What is a function in the context of Cambridge International A Levels Pure Mathematics 1?

In Cambridge International A Levels Pure Mathematics 1, a function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. Functions are often represented as f(x), where x is the input variable, and f(x) is the output. Understanding functions is crucial as they form the foundation for many mathematical concepts and applications.

How do you determine if a relation is a function?

To determine if a relation is a function, each input value must correspond to exactly one output value. This can be checked using the vertical line test on a graph: if a vertical line intersects the graph at more than one point, the relation is not a function. This ensures that for every x-value, there is only one corresponding y-value.

What are the different types of functions studied in Pure Mathematics 1?

In Pure Mathematics 1, students study various types of functions including linear functions, quadratic functions, polynomial functions, rational functions, and exponential functions. Each type has distinct characteristics and equations, and understanding these differences is essential for solving problems related to functions in the Cambridge International A Levels curriculum.

How do you find the inverse of a function?

To find the inverse of a function, you need to swap the roles of the input and output variables and solve for the new output variable. For a function f(x), its inverse is denoted as f^(-1)(x). The process involves replacing f(x) with y, swapping x and y, and then solving for y. It's important to verify that the inverse is indeed a function by checking that it passes the horizontal line test.

What is the significance of domain and range in functions?

The domain of a function is the set of all possible input values, while the range is the set of all possible output values. Understanding the domain and range is crucial because it defines the limits within which the function operates. In Cambridge International A Levels Pure Mathematics 1, students learn to determine the domain and range from a function's equation or graph, which is essential for correctly interpreting and solving function-related problems.

Why is "Applying $f$ before $g$ in $fg(x)$" a common mistake in Functions?

Why it happens: Reading left to right. How to avoid it: fg(x) = f(g(x)) — INSIDE function applied FIRST. Think of it as f acting on the output of g. Source: 9709 Examiner Reports 2022-2024

Why is "Forgetting to swap domain and range for inverse" a common mistake in Functions?

Why it happens: Domain of f feels natural. How to avoid it: Domain of f^-1 = RANGE of f. Range of f^-1 = DOMAIN of f. Swap them. Source: 9709 Examiner Reports 2022-2024

Why is "Inverse of $f(x) = x^2$ on entire $\mathbb{R}$" a common mistake in Functions?

Why it happens: Forgetting one-to-one requirement. How to avoid it: f(x) = x^2 is not one-to-one on R — no inverse. Restrict domain (e.g. x ≥ 0) to make it invertible. Source: 9709 Examiner Reports 2022-2024

Pure Mathematics 1Cambridge International A Levels Mathematics (9709)

Functions

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Functions Study Notes — Cambridge International A Level Mathematics 9709 P1 (2024-2026 syllabus)

Domain, range, one-to-one, inverses, composites. The vocabulary and tools for everything that follows.

What you’ll learn

Mapped to the Cambridge IGCSE 9709 syllabus (2024-2026).

P1.2.1 — Use functional notation; identify domain and range.
P1.2.2 — Determine if function is one-to-one.
P1.2.3 — Find inverse functions.
P1.2.4 — Form composite functions.

Domain and range

Inputs and outputs.

Domain. Set of allowed inputs. Often $\mathbb{R}$ or restricted (e.g. $x \geq 0$ for square root).

Range. Set of all outputs achievable. Depends on BOTH the rule AND the domain.

Example. $f(x) = x^2 - 4x + 5$ for $x \geq 2$ .

Complete the square: $f(x) = (x-2)^2 + 1$ .
Vertex at $x = 2$ , $y = 1$ .
For $x \geq 2$ (right side of parabola), $f(x)$ ranges from $1$ (at $x = 2$ ) upwards.
Range: $f(x) \geq 1$ .

Notation. $x \in \mathbb{R}$ means $x$ is a real number. $x \in [a, b]$ is closed interval; $(a, b)$ open.

Cambridge tip. Find vertex via completing-the-square — fastest route to range.

Domain = inputs.
Range = outputs.
Both depend on rule and domain.

See the full worked example for functions →

Inverse functions

Swap, solve.

Inverse $f^{-1}$ undoes $f$ : $f^{-1}(f(x)) = x$ .

Exists if and only if $f$ is one-to-one — each output from exactly one input. For $f(x) = x^2$ on $\mathbb{R}$ , NO inverse (both $2$ and $-2$ give $4$ ). Restrict to $x \geq 0$ to make invertible.

Method.

Let $y = f(x)$ .
Swap $x$ and $y$ .
Solve for $y$ .
Replace $y$ with $f^{-1}(x)$ .

Example. $f(x) = 2x - 3$ .

$y = 2x - 3$ → $x = 2y - 3$ → $y = \frac{x+3}{2}$ .
$f^{-1}(x) = \frac{x+3}{2}$ .

Domain/range swap.

Domain of $f^{-1}$ = range of $f$ .
Range of $f^{-1}$ = domain of $f$ .

Graphical view. $y = f^{-1}(x)$ is reflection of $y = f(x)$ in line $y = x$ .

Cambridge tip. Always state domain and range explicitly when finding inverse.

Swap $x$ and $y$ , solve.
Domain/range swap.
Reflection in $y = x$ .

See the full worked example for functions →

Composite functions

$fg(x) = f(g(x))$ .

Composite $f \circ g$ . Apply $g$ first, then $f$ . So $fg(x) = f(g(x))$ .

Order matters. $fg \ne gf$ in general.

Example. $f(x) = x^2 + 1$ , $g(x) = 2x - 1$ .

$fg(x) = f(g(x)) = f(2x - 1) = (2x-1)^2 + 1 = 4x^2 - 4x + 2$ .
$gf(x) = g(f(x)) = g(x^2 + 1) = 2(x^2 + 1) - 1 = 2x^2 + 1$ .

Domain of composite. $fg(x)$ is defined when $x \in \text{dom}(g)$ AND $g(x) \in \text{dom}(f)$ .

Cambridge tip. Inside function FIRST. Read $fg$ as " $f$ AFTER $g$ ".

Apply inner function first.
$fg \ne gf$ generally.
Read right-to-left.

See the full worked example for functions →

How it’s examined

Functions appears every P1 — usually Q1 or Q2 on 9709/12. Most-tested: inverse (5 marks), composite (6 marks), range with restricted domain (5 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/12 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Functions

Step-by-step solutions to past-paper-style questions on functions, written exactly the way a tutor would explain them at the board.

1Finding an inverse function (5 marks)
Extended• Adapted from 9709/12 May/Jun 2024• inverse
Question
The function $f$ is defined by $f(x) = 2x - 3$ for $x \in \mathbb{R}$ . Find $f^{-1}(x)$ and state its domain and range. (5 marks)
Step-by-step solution
1. Step 1
  Let $y = f(x)$ .
  $y = 2x - 3$
2. Step 2
  Swap $x$ and $y$ , then solve for $y$ .
  $x = 2y - 3 \Rightarrow y = \frac{x+3}{2}$
3. Step 3
  Replace $y$ with $f^{-1}(x)$ .
  $f^{-1}(x) = \dfrac{x+3}{2}$
4. Step 4
  Domain of $f^{-1}$ is the range of $f$ . Since $f$ has range $\mathbb{R}$ , $f^{-1}$ has domain $\mathbb{R}$ .
5. Step 5
  Range of $f^{-1}$ is the domain of $f$ . Range of $f^{-1}$ is $\mathbb{R}$ .
Answer
$f^{-1}(x) = \frac{x+3}{2}$ , domain and range both $\mathbb{R}$ .
2Composite functions (6 marks)
Extended• composite
Question
Functions $f$ and $g$ are defined by $f(x) = x^2 + 1$ for $x \geq 0$ and $g(x) = 2x - 1$ for $x \in \mathbb{R}$ . Find $fg(x)$ and $gf(x)$ . (6 marks)
Step-by-step solution
1. Step 1
  $fg(x) = f(g(x))$ . Apply $g$ first, then $f$ .
  $fg(x) = f(2x-1) = (2x-1)^2 + 1$
2. Step 2
  Expand.
  $fg(x) = 4x^2 - 4x + 1 + 1 = 4x^2 - 4x + 2$
3. Step 3
  $gf(x) = g(f(x))$ . Apply $f$ first, then $g$ .
  $gf(x) = g(x^2 + 1) = 2(x^2 + 1) - 1 = 2x^2 + 1$
4. Step 4
  Note. $fg \ne gf$ in general — composition is not commutative.
Answer
$fg(x) = 4x^2 - 4x + 2$ ; $gf(x) = 2x^2 + 1$ .
3Range with restricted domain (5 marks)
Extended• range
Question
The function $f$ is defined by $f(x) = x^2 - 4x + 5$ for $x \geq 2$ . Find the range of $f$ . (5 marks)
Step-by-step solution
1. Step 1
  Complete the square to find vertex.
  $f(x) = (x-2)^2 + 1$
2. Step 2
  Vertex at $x = 2$ . Since domain starts at $x = 2$ , vertex is INCLUDED.
3. Step 3
  At $x = 2$ , $f(x) = 1$ .
4. Step 4
  For $x > 2$ , $f(x)$ increases. Curve opens upwards.
5. Step 5
  Range: $f(x) \geq 1$ .
Answer
$f(x) \geq 1$ .
Examiner tip
Top-band candidates state the range using the SAME form as the question (i.e. $f(x) \geq 1$ , not $y \geq 1$ ).

Key Formulae — Functions

The formulae you need to memorise for functions on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

Inverse function
$f \circ f^{-1}(x) = x = f^{-1} \circ f(x)$
$f^{-1}$
inverse function
When to use
Verifying or constructing inverse. Domain of $f^{-1}$ = range of $f$ ; range of $f^{-1}$ = domain of $f$ .
Composite function
$(f \circ g)(x) = f(g(x))$
$f \circ g$
apply $g$ first, then $f$
When to use
When composing two functions. Order matters: $f \circ g \ne g \circ f$ in general.

Key Definitions and Keywords — Functions

Definitions to memorise and the exact keywords mark schemes credit for functions answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9709 sitting.

Function
Examiner keyword
A rule assigning to each input from the domain EXACTLY ONE output in the range.
Domain
Examiner keyword
Set of allowed inputs for a function. Often $\mathbb{R}$ or a subset.
Range
Examiner keyword
Set of all outputs the function can produce. Depends on both rule AND domain.
One-to-one
Examiner keyword
Each output comes from EXACTLY one input. Required for an inverse to exist.
Inverse function
Examiner keyword
Function $f^{-1}$ such that $f \circ f^{-1} = f^{-1} \circ f = \text{identity}$ .
Composite function
Examiner keyword
Function formed by applying one function to another's output. $(f \circ g)(x) = f(g(x))$ .

Common Mistakes and Misconceptions — Functions

The traps other students keep falling into on functions questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Applying $f$ before $g$ in $fg(x)$
9709 Examiner Reports 2022-2024
Why it happens
Reading left to right.
How to avoid it
$fg(x) = f(g(x))$ — INSIDE function applied FIRST. Think of it as $f$ acting on the output of $g$ .
✕Forgetting to swap domain and range for inverse
9709 Examiner Reports 2022-2024
Why it happens
Domain of $f$ feels natural.
How to avoid it
Domain of $f^{-1}$ = RANGE of $f$ . Range of $f^{-1}$ = DOMAIN of $f$ . Swap them.
✕Inverse of $f(x) = x^2$ on entire $\mathbb{R}$
9709 Examiner Reports 2022-2024
Why it happens
Forgetting one-to-one requirement.
How to avoid it
$f(x) = x^2$ is not one-to-one on $\mathbb{R}$ — no inverse. Restrict domain (e.g. $x \geq 0$ ) to make it invertible.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Functions

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Finding an inverse function (5 marks)

2Composite functions (6 marks)

3Range with restricted domain (5 marks)

Inverse function

Composite function

Function

Domain

Range

One-to-one

Inverse function

Composite function

✕Applying fff before ggg in fg(x)fg(x)fg(x)

✕Forgetting to swap domain and range for inverse

✕Inverse of f(x)=x2f(x) = x^2f(x)=x2 on entire R\mathbb{R}R

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Functions — frequently asked questions

✕Applying $f$ before $g$ in $fg(x)$

✕Inverse of $f(x) = x^2$ on entire $\mathbb{R}$