Functions: Domain, Range & Mapping

This lesson covers functions, domain, range, and mappings as required by the AQA A-Level Mathematics specification (7357). A solid understanding of functions is essential for the rest of the course — you will need to know what makes a relation a function, how to determine domain and range, and how restricting the domain can make a function invertible.

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What Is a Function?

A function is a mapping that takes each element of a set (called the domain) and assigns it to exactly one element of another set (called the codomain). We write f : A → B to mean "f is a function from set A to set B".

The key requirement is: every input produces exactly one output.

Function vs Relation

A relation is any rule that connects elements of one set to elements of another. Not every relation is a function.

Example: The relation y² = x is not a function of x, because for each positive x there are two values of y (e.g. if x = 4, then y = 2 or y = −2).

The vertical line test can be used on a graph: if any vertical line crosses the graph more than once, the relation is not a function.

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Types of Mapping

Mapping Type	Description	Example
One-to-one	Each output comes from exactly one input	f(x) = 2x + 1
Many-to-one	Different inputs can give the same output	f(x) = x²
One-to-many	One input gives multiple outputs	y² = x (not a function!)

At A-Level, a function must be either one-to-one or many-to-one. A one-to-many mapping is not a function.

The Horizontal Line Test

To determine whether a function is one-to-one or many-to-one, use the horizontal line test:

• If every horizontal line intersects the graph at most once, the function is one-to-one.
• If any horizontal line intersects the graph more than once, the function is many-to-one.

Example: f(x) = x³ is one-to-one (every horizontal line meets the curve once).

Example: f(x) = x² is many-to-one (the horizontal line y = 4 meets the curve at x = 2 and x = −2).

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Domain and Range

Domain

The domain of a function is the set of all allowed input values (x-values). When defining a function, you should always state its domain.

Common restrictions on domain:

• Square roots: the expression under the root must be ≥ 0.
• Fractions: the denominator must not be zero.
• Logarithms: the argument must be > 0.

Example: f(x) = √(x − 3). The domain is x ≥ 3 or [3, ∞) because x − 3 ≥ 0.

Example: f(x) = 1/(x − 2). The domain is x ∈ ℝ, x ≠ 2 because the denominator cannot be zero.

Range

The range of a function is the set of all possible output values (y-values). The range depends on both the function rule and the domain.

Example: f(x) = x² with domain x ∈ ℝ. The range is f(x) ≥ 0 or [0, ∞).

Example: f(x) = x² with domain x ≥ 3. The range is f(x) ≥ 9 or [9, ∞).

Example: f(x) = 3 − 2x with domain x ∈ ℝ. The range is f(x) ∈ ℝ (all real numbers).

Exam Tip: To find the range, consider the graph of the function. Identify the minimum and maximum values of the output. If the function has a restricted domain, evaluate the function at the endpoints and at any turning points within the domain.

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Restricting the Domain

A many-to-one function cannot have an inverse function (because the inverse would be one-to-many, which is not a function). However, by restricting the domain, we can make a many-to-one function into a one-to-one function, thereby enabling it to have an inverse.

Example: f(x) = x² with domain x ∈ ℝ is many-to-one (e.g. f(2) = f(−2) = 4).

If we restrict the domain to x ≥ 0, then f(x) = x² is now one-to-one, and it has an inverse: f⁻¹(x) = √x.

If instead we restrict the domain to x ≤ 0, then f⁻¹(x) = −√x.

Example: g(x) = (x − 1)² + 3 has vertex at (1, 3).

• Restricting to x ≥ 1 gives a one-to-one function with range g(x) ≥ 3.
• Restricting to x ≤ 1 gives a one-to-one function with range g(x) ≥ 3.

Choosing a Domain Restriction

To make a quadratic function one-to-one, restrict the domain to one side of the vertex. The standard choice (unless told otherwise) is to include the vertex:

• For f(x) = (x − a)² + b, restrict to x ≥ a (or x ≤ a).

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Piecewise Functions

A piecewise function is defined by different rules on different parts of its domain.

Example:

f(x) = { 2x + 1   if x < 0
        { x²       if x ≥ 0

For x = −3: f(−3) = 2(−3) + 1 = −5. For x = 2: f(2) = 4. For x = 0: f(0) = 0 (using the x ≥ 0 rule).

The domain is x ∈ ℝ. The range requires checking both pieces:

• For x < 0: 2x + 1 < 1, so this piece gives (−∞, 1).
• For x ≥ 0: x² ≥ 0, so this piece gives [0, ∞).
• Combined range: (−∞, 1) ∪ [0, ∞) = (−∞, ∞) = ℝ (since (−∞, 1) and [0, ∞) overlap).

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Function Notation and Evaluation

Example: Given f(x) = 3x² − 2x + 1, find:

(a) f(2) = 3(4) − 4 + 1 = 9

(b) f(−1) = 3(1) + 2 + 1 = 6

(c) f(a + 1) = 3(a + 1)² − 2(a + 1) + 1 = 3a² + 6a + 3 − 2a − 2 + 1 = 3a² + 4a + 2

Solving f(x) = k

Example: Given f(x) = x² − 4x + 7 with domain x ≥ 2, solve f(x) = 3.

x² − 4x + 7 = 3
x² − 4x + 4 = 0
(x − 2)² = 0
x = 2

Check: x = 2 is in the domain x ≥ 2. ✓

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Summary

• A function maps each input to exactly one output.
• Functions can be one-to-one or many-to-one; one-to-many mappings are not functions.
• The domain is the set of allowed inputs; the range is the set of possible outputs.
• Restricting the domain of a many-to-one function can make it one-to-one, enabling an inverse.
• Use the vertical line test to check whether a relation is a function.
• Use the horizontal line test to check whether a function is one-to-one.
• Always state the domain and range when defining a function.

Exam Tip: When asked for the domain or range, express your answer in set notation or interval notation. Always check for restrictions such as division by zero, square roots of negatives, or logarithms of non-positive numbers. When restricting a domain to make a function invertible, state clearly which restriction you are using and why.

A-Level Deep Dive: Functions, Domain and Range

Spec mapping

AQA 7357 specification, Pure Mathematics Section B — Algebra and Functions. The function concept is the connective tissue of A-Level Pure: every later topic — composite functions, inverse functions, transformations, calculus — assumes confident handling of $f: A \to B$f:A→B, the domain (the set of permissible inputs), and the range (the actual set of outputs that the rule produces). The specification expects students to use formal function notation, distinguish one-to-one (injective) from many-to-one mappings, and recognise that a function only has an inverse on a domain where it is one-to-one. Domain/range reasoning is examined directly on Paper 2 (Pure) and synoptically on Paper 1 (Pure) wherever transformations or calculus appear, and indirectly on Paper 3 (Statistics & Mechanics) where probability density functions and kinematics functions both rely on domain-restriction logic. The AQA formula booklet does not list any function definitions — they must be carried in working memory.

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Worked example with full mark scheme

Question (8 marks):

The function $f$f is defined by $f(x) = (x - 2)^2 + 5$f(x)=(x−2)2+5 for $x \geq 0$x≥0.

(a) State the range of $f$f. (3)

(b) Determine, with reasons, whether $f$f is one-to-one on the given domain. (3)

(c) State the largest sub-domain containing $x = 0$x=0 on which $f$f has an inverse, and give the range of $f$f on that sub-domain. (2)

Solution with mark scheme:

(a) Step 1 — recognise the structure.

The expression $(x - 2)^2 + 5$(x−2)2+5 is in completed-square form, with vertex at $(2, 5)$(2,5). The squared term $(x - 2)^2 \geq 0$(x−2)2≥0 for all real $x$x, with equality at $x = 2$x=2.

M1 — identifying the completed-square form and noting that the minimum of $(x - 2)^2$(x−2)2 over the reals is 0, attained at $x = 2$x=2.

Step 2 — check whether the vertex lies in the domain.

The domain is $x \geq 0$x≥0, and $x = 2$x=2 satisfies $x \geq 0$x≥0, so the vertex is inside the domain. The minimum value of $f$f on the domain is therefore $f(2) = 0 + 5 = 5$f(2)=0+5=5.

M1 — verifying that the unconstrained minimum point lies inside the restricted domain (this is the step most students skip).

Step 3 — examine the boundary and behaviour.

At the left boundary $x = 0$x=0: $f(0) = (-2)^2 + 5 = 9$f(0)=(−2)2+5=9. As $x \to \infty$x→∞, $f(x) \to \infty$f(x)→∞. The parabola decreases from $f(0) = 9$f(0)=9 down to $f(2) = 5$f(2)=5, then increases without bound.

So the range is $f(x) \geq 5$f(x)≥5, or in interval notation $[5, \infty)$[5,∞).

A1 — correct range $[5, \infty)$[5,∞) or $f(x) \geq 5$f(x)≥5, written using a closed lower bound (the minimum is attained).

(b) Step 1 — test for one-to-one.

A function is one-to-one on a domain if every output is produced by at most one input. Equivalently, no horizontal line crosses the graph more than once on that domain.

Step 2 — find a counterexample.

Consider $f(0) = 9$f(0)=9 and $f(4) = (4 - 2)^2 + 5 = 4 + 5 = 9$f(4)=(4−2)2+5=4+5=9. Both $x = 0$x=0 and $x = 4$x=4 lie in the domain $x \geq 0$x≥0, and $f(0) = f(4) = 9$f(0)=f(4)=9.

M1 — attempting to find two distinct domain values with the same image.

A1 — a correct concrete pair such as $(0, 4)$(0,4) or $(1, 3)$(1,3) (both give image 9 and 6 respectively).

Step 3 — conclude.

Since two distinct inputs share an image, $f$f is many-to-one on $x \geq 0$x≥0, hence not one-to-one.

A1 — explicit conclusion stating "not one-to-one" or "many-to-one", referencing the counterexample.

(c) The vertex sits at $x = 2$x=2, splitting the domain into a decreasing branch $0 \leq x \leq 2$0≤x≤2 and an increasing branch $x \geq 2$x≥2. The branch containing $x = 0$x=0 is $[0, 2]$[0,2].

M1 — identifying $[0, 2]$[0,2] as the largest sub-domain containing 0 on which $f$f is monotonic (here, monotonically decreasing).

On $[0, 2]$[0,2], $f$f runs from $f(0) = 9$f(0)=9 down to $f(2) = 5$f(2)=5, so the range is $[5, 9]$[5,9].

A1 — range $[5, 9]$[5,9], with both endpoints attained.

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the AQA Paper 2 format

Question (6 marks): The function $g$g is defined by $g(x) = \dfrac{1}{x - 3}$g(x)=x−31​ for $x \in \mathbb{R}$x∈R, $x \neq 3$x=3.

(a) State the domain and range of $g$g. (3)

(b) Explain why $g$g is one-to-one on its domain. (3)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — domain stated as $x \in \mathbb{R}$x∈R, $x \neq 3$x=3, or equivalently $\mathbb{R} \setminus \{3\}$R∖{3}.
• M1 (AO1.1b) — recognising that $\dfrac{1}{x - 3}$x−31​ takes every non-zero value as $x$x varies over its domain (consider solving $y = \dfrac{1}{x - 3}$y=x−31​ for $x$x — this gives $x = 3 + \dfrac{1}{y}$x=3+y1​, defined for every $y \neq 0$y=0).
• A1 (AO2.5) — range stated as $g(x) \in \mathbb{R}$g(x)∈R, $g(x) \neq 0$g(x)=0, or equivalently $\mathbb{R} \setminus \{0\}$R∖{0}.

(b)

• M1 (AO2.1) — setting $g(x_1) = g(x_2)$g(x1​)=g(x2​) and attempting algebraic manipulation: $\dfrac{1}{x_1 - 3} = \dfrac{1}{x_2 - 3}$x1​−31​=x2​−31​.
• M1 (AO2.1) — deducing $x_1 - 3 = x_2 - 3$x1​−3=x2​−3 by taking reciprocals (valid because both sides are non-zero on the domain).
• A1 (AO2.5) — concluding $x_1 = x_2$x1​=x2​, hence $g$g is one-to-one. Reference to the horizontal-line test on the rectangular hyperbola is acceptable but earns full marks only with explicit reasoning.

Total: 6 marks split AO1 = 1, AO2 = 5. This is unusually AO2-heavy for a Pure question — AQA uses domain/range items to probe reasoning, not just procedure.

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Synoptic links

Connects to:

1. Composite functions $(g \circ f)(x) = g(f(x))$(g∘f)(x)=g(f(x)): the domain of $g \circ f$g∘f is the set of $x$x in $\text{dom}(f)$dom(f) for which $f(x) \in \text{dom}(g)$f(x)∈dom(g). So composite-function domain questions reduce to a range vs domain compatibility check. If $f(x) = \sqrt{x}$f(x)=x​ (range $[0, \infty)$[0,∞)) and $g(x) = \dfrac{1}{x - 4}$g(x)=x−41​ (domain $x \neq 4$x=4), then $g(f(x))$g(f(x)) requires $\sqrt{x} \neq 4$x​=4, i.e. $x \neq 16$x=16 — and of course $x \geq 0$x≥0 for $f$f itself.

2. Inverse functions $f^{-1}$f−1: an inverse exists only when $f$f is one-to-one. If $f$f is many-to-one, you must restrict the domain first. Crucially, the domain of $f^{-1}$f−1 is the range of $f$f, and the range of $f^{-1}$f−1 is the domain of $f$f — they swap. Forgetting this swap is the single most common A-Level error.

3. Transformations of graphs: translations $y = f(x) + a$y=f(x)+a and $y = f(x + a)$y=f(x+a) act on the range and domain respectively. A vertical stretch $y = kf(x)$y=kf(x) scales the range; a horizontal stretch $y = f(x/k)$y=f(x/k) scales the domain. The exam-favourite question "describe the transformation that maps $y = f(x)$y=f(x) to $y = 2f(x - 3) + 1$y=2f(x−3)+1" is fundamentally a domain/range manipulation.

4. Calculus on functions: differentiability requires the function to be defined on an open interval around the point. $f(x) = \sqrt{x}$f(x)=x​ is differentiable on $(0, \infty)$(0,∞) but not at $x = 0$x=0 — the derivative $\dfrac{1}{2\sqrt{x}}$2x​1​ has the smaller domain $(0, \infty)$(0,∞). Range questions resurface in optimisation: the maximum value of $f$f on a closed interval is the largest element of its range.

5. Modelling and validity intervals: real-world models impose physical domains. A projectile-height function $h(t) = -4.9t^2 + 20t + 1.5$h(t)=−4.9t2+20t+1.5 is mathematically defined for all $t \in \mathbb{R}$t∈R, but the modelling domain is $0 \leq t \leq T$0≤t≤T where $T$T is the time of impact. AQA mark schemes deduct marks for using values outside the modelling domain.

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Mark-scheme literacy

Domain/range questions on AQA 7357 split AO marks unusually across the three objectives: