Solutions — Stationary Points and the First Derivative Test

Q1 — Finding stationary points

(a) y = x² − 6x + 5:
y′ = 2x − 6 = 0 → x = 3. y(3) = 9 − 18 + 5 = −4. Stationary point: (3, −4).

(b) y = x³ − 12x:
y′ = 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2) = 0 → x = ±2.
y(2) = 8 − 24 = −16. y(−2) = −8 + 24 = 16.
Stationary points: (2, −16) and (−2, 16).

(c) y = 2x³ − 3x²:
y′ = 6x² − 6x = 6x(x − 1) = 0 → x = 0 or x = 1.
y(0) = 0. y(1) = 2 − 3 = −1.
Stationary points: (0, 0) and (1, −1).

(d) y = x⁴ − 8x²:
y′ = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2) = 0 → x = 0, ±2.
y(0) = 0. y(2) = 16 − 32 = −16. y(−2) = −16.
Stationary points: (0, 0), (2, −16), and (−2, −16).

Q2 — Classification using first derivative test

(a) y = x² − 6x + 5, x = 3:
Test x = 2: y′ = −2 < 0 (−). Test x = 4: y′ = 2 > 0 (+). Sign changes −→+. Local minimum at (3, −4).

(b) y = x³ − 12x:
x = −2: Test x = −3: y′ = 3(9)−12 = 15 > 0 (+). Test x = 0: y′ = −12 < 0 (−). Sign +→−. Local maximum at (−2, 16).
x = 2: Test x = 0: y′ = −12 < 0 (−). Test x = 3: y′ = 27−12 = 15 > 0 (+). Sign −→+. Local minimum at (2, −16).

(c) y = 2x³ − 3x²:
x = 0: Test x = −1: y′ = 6(1)+6 = 12 > 0 (+). Test x = 0.5: y′ = 6(0.25)−3 = −1.5 < 0 (−). Sign +→−. Local maximum at (0, 0).
x = 1: Test x = 0.5: y′ = −1.5 < 0 (−). Test x = 2: y′ = 24−12 = 12 > 0 (+). Sign −→+. Local minimum at (1, −1).

(d) y = x⁴ − 8x²:
x = −2: Test x = −3: y′ = 4(−27)−16(−3) = −108+48 = −60 < 0. Test x = −1: y′ = −4+16 = 12 > 0. Sign −→+. Local minimum at (−2, −16).
x = 0: Test x = −1: y′ = 12 > 0 (+). Test x = 1: y′ = 4−16 = −12 < 0 (−). Sign +→−. Local maximum at (0, 0).
x = 2: Test x = 1: y′ = −12 < 0 (−). Test x = 3: y′ = 108−48 = 60 > 0 (+). Sign −→+. Local minimum at (2, −16).

Q3 — Increasing and decreasing intervals

(a) y = x² − 6x + 5: Decreasing on (−∞, 3); increasing on (3, ∞).

(b) y = x³ − 12x: Increasing on (−∞, −2) ∪ (2, ∞); decreasing on (−2, 2).

(c) y = 2x³ − 3x²: Increasing on (−∞, 0) ∪ (1, ∞); decreasing on (0, 1).

(d) y = x⁴ − 8x²: Increasing on (−2, 0) ∪ (2, ∞); decreasing on (−∞, −2) ∪ (0, 2).

Q4 — Stationary points from f′(x) = x(x − 2)(x + 3)

Stationary points at f′ = 0: x = −3, x = 0, x = 2.

Sign chart (test values x = −4, −1, 1, 3):

x = −4: (−4)(−6)(−1) = −24 < 0 → −
x = −1: (−1)(−3)(2) = 6 > 0 → +
x = 1: (1)(−1)(4) = −4 < 0 → −
x = 3: (3)(1)(6) = 18 > 0 → +

x = −3: −→+ → local minimum.

x = 0: +→− → local maximum.

x = 2: −→+ → local minimum.

Q5 — f(x) = 2x³ − 9x² + 12x − 4

f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2)

Stationary points at x = 1 and x = 2.

f(1) = 2 − 9 + 12 − 4 = 1. f(2) = 16 − 36 + 24 − 4 = 0.

Sign test: f′(0) = 12 > 0; f′(1.5) = 6(0.5)(−0.5) = −1.5 < 0; f′(3) = 6(2)(1) = 12 > 0.

x = 1: +→− → local maximum, f(1) = 1.

x = 2: −→+ → local minimum, f(2) = 0.

Q6 — y = x³ + x always increasing

y′ = 3x² + 1.

Since x² ≥ 0 for all real x, we have 3x² ≥ 0, so y′ = 3x² + 1 ≥ 1 > 0 for all real x.

Because y′ > 0 everywhere, y′ is never zero and the function has no stationary points. Since y′ > 0 for all x, the function is always increasing. □

Q7 — f(x) = x⁴ − 4x³

(a) f′(x) = 4x³ − 12x² = 4x²(x − 3). f′ = 0 at x = 0 and x = 3.
f(0) = 0. f(3) = 81 − 108 = −27. Stationary points: (0, 0) and (3, −27).

(b) Sign chart:
x = −1: f′(−1) = 4(1)(−4) = −16 < 0 (−).
x = 1: f′(1) = 4(1)(−2) = −8 < 0 (−).
x = 4: f′(4) = 4(16)(1) = 64 > 0 (+).
x = 0: sign is − on both sides → stationary point of inflection at (0, 0).
x = 3: sign −→+ → local minimum at (3, −27).

(c) Increasing on (3, ∞); decreasing on (−∞, 0) ∪ (0, 3).

Q8 — f(x) = ax³ + bx² + 3x − 2, stationary point at x = 1, f(1) = 0

f′(x) = 3ax² + 2bx + 3.

Condition 1 — stationary point at x = 1:
f′(1) = 3a + 2b + 3 = 0 → 3a + 2b = −3 …(1)

Condition 2 — f(1) = 0:
a + b + 3 − 2 = 0 → a + b = −1 …(2)

From (2): a = −1 − b. Substitute into (1): 3(−1 − b) + 2b = −3 → −3 − 3b + 2b = −3 → −b = 0 → b = 0.

Then a = −1 − 0 = −1.

Answer: a = −1, b = 0.

Verify: f(x) = −x³ + 3x − 2. f(1) = −1 + 3 − 2 = 0 ✓. f′(1) = −3 + 3 = 0 ✓.

Q9 — f(x) = x³ − 3kx + 2: no stationary points

f′(x) = 3x² − 3k.

f has no stationary points when f′(x) = 0 has no real solutions.

3x² − 3k = 0 → x² = k. This has no real solutions when k < 0.

Answer: k < 0.

Note: When k = 0, f′(x) = 3x² = 0 only at x = 0, which is a stationary point of inflection. For no stationary points at all, we need k < 0.

Q10 — Maximum area A(x) = 3x² − x³⁄3

(a) A′(x) = 6x − x² = x(6 − x). Setting A′(x) = 0: x = 0 or x = 6.

(b) Sign chart: Test x = 3 (between 0 and 6): A′(3) = 3(6−3) = 9 > 0 (+). Test x = 7: A′(7) = 7(6−7) = −7 < 0 (−).
x = 0 is a minimum (or boundary); x = 6: sign changes +→− → local maximum. ✓

(c) A(6) = 3(36) − 216⁄3 = 108 − 72 = 36 km².