Mathematics Advanced • Year 11 • Module 1 • Lesson 2
Function Notation & Evaluation
Apply function notation and the difference quotient to taxi fares, drone flights, audio signals and physics.
Problem 1 — Taxi fare modelling
A taxi charges C(d) = 5 + 1.8d, where C is cost in dollars and d is distance in kilometres.
Set up: What are we solving for?
(i) Find C(15) and C(30), showing brackets at the substitution step. 2 marks
(ii) Interpret C(0) in context. Then use C(d) to find the distance for a $50 fare. 3 marks
(iii) Compute the difference quotient [C(d + h) − C(d)] / h in simplest form. What does this number represent in dollars-per-kilometre? 3 marks
Stuck? Revisit lesson § Worked Example 2 — Function in Context.Problem 2 — Drone altitude
A delivery drone flies in a vertical plane. Its altitude (in metres) t seconds after take-off is modelled by h(t) = −t² + 20t for 0 ≤ t ≤ 20.
Set up: What are we solving for?
(i) Find h(0), h(5), h(10) and h(20). Interpret h(20). 3 marks
(ii) Solve h(t) = 75 algebraically. What do the two solutions represent physically? 3 marks
(iii) Compute the difference quotient [h(t + Δt) − h(t)] / Δt in simplest form, and use it to find the average vertical speed between t = 4 and t = 6. 3 marks
Problem 3 — Electrical power
The instantaneous power (in watts) dissipated in a resistor is modelled by P(I) = R · I², where R = 8 ohms.
Set up: What are we solving for?
(i) Write the explicit rule P(I), then evaluate P(2) and P(3), showing brackets. 2 marks
(ii) A safety standard caps power at 200 W. Solve P(I) ≤ 200 to find the maximum permissible current. 3 marks
(iii) Compute and simplify [P(I + h) − P(I)] / h for a general h. Interpret this expression as h → 0 in terms of physical quantities. 3 marks
Stuck on (iii)? Expand (I + h)² = I² + 2Ih + h² carefully.Problem 4 — Audio signal in dB
The intensity of an audio signal in decibels is L(I) = 10 log₁₀(I / 10⁻¹²), where I is the intensity in W/m² and 10⁻¹² is the threshold of hearing.
Set up: What are we solving for?
(i) Evaluate L(10⁻¹²) and L(10⁻⁶) without a calculator. 3 marks
(ii) Show that doubling the intensity (I → 2I) increases the dB reading by exactly 10 log₁₀ 2 ≈ 3 dB, independent of the starting intensity. 3 marks
(iii) Concert speakers measure 100 dB. Use L(I) to find I in W/m². 2 marks
Problem 5 — Revenue function
A bakery's daily revenue (in dollars) from selling x boxes of croissants is R(x) = 8x − 0.05x², for 0 ≤ x ≤ 160.
Set up: What are we solving for?
(i) Evaluate R(50) and R(100), and interpret in dollars. 2 marks
(ii) Find R(x + 1) − R(x) in simplest form. What does this expression represent in business terms? 3 marks
(iii) Use part (ii) to find the value of x at which the marginal revenue R(x + 1) − R(x) becomes zero. What does this tell the bakery about its optimal production volume? 3 marks
Stuck on (iii)? Set the expression from (ii) equal to zero and solve linearly for x.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Taxi fare
Set up. We use function notation to compute fares, interpret a fixed value, and read the slope as a rate.
(i) C(15) = 5 + 1.8(15) = 5 + 27 = $32. C(30) = 5 + 1.8(30) = 5 + 54 = $59.
(ii) C(0) = $5 — the flag-fall (booking fee) charged before any distance. For $50: 5 + 1.8d = 50 ⇒ 1.8d = 45 ⇒ d = 25 km.
(iii) C(d + h) − C(d) = (5 + 1.8(d + h)) − (5 + 1.8d) = 1.8h. Dividing by h: 1.8. This is the per-kilometre charge ($1.80/km) — the marginal cost of one extra kilometre, independent of distance already travelled.
Problem 2 — Drone altitude
Set up. We evaluate a quadratic altitude model, solve for time at a given height, and compute average velocity over an interval.
(i) h(0) = 0 m (lifting off). h(5) = −25 + 100 = 75 m. h(10) = −100 + 200 = 100 m (peak). h(20) = −400 + 400 = 0 m: the drone has landed.
(ii) −t² + 20t = 75 ⇒ t² − 20t + 75 = 0 ⇒ (t − 5)(t − 15) = 0 ⇒ t = 5 or t = 15. The drone is at 75 m height twice: once on the way up (t = 5 s) and once on the way down (t = 15 s).
(iii) h(t + Δt) = −(t + Δt)² + 20(t + Δt) = −t² − 2tΔt − (Δt)² + 20t + 20Δt. Then h(t + Δt) − h(t) = −2tΔt − (Δt)² + 20Δt. Dividing by Δt: −2t − Δt + 20. Between t = 4 and t = 6, Δt = 2: average vertical speed = −2(4) − 2 + 20 = 10 m/s upward.
Problem 3 — Power
Set up. We work with a quadratic power function, solve an inequality for safe operation, and prepare the calculus expression for power vs current.
(i) P(I) = 8I². P(2) = 8(2)² = 32 W. P(3) = 8(3)² = 72 W.
(ii) 8I² ≤ 200 ⇒ I² ≤ 25 ⇒ −5 ≤ I ≤ 5. In a physical resistor, I ≥ 0, so the safety cap is I ≤ 5 A.
(iii) P(I + h) − P(I) = 8(I + h)² − 8I² = 8(I² + 2Ih + h²) − 8I² = 16Ih + 8h². Dividing by h: 16I + 8h. As h → 0 this tends to 16I W/A — the instantaneous rate at which power changes with current (in Year 12 it will be dP/dI = 16I).
Problem 4 — Audio dB
Set up. We evaluate a log-based intensity function and use log laws to show doubling = +3 dB.
(i) L(10⁻¹²) = 10 log₁₀(10⁻¹² / 10⁻¹²) = 10 log₁₀(1) = 0 dB. L(10⁻⁶) = 10 log₁₀(10⁻⁶ / 10⁻¹²) = 10 log₁₀(10⁶) = 10 · 6 = 60 dB.
(ii) L(2I) = 10 log₁₀(2I / 10⁻¹²) = 10 [log₁₀ 2 + log₁₀(I / 10⁻¹²)] = 10 log₁₀ 2 + L(I). So L(2I) − L(I) = 10 log₁₀ 2 ≈ 3.01 dB, independent of I. (Each doubling of intensity adds about 3 dB.)
(iii) L(I) = 100 ⇒ 10 log₁₀(I / 10⁻¹²) = 100 ⇒ log₁₀(I / 10⁻¹²) = 10 ⇒ I / 10⁻¹² = 10¹⁰ ⇒ I = 10⁻² W/m² (= 0.01 W/m²).
Problem 5 — Bakery revenue
Set up. We use function evaluation, finite differences (marginal revenue), and solve for the production point at which extra units stop adding revenue.
(i) R(50) = 8(50) − 0.05(50)² = 400 − 125 = $275. R(100) = 8(100) − 0.05(100)² = 800 − 500 = $300.
(ii) R(x + 1) = 8(x + 1) − 0.05(x + 1)² = 8x + 8 − 0.05(x² + 2x + 1) = 8x + 8 − 0.05x² − 0.1x − 0.05. Subtracting R(x): R(x + 1) − R(x) = 8 − 0.1x − 0.05 = 7.95 − 0.1x. This is the marginal revenue: the extra dollars from selling the (x + 1)-th box.
(iii) 7.95 − 0.1x = 0 ⇒ x = 79.5. Selling one more box only stops adding revenue around 79-80 boxes. Beyond that, total revenue is declining — the bakery should produce at most ~80 boxes/day.