Mathematics Advanced • Year 11 • Module 1 • Lesson 3
Domain & Range
Practise HSC-style writing on domain restrictions, interval notation, and combined-restriction problems.
1. Short-answer questions
1.1 Find the natural domain of f(x) = √(2x + 6), giving your answer in interval notation. Justify your choice of bracket at the endpoint. 2 marks Band 3
1.2 Find the domain and range of f(x) = x² − 6x + 11. Use the vertex form (or complete the square) to justify the range. 3 marks Band 3-4
1.3 Find the domain of g(x) = √(x² − x − 6) / (x − 4). Show how you combined the two restrictions. 4 marks Band 4-5
Stuck on 1.3? Factor x² − x − 6 = (x − 3)(x + 2), then determine where this is ≥ 0 and combine with x ≠ 4.2. Extended response
2.1 Consider the function f(x) = 1 / √(x² − 9).
(a) Determine the natural domain of f, showing your inequality and the test of signs.
(b) Find the range of f.
(c) Sketch a clean graph of f for x > 3 only, labelling the asymptote(s) and the behaviour as x → 3⁺ and as x → ∞.
(d) Hence describe in one sentence how the domain and range of f compare with those of the related function g(x) = √(x² − 9). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — combines the two restrictions: radicand > 0 (strictly, because the √ is in a denominator), giving x² − 9 > 0.
• 1 mark — factors (x − 3)(x + 3) > 0 and uses signs / number line / quadratic graph.
• 1 mark — writes domain in interval notation: (−∞, −3) ∪ (3, ∞).
Part (b) — 2 marks
• 1 mark — argues f(x) > 0 (positive denominator).
• 1 mark — describes behaviour at endpoints: as x → 3⁺ or x → −3⁻, denominator → 0⁺ so f → ∞; as |x| → ∞, f → 0. Range = (0, ∞).
Part (c) — 2 marks
• 1 mark — sketch shows curve diving to ∞ near x = 3 and dropping toward y = 0 as x → ∞ (decreasing branch).
• 1 mark — labels vertical asymptote x = 3 and horizontal asymptote y = 0.
Part (d) — 1 mark
• 1 mark — explicit contrast: g has same domain but range [0, ∞) (including 0 at x = ±3, growing to ∞); f has same domain but range (0, ∞) (positive, never 0, growing to ∞ near boundaries).
Your response:
Stuck on (b)? Ask: as x slides from just above 3 toward ∞, does f get bigger or smaller? What about as x → 3⁺?How did this worksheet feel?
What I'll revisit before next class:
1.1 — Domain of f(x) = √(2x + 6) (2 marks)
Sample response. The radicand must be ≥ 0: 2x + 6 ≥ 0 ⇒ x ≥ −3. At x = −3 the radicand is exactly 0, so √0 = 0 is defined and the endpoint is included. Hence the bracket at −3 is square. Domain = [−3, ∞).
Marking notes. 1 mark — correct inequality and solving. 1 mark — explicit justification for square bracket at −3 (or round bracket at ∞). Common error: writing (−3, ∞) and forgetting that √0 is defined: loses 1 mark.
1.2 — Domain and range of f(x) = x² − 6x + 11 (3 marks)
Sample response. Completing the square: f(x) = (x − 3)² + 2. Domain: polynomial, all reals ⇒ (−∞, ∞). Because (x − 3)² ≥ 0 with minimum 0 at x = 3, f(x) ≥ 2 with minimum 2 attained. Range = [2, ∞).
Marking notes. 1 mark — correct completed-square form (or correct vertex via x = −b/2a = 3, f(3) = 2). 1 mark — correctly states domain = (−∞, ∞). 1 mark — correct range with [ at 2 (because the vertex is attained).
1.3 — Domain of g(x) = √(x² − x − 6) / (x − 4) (4 marks)
Sample response. Two restrictions:
(a) Radicand ≥ 0: x² − x − 6 = (x − 3)(x + 2) ≥ 0. Sign chart: positive outside the roots ⇒ x ≤ −2 or x ≥ 3.
(b) Denominator ≠ 0: x − 4 ≠ 0 ⇒ x ≠ 4.
Combine: (a) gives (−∞, −2] ∪ [3, ∞). Removing x = 4 from this gives
Domain = (−∞, −2] ∪ [3, 4) ∪ (4, ∞).
Marking notes. 1 mark — factorising the radicand. 1 mark — sign-test giving x ≤ −2 or x ≥ 3. 1 mark — identifying x ≠ 4. 1 mark — correctly stitched final interval notation with [ at −2 and 3 (radicand can be 0) and a hole at 4. Common error: forgetting to exclude x = 4 (loses 1 mark); writing the radicand restriction as strict > 0 (loses 1 mark; only the denominator's nested factor needs strict).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). f(x) = 1 / √(x² − 9). The expression √(x² − 9) appears in the denominator, so we need both
x² − 9 ≥ 0 (for the square root to be real), and
√(x² − 9) ≠ 0 (no division by zero).
Together these force the strict inequality x² − 9 > 0. [1 mark — combined restrictions.]
Factor: (x − 3)(x + 3) > 0. Sign analysis: the product is positive when both factors share the same sign, i.e. x < −3 or x > 3. [1 mark — sign analysis / factor inequality.]
Domain = (−∞, −3) ∪ (3, ∞). [1 mark — interval notation, round brackets at ±3.]
Part (b). Since √(x² − 9) > 0 on the domain, f(x) = 1 / (positive) is strictly positive. [1 mark — argues positivity.]
As x → 3⁺ (or x → −3⁻), the radicand → 0⁺ and √(...) → 0⁺, so f → +∞. As |x| → ∞, x² − 9 → ∞ and 1/√(...) → 0⁺. By continuity on each branch, every positive value is attained. Range = (0, ∞). [1 mark — endpoint behaviour and range.]
Part (c). Sketch on x > 3:
y | | vertical asymptote at x = 3 | | | |\ | | \____ | | \________ |__|________________________ y → 0 3 x
Vertical asymptote x = 3 (curve shoots up). Horizontal asymptote y = 0 (curve drops toward zero). [1 mark — shape; 1 mark — both asymptotes labelled.]
Part (d). g(x) = √(x² − 9) shares the wider closed-root domain (−∞, −3] ∪ [3, ∞) (since √0 is fine), and its range is [0, ∞) (touches 0 at x = ±3). f has the strictly open domain (−∞, −3) ∪ (3, ∞) and range (0, ∞) — it never reaches 0, and blows up where g touches 0. [1 mark — explicit contrast of both domain and range, naming the open/closed difference at ±3.]
Total: 8/8.
Band descriptors for marker.
Band 3: Solves x² − 9 ≥ 0 (loses the strict point) and writes [−3, −3] ∪ [3, ∞) wrong endpoints. Range vague (e.g. "positive numbers"). ≈ 3-4 marks.
Band 4: Correct domain with strict brackets, range stated as (0, ∞) but without endpoint analysis. Sketch missing asymptote labels. ≈ 5-6 marks.
Band 5: Correct (a) and (b) with limit reasoning, sketch present and labelled. (d) attempted but not explicit. ≈ 7 marks.
Band 6: All four parts complete, with strict-vs-closed inequality reasoning, asymptote labels, and an explicit one-sentence contrast in (d). 8/8.