This section provides comprehensive practice with the definition and formulas of differentiation, tangent lines, and increase and decrease.
Example ① (Maximum and Minimum)
Find the maximum and minimum values of f(x) = x³ − 6x² + 9x + 1 for 0 ≤ x ≤ 4.
Answer:f'(x) = 3(x−1)(x−3). Local maximum f(1) = 5, local minimum f(3) = 1.
Endpoints: f(0) = 1, f(4) = 5. Maximum value 5 (x=1,4), minimum value 1 (x=0,3)
Find the maximum and minimum values of f(x) = x³ − 6x² + 9x + 1 for 0 ≤ x ≤ 4.
Answer:f'(x) = 3(x−1)(x−3). Local maximum f(1) = 5, local minimum f(3) = 1.
Endpoints: f(0) = 1, f(4) = 5. Maximum value 5 (x=1,4), minimum value 1 (x=0,3)
Example ②
If f(x) = ax³ + bx has an extreme value of −2 at x = 1, find a and b.
Answer:f'(x) = 3ax² + b. f'(1) = 0 → 3a + b = 0. f(1) = a + b = −2.
Simultaneous equations: 3a + b = 0, a + b = −2 → 2a = 2 → a = 1, b = −3.
If f(x) = ax³ + bx has an extreme value of −2 at x = 1, find a and b.
Answer:f'(x) = 3ax² + b. f'(1) = 0 → 3a + b = 0. f(1) = a + b = −2.
Simultaneous equations: 3a + b = 0, a + b = −2 → 2a = 2 → a = 1, b = −3.
Review
- Definition of the derivative: lim(f(x+h)−f(x))/h
- Basic formula: (xⁿ)' = nxⁿ⁻¹
- Tangent line: y−f(a) = f'(a)(x−a)
- Maximum and minimum on a closed interval: compare the extreme values and endpoint values
Practice Problems
- Find the maximum and minimum values of f(x) = −x³ + 3x² + 1 for −1 ≤ x ≤ 3.
- Find the local maximum and local minimum values of y = x³ − 3x + 1, and describe the general shape of its graph.
- If f(x) = x³ + ax + b has an extreme value of −5 at x = 2, find a and b.
Answers and Explanations
- Answer:f'=−3x²+6x=−3x(x−2). Local maximum f(2)=5, local minimum f(0)=1. Endpoint values f(−1)=−3, f(3)=1. Maximum value 5 (x=2), minimum value −3 (x=−1)
Explanation: Compare all extreme values and endpoint values in the closed interval. - Answer:f'=3x²−3=3(x+1)(x−1). Local maximum f(−1)=3, local minimum f(1)=−1
Explanation: The graph has a peak at x=−1 and a valley at x=1. As x→±∞, y→±∞. - Answer:a = −12, b = 11
Explanation: f'=3x²+a. f'(2)=12+a=0→a=−12. f(2)=8+2a+b=8−24+b=−5→b=11.