm{f}(x) = 3x^2 - 12x + 14$ is one-one on $x \geqslant k$
Find $\mathrm{f}^{-1}(x)$ when $x$ appears twice, e.g. $\mathrm{f}(x) = \dfrac{2x+1}{x-3}$
Invert a restricted quadratic and justify which square-root branch you kept
Write the domain of $\mathrm{f}^{-1}$ without doing any new algebra
Write down $(\mathrm{g}\mathrm{f})^{-1}$ in terms of $\mathrm{f}^{-1}$ and $\mathrm{g}^{-1}$ , in the correct order
Sketch $y = \mathrm{f}^{-1}(x)$ from a given curve, with the mirror line drawn and the asymptotes swapped
Find where $y = \mathrm{f}(x)$ meets $y = \mathrm{f}^{-1}(x)$ without finding $\mathrm{f}^{-1}$
Describe fully the sequence taking $y = \sqrt{x}$ to $y = 3\sqrt{x+2} - 5$ , in order
Say which axis $y = -\mathrm{f}(x)$ is reflected in, and which axis $y = \mathrm{f}(-x)$ is reflected in
Apply a printed sequence of three transformations to an equation, one step at a time
Write "stretch $\times 3$ in $y$ , then reflect in the $y$ -axis" as an equation in terms of $\mathrm{g}$
Describe $y = \mathrm{f}(2x + 6)$ both ways round, and write a graph change as $a\mathrm{f}(bx + c)$
Follow a single point, such as a minimum, through a sequence of transformations
Invert $1 + 2\sin x$ for $-\tfrac12\pi \leqslant x \leqslant \tfrac12\pi$ using $\sin^{-1}$

Now do the questions

162 real Paper 1 parts from 2021–2025, sorted by difficulty, with mark schemes

Quadratics

Coordinate Geometry

Functions

The language, used precisely

Reading a function definition

One-one and many-one

The quickest one-one argument: always going the same way

Finding the range

Composite functions

Can you do all of these?

The language, used precisely

Reading a function definition

One-one and many-one

The quickest one-one argument: always going the same way

The rest of this note

Can you do all of these?