Fix the mistake: Applications of Differentiation ② (Maximum and Minimum Values, Velocity and Acceleration, Approximat
Problems on Maximum and Minimum Values
A function continuous on the closed interval [a, b] always has a maximum and a minimum value (the extreme value theorem). The candidates are
① Extreme values (points where f'(x)=0)
② The endpoints of the interval, x=a and x=b
Example ① Find the maximum and minimum of f(x) = x³ − 3x on [−2, 2].
Solution:f'=3x²−3=0 ⇒ x=±1。
f(−2)=−2、f(−1)=2（local maximum）、f(1)=−2（local minimum）、f(2)=2。
Maximum value 2 (x=−1, 2), minimum value −2 (x=−2, 1)
Solution:f'=3x²−3=0 ⇒ x=±1。
f(−2)=−2、f(−1)=2（local maximum）、f(1)=−2（local minimum）、f(2)=2。
Maximum value 2 (x=−1, 2), minimum value −2 (x=−2, 1)
Key Point
- Compare the values at the candidate points and the endpoints
- Pay attention to open intervals and the domain
Practice Problems
- Find the maximum value of f(x) = x·e^{−x} for x ≧ 0。
Solutions and Explanations
- Solution:1/e at x=1
f'=(1−x)e^{−x}。
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