Mathematics Advanced • Year 11 • Module 1 • Lesson 4
Piecewise & Absolute Value Functions
HSC-style writing on piecewise definitions, absolute value equations, and continuity at a boundary.
1. Short-answer questions
1.1 Solve |3x − 6| = 9, showing both cases and a check of each solution. 2 marks Band 3
1.2 A theme park charges $40 for entry if a person is under 16 years old, and $60 if they are 16 or older. Let A(x) be the admission cost in dollars for a person of age x years (x ≥ 0).
(a) Write A(x) as a piecewise function.
(b) State A(15) and A(16) and explain in one sentence why a single linear formula cannot model this pricing structure. 3 marks Band 3-4
1.3 Consider f(x) = { x + 2 if x < 1 ; x² + 2 if x ≥ 1 }.
(a) Evaluate f(0), f(1) and f(3), stating the piece used.
(b) By comparing the two one-sided values at the boundary, decide whether f is continuous at x = 1, and justify in one line. 4 marks Band 4
2. Extended response
2.1 A delivery company has the following pricing rule: deliveries up to and including 3 km cost $6 (flat), and longer deliveries cost $6 plus $2.50 per kilometre beyond 3 km. Let C(d) be the cost in dollars for a delivery of d kilometres (d > 0).
(a) Write C(d) as a piecewise function and simplify the second piece.
(b) Evaluate C(3) and C(8), stating the piece used for each.
(c) A customer is quoted $16.00 for their delivery. Determine the trip distance, and verify that the result satisfies the condition for the piece used.
(d) Show that C is continuous at the boundary d = 3 by comparing the two one-sided values, and explain in one sentence what would change physically if it were not continuous. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct conditions and pieces: $6 for 0 < d ≤ 3 and 6 + 2.5(d − 3) for d > 3.
• 1 mark — simplified second piece: 2.5d − 1.5 (or 2.5(d − 3) + 6 stated explicitly).
Part (b) — 1 mark
• 1 mark — both values correct with the piece stated: C(3) = $6 (first piece, 3 ≤ 3), C(8) = $18.50 (second piece, 8 > 3).
Part (c) — 2 marks
• 1 mark — identifies that $16 > $6 (first-piece max), so the second piece applies; solves 2.5d − 1.5 = 16 ⇒ d = 7 km.
• 1 mark — verifies 7 > 3, so the condition for the second piece is satisfied.
Part (d) — 2 marks
• 1 mark — computes both one-sided values at d = 3: from the first piece C(3) = 6; from the second piece, as d → 3⁺, C(d) → 2.5(3) − 1.5 = 6.
• 1 mark — states both pieces meet at value 6, so C is continuous; and explains in one sentence that a jump would mean a customer at d = 3 km could be charged a different price depending on rounding, which would be unfair/inconsistent.
Your response:
Stuck on (d)? Compare the value the first piece gives at d = 3 with the limit of the second piece as d → 3⁺.How did this worksheet feel?
What I'll revisit before next class:
1.1 — |3x − 6| = 9 (2 marks)
Sample response. Case 1: 3x − 6 = 9 ⇒ 3x = 15 ⇒ x = 5. Case 2: 3x − 6 = −9 ⇒ 3x = −3 ⇒ x = −1. Check: |3(5) − 6| = |9| = 9 ✓; |3(−1) − 6| = |−9| = 9 ✓. Solutions: x = 5 or x = −1.
Marking notes. 1 mark — sets up both cases A = B and A = −B (not just one). 1 mark — both solutions correct with at least one shown check, or both values found and equation solved correctly. A response that gives only x = 5 (the most common partial-credit answer) scores 1/2.
1.2 — Theme park admission (3 marks)
(a) Sample response. A(x) = { 40 if 0 ≤ x < 16; 60 if x ≥ 16 }.
(b) Sample response. A(15) = $40 (first piece, since 15 < 16). A(16) = $60 (second piece, since 16 ≥ 16). A single linear formula would produce a gradual increase in cost with age, but the actual cost jumps abruptly at exactly age 16 — the rate of change is zero everywhere except at one point — which only a piecewise (step) function can model.
Marking notes. (a) 1 mark — both pieces with strict/non-strict inequalities matching the description (the boundary x = 16 belongs to the $60 piece). (b) 1 mark — both numerical values correct. 1 mark — explanation invokes "step" or "jump" behaviour that a single linear formula cannot reproduce. Common error: writing the inequalities the wrong way at the boundary (so A(16) = 40) scores half the (a) mark.
1.3 — Piecewise continuity (4 marks)
(a) Sample response. f(0): 0 < 1, first piece ⇒ f(0) = 0 + 2 = 2. f(1): 1 ≥ 1, second piece ⇒ f(1) = (1)² + 2 = 3. f(3): 3 ≥ 1, second piece ⇒ f(3) = (3)² + 2 = 11.
(b) Sample response. From the left (x → 1⁻, first piece): f → 1 + 2 = 3. At x = 1 and from the right (second piece, since x ≥ 1 includes x = 1): f(1) = 1² + 2 = 3. Both one-sided values agree (3 = 3), so f is continuous at x = 1.
Marking notes. (a) 1 mark — all three values correct with the piece stated for each. (b) 1 mark — computes the left limit using the first piece. 1 mark — computes the right value using the second piece. 1 mark — states "continuous" with the correct justification (both pieces meet at value 3). Common error: using x = 1 in the first piece's formula because of a strict/non-strict mix-up.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Piecewise function.
For 0 < d ≤ 3: flat rate, so C(d) = 6. For d > 3: C(d) = 6 + 2.5(d − 3) = 6 + 2.5d − 7.5 = 2.5d − 1.5. [1 mark — pieces and conditions correct.]
C(d) = { 6 if 0 < d ≤ 3; 2.5d − 1.5 if d > 3 }. [1 mark — second piece simplified.]
(b) Evaluate.
C(3): 3 ≤ 3, use first piece ⇒ C(3) = $6. C(8): 8 > 3, use second piece ⇒ C(8) = 2.5(8) − 1.5 = 20 − 1.5 = $18.50. [1 mark — both with piece stated.]
(c) Reverse the rule for $16.
$16 > $6 (the maximum cost from the first piece), so the second piece applies. [1 mark — justifies the choice of piece.]
Solve 2.5d − 1.5 = 16 ⇒ 2.5d = 17.5 ⇒ d = 7 km. Check: 7 > 3 ✓, so the second-piece condition is satisfied. [1 mark — correct distance with the condition verified.]
(d) Continuity at d = 3.
From the first piece: C(3) = 6. From the second piece, as d → 3⁺: C(d) → 2.5(3) − 1.5 = 7.5 − 1.5 = 6. [1 mark — both one-sided values calculated.]
Both pieces meet at the value 6, so C is continuous at d = 3 — there is no jump in price exactly at the boundary. If C were discontinuous, a 3.000 km trip and a 3.001 km trip would be priced very differently, leading to unfair pricing that customers would (correctly) dispute. [1 mark — states continuous, justifies, and gives the practical consequence.]
Total: 7/7.
Band descriptors for marker.
Band 3: Writes the piecewise function with at most one of the two pieces correct, or has the wrong boundary inequality. Attempts (b) and (c) but uses only one piece. ≈ 2-3 marks.
Band 4: Both pieces correctly defined and simplified; (b) correct; (c) gives d = 7 km but skips the "which piece" justification. Does not attempt (d). ≈ 4-5 marks.
Band 5: All of (a), (b), (c) correct with piece justification. Attempts (d) but only computes one side (often forgets the limit from the right). ≈ 5-6 marks.
Band 6: Complete response. Both one-sided values calculated and equated; explicit "continuous" conclusion; practical-consequence sentence in (d). All conditions and inequalities stated correctly. 7/7.