Mathematics Advanced • Year 11 • Module 1 • Lesson 8
Working with Functions — Synthesis
Practise HSC-style writing on integrated function problems — composite, inverse, domain, and a structured evaluation response.
1. Short-answer questions
1.1 Let f(x) = 2x + 5 and g(x) = x − 3.
(a) Find (f ∘ g)(x).
(b) Find (f ∘ g)⁻¹(x).
(c) Verify that (f ∘ g)⁻¹(11) = 6 by showing your substitution. 4 marks Band 3-4
1.2 Consider f(x) = √(x − 1) and g(x) = 2/x.
(a) Find (f ∘ g)(x) in simplified surd form.
(b) Find the domain of (f ∘ g)(x), showing both the inner-function condition and the outer-function inequality, with the final domain in interval notation. 4 marks Band 4-5
1.3 The function p(x) = x² + 4 is given on the unrestricted real line.
(a) Explain in one sentence why p has no inverse on the unrestricted real line.
(b) State a suitable restriction on the domain and find the corresponding inverse function, including its domain in interval notation. 3 marks Band 4
2. Extended response
2.1 A student is given f(x) = x² + 2x. They claim: "This function does not have an inverse because it is a quadratic."
(a) Evaluate the student's claim. State whether they are entirely correct, partially correct, or incorrect, and justify in one sentence.
(b) By completing the square, write f in vertex form and state the coordinates of the vertex.
(c) State a domain restriction on f under which an inverse does exist, and find f⁻¹(x) explicitly using swap-and-solve.
(d) State the domain and range of your f⁻¹, and verify your inverse by showing f(f⁻¹(x)) = x. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — identifies the claim as partially correct (without restriction f is not one-to-one) and notes that a restricted-domain inverse does exist.
Part (b) — 1 mark
• 1 mark — f(x) = (x + 1)² − 1, vertex (−1, −1).
Part (c) — 3 marks
• 1 mark — states a valid restriction, e.g. x ≥ −1 (or x ≤ −1) with reason "so f is one-to-one".
• 1 mark — applies swap-and-solve correctly: y = (x + 1)² − 1 → x = (y + 1)² − 1 → y + 1 = ±√(x + 1).
• 1 mark — chooses the correct branch consistent with the restriction (positive root if x ≥ −1) and writes f⁻¹(x) = √(x + 1) − 1.
Part (d) — 2 marks
• 1 mark — Domain of f⁻¹ = range of restricted f = [−1, ∞); Range of f⁻¹ = restricted domain of f = [−1, ∞).
• 1 mark — Verification f(f⁻¹(x)) = (√(x+1) − 1 + 1)² − 1 = (√(x+1))² − 1 = (x + 1) − 1 = x. ✓
Your response:
Stuck on (d)? The domain of f⁻¹ is always the range of (restricted) f — find the y-values f can take when x ≥ −1.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Composite then inverse (4 marks)
Sample response. (a) (f ∘ g)(x) = f(x − 3) = 2(x − 3) + 5 = 2x − 1.
(b) y = 2x − 1 ⇒ swap: x = 2y − 1 ⇒ y = (x + 1)/2. So (f ∘ g)⁻¹(x) = (x + 1)/2.
(c) (f ∘ g)⁻¹(11) = (11 + 1)/2 = 12/2 = 6. ✓
Marking notes. (a) 1 mark — correct composite shown with intermediate 2(x − 3) + 5. (b) 1 mark — correct swap step; 1 mark — final (x + 1)/2. (c) 1 mark — substitution and conclusion. Common error: solving y = 2x − 1 without swapping (gives the same expression by coincidence here, but the method is wrong).
1.2 — Composite with full domain analysis (4 marks)
Sample response. (a) (f ∘ g)(x) = f(2/x) = √((2/x) − 1) = √((2 − x)/x).
(b) Inner condition: x ≠ 0 (so 2/x is defined). Outer condition: (2/x) − 1 ≥ 0 ⇒ (2 − x)/x ≥ 0. Critical values at x = 0 and x = 2. Sign chart: on (−∞, 0) the ratio is (+)/(−) = −, fails; on (0, 2] the ratio is (≥ 0)/(+) = ≥ 0, satisfies; on (2, ∞) the ratio is (−)/(+) = −, fails. Combined with x ≠ 0: Domain = (0, 2].
Marking notes. (a) 1 mark — correct composite in simplified surd form. (b) 1 mark — both conditions stated separately. 1 mark — sign analysis of (2 − x)/x carried out and correct interval (0, 2] identified. 1 mark — final interval notation includes the right bracket at 2 (closed because numerator = 0 is allowed) and the open bracket at 0 (excluded by inner condition). Common error: writing [0, 2] or (0, 2) — both lose the final mark.
1.3 — Restricted quadratic inverse (3 marks)
Sample response. (a) p is not one-to-one on the unrestricted real line because p(a) = p(−a) for every a (e.g. p(1) = p(−1) = 5), so no function can undo p uniquely.
(b) Restrict to x ≥ 0 (then p is increasing and one-to-one). y = x² + 4 ⇒ x = y² + 4 ⇒ y² = x − 4 ⇒ y = √(x − 4) (positive root, matching the restriction). p⁻¹(x) = √(x − 4), with domain [4, ∞).
Marking notes. (a) 1 mark — explicit reason (not one-to-one, with one supporting example). (b) 1 mark — chooses a valid restriction (x ≥ 0 or x ≤ 0 both acceptable); 1 mark — correct inverse and correct domain in interval notation. Accept x ≤ 0 with inverse p⁻¹(x) = −√(x − 4) and same domain.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). The student's claim is partially correct. It is true that f(x) = x² + 2x has no inverse function on the unrestricted real line, because f is not one-to-one (it takes the same value at pairs of x-values, e.g. f(0) = f(−2) = 0). However, restricting the domain so that f becomes one-to-one produces a valid inverse, so the claim is too sweeping. [1 mark — partial correctness identified.]
Part (b). Completing the square: f(x) = x² + 2x = (x² + 2x + 1) − 1 = (x + 1)² − 1. Vertex at (−1, −1). [1 mark.]
Part (c). Restrict to x ≥ −1, the right-hand branch of the parabola, so f is increasing and one-to-one on this domain. [1 mark — restriction stated with justification.]
Apply swap-and-solve to y = (x + 1)² − 1:
x = (y + 1)² − 1 ⇒ (y + 1)² = x + 1 ⇒ y + 1 = ±√(x + 1). [1 mark — swap and solve to ±√.]
Since the restricted domain has y + 1 ≥ 0 (because x ≥ −1 ⇒ x + 1 ≥ 0 here applies to the original input, and the inverse outputs match — choose the positive branch), f⁻¹(x) = √(x + 1) − 1. [1 mark — correct branch and final expression.]
Part (d). Domain of f⁻¹ = range of restricted f. On x ≥ −1, f starts at the minimum f(−1) = −1 and increases without bound, so range of f is [−1, ∞). Hence domain of f⁻¹ = [−1, ∞); range of f⁻¹ = restricted domain of f = [−1, ∞). [1 mark.]
Verification. f(f⁻¹(x)) = (√(x + 1) − 1 + 1)² − 1 = (√(x + 1))² − 1 = (x + 1) − 1 = x. ✓ [1 mark — verification reaches x with all steps shown.] ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Identifies the claim as incomplete in part (a) and completes the square in (b), but gets stuck at the ± in (c) or omits the verification in (d). ≈ 2-3 marks.
Band 4: Selects a valid restriction in (c) and finds f⁻¹, but does not consistently state the domain and range of f⁻¹ in (d) or skips the verification. ≈ 4-5 marks.
Band 5: All four parts attempted with correct algebra; domain/range of f⁻¹ correct; verification given but not fully simplified. ≈ 5-6 marks.
Band 6: Full response with explicit one-to-one reasoning in (a), justified branch choice in (c), domain and range stated for f⁻¹ in (d), and a fully simplified verification ending in "= x ✓". 7/7.