Mathematics Advanced • Year 11 • Module 1 • Lesson 3
Domain & Range
Apply domain & range reasoning to real models — heights, costs, areas — and the natural/contextual domain distinction.
Problem 1 — Projectile height
A ball is thrown upward from a 1-metre platform. Its height (in metres) after t seconds is h(t) = −5t² + 10t + 1.
Set up: What are we solving for?
(i) Find the natural domain of h algebraically (no context). Then find the contextual domain for the flight — from launch to hitting the ground. 3 marks
(ii) Find the range of h on the contextual domain. (Find the maximum height and use the launch height.) 3 marks
(iii) Explain in 1-2 sentences why the contextual domain is shorter than the natural domain, and what would go wrong if a student used (−∞, ∞) here. 2 marks
Stuck on (i)? Solve h(t) = 0 with the quadratic formula and take the positive root.Problem 2 — Rectangle area as a function of width
A farmer has 40 m of fencing for a rectangular pen. If the width is w metres, the length is (20 − w) m. The area is A(w) = w(20 − w).
Set up: What are we solving for?
(i) State the natural domain of A and the contextual (geometric) domain. Justify the contextual bounds with one sentence each. 3 marks
(ii) Expand A(w) into ax² + bx + c form. Find the vertex and hence the range of A on the contextual domain. 3 marks
(iii) Without solving, explain why the endpoint area A(0) = 0 must be excluded from the range if the problem asks for a "valid pen". 2 marks
Problem 3 — Sound intensity model
The sound intensity (in W/m²) at distance r metres from a loudspeaker is I(r) = 0.5 / r².
Set up: What are we solving for?
(i) Find the natural domain of I as a function of r ∈ ℝ. Then find the contextual domain (people stand at distances r > 0). 2 marks
(ii) Find the range of I on the contextual domain (0, ∞). What value does I approach as r → ∞ and as r → 0⁺? 3 marks
(iii) Audiologists recommend listeners stay below 0.02 W/m². Solve I(r) ≤ 0.02 to find the minimum safe distance. 3 marks
Stuck on (iii)? 0.5/r² ≤ 0.02 gives r² ≥ 25.Problem 4 — Wheelchair ramp profile
A ramp's vertical rise (in cm) over a horizontal run x cm is modelled by f(x) = √(900 − x²) − 30, valid for 0 ≤ x ≤ 30 cm (the curved top of a quarter-circle arc base).
Set up: What are we solving for?
(i) Find the natural domain of the algebraic expression √(900 − x²) − 30 over ℝ, then state the contextual domain for the ramp. 2 marks
(ii) Evaluate f(0), f(15), f(30). Hence find the range of f on the contextual domain. 3 marks
(iii) A building code requires the ramp rise to be less than 10 cm anywhere. Explain whether this ramp satisfies the code, citing values from (ii). 2 marks
Problem 5 — Investment value
A $1000 investment compounded monthly at 6% p.a. nominal has value V(t) = 1000 (1.005)^(12t), where t is years.
Set up: What are we solving for?
(i) State the natural domain of V over ℝ and the contextual domain. Justify each. 2 marks
(ii) Find V(0) and V(10). Use these (and the function's behaviour) to state the range of V on the contextual domain. 3 marks
(iii) Use V(t) to find the time required for the investment to double. Express the answer in terms of log₁₀ or ln, then evaluate to 2 dp. 3 marks
Stuck on (iii)? Solve (1.005)^(12t) = 2 by taking logs of both sides.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Projectile
Set up. We distinguish the algebraic (natural) domain of a polynomial from the physical (contextual) domain of a flight, then compute the range over the physical interval.
(i) Natural domain of h(t) = −5t² + 10t + 1 is (−∞, ∞) (polynomial). Contextual domain: t ≥ 0 (time after launch) and t ≤ first positive root of h(t) = 0. Solving: t = (10 + √(100 + 20))/(10) = (10 + √120)/10 = 1 + √(30)/5 ≈ 2.095. Contextual domain ≈ [0, 1 + √30/5] ≈ [0, 2.10] s.
(ii) Vertex at t = −10/(2·−5) = 1; h(1) = −5 + 10 + 1 = 6. h(0) = 1, h(t_ground) = 0. Over the flight, h takes all values from 0 (touching down) up to the maximum 6. Range = [0, 6] m.
(iii) The algebraic rule produces meaningful real numbers for all t (including negative t and t after landing), but those values do not correspond to a real flight. Using (−∞, ∞) would give absurd "negative heights" and "heights before launch" that don't model the physical motion.
Problem 2 — Rectangle area
Set up. We model an enclosed pen with two domain layers: what the formula allows, and what the geometry insists on.
(i) Natural domain of A(w) = w(20 − w) is ℝ (polynomial). Contextually we need w > 0 (positive width) and 20 − w > 0 ⇒ w < 20. Contextual domain = (0, 20).
(ii) A(w) = −w² + 20w. Vertex at w = 10, A(10) = 100. Range on (0, 20) ⇒ values from just above 0 up to and including 100. Range = (0, 100]. (The maximum 100 is attained at w = 10.)
(iii) A(0) = 0 would correspond to width 0 (no rectangle). For a "valid pen" both length and width must be strictly positive, so we exclude the open endpoints and their image A = 0.
Problem 3 — Sound intensity
Set up. We treat I(r) = 0.5/r² as a rational function with one excluded denominator value and read off limits for the range.
(i) Natural domain: r ≠ 0 ⇒ (−∞, 0) ∪ (0, ∞). Contextual: distance must be positive ⇒ (0, ∞).
(ii) As r → 0⁺, I → +∞ (very close to speaker, infinitely loud). As r → ∞, I → 0 (vanishingly quiet). I is continuous and decreasing on (0, ∞), taking every positive value. Range = (0, ∞).
(iii) 0.5/r² ≤ 0.02 ⇒ r² ≥ 0.5/0.02 = 25 ⇒ r ≥ 5 (taking the positive root). Minimum safe distance = 5 m.
Problem 4 — Ramp profile
Set up. We use a quarter-circle-based rule and pin down what values it takes on the design interval.
(i) Natural domain of √(900 − x²) − 30 over ℝ is 900 − x² ≥ 0 ⇒ −30 ≤ x ≤ 30 ⇒ [−30, 30]. Contextual ramp domain: [0, 30] cm.
(ii) f(0) = √900 − 30 = 30 − 30 = 0 cm (level at the base). f(15) = √(900 − 225) − 30 = √675 − 30 ≈ 25.98 − 30 ≈ −4.02 cm. f(30) = √0 − 30 = −30 cm. On [0, 30], f decreases from 0 to −30. Range = [−30, 0] cm.
(iii) The ramp drops as x increases (negative values), so its absolute "rise" magnitude grows from 0 to 30 cm. The maximum vertical change is 30 cm, which exceeds the 10 cm code limit. The ramp does not satisfy the code (e.g. |f(30)| = 30 > 10).
Problem 5 — Investment
Set up. We treat compound interest as an exponential function, distinguish algebraic from financial domain, and invert at a target.
(i) Natural domain: all reals (1.005^x defined everywhere). Contextual: t ≥ 0 (cannot have negative elapsed time after investment). Contextual domain = [0, ∞).
(ii) V(0) = 1000(1.005)⁰ = $1000. V(10) = 1000(1.005)¹²⁰ ≈ 1000 × 1.8194 ≈ $1819.40. V is increasing and unbounded on [0, ∞). Range = [1000, ∞).
(iii) Set V(t) = 2000: 1000(1.005)¹²ᵗ = 2000 ⇒ (1.005)¹²ᵗ = 2 ⇒ 12t · ln(1.005) = ln 2 ⇒ t = ln 2 / (12 ln 1.005). Evaluating: ln 2 ≈ 0.69315, ln 1.005 ≈ 0.004988, so t ≈ 0.69315 / (12 × 0.004988) ≈ 0.69315 / 0.05985 ≈ 11.58 years.