Mathematics Advanced • Year 11 • Module 1 • Lesson 7
Composite Functions
Practise HSC-style writing on composition — algebraic expressions, domain restrictions, and a structured proof comparing two orderings.
1. Short-answer questions
1.1 Let f(x) = 2x − 1 and g(x) = x² + 3. Find (f ∘ g)(x) in simplest form, showing the substitution step in brackets. 2 marks Band 3
1.2 Let f(x) = √(x + 2) and g(x) = x − 3.
(a) Write (f ∘ g)(x) in simplified surd form.
(b) Solve the inequality required for the domain and write the domain in interval notation. 3 marks Band 3-4
1.3 Let f(x) = 2x + 1 and g(x) = x² − 3.
(a) Find (f ∘ g)(x) and (g ∘ f)(x) in expanded form.
(b) Show, by giving the value of x for which (f ∘ g)(x) − (g ∘ f)(x) = 0 (if any), that the two composites are equal at most at isolated x-values. 4 marks Band 4
2. Extended response
2.1 Let f(x) = ax + b and g(x) = cx + d, where a, b, c, d are real constants and a, c ≠ 0.
(a) Find (f ∘ g)(x) and (g ∘ f)(x), simplifying each to the form mx + n.
(b) Prove that (f ∘ g)(x) = (g ∘ f)(x) for every x if and only if a · d + b = b · c + d.
(c) Use part (b) to find all values of b for which the linear functions f(x) = 3x + b and g(x) = 2x + 1 commute under composition.
(d) Hence give a single concrete example of a pair of distinct, non-inverse linear functions f and g for which f ∘ g = g ∘ f, and verify by direct substitution. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correctly substitutes g(x) into f and simplifies: (f ∘ g)(x) = a(cx + d) + b = acx + ad + b.
• 1 mark — correctly substitutes f(x) into g and simplifies: (g ∘ f)(x) = c(ax + b) + d = acx + bc + d.
Part (b) — 2 marks
• 1 mark — observes that the x-coefficients are equal (both ac) for any a, b, c, d, so equality of composites reduces to equality of constants.
• 1 mark — concludes ad + b = bc + d, with an "if and only if" clearly stated.
Part (c) — 2 marks
• 1 mark — substitutes a = 3, c = 2, d = 1: 3·1 + b = b·2 + 1, i.e. 3 + b = 2b + 1.
• 1 mark — solves to give b = 2, with a closing sentence "so f(x) = 3x + 2".
Part (d) — 1 mark
• 1 mark — gives any explicit pair satisfying ad + b = bc + d with f ≠ g and f, g not inverses (e.g. f(x) = 3x + 2 and g(x) = 2x + 1 from part (c)), then verifies by direct computation that both composites equal the same linear expression in x.
Your response:
Stuck on (d)? Use the b you found in (c) — that is an explicit non-inverse pair that commutes.How did this worksheet feel?
What I'll revisit before next class:
1.1 — (f ∘ g)(x) with f(x) = 2x − 1, g(x) = x² + 3 (2 marks)
Sample response. (f ∘ g)(x) = f(g(x)) = f(x² + 3) = 2(x² + 3) − 1 = 2x² + 6 − 1 = 2x² + 5.
Marking notes. 1 mark — correct substitution shown with brackets: 2(x² + 3) − 1. 1 mark — correct simplification to 2x² + 5. A response that writes only "2x² + 5" without the substitution line scores 1/2.
1.2 — Composite with domain restriction (3 marks)
Sample response. (a) (f ∘ g)(x) = f(x − 3) = √((x − 3) + 2) = √(x − 1).
(b) Need x − 1 ≥ 0 ⇒ x ≥ 1. Domain: [1, ∞).
Marking notes. (a) 1 mark — correct composite √(x − 1), with the intermediate step √((x − 3) + 2) shown (brackets around the inner expression). (b) 1 mark — correct inequality x − 1 ≥ 0; 1 mark — correct interval notation [1, ∞). Common error: writing (1, ∞) instead of [1, ∞), losing the endpoint mark.
1.3 — Comparing (f ∘ g) and (g ∘ f) (4 marks)
Sample response. (a) (f ∘ g)(x) = f(x² − 3) = 2(x² − 3) + 1 = 2x² − 5. (g ∘ f)(x) = g(2x + 1) = (2x + 1)² − 3 = 4x² + 4x − 2.
(b) Set (f ∘ g)(x) − (g ∘ f)(x) = 0: (2x² − 5) − (4x² + 4x − 2) = −2x² − 4x − 3 = 0, i.e. 2x² + 4x + 3 = 0. Discriminant = 16 − 24 = −8 < 0, so there are no real x for which the two composites are equal. Hence (f ∘ g)(x) ≠ (g ∘ f)(x) for every real x.
Marking notes. (a) 1 mark for each correct expanded composite (2 marks total). (b) 1 mark — sets up the equation (f ∘ g)(x) − (g ∘ f)(x) = 0 and reaches a correct quadratic in x. 1 mark — uses the discriminant (or completing the square) to conclude there are no real solutions, and states the conclusion cleanly. Accept the alternative form: "The leading coefficients differ (2 vs 4), so the two polynomials cannot be identical, and a real intersection would require the discriminant of their difference to be ≥ 0 — which it isn't."
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a).
(f ∘ g)(x) = f(g(x)) = f(cx + d) = a(cx + d) + b = acx + (ad + b). [1 mark — substitution and simplification for f ∘ g.]
(g ∘ f)(x) = g(f(x)) = g(ax + b) = c(ax + b) + d = acx + (bc + d). [1 mark — substitution and simplification for g ∘ f.]
Part (b). Comparing the two linear expressions: both have x-coefficient ac, so they are equal for every x if and only if their constant terms are equal: [1 mark — observes coefficients of x already match.]
acx + (ad + b) = acx + (bc + d) ⇔ ad + b = bc + d. [1 mark — iff statement with the correct equation.]
Part (c). Substituting a = 3, c = 2, d = 1 into ad + b = bc + d:
3·1 + b = b·2 + 1 ⇒ 3 + b = 2b + 1 ⇒ 2 = b. [1 mark — correct equation, 1 mark — solves to b = 2.]
So f(x) = 3x + 2 commutes with g(x) = 2x + 1 under composition.
Part (d). Take f(x) = 3x + 2 and g(x) = 2x + 1 (from part (c)). These are distinct (different gradients), and not inverses (the inverse of g is (x − 1)/2 ≠ 3x + 2). Verification: [1 mark — explicit pair + verification.]
(f ∘ g)(x) = 3(2x + 1) + 2 = 6x + 3 + 2 = 6x + 5.
(g ∘ f)(x) = 2(3x + 2) + 1 = 6x + 4 + 1 = 6x + 5. ✓ Both equal 6x + 5, so f ∘ g = g ∘ f. ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Computes (f ∘ g) and (g ∘ f) correctly for part (a) but does not abstract to general a, b, c, d, or stops at the iff equation without solving anything in (c). ≈ 2-3 marks.
Band 4: Completes parts (a) and (b) and finds b = 2 in part (c), but does not give an explicit pair in (d) or omits the verification step. ≈ 4-5 marks.
Band 5: Completes all four parts but the iff in (b) is reasoned in one direction only, or the verification in (d) shows just one composite. ≈ 5-6 marks.
Band 6: All parts fully justified, explicit iff statement, explicit verification of both composites in (d), and a closing remark identifying the pair as commuting but not inverses. 7/7.