Mathematics Advanced • Year 11 • Module 3 • Lesson 7
The Second Derivative
Practise HSC-style writing on second derivatives, concavity and inflection — culminating in a structured proof / explanation.
1. Short-answer questions
1.1 For f(x) = 2x³ − 9x² + 12x, find f″(x) and determine the values of x for which the curve is concave up. 3 marks Band 3-4
1.2 A particle moves with displacement s(t) = t³ − 6t² + 9t metres for t ≥ 0. Find the time at which the acceleration is zero, and the velocity at that instant. 4 marks Band 4
1.3 Show that the curve y = x⁴ − 4x³ has a point of inflection at x = 2. Use a sign-of-y″ argument; do not rely solely on y″(2) = 0. 3 marks Band 4
Stuck on 1.3? Compute y″ at one point each side of x = 2 and compare signs.2. Extended response
2.1 Consider the curve C with equation y = x³ − 3x² − 9x + 5.
(a) Find dy/dx and d²y/dx².
(b) Find all stationary points of C and classify each using the second derivative test.
(c) Find the coordinates of the point of inflection and the values of x for which the curve is concave up.
(d) Hence sketch the curve, labelling all stationary points, the inflection point, and the y-intercept. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct dy/dx = 3x² − 6x − 9 and d²y/dx² = 6x − 6.
Part (b) — 3 marks
• 1 mark — solves 3x² − 6x − 9 = 0 (factor 3(x − 3)(x + 1)) to get x = 3, x = −1; finds y-coords (3, −22) and (−1, 10).
• 1 mark — applies second derivative test at x = 3: y″(3) = 12 > 0 ⇒ local min at (3, −22).
• 1 mark — applies test at x = −1: y″(−1) = −12 < 0 ⇒ local max at (−1, 10).
Part (c) — 2 marks
• 1 mark — solves d²y/dx² = 0, i.e. 6x − 6 = 0 ⇒ x = 1; y(1) = 1 − 3 − 9 + 5 = −6; verifies sign change of y″ around x = 1.
• 1 mark — concave up for x > 1 (since 6x − 6 > 0).
Part (d) — 2 marks
• 1 mark — sketch shows the correct cubic shape with both extrema in the right positions.
• 1 mark — labels: stationary points (−1, 10) and (3, −22), inflection (1, −6), y-intercept (0, 5).
Your response (use a separate page or large grid for the sketch):
Stuck on (d)? After (b) you know where the curve turns; after (c) you know where it changes concavity. Draw the cubic so it passes through both points and bends correctly.How did this worksheet feel?
What I'll revisit before next class:
1.1 — f(x) = 2x³ − 9x² + 12x (3 marks)
Sample response. f′(x) = 6x² − 18x + 12, so f″(x) = 12x − 18. The curve is concave up where f″(x) > 0, i.e. 12x − 18 > 0, giving x > 3/2.
Marking notes. 1 — correct f′(x). 1 — correct f″(x). 1 — solves f″(x) > 0 and writes the inequality (or set) x > 3/2. Writing x ≥ 3/2 is acceptable, but x < 3/2 (the opposite) loses the final mark.
1.2 — Particle s(t) = t³ − 6t² + 9t (4 marks)
Sample response. v(t) = s′(t) = 3t² − 12t + 9. a(t) = s″(t) = 6t − 12.
Set a(t) = 0: 6t − 12 = 0 ⇒ t = 2 s. At t = 2: v(2) = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3 m/s.
Marking notes. 1 — correct v(t). 1 — correct a(t). 1 — solves a(t) = 0 for t. 1 — substitutes back to find v(2) with correct sign. A response that solves v(t) = 0 by mistake (giving t = 1, 3) scores at most 1/4.
1.3 — y = x⁴ − 4x³, inflection at x = 2 (3 marks)
Sample response. y′ = 4x³ − 12x², so y″ = 12x² − 24x = 12x(x − 2). At x = 2, y″(2) = 0. ✓
Sign of y″ around x = 2: at x = 1, y″ = 12 − 24 = −12 (negative); at x = 3, y″ = 108 − 72 = 36 (positive). Sign changes from − to +.
Since y″ changes sign at x = 2, the concavity changes there, so x = 2 is a point of inflection.
Marking notes. 1 — correct y″. 1 — substitutes test values either side of x = 2. 1 — concludes inflection with explicit reference to sign change. A response that just says "y″(2) = 0 so it's an inflection" scores at most 1/3, because that condition alone is insufficient (e.g. y = x⁴ at x = 0).
2.1 — Curve sketch of y = x³ − 3x² − 9x + 5 (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). dy/dx = 3x² − 6x − 9; d²y/dx² = 6x − 6. [1 mark.]
Part (b). Stationary points where dy/dx = 0:
3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1) = 0 ⇒ x = 3 or x = −1.
y(−1) = −1 − 3 + 9 + 5 = 10; y(3) = 27 − 27 − 27 + 5 = −22. [1 mark — both stationary points (−1, 10) and (3, −22).]
Second-derivative test: y″(−1) = −6 − 6 = −12 < 0 ⇒ local maximum at (−1, 10). [1 mark.]
y″(3) = 18 − 6 = 12 > 0 ⇒ local minimum at (3, −22). [1 mark.]
Part (c). Inflection: y″ = 0 ⇒ 6x = 6 ⇒ x = 1. y(1) = 1 − 3 − 9 + 5 = −6. Sign check: y″(0) = −6 (−); y″(2) = 6 (+). Sign change ✓ Point of inflection at (1, −6). [1 mark — inflection.]
Concave up when 6x − 6 > 0, i.e. x > 1. [1 mark — concavity interval.]
Part (d). Sketch: a cubic with positive leading coefficient, passing through y-intercept (0, 5); local max (−1, 10); inflection (1, −6); local min (3, −22). The curve is concave down for x < 1 (going through the max and dropping toward the inflection) and concave up for x > 1 (continuing down to the min then rising). [1 mark — correct cubic shape; 1 mark — all four features labelled.]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes both derivatives but stops at finding x-coordinates of stationary points; no classification; no inflection or sketch. ≈ 2-3 marks.
Band 4: Both stationary points found and classified by inspection of y″ but no inflection point, or inflection point found without verifying sign change. ≈ 4-5 marks.
Band 5: All algebra complete (parts a–c); sketch attempted but missing labels or has the wrong concavity at one section. ≈ 6-7 marks.
Band 6: All parts complete; uses second derivative test correctly to classify; sketch is correctly oriented with all four labelled features; concavity intervals stated. 8/8.