Module 5 Practice Problems: Applications of Derivatives II
Instructions: Work through each problem on paper first, then reveal the solution.
Problem 1
Find the intervals where f(x) = x³ − 6x² + 9x + 1 is increasing and decreasing.
Solution
f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). f' > 0 on (−∞, 1) and (3, ∞): increasing. f' < 0 on (1, 3): decreasing.
Problem 2
Use the Second Derivative Test to classify the critical points of f(x) = x³ − 3x.
Solution
f'(x) = 3x² − 3, crit pts x = ±1. f''(x) = 6x. f''(1) = 6 > 0: local min. f''(−1) = −6 < 0: local max.
Problem 3
Find the inflection points of f(x) = x4 − 6x² + 8x.
Solution
f''(x) = 12x² − 12 = 12(x − 1)(x + 1) = 0 at x = ±1. Sign changes at both. Inflection points: (−1, 1) and (1, 3).
Problem 4
Sketch f(x) = x/(x − 1). List domain, asymptotes, intercepts, and monotonicity.
Solution
Domain: x ≠ 1. VA: x = 1. HA: y = 1. x-int: (0, 0). y-int: (0, 0). f'(x) = −1/(x − 1)² < 0 always: decreasing on (−∞, 1) and (1, ∞). No extrema.
Problem 5
Find two positive numbers that sum to 50 and have the largest possible product.
Solution
P = x(50 − x). P' = 50 − 2x = 0, x = 25. Both numbers: 25 and 25. Max product = 625.
Problem 6
An open-top box from a 20×20 sheet. Find the cut size x that maximizes volume.
Solution
V = x(20 − 2x)². V' = (20 − 2x)(20 − 6x) = 0. x = 10/3 (x = 10 out of domain). Max volume: (10/3)(40/3)² = 16000/27 ≈ 592.6 cubic units.
Problem 7
Find where f(x) = x2/3(x − 5) is concave up.
Solution
f'(x) = (5/3)x2/3 − (10/3)x−1/3 = (5x − 10)/(3x1/3). f''(x) = (10x + 10)/(9x4/3). f'' > 0 when x > 0 or x < −1. Concave up on (−∞, −1) and (0, ∞).
Problem 8
Find the point on y = x² closest to (0, 3).
Solution
Minimize D² = x² + (x² − 3)² = x4 − 5x² + 9. d(D²)/dx = 4x³ − 10x = 2x(2x² − 5) = 0. x = 0 or x = ±√(5/2). D²(0) = 9, D²(√(5/2)) = 11/4. Closest: (±√(5/2), 5/2).
Problem 9
Classify all critical points of f(x) = 3x5 − 5x³.
Solution
f'(x) = 15x4 − 15x² = 15x²(x² − 1). Crit pts: x = 0, ±1. FDT: f' changes + to − at x = −1 (local max), − to + at x = 1 (local min), no change at x = 0 (neither).
Problem 10
A rectangle has perimeter 40. What dimensions maximize the area?
Solution
2l + 2w = 40, so l = 20 − w. A = w(20 − w). A' = 20 − 2w = 0, w = 10, l = 10. Square with side 10. Max area = 100.