Use differentiation to solve problems such as proving “f(x) ≥ 0” and finding minimum values.
Example ① (Proof of an Inequality)
Show that x³ + 2 > 3x when x > 0.
Answer:Let f(x) = x³ − 3x + 2. f'(x) = 3x²−3 = 3(x+1)(x−1).
For x > 0, the sign of f'(x) is as follows: f'<0 for 0 < x < 1 (decreasing), and f'>0 for x > 1 (increasing).
f(1) = 1−3+2 = 0 is the minimum.
For x > 0, f(x) ≥ 0, with equality at x = 1. Therefore x³+2 ≥ 3x. (Except for equality at x=1, x³+2 > 3x... equality holds at x=1.)
Thus, x > 0, x ≠ 1 gives x³+2 > 3x, while equality holds at x=1. If the problem states “>”, add: the inequality “>” does not hold when x=1 because equality holds. Correction: x³+2 ≥ 3x (x > 0) is precise.
Show that x³ + 2 > 3x when x > 0.
Answer:Let f(x) = x³ − 3x + 2. f'(x) = 3x²−3 = 3(x+1)(x−1).
For x > 0, the sign of f'(x) is as follows: f'<0 for 0 < x < 1 (decreasing), and f'>0 for x > 1 (increasing).
f(1) = 1−3+2 = 0 is the minimum.
For x > 0, f(x) ≥ 0, with equality at x = 1. Therefore x³+2 ≥ 3x. (Except for equality at x=1, x³+2 > 3x... equality holds at x=1.)
Thus, x > 0, x ≠ 1 gives x³+2 > 3x, while equality holds at x=1. If the problem states “>”, add: the inequality “>” does not hold when x=1 because equality holds. Correction: x³+2 ≥ 3x (x > 0) is precise.
Example ② (Application of Maximum and Minimum)
For a cylinder with base radius r and height h, when r + h = 6 (constant), find the maximum value of the volume V = πr²h.
Answer:h = 6−r (0 < r < 6). V = πr²(6−r) = π(6r²−r³).
dV/dr = π(12r−3r²) = 3πr(4−r) = 0 → r = 4 (0 < r < 6).
The maximum occurs at r = 4. V = π × 16 × 2 = 32π
For a cylinder with base radius r and height h, when r + h = 6 (constant), find the maximum value of the volume V = πr²h.
Answer:h = 6−r (0 < r < 6). V = πr²(6−r) = π(6r²−r³).
dV/dr = π(12r−3r²) = 3πr(4−r) = 0 → r = 4 (0 < r < 6).
The maximum occurs at r = 4. V = π × 16 × 2 = 32π
Key Points
- Let g(x) = f(x)−k and find the minimum value of g(x) ≥ 0
- For conditional maximum and minimum problems: reduce the number of variables to one and differentiate
Practice Problems
- Show that x³ ≥ 3x − 2 when x ≥ 0.
- When a + b = 10 (a > 0, b > 0), find the maximum value of ab².
- Using differentiation, show that x + 4/x ≥ 4 when x > 0 (or find the minimum value).
Answers and Explanations
- Answer (Outline):f(x)=x³−3x+2. f'=3(x−1)(x+1). For x≥0, x=1 is the local and absolute minimum: f(1)=0≥0. Therefore x³≥3x−2. Equality holds at x=1.
Explanation: f(0)=2>0, f(1)=0 (minimum). - Answer:Maximum value 2000/3 (when a=10/3, b=20/3)
Explanation: b=10−a. f(a)=a(10−a)². f'=(10−a)²+a×2(10−a)(−1)=(10−a)(10−3a)=0. a=10/3. - Answer:f(x)=x+4/x. f'=1−4/x²=0→x=2 (x>0). f(2)=4 is the minimum value. Therefore x+4/x≥4.
Explanation: The local minimum at x=2 is also the absolute minimum. The same conclusion can be reached using the arithmetic-geometric mean inequality.