Composite & Inverse Functions

This lesson covers composite functions and inverse functions as required by the AQA A-Level Mathematics specification (7357). Composing functions means applying one function after another, while an inverse function reverses the effect of the original function. These concepts are fundamental to A-Level mathematics and appear frequently in exam papers.

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Composite Functions

A composite function is formed by applying one function and then another. The notation fg(x) or f ∘ g(x) means "apply g first, then apply f to the result".

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

Important: fg(x) means apply g first, then f. The order matters — in general, fg(x) ≠ gf(x).

Evaluating Composite Functions

Example: Given f(x) = 2x + 3 and g(x) = x², find:

(a) fg(2):

g(2) = 4
fg(2) = f(4) = 2(4) + 3 = 11

(b) gf(2):

f(2) = 7
gf(2) = g(7) = 49

(c) fg(x):

fg(x) = f(g(x)) = f(x²) = 2x² + 3

(d) gf(x):

gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)²

Notice that fg(x) = 2x² + 3 and gf(x) = (2x + 3)² are different — composition is not commutative.

Domain of a Composite Function

The domain of fg is the set of all x in the domain of g such that g(x) is in the domain of f.

Example: f(x) = √x (domain x ≥ 0) and g(x) = 2x − 6 (domain x ∈ ℝ).

fg(x) = √(2x − 6). For this to be defined, we need 2x − 6 ≥ 0, so x ≥ 3.

The domain of fg is x ≥ 3 or [3, ∞).

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Composing a Function with Itself

Example: Given f(x) = 3x − 1, find f²(x) = ff(x).

ff(x) = f(f(x)) = f(3x − 1) = 3(3x − 1) − 1 = 9x − 4

Example: Given f(x) = (x + 1)/(x − 1) for x ≠ 1, find ff(x).

ff(x) = f((x + 1)/(x − 1))
      = ((x + 1)/(x − 1) + 1) / ((x + 1)/(x − 1) − 1)
      = ((x + 1 + x − 1)/(x − 1)) / ((x + 1 − x + 1)/(x − 1))
      = (2x/(x − 1)) / (2/(x − 1))
      = 2x/2 = x

So ff(x) = x — the function is its own inverse! Such a function is called a self-inverse function (or involution).

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Inverse Functions

The inverse function f⁻¹ reverses the effect of f. If f maps a to b, then f⁻¹ maps b back to a.

f⁻¹(f(x)) = x   and   f(f⁻¹(x)) = x

Conditions for an Inverse to Exist

A function has an inverse if and only if it is one-to-one (injective). Many-to-one functions do not have inverses unless the domain is restricted.

Finding the Inverse Algebraically

1. Write y = f(x).
2. Swap x and y (i.e., replace y with x and x with y).
3. Rearrange to make y the subject — this gives f⁻¹(x).

Example: Find the inverse of f(x) = 3x − 7.

y = 3x − 7
x = 3y − 7    (swap x and y)
3y = x + 7
y = (x + 7)/3

So f⁻¹(x) = (x + 7)/3.

Check: f(f⁻¹(x)) = 3 × (x + 7)/3 − 7 = x + 7 − 7 = x. ✓

Example: Find the inverse of f(x) = (2x + 1)/(x − 3), x ≠ 3.

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

So f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2.

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Domain and Range of the Inverse

The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.

Example: f(x) = x² + 1 with domain x ≥ 0.

• Domain of f: x ≥ 0
• Range of f: f(x) ≥ 1

Therefore:

• Domain of f⁻¹: x ≥ 1
• Range of f⁻¹: f⁻¹(x) ≥ 0

To find f⁻¹: y = x² + 1, swap: x = y² + 1, so y² = x − 1, y = √(x − 1) (taking the positive root since the range of f⁻¹ must be ≥ 0).

f⁻¹(x) = √(x − 1), x ≥ 1.

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Graphical Relationship

The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

This is because if (a, b) lies on y = f(x), then (b, a) lies on y = f⁻¹(x) — and reflecting (a, b) in the line y = x gives (b, a).

Key properties:

• The graphs of f and f⁻¹ are symmetrical about y = x.
• The graphs intersect on the line y = x (at points where f(x) = x).
• A self-inverse function has a graph that is symmetrical about y = x.

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Self-Inverse Functions

A function f is self-inverse if f⁻¹ = f, meaning ff(x) = x for all x in the domain.

Examples of self-inverse functions:

• f(x) = −x
• f(x) = 1/x (x ≠ 0)
• f(x) = (x + 1)/(x − 1) (x ≠ 1)
• f(x) = a − x (for any constant a)

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Summary

• Composite functions: fg(x) = f(g(x)) — apply g first, then f.
• Composition is not commutative: fg(x) ≠ gf(x) in general.
• The domain of fg requires g(x) to be in the domain of f.
• An inverse function f⁻¹ reverses f: f⁻¹(f(x)) = x.
• A function has an inverse only if it is one-to-one.
• To find f⁻¹: write y = f(x), swap x and y, rearrange for y.
• Domain of f⁻¹ = range of f and range of f⁻¹ = domain of f.
• The graph of f⁻¹ is the reflection of f in the line y = x.
• A self-inverse function satisfies ff(x) = x.

Exam Tip: In the exam, always state the domain of the inverse function. When finding an inverse, check your answer by verifying that f(f⁻¹(x)) = x. Remember that fg means "apply g first" — a very common source of errors. When sketching f⁻¹, reflect the graph of f in y = x and mark corresponding points.

A-Level Deep Dive: Composite and Inverse Functions

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section B (Algebra and Functions) covers composite and inverse functions, including domain and range (refer to the official specification document for exact wording). Composite-and-inverse-function content sits within the Year 2 portion of AQA Pure and is examined across Paper 1 and Paper 2. The topic relies on confident handling of $f: A \to B$f:A→B notation, domain restrictions, and graph reflection in $y = x$y=x. Composite construction $(f \circ g)(x) = f(g(x))$(f∘g)(x)=f(g(x)) and the inverse-existence condition (one-to-one) are stated directly in the specification, while their interaction with logarithms (as inverse exponentials) and the chain rule (for differentiating composites) is examined synoptically. The AQA formula booklet does not list composite or inverse rules — they must be memorised, including the order convention.

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Worked example with full mark scheme

Question (8 marks):

Let $f(x) = 2x - 3$f(x)=2x−3 for $x \in \mathbb{R}$x∈R and $g(x) = x^2 + 1$g(x)=x2+1 for $x \geq 0$x≥0.

(a) Find $(fg)(x)$(fg)(x) and state its range. (3)

(b) Show that $f$f has an inverse, and find $f^{-1}(x)$f−1(x). (2)

(c) The function $h$h is defined by $h(x) = x^2 - 4x + 7$h(x)=x2−4x+7 for $x \geq 2$x≥2. Show that $h$h is one-to-one on this domain, and find $h^{-1}(x)$h−1(x) stating its domain. (3)

Solution with mark scheme:

(a) Step 1 — apply the order convention.

$(fg)(x) = f(g(x)) = f(x^2 + 1) = 2(x^2 + 1) - 3 = 2x^2 - 1$(fg)(x)=f(g(x))=f(x2+1)=2(x2+1)−3=2x2−1.

M1 — correct order: substitute $g$g into $f$f, not the other way around. Many candidates compute $g(f(x)) = (2x - 3)^2 + 1$g(f(x))=(2x−3)2+1 by mistake. The convention $(fg)(x) = f(g(x))$(fg)(x)=f(g(x)) means "do $g$g first, then $f$f" — read right to left.

A1 — correct simplified expression $2x^2 - 1$2x2−1.

Step 2 — range. Since $x \geq 0$x≥0 is the domain of $g$g, $x^2 \geq 0$x2≥0, so $2x^2 \geq 0$2x2≥0, hence $2x^2 - 1 \geq -1$2x2−1≥−1. The range is $(fg)(x) \geq -1$(fg)(x)≥−1, i.e. $[-1, \infty)$[−1,∞).

B1 — correct range stated using the domain restriction. Candidates who forget that $g$g's domain is $x \geq 0$x≥0 often write the range as all reals $\geq -1$≥−1, which is correct here only because $x^2 \geq 0$x2≥0 for all real $x$x anyway — but the reasoning must reference the restricted domain.

(b) Step 1 — establish one-to-one.

$f(x) = 2x - 3$f(x)=2x−3 is linear with non-zero gradient $2$2, so it is strictly increasing on $\mathbb{R}$R, hence one-to-one and invertible.

B1 — explicit justification of one-to-one. Stating "linear so invertible" is too brief; the gradient must be referenced, since a constant function (gradient $0$0) is linear but not one-to-one.

Step 2 — invert.

Let $y = 2x - 3$y=2x−3. Swap and solve: $x = 2y - 3 \implies y = \dfrac{x + 3}{2}$x=2y−3⟹y=2x+3​. So $f^{-1}(x) = \dfrac{x + 3}{2}$f−1(x)=2x+3​.

B1 — correct inverse expression with $f^{-1}(x)$f−1(x) notation (not $y$y or $x$x).

(c) Step 1 — complete the square.

$h(x) = x^2 - 4x + 7 = (x - 2)^2 + 3$h(x)=x2−4x+7=(x−2)2+3.

M1 — completed-square form. This is the key insight: the vertex is at $x = 2$x=2, and the domain restriction $x \geq 2$x≥2 gives the right half of the parabola, on which $h$h is strictly increasing — hence one-to-one.

Step 2 — invert.

Let $y = (x - 2)^2 + 3$y=(x−2)2+3. Then $(x - 2)^2 = y - 3$(x−2)2=y−3, so $x - 2 = \pm\sqrt{y - 3}$x−2=±y−3​. Since $x \geq 2$x≥2, take the positive root: $x = 2 + \sqrt{y - 3}$x=2+y−3​.

So $h^{-1}(x) = 2 + \sqrt{x - 3}$h−1(x)=2+x−3​.

A1 — correct inverse with positive root selected, justified by the domain restriction.

Step 3 — domain of $h^{-1}$h−1. The domain of $h^{-1}$h−1 is the range of $h$h. Since $h(2) = 3$h(2)=3 and $h$h is increasing on $[2, \infty)$[2,∞), the range of $h$h is $[3, \infty)$[3,∞). Hence the domain of $h^{-1}$h−1 is $x \geq 3$x≥3.

B1 — domain of inverse stated as $x \geq 3$x≥3, with the connection "domain of $h^{-1}$h−1 = range of $h$h" implicit in the working.

Total: 8 marks.

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): The functions $p$p and $q$q are defined by $p(x) = \ln(x - 1)$p(x)=ln(x−1) for $x > 1$x>1 and $q(x) = e^{2x} + 1$q(x)=e2x+1 for $x \in \mathbb{R}$x∈R.

(a) Find $(pq)(x)$(pq)(x), simplifying your answer. (3)

(b) State the range of $(pq)(x)$(pq)(x). (1)

(c) Find $p^{-1}(x)$p−1(x) and state its domain. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — substituting $q$q into $p$p: $(pq)(x) = p(q(x)) = \ln(e^{2x} + 1 - 1) = \ln(e^{2x})$(pq)(x)=p(q(x))=ln(e2x+1−1)=ln(e2x).
• M1 (AO1.1b) — applying $\ln(e^k) = k$ln(ek)=k: $\ln(e^{2x}) = 2x$ln(e2x)=2x.
• A1 (AO1.1b) — final answer $(pq)(x) = 2x$(pq)(x)=2x.

(b)

• B1 (AO2.5) — range is $\mathbb{R}$R, since $2x$2x takes all real values for $x \in \mathbb{R}$x∈R.

(c)

• M1 (AO1.1b) — let $y = \ln(x - 1)$y=ln(x−1), so $e^y = x - 1$ey=x−1, giving $x = e^y + 1$x=ey+1, hence $p^{-1}(x) = e^x + 1$p−1(x)=ex+1.
• A1 (AO2.5) — domain of $p^{-1}$p−1 is $x \in \mathbb{R}$x∈R (the range of $p$p, since $\ln$ln produces all real outputs).

Total: 6 marks split AO1 = 4, AO2 = 2. This question rewards two synoptic insights: that logarithm and exponential are mutual inverses, and that the domain of an inverse equals the range of the original.

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Synoptic links

Connects to:

1. Functions (domain and range): every composite $(fg)(x)$(fg)(x) requires the range of $g$g to lie within the domain of $f$f, otherwise $(fg)$(fg) is undefined for some $x$x. AQA examiners specifically test this by giving a $g$g whose range partially exits $f$f's domain — candidates who blindly substitute lose marks.

2. Graph transformations: the inverse $f^{-1}$f−1 is the reflection of $f$f in the line $y = x$y=x. This geometric fact justifies why $(f \circ f^{-1})(x) = x$(f∘f−1)(x)=x — composing a function with its reflection gives the identity. Students who already understand reflection transformations can deploy this as a sanity check.

3. Chain rule (Year 2 differentiation): $\dfrac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$dxd​f(g(x))=f′(g(x))⋅g′(x) is composite-function differentiation. Without confident composite-function notation, the chain rule is impossible to apply correctly; the order $f'(g(x))$f′(g(x)) mirrors the composite order $f(g(x))$f(g(x)).

4. Logarithms as inverse exponentials: $\ln x$lnx is defined as the inverse of $e^x$ex. The identities $\ln(e^x) = x$ln(ex)=x and $e^{\ln x} = x$elnx=x are exactly the composite-with-inverse identity $(f \circ f^{-1})(x) = x$(f∘f−1)(x)=x specialised to $f = e^{(\cdot)}$f=e(⋅). Recognising this collapses a whole class of "simplify $\ln(e^{2x} - 1)$ln(e2x−1)"-style questions to algebra.

5. Trigonometric inverses: $\arcsin$arcsin, $\arccos$arccos, $\arctan$arctan exist only because $\sin$sin, $\cos$cos, $\tan$tan are restricted to one-to-one intervals ($[-\pi/2, \pi/2]$[−π/2,π/2] for $\sin$sin, $[0, \pi]$[0,π] for $\cos$cos, $(-\pi/2, \pi/2)$(−π/2,π/2) for $\tan$tan). The same domain-restriction trick used for $h(x) = (x - 2)^2 + 3$h(x)=(x−2)2+3 above appears here at scale.

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Mark-scheme literacy

Composite/inverse questions on AQA 7357 split AO marks roughly as follows: