Solutions — Second Derivative Test for Stationary Points

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f(x) = x² − 6x + 5
Step 1 — Find stationary point:
f′(x) = 2x − 6 = 0 ⇒ x = 3

Step 2 — Second derivative test:
f″(x) = 2; f″(3) = 2 > 0 ⇒ concave up ⇒ local minimum

Step 3 — Coordinates:
f(3) = 9 − 18 + 5 = −4

Local minimum at (3, −4)
f(x) = −x² + 4x + 1
Step 1 — Find stationary point:
f′(x) = −2x + 4 = 0 ⇒ x = 2

Step 2 — Second derivative test:
f″(x) = −2; f″(2) = −2 < 0 ⇒ concave down ⇒ local maximum

Step 3 — Coordinates:
f(2) = −4 + 8 + 1 = 5

Local maximum at (2, 5)
f(x) = x³ − 12x
Step 1 — Find stationary points:
f′(x) = 3x² − 12 = 3(x² − 4) = 3(x−2)(x+2) = 0
x = 2 or x = −2

Step 2 — Second derivative:
f″(x) = 6x
f″(2) = 12 > 0 ⇒ concave up ⇒ local minimum; f(2) = 8 − 24 = −16
f″(−2) = −12 < 0 ⇒ concave down ⇒ local maximum; f(−2) = −8 + 24 = 16

Local maximum at (−2, 16); local minimum at (2, −16)
f(x) = 3x⁴ − 4x³
Step 1 — Find stationary points:
f′(x) = 12x³ − 12x² = 12x²(x − 1) = 0
x = 0 or x = 1

Step 2 — Second derivative:
f″(x) = 36x² − 24x
f″(1) = 36 − 24 = 12 > 0 ⇒ local minimum at (1, −1)
f(1) = 3 − 4 = −1 ✓

f″(0) = 0 ⇒ inconclusive. Use first derivative sign test:
f′(−0.5) = 12(0.25)(−1.5) = −4.5 < 0
f′(0.5) = 12(0.25)(−0.5) = −1.5 < 0
No sign change in f′ ⇒ stationary inflection point at (0, 0)
f(x) = 2x³ − 9x² + 12x − 4
Step 1 — Find stationary points:
f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x−1)(x−2) = 0
x = 1 or x = 2

Step 2 — Second derivative:
f″(x) = 12x − 18
f″(1) = 12 − 18 = −6 < 0 ⇒ local maximum; f(1) = 2 − 9 + 12 − 4 = 1
f″(2) = 24 − 18 = 6 > 0 ⇒ local minimum; f(2) = 16 − 36 + 24 − 4 = 0

Local maximum at (1, 1); local minimum at (2, 0)
f(x) = x⁴ − 8x²
Step 1 — Find stationary points:
f′(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x+2)(x−2) = 0
x = 0, x = 2, or x = −2

Step 2 — Second derivative:
f″(x) = 12x² − 16
f″(0) = −16 < 0 ⇒ local maximum; f(0) = 0
f″(2) = 48 − 16 = 32 > 0 ⇒ local minimum; f(2) = 16 − 32 = −16
f″(−2) = 48 − 16 = 32 > 0 ⇒ local minimum; f(−2) = 16 − 32 = −16

Local maximum at (0, 0); local minima at (2, −16) and (−2, −16)
f(x) = x³ + ax² + bx
f′(x) = 3x² + 2ax + b

Since x = −3 and x = 1 are the stationary points:
f′(x) = 3(x − (−3))(x − 1) = 3(x + 3)(x − 1)
= 3(x² + 2x − 3)
= 3x² + 6x − 9

Comparing with 3x² + 2ax + b:
2a = 6 ⇒ a = 3
b = −9 ⇒ b = −9

Verification with second derivative:
f″(x) = 6x + 6
f″(−3) = −12 < 0 ⇒ local maximum at x = −3 ✓
f″(1) = 12 > 0 ⇒ local minimum at x = 1 ✓
f(x) = x⁵
f′(x) = 5x⁴ = 0 ⇒ x = 0
f″(x) = 20x³; f″(0) = 0

Second derivative test is inconclusive because f″(0) = 0.

First derivative sign test:
f′(−1) = 5(−1)⁴ = 5 > 0 (increasing)
f′(1) = 5(1)⁴ = 5 > 0 (increasing)

No sign change in f′: positive on both sides of x = 0.
(0, 0) is a stationary inflection point — the curve is momentarily flat but does not turn.
f(x) = xe^−x

(a) Stationary points:
Using the product rule: f′(x) = 1 · e^−x + x · (−e^−x) = e^−x(1 − x)
e^−x > 0 always, so f′(x) = 0 ⇒ 1 − x = 0 ⇒ x = 1

f″(x): differentiate e^−x(1−x) using product rule:
f″(x) = −e^−x(1−x) + e^−x(−1) = e^−x[−(1−x) − 1] = e^−x(x − 2)
f″(1) = e⁻¹(1 − 2) = −e⁻¹ < 0 ⇒ local maximum at (1, e⁻¹)

(b) Inflection points:
f″(x) = e^−x(x − 2) = 0 ⇒ x = 2
f″(1.5) = e^−1.5(1.5 − 2) = −0.5e^−1.5 < 0
f″(3) = e⁻³(3 − 2) = e⁻³ > 0
Sign changes ⇒ inflection point at (2, 2e⁻²)
(a) Classify stationary points:
f′(x) = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2
f″(x) = 6x − 6
f″(0) = −6 < 0 ⇒ local maximum at x = 0
f″(2) = 6 > 0 ⇒ local minimum at x = 2

(b) Find f(x):
f(x) = ∫(3x² − 6x) dx = x³ − 3x² + c
f(0) = 0 − 0 + c = 4 ⇒ c = 4
f(x) = x³ − 3x² + 4

Global minimum on [−1, 3]:
Evaluate at stationary points and endpoints:
f(−1) = −1 − 3 + 4 = 0
f(0) = 0 − 0 + 4 = 4 (local max)
f(2) = 8 − 12 + 4 = 0 (local min)
f(3) = 27 − 27 + 4 = 4

The global minimum value is 0, achieved at x = −1 and x = 2.