Functions

This lesson covers the formal definition of functions, domain and range, composite functions, and inverse functions — the language and machinery of A-Level mathematics.

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What Is a Function?

A function is a mapping that takes each element from a set (the domain) to exactly one element in another set (the codomain). The set of actual output values is the range.

Notation

f: x ↦ 2x + 1 means “f maps x to 2x + 1”.

Equivalently: f(x) = 2x + 1.

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Domain and Range

Term	Meaning
Domain	The set of allowed input values
Range	The set of all output values

Worked Example 1

f(x) = √(x − 3). Find the domain and range.

Domain: x − 3 ≥ 0 → x ≥ 3, so domain is [3, ∞).

Range: √(x − 3) ≥ 0, so range is [0, ∞).

Worked Example 2

g(x) = 1/(x − 2), x ≠ 2.

Domain: all real numbers except x = 2. Range: all real numbers except y = 0 (since 1/(x − 2) can never equal zero).

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Types of Mapping

Type	Description
One-to-one	Each output corresponds to exactly one input
Many-to-one	More than one input can give the same output (e.g. f(x) = x²)

Only one-to-one functions have inverses that are also functions. A many-to-one function can be made one-to-one by restricting its domain.

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Composite Functions

The composite function fg(x) means “apply g first, then f”.

fg(x) = f(g(x))

Worked Example 3

f(x) = 2x + 1 and g(x) = x².

fg(x) = f(g(x)) = f(x²) = 2x² + 1

gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)²

Note: fg(x) ≠ gf(x) in general — order matters.

Worked Example 4

f(x) = eˣ and g(x) = 3x − 1. Find fg(2).

g(2) = 5, so fg(2) = f(5) = e⁵ ≈ 148.41

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Domain of Composite Functions

For fg(x) to exist, the range of g must be contained in (or overlap with) the domain of f.

Worked Example 5

f(x) = ln x (domain: x > 0) and g(x) = x² + 1 (range: y ≥ 1).

Since the range of g is [1, ∞) which is a subset of the domain of f (x > 0), fg(x) = ln(x² + 1) is defined for all real x.

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Inverse Functions

If f is a one-to-one function, its inverse f⁻¹ reverses the mapping.

How to Find an Inverse

1. Write y = f(x).
2. Swap x and y.
3. Rearrange to make y the subject.
4. Write f⁻¹(x) = …

Worked Example 6

Find the inverse of f(x) = (3x − 1)/2.

y = (3x − 1)/2 → 2y = 3x − 1 → 3x = 2y + 1 → x = (2y + 1)/3

Swap: f⁻¹(x) = (2x + 1)/3

Worked Example 7

Find the inverse of f(x) = e^(2x) + 3.

y = e^(2x) + 3 → y − 3 = e^(2x) → 2x = ln(y − 3) → x = ln(y − 3)/2

f⁻¹(x) = ln(x − 3)/2, domain: x > 3

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Graphical Properties

• The graph of f⁻¹ is the reflection of the graph of f in the line y = x.
• The domain of f⁻¹ equals the range of f, and vice versa.
• f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.

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Self-Inverse Functions

A function is self-inverse if f(f(x)) = x, meaning f⁻¹ = f.

Worked Example 8

Show that f(x) = 1/x is self-inverse.

f(f(x)) = f(1/x) = 1/(1/x) = x ✓

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Exam Tips

• Always state the domain and range clearly.
• When finding inverses, swap x and y then solve — do not just rearrange.
• The graph of f⁻¹ is a reflection in y = x — use this for quick sketches.
• For composite functions, apply the inner function first.
• Check whether a function is one-to-one before attempting to find an inverse.

A-Level Deep Dive: Functions

Spec mapping

AQA 7357 Pure section B (Year 2) — Algebra and functions covers the definition of a function; domain and range of functions; composition of functions; inverse functions and their graphs (refer to the official specification document for exact wording). Functions sit at the centre of the Year 2 Pure paper and reach across the specification synoptically. Notation matters: AQA uses both $f(x)$f(x) and the mapping form $f: x \mapsto \ldots$f:x↦… (also written $f: x \to \ldots$f:x→…), and expects candidates to switch between them. Synoptic links to track: graph transformations (the inverse $f^{-1}$f−1 is a reflection of $f$f in $y = x$y=x), modulus functions (where $|f(x)|$∣f(x)∣ is examined alongside $f(|x|)$f(∣x∣)), calculus on composite functions (the chain rule $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$dxd​f(g(x))=f′(g(x))g′(x)), and set notation (domains expressed as $\{x \in \mathbb{R} : x \geq 0\}${x∈R:x≥0}). The AQA formula booklet does not list the chain rule or the inverse-function condition — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

The functions $f$f and $g$g are defined by $f: x \mapsto 2x - 3$f:x↦2x−3 for $x \in \mathbb{R}$x∈R and $g: x \mapsto \sqrt{x + 4}$g:x↦x+4​ for $x \geq -4$x≥−4.

(a) Find an expression for $fg(x)$fg(x) and state its domain. (3)

(b) Find $f^{-1}(x)$f−1(x) and state its domain and range. (3)

(c) Solve the equation $fg(x) = f^{-1}(x)$fg(x)=f−1(x). (2)

Solution with mark scheme:

(a) Apply $g$g first, then $f$f.

$fg(x) = f(g(x)) = f(\sqrt{x + 4}) = 2\sqrt{x + 4} - 3$fg(x)=f(g(x))=f(x+4​)=2x+4​−3

M1 — correct order of composition (inner function $g$g substituted into $f$f). A common slip is to compute $gf(x) = \sqrt{2x - 3 + 4} = \sqrt{2x + 1}$gf(x)=2x−3+4​=2x+1​, the wrong way round.

A1 — correct simplified expression $2\sqrt{x + 4} - 3$2x+4​−3.

B1 — domain $x \geq -4$x≥−4 (the domain of $fg$fg is the set of inputs for which the inner function $g$g is defined and produces an output in the domain of $f$f; here $f$f is defined on all reals so the constraint is just $x + 4 \geq 0$x+4≥0).

(b) Set $y = 2x - 3$y=2x−3 and rearrange for $x$x:

$y + 3 = 2x \implies x = \dfrac{y + 3}{2}$y+3=2x⟹x=2y+3​

Swap variables to express the inverse as a function of $x$x:

$f^{-1}(x) = \dfrac{x + 3}{2}$f−1(x)=2x+3​

M1 — correct rearrangement and variable swap.

A1 — correct inverse expression.

B1 — domain $x \in \mathbb{R}$x∈R, range $f^{-1}(x) \in \mathbb{R}$f−1(x)∈R (since $f$f is defined on all reals and is one-to-one, $f^{-1}$f−1 inherits the full real line).

(c) Equate:

$2\sqrt{x + 4} - 3 = \dfrac{x + 3}{2}$2x+4​−3=2x+3​

Multiply through by 2 and isolate the surd:

$4\sqrt{x + 4} - 6 = x + 3 \implies 4\sqrt{x + 4} = x + 9$4x+4​−6=x+3⟹4x+4​=x+9

Square both sides:

$16(x + 4) = (x + 9)^2 \implies 16x + 64 = x^2 + 18x + 81 \implies x^2 + 2x + 17 = 0$16(x+4)=(x+9)2⟹16x+64=x2+18x+81⟹x2+2x+17=0

The discriminant is $4 - 68 < 0$4−68<0, so no real solutions exist.

M1 — correct setup and squaring step (with the implicit checking that any solution must satisfy $x + 9 \geq 0$x+9≥0 for the original equation).

A1 — correct conclusion: no real solutions, with discriminant justification.

Total: 8 marks (M3 A3 B2).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): The function $h$h is defined by $h: x \mapsto x^2 - 6x + 11$h:x↦x2−6x+11 for $x \geq 3$x≥3.

(a) Show that $h$h is one-to-one on this domain. (2)

(b) Find $h^{-1}(x)$h−1(x) and state its domain. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO2.1) — completing the square: $h(x) = (x - 3)^2 + 2$h(x)=(x−3)2+2, identifying the minimum at $x = 3$x=3.
• A1 (AO2.4) — argument that on $x \geq 3$x≥3 the parabola is monotonically increasing, hence one-to-one.

(b)

• M1 (AO1.1a) — set $y = (x - 3)^2 + 2$y=(x−3)2+2 and rearrange: $(x - 3)^2 = y - 2$(x−3)2=y−2.
• M1 (AO1.1b) — take the positive square root (since $x \geq 3$x≥3 means $x - 3 \geq 0$x−3≥0): $x - 3 = \sqrt{y - 2}$x−3=y−2​.
• A1 (AO1.1b) — $h^{-1}(x) = 3 + \sqrt{x - 2}$h−1(x)=3+x−2​.
• A1 (AO2.5) — domain of $h^{-1}$h−1 is $x \geq 2$x≥2 (the range of the original $h$h).

Total: 6 marks split AO1 = 3, AO2 = 3. AQA balances procedure and reasoning roughly evenly on inverse-function questions; the AO2 marks reward explicit justification of the sign choice when taking the square root.

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Synoptic links

1. Graph transformations: the inverse $f^{-1}$f−1 is the reflection of $f$f in the line $y = x$y=x. This connects to the broader transformation toolkit ($y = f(x) + a$y=f(x)+a, $y = f(x - a)$y=f(x−a), $y = af(x)$y=af(x), $y = f(ax)$y=f(ax)) and gives a fast geometric route to sketching inverses without algebra.

2. Modulus functions: $y = |f(x)|$y=∣f(x)∣ reflects negative $y$y-values above the $x$x-axis, while $y = f(|x|)$y=f(∣x∣) reflects the right-hand portion of $f$f to the left. Composing modulus with restricted-domain functions to enforce one-to-one behaviour is a recurring AQA Year 2 motif.

3. Differentiation via the chain rule: for composite $y = f(g(x))$y=f(g(x)), $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$dxdy​=f′(g(x))⋅g′(x). Implicit in this is that $g$g is differentiable on the domain of the composite — exactly the same domain check needed for $fg$fg to exist.

4. Set notation and inequalities: domains and ranges are written using interval or set-builder notation, e.g. $\{x \in \mathbb{R} : x \geq -4\}${x∈R:x≥−4} or $[-4, \infty)$[−4,∞). Confidence with this notation feeds directly into Statistics (sample spaces) and Mechanics (validity intervals on motion equations).

5. Exponentials and logarithms as inverses: $f(x) = e^x$f(x)=ex and $g(x) = \ln x$g(x)=lnx are mutual inverses, with $f(g(x)) = x$f(g(x))=x for $x > 0$x>0 and $g(f(x)) = x$g(f(x))=x for $x \in \mathbb{R}$x∈R. The asymmetry in domains is itself an instance of inverse-function domain logic.

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Mark-scheme literacy

Functions questions on AQA 7357 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Correctly composing functions in the right order, rearranging for inverses, applying domain and range definitions
AO2 (reasoning / interpretation)	30–40%	Justifying one-to-one behaviour, choosing the correct square-root sign on a restricted domain, presenting domain restrictions in valid set notation
AO3 (problem-solving)	5–15%	Using inverse-function or composition relationships to solve unfamiliar equations, often combined with calculus or modulus content