Mathematics Advanced • Year 11 • Module 3 • Lesson 7
The Second Derivative
Build procedural fluency in computing f″(x), interpreting concavity, and finding points of inflection.
1. Quick recall
Answer each in the space provided. 1 mark each
Q1.1 Complete each statement using >, <, or =.
A curve is concave up when f″(x) ______ 0.
A curve is concave down when f″(x) ______ 0.
A point of inflection requires f″(x) ______ 0 and a sign change.
Q1.2 Match each notation with what it means:
f′(x) ↔ ____________________________
f″(x) ↔ ____________________________
d²y/dx² ↔ ____________________________
Q1.3 In kinematics: if s(t) is position, then s′(t) represents ____________ and s″(t) represents ____________.
2. Worked example — f(x) = 3x⁴ − 2x³ + x
Differentiate twice, then factorise to find candidates for inflection points.
Problem. Find f″(x) for f(x) = 3x⁴ − 2x³ + x.
Step 1 — Differentiate term by term to get f′(x).
f′(x) = 12x³ − 6x² + 1
Reason: power rule on each term; the constant term in f drops to 0 (no contribution).
Step 2 — Differentiate f′(x) to get f″(x).
f″(x) = 36x² − 12x
Reason: differentiate the first derivative; the +1 drops out.
Step 3 — Factorise to find where f″(x) = 0.
f″(x) = 12x(3x − 1) = 0 ⇒ x = 0 or x = 1/3
Reason: factoring exposes the two candidate inflection points (sign change still needs to be checked).
Conclusion. f″(x) = 12x(3x − 1); candidates for inflection are x = 0 and x = 1/3.
3. Faded example — fill in the missing steps
Show that y = x³ − 6x² + 9x has a point of inflection and find its coordinates. Fill in each blank. 4 marks
Step 1 — First derivative:
dy/dx = ______________________
Step 2 — Second derivative:
d²y/dx² = ______________________
Step 3 — Set d²y/dx² = 0 and solve:
______________________ = 0 ⇒ x = ______________
Step 4 — Check sign change of d²y/dx² across x = ______:
For x just below: d²y/dx² is ______; for x just above: d²y/dx² is ______. Sign change confirmed.
Step 5 — y-coordinate: at x = ______, y = ______________________.
Conclusion. Point of inflection at ( ______ , ______ ).
4. Graduated practice — second derivative work
For each function, compute the requested derivative and answer concisely.
Foundation — pure derivative computation (4 questions)
| Q | Function f(x) | f′(x) | f″(x) |
|---|---|---|---|
| 4.1 1 | f(x) = x⁵ − 4x³ + 2x | ||
| 4.2 1 | f(x) = sin x | ||
| 4.3 1 | f(x) = cos x | ||
| 4.4 1 | f(x) = ex |
Standard — concavity and inflection (6 questions)
4.5 Find d²y/dx² for y = sin x, then state where d²y/dx² < 0 in the interval 0 ≤ x ≤ 2π. 2 marks
4.6 Determine the concavity of y = x⁴ − 2x² at x = 1. 2 marks
4.7 Find the point of inflection of y = x³ − 3x². 2 marks
4.8 For f(x) = 2x³ − 9x² + 12x, find f″(x) and find the x-values where the curve is concave up. 2 marks
4.9 If s(t) = 2t³ − t² metres, find the acceleration at t = 1 s. 2 marks
4.10 Show that y = x⁴ has f″(0) = 0 but x = 0 is not a point of inflection. 2 marks
Extension — combine ideas (2 questions)
4.11 Show that y = x⁴ − 4x³ has two points of inflection, finding both x-values and the corresponding y-coordinates. 3 marks
4.12 A particle's position is s(t) = t³ − 6t² + 9t metres for t ≥ 0. Find the time at which the acceleration is zero, and the velocity at that instant. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your work.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Concavity conditions
Concave up: f″(x) > 0. Concave down: f″(x) < 0. Inflection: f″(x) = 0 (and sign change).
Q1.2 — Notation
f′(x): first derivative (rate of change / gradient). f″(x): second derivative (rate of change of the gradient / concavity / acceleration). d²y/dx²: Leibniz notation for the second derivative (single symbol, not a fraction to cancel).
Q1.3 — Kinematics
s′(t) = velocity. s″(t) = acceleration.
Q3 — Faded example y = x³ − 6x² + 9x
Step 1: dy/dx = 3x² − 12x + 9. Step 2: d²y/dx² = 6x − 12. Step 3: 6x − 12 = 0 ⇒ x = 2. Step 4: at x = 1, d²y/dx² = −6 (negative); at x = 3, d²y/dx² = 6 (positive); sign change confirmed. Step 5: y(2) = 8 − 24 + 18 = 2. Conclusion: point of inflection at (2, 2).
Q4.1 — f(x) = x⁵ − 4x³ + 2x
f′(x) = 5x⁴ − 12x² + 2. f″(x) = 20x³ − 24x.
Q4.2 — f(x) = sin x
f′(x) = cos x. f″(x) = −sin x.
Q4.3 — f(x) = cos x
f′(x) = −sin x. f″(x) = −cos x.
Q4.4 — f(x) = ex
f′(x) = ex. f″(x) = ex. The exponential function is its own derivative at every order.
Q4.5 — d²y/dx² for y = sin x; concavity on [0, 2π]
d²y/dx² = −sin x. We need −sin x < 0, i.e. sin x > 0. On [0, 2π] this happens for 0 < x < π.
Q4.6 — concavity of y = x⁴ − 2x² at x = 1
dy/dx = 4x³ − 4x; d²y/dx² = 12x² − 4. At x = 1: d²y/dx² = 12 − 4 = 8 > 0, so the curve is concave up at x = 1.
Q4.7 — point of inflection of y = x³ − 3x²
dy/dx = 3x² − 6x; d²y/dx² = 6x − 6 = 0 ⇒ x = 1. Sign change: at x = 0, d²y/dx² = −6 (−); at x = 2, d²y/dx² = 6 (+). ✓ y(1) = 1 − 3 = −2. Point of inflection at (1, −2).
Q4.8 — f(x) = 2x³ − 9x² + 12x
f′(x) = 6x² − 18x + 12; f″(x) = 12x − 18. Concave up when f″(x) > 0: 12x − 18 > 0 ⇒ x > 3/2.
Q4.9 — acceleration for s(t) = 2t³ − t² at t = 1
v(t) = 6t² − 2t; a(t) = 12t − 2. a(1) = 12 − 2 = 10 m/s².
Q4.10 — y = x⁴ at x = 0
y′ = 4x³; y″ = 12x². At x = 0, y″(0) = 0. But for x < 0, y″ = 12x² > 0, and for x > 0, y″ = 12x² > 0 also. There is no sign change, so x = 0 is not a point of inflection (the curve is concave up on both sides — it's actually a minimum).
Q4.11 — inflection points of y = x⁴ − 4x³
y′ = 4x³ − 12x²; y″ = 12x² − 24x = 12x(x − 2) = 0 ⇒ x = 0 or x = 2.
At x = 0: for x = −1, y″ = 36 (+); for x = 1, y″ = −12 (−). Sign change ✓ y(0) = 0.
At x = 2: for x = 1, y″ = −12 (−); for x = 3, y″ = 36 (+). Sign change ✓ y(2) = 16 − 32 = −16.
Inflection points at (0, 0) and (2, −16).
Q4.12 — s(t) = t³ − 6t² + 9t: when a = 0
v(t) = 3t² − 12t + 9; a(t) = 6t − 12 = 0 ⇒ t = 2 s. v(2) = 12 − 24 + 9 = −3 m/s (particle moving in the negative direction at the instant acceleration vanishes).