Mathematics Advanced • Year 11 • Module 1 • Lesson 4
Piecewise & Absolute Value Functions
Build fluency in evaluating piecewise functions, solving |ax + b| = c, and the open/closed dot convention.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 Write the piecewise definition of |x|:
|x| = { ____________ if x ≥ 0 ; ____________ if x < 0 }
Q1.2 If |A| = B and B ≥ 0, the two equations to solve are: A = ______ or A = ______.
Q1.3 On a piecewise graph, an open circle at a boundary point means the value ____________ included; a closed (filled) circle means the value ____________ included.
2. Worked example — solving |3x − 6| = 9
Follow each line.
Problem. Solve |3x − 6| = 9.
Step 1 — Set up the two cases.
3x − 6 = 9 OR 3x − 6 = −9
Step 2 — Solve Case 1.
3x = 15 ⇒ x = 5
Step 3 — Solve Case 2.
3x = −3 ⇒ x = −1
Step 4 — Check both in the original equation.
|3(5) − 6| = |9| = 9 ✓ |3(−1) − 6| = |−9| = 9 ✓
Conclusion. x = 5 or x = −1.
3. Faded example — evaluating a piecewise function
For the piecewise function below, evaluate f(−1), f(2) and f(4). Fill in each blank. 3 marks
f(x) = { x² + 1, x ≤ 2 ; 5x − 3, x > 2 }
f(−1): Since −1 ____ 2, we use the ____________ rule.
f(−1) = ( ______ )² + 1 = ______ + 1 = ______
f(2): Since 2 ____ 2 (boundary), we use the ____________ rule (because the boundary is included).
f(2) = ( ______ )² + 1 = ______ + 1 = ______
f(4): Since 4 ____ 2, we use the ____________ rule.
f(4) = 5( ______ ) − 3 = ______
4. Graduated practice
Show working in the space below each part. Use brackets when substituting.
Foundation — single evaluations (4 questions)
| Q | Task | Answer |
|---|---|---|
| 4.1 1 | Evaluate |−7|. | |
| 4.2 1 | Evaluate |3 − 8|. | |
| 4.3 1 | Solve |x| = 6. | |
| 4.4 1 | For f(x) = { 2x if x ≥ 0 ; −x if x < 0 }, find f(−4). |
Standard — typical HSC difficulty (6 questions)
4.5 Solve |2x + 1| = 7. 2 marks
4.6 Solve |x − 5| = 3. State both solutions and check one. 2 marks
4.7 Given f(x) = { 3x + 1 if x < 0 ; x² if x ≥ 0 }, evaluate f(−2), f(0) and f(3). 2 marks
4.8 A ride-share charges $2/km for the first 5 km, then $1.50/km after. Write the cost C(d) as a piecewise function for d ≥ 0. 2 marks
4.9 Solve |x + 4| = −3. Explain your answer in one line. 2 marks
4.10 A taxi charges a $5 flat rate for trips up to 2 km, then $5 plus $1.50 per extra km after 2 km. Write the rule C(d) as a piecewise function. 2 marks
Extension — combined / harder (2 questions)
4.11 Solve |2x − 1| = |x + 4|. (Hint: equate the inside expressions for case A and the negative of one expression for case B.) 3 marks
4.12 Determine whether f(x) = { x + 2 if x < 1 ; x² + 2 if x ≥ 1 } is continuous at x = 1. Justify by comparing the two one-sided values. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Piecewise definition of |x|
|x| = { x if x ≥ 0; −x if x < 0 }.
Q1.2 — Equations from |A| = B
A = B or A = −B.
Q1.3 — Open vs closed dots
Open circle: value is not included. Closed circle: value is included.
Q3 — Faded piecewise example
f(−1): −1 ≤ 2, use the first rule. f(−1) = (−1)² + 1 = 1 + 1 = 2.
f(2): 2 = 2 (boundary included in first rule), use the first rule. f(2) = (2)² + 1 = 4 + 1 = 5.
f(4): 4 > 2, use the second rule. f(4) = 5(4) − 3 = 20 − 3 = 17.
Q4.1 — |−7|
|−7| = 7 (distance from 0).
Q4.2 — |3 − 8|
|3 − 8| = |−5| = 5.
Q4.3 — |x| = 6
x = 6 or x = −6.
Q4.4 — f(−4) where f(x) = 2x if x ≥ 0, −x if x < 0
−4 < 0 ⇒ use −x branch. f(−4) = −(−4) = 4.
Q4.5 — |2x + 1| = 7
Case 1: 2x + 1 = 7 ⇒ x = 3. Case 2: 2x + 1 = −7 ⇒ 2x = −8 ⇒ x = −4.
Q4.6 — |x − 5| = 3
x − 5 = 3 ⇒ x = 8; x − 5 = −3 ⇒ x = 2. Check x = 8: |8 − 5| = 3 ✓.
Q4.7 — Piecewise f(x) = 3x + 1 if x < 0; x² if x ≥ 0
f(−2) = 3(−2) + 1 = −5. f(0) = (0)² = 0. f(3) = (3)² = 9.
Q4.8 — Ride-share cost
For 0 ≤ d ≤ 5: C(d) = 2d. For d > 5: C(d) = 2(5) + 1.5(d − 5) = 10 + 1.5(d − 5) = 1.5d + 2.5.
C(d) = { 2d if 0 ≤ d ≤ 5; 1.5d + 2.5 if d > 5 }.
Q4.9 — |x + 4| = −3
No solution. Absolute value is always ≥ 0, so it can never equal a negative number.
Q4.10 — Taxi piecewise
For 0 ≤ d ≤ 2: C(d) = 5. For d > 2: C(d) = 5 + 1.5(d − 2) = 2 + 1.5d.
C(d) = { 5 if 0 ≤ d ≤ 2; 2 + 1.5d if d > 2 }.
Q4.11 — |2x − 1| = |x + 4|
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: |9| = |9| ✓. Check x = −1: |−3| = |3| ✓. Solutions: x = 5 or x = −1.
Q4.12 — Continuity at x = 1
From the left (x → 1⁻, first rule): f → 1 + 2 = 3.
From the right and at x = 1 (second rule, since x ≥ 1 includes x = 1): f(1) = (1)² + 2 = 3.
Both pieces meet at the boundary value 3, so f is continuous at x = 1 (no jump or gap).