Mathematics Advanced • Year 11 • Module 1 • Lesson 4
Piecewise & Absolute Value Functions
Apply piecewise and absolute-value rules to realistic pricing, motion, and tolerance contexts.
Problem 1 — Ride-share pricing (piecewise from context)
A ride-share app charges $2 per km for the first 5 km, then $1.50 per km for every additional km after that. Let C(d) be the cost in dollars of a trip of d km, for d ≥ 0.
Set up: What are we solving for?
(i) Write C(d) as a piecewise function (simplify the second piece). 3 marks
(ii) Calculate the cost of a 3 km trip and an 8 km trip. State which piece you used for each. 2 marks
(iii) A rider reports paying $14.50. Use C(d) to determine the trip distance, showing why the result must come from the second piece. 2 marks
Stuck? Revisit lesson § Piecewise Functions and Worked Example 3.Problem 2 — Two-band income tax (piecewise evaluate & reverse)
A simplified two-band tax rule charges 0% tax on annual income up to $18,200, and 20 cents on each dollar earned above $18,200. Let T(x) be the tax in dollars on an annual income of x dollars.
Set up: What are we solving for?
(i) Write T(x) as a piecewise function for x ≥ 0. 2 marks
(ii) Calculate T(15 000) and T(45 000), stating which piece you used. 2 marks
(iii) A worker pays $3 200 in tax. Find their annual income. 2 marks
Stuck on (iii)? Set the second piece equal to 3200 and solve for x.Problem 3 — Machine tolerance (absolute value as distance)
A factory cuts steel rods to a target length of 250 mm. A rod is "in spec" when its measured length L (in mm) satisfies |L − 250| ≤ 0.4.
Set up: What are we solving for?
(i) Translate the spec into the two boundary equations |L − 250| = 0.4 and solve for the maximum and minimum acceptable lengths. 2 marks
(ii) Three rods measure 249.7 mm, 250.5 mm and 250.4 mm. Decide which are in spec by checking the absolute value condition for each. 2 marks
(iii) Explain in one sentence why the equation |L − 250| = −0.4 has no solution, and what this would mean physically if a quality inspector wrote it. 2 marks
Stuck? Recall |x| ≥ 0 always — revisit lesson § Absolute Value Functions.Problem 4 — Courier company piecewise rate
A courier company charges $5 for any delivery of 2 km or less, and $5 plus $1.50 per km beyond 2 km for longer deliveries. Let C(d) be the cost in dollars for a delivery of d km.
Set up: What are we solving for?
(i) Write C(d) as a piecewise function with the second piece in simplified form. 2 marks
(ii) Calculate C(2) and C(7), and explain whether the boundary point d = 2 belongs to the first or second piece. 2 marks
(iii) A customer is quoted $15.50 for a delivery. Find the trip distance, then check that the result satisfies the condition for the piece used. 3 marks
Stuck? Revisit lesson § Worked Example 3 — Writing a Piecewise Rule from Context.Problem 5 — Position relative to a target (absolute value equation)
A drone hovers at horizontal position x metres along a corridor (the target is at x = 8 m). Its "miss distance" is given by m(x) = |x − 8|.
Set up: What are we solving for?
(i) Solve m(x) = 3 to find the two possible positions of the drone. Interpret each geometrically as a distance from the target. 2 marks
(ii) Solve m(x) = 0. What does this single solution tell us about absolute value equations of the form |A| = 0? 2 marks
(iii) Solve |2x − 1| = |x + 4| for the drone's horizontal position when its miss distance from two targets at x = ½ (target A) and x = −4 (target B, scaled) are equal. Show both cases (A = B and A = −B), then check each solution in the original equation. 3 marks
Stuck on (iii)? Two cases: 2x − 1 = x + 4 and 2x − 1 = −(x + 4).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ride-share pricing
Set up. We need a rule for cost that changes at d = 5 km, then to evaluate and invert it.
(i) For 0 ≤ d ≤ 5: C(d) = 2d. For d > 5: C(d) = 2(5) + 1.5(d − 5) = 10 + 1.5d − 7.5 = 1.5d + 2.5.
C(d) = { 2d if 0 ≤ d ≤ 5; 1.5d + 2.5 if d > 5 }.
(ii) 3 km: 3 ≤ 5, use first piece. C(3) = 2(3) = $6.00. 8 km: 8 > 5, use second piece. C(8) = 1.5(8) + 2.5 = 12 + 2.5 = $14.50.
(iii) Test the first-piece maximum: C(5) = $10, so any cost above $10 must come from the second piece. Solve 1.5d + 2.5 = 14.5 ⇒ 1.5d = 12 ⇒ d = 8 km. Confirm 8 > 5 ✓.
Problem 2 — Two-band income tax
Set up. Tax-free up to $18 200, then 20¢ per dollar above. Express piecewise, evaluate two incomes, then invert.
(i) T(x) = { 0 if 0 ≤ x ≤ 18 200; 0.20(x − 18 200) if x > 18 200 }.
(ii) T(15 000): 15 000 ≤ 18 200, use first piece ⇒ T(15 000) = $0. T(45 000): 45 000 > 18 200, use second piece ⇒ T(45 000) = 0.20 × (45 000 − 18 200) = 0.20 × 26 800 = $5 360.
(iii) 3 200 > 0 so the second piece applies. 0.20(x − 18 200) = 3 200 ⇒ x − 18 200 = 16 000 ⇒ x = $34 200.
Problem 3 — Machine tolerance
Set up. The absolute value measures how far L is from 250 mm. The spec window is ±0.4 mm.
(i) |L − 250| = 0.4 gives L − 250 = 0.4 or L − 250 = −0.4, so L = 250.4 mm (max) or L = 249.6 mm (min). The in-spec window is 249.6 ≤ L ≤ 250.4.
(ii) 249.7: |249.7 − 250| = 0.3 ≤ 0.4 ✓ in spec. 250.5: |250.5 − 250| = 0.5 > 0.4 ✗ out of spec. 250.4: |250.4 − 250| = 0.4 ≤ 0.4 ✓ in spec (on the boundary).
(iii) |L − 250| is a distance, so it is always ≥ 0 and can never equal a negative number. Writing |L − 250| = −0.4 would be a nonsense statement — the inspector has confused a sign convention; no measurement can have negative distance from the target.
Problem 4 — Courier piecewise
Set up. Flat rate up to and including 2 km; extra $1.50/km beyond. Mirrors the lesson's Worked Example 3.
(i) For 0 ≤ d ≤ 2: C(d) = 5. For d > 2: C(d) = 5 + 1.5(d − 2) = 5 + 1.5d − 3 = 1.5d + 2. C(d) = { 5 if 0 ≤ d ≤ 2; 1.5d + 2 if d > 2 }.
(ii) C(2): 2 ≤ 2, use first piece ⇒ $5. The boundary d = 2 belongs to the first piece because the condition is d ≤ 2 (closed dot at d = 2 on the flat-rate side). C(7): 7 > 2, use second piece ⇒ 1.5(7) + 2 = $12.50.
(iii) $15.50 > $5 (the first-piece maximum), so the cost must come from the second piece. Solve 1.5d + 2 = 15.50 ⇒ 1.5d = 13.50 ⇒ d = 9 km. Check d = 9 > 2 ✓ — the condition for the second piece is satisfied.
Problem 5 — Drone miss distance
Set up. Apply the two-case rule for |A| = B (with B ≥ 0).
(i) |x − 8| = 3 gives x − 8 = 3 or x − 8 = −3, so x = 11 or x = 5. The drone is either 3 m to the right of the target (x = 11) or 3 m to the left (x = 5) — both are 3 m of "miss distance".
(ii) |x − 8| = 0 gives x − 8 = 0, so x = 8 (a single solution). In general, |A| = 0 has exactly one solution, A = 0, because zero is the only number whose distance from zero is zero.
(iii) Case A: 2x − 1 = x + 4 ⇒ x = 5. Case B: 2x − 1 = −(x + 4) ⇒ 2x − 1 = −x − 4 ⇒ 3x = −3 ⇒ x = −1. Check x = 5: |2(5) − 1| = |9| = 9 and |5 + 4| = 9 ✓. Check x = −1: |2(−1) − 1| = |−3| = 3 and |−1 + 4| = 3 ✓. Solutions: x = 5 or x = −1.