Mathematics Advanced • Year 11 • Module 1 • Lesson 7
Composite Functions
Apply composition to multi-stage processes — pricing, manufacturing, conversion chains and inverse pairs.
Problem 1 — Order matters in retail (price + GST)
A store discounts the marked price x dollars by 20%, giving d(x) = 0.8 x. A separate function adds 10% GST: g(p) = 1.1 p. Two cashiers disagree about the order of operations.
d(x) = 0.8 x g(p) = 1.1 p
Set up: What are we solving for?
(i) Cashier A computes (g ∘ d)(x) = "discount first, then add GST". Find this composite as a simplified linear expression in x. 2 marks
(ii) Cashier B computes (d ∘ g)(x) = "add GST first, then discount". Find this composite as a simplified linear expression in x. 2 marks
(iii) Compare your answers from (i) and (ii). Explain in one sentence whether the customer pays the same amount either way, and what this tells us about commutativity of composition for linear functions through 0. 1 mark
Stuck? Revisit lesson § Order Matters.Problem 2 — Supply-chain composition (chained processes)
Raw material x kilograms is processed by Factory A into components, weight A(x) = 0.9 x − 2 kg (some material is lost as waste). Factory B then assembles, output B(c) = 1.5 c + 4 kg (packaging adds weight).
Set up: What are we solving for?
(i) Find the composite (B ∘ A)(x) giving the final packaged weight as a function of raw input x. 3 marks
(ii) If 50 kg of raw material enters Factory A, find the final packaged weight. 1 mark
(iii) Factory A's output cannot be negative (it would mean zero or sub-zero kilograms entering Factory B). Find the smallest value of x for which (B ∘ A)(x) is physically meaningful, i.e. for which A(x) ≥ 0. 2 marks
Problem 3 — Composite with a domain restriction (engineering signal)
A sensor measures temperature x °C, then a calibration step converts it via g(x) = x − 7, then a square-root processing stage f(t) = √t produces the displayed reading.
f(t) = √t g(x) = x − 7
Set up: What are we solving for?
(i) Find (f ∘ g)(x) and state the inequality that x must satisfy for the display reading to exist (i.e. the domain). Give the domain in interval notation. 3 marks
(ii) Explain in one sentence what physically happens if the sensor reads x = 5 °C — referring to which of the two domain conditions fails. 1 mark
(iii) The engineer wants the display to read exactly 4. Find the temperature x that produces this. 2 marks
Stuck on (iii)? Set √(x − 7) = 4 and square both sides.Problem 4 — When composition is commutative (inverses)
A student wonders whether (f ∘ g)(x) can ever equal (g ∘ f)(x). Investigate using f(x) = 3x − 5 and g(x) = (x + 5)/3.
Set up: What are we solving for?
(i) Find (f ∘ g)(x), simplifying fully. 2 marks
(ii) Find (g ∘ f)(x), simplifying fully. 2 marks
(iii) State the relationship between f and g that makes both composites equal x, and write a one-sentence rule the student can use to recognise such pairs in future. 2 marks
Problem 5 — Triple composite (currency-conversion chain)
A traveller converts AUD → USD using u(x) = 0.65 x, then USD → EUR using e(u) = 0.92 u, then pays a 3% card fee using c(p) = 1.03 p (the fee is added on top of the EUR amount).
Set up: What are we solving for?
(i) Write the final cost in EUR as a triple composite (c ∘ e ∘ u)(x) and simplify the constant. 3 marks
(ii) Use your result from (i) to find the final EUR cost of an AUD 500 purchase, to the nearest cent. 1 mark
(iii) If the bank reorders the steps and charges the card fee first (in AUD, then converts), would the customer be better off, worse off, or pay the same? Justify in one line. 2 marks
Stuck on (iii)? Compare (c ∘ e ∘ u)(x) with (e ∘ u ∘ c)(x) — both are scalar multiples of x, look at the constants.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Tax and discount
Set up. Comparing two orderings of discount and GST on the same marked price.
(i) (g ∘ d)(x) = g(0.8 x) = 1.1 · (0.8 x) = 0.88 x.
(ii) (d ∘ g)(x) = d(1.1 x) = 0.8 · (1.1 x) = 0.88 x.
(iii) Both composites equal 0.88 x, so the customer pays the same either way. Linear scaling functions through 0 (multiplications by a constant) commute under composition, because the product of the scale factors does not depend on order: 0.8 · 1.1 = 1.1 · 0.8 = 0.88.
Problem 2 — Supply chain
Set up. Chaining two factory transformations to express final packaged weight as a function of raw input.
(i) (B ∘ A)(x) = B(A(x)) = B(0.9 x − 2) = 1.5 (0.9 x − 2) + 4 = 1.35 x − 3 + 4 = 1.35 x + 1 kg.
(ii) (B ∘ A)(50) = 1.35 (50) + 1 = 67.5 + 1 = 68.5 kg.
(iii) Need A(x) ≥ 0: 0.9 x − 2 ≥ 0 ⇒ x ≥ 2/0.9 ⇒ x ≥ 20/9 ≈ 2.22 kg. Below this raw weight, Factory A's output is non-positive and the composite is not physically meaningful.
Problem 3 — Domain restriction
Set up. Two-stage sensor pipeline whose composite has a domain set by the square-root stage.
(i) (f ∘ g)(x) = f(x − 7) = √(x − 7). Need x − 7 ≥ 0 ⇒ x ≥ 7. Domain: [7, ∞).
(ii) At x = 5, g(5) = −2, which is not in the domain of f (square root needs non-negative input). The second condition (g(x) must lie in the domain of f) fails, so the composite is undefined and the display shows no reading.
(iii) Solve √(x − 7) = 4 ⇒ x − 7 = 16 ⇒ x = 23 °C.
Problem 4 — Inverse pair
Set up. Testing whether a particular f and g satisfy (f ∘ g)(x) = (g ∘ f)(x) = x.
(i) (f ∘ g)(x) = f((x + 5)/3) = 3 · (x + 5)/3 − 5 = (x + 5) − 5 = x.
(ii) (g ∘ f)(x) = g(3x − 5) = ((3x − 5) + 5)/3 = 3x/3 = x.
(iii) f and g are inverses of each other — applying one and then the other returns the original input. Rule: whenever (f ∘ g)(x) = x and (g ∘ f)(x) = x for every x in the relevant domain, f and g are inverse functions; this is the prototypical case in which composition commutes.
Problem 5 — Triple composite
Set up. Chaining three scale-by-a-constant conversions.
(i) (c ∘ e ∘ u)(x) = c(e(u(x))) = c(e(0.65 x)) = c(0.92 · 0.65 x) = c(0.598 x) = 1.03 · 0.598 x = 0.61594 x EUR.
(ii) (c ∘ e ∘ u)(500) = 0.61594 × 500 = EUR 307.97 (to the nearest cent).
(iii) Same. All three steps are scalar multiplications, and the product 0.65 × 0.92 × 1.03 = 0.61594 does not depend on the order of the factors, so any reorder gives the same total.