Questions: Absolute Extrema and the Closed Interval Method
4 questions to test your understanding
Score: 0 / 4
Question 1 Multiple Choice
Find the absolute maximum value of f(x) = x³ - 3x on [0, 2].
A-2
B0
C2
D8
f'(x) = 3x² - 3 = 0 gives x = ±1, and only x = 1 lies in [0, 2]. The candidates are f(0) = 0, f(1) = -2 and f(2) = 2. The largest is 2, at the endpoint x = 2. The value -2 is the absolute minimum.
Question 2 True / False
If f is continuous on [a, b], its absolute maximum on [a, b] must occur at a point where f'(x) = 0.
TTrue
FFalse
Answer: False
The maximum can occur at an endpoint, or at an interior point where f' does not exist. In the previous question the maximum occurs at the endpoint x = 2, where f'(2) = 9 is not zero.
Question 3 Short Answer
Give an example of a function on an interval that has no absolute maximum, and say which hypothesis of the Extreme Value Theorem fails.
Think about your answer, then reveal below.
Model answer: f(x) = x on the open interval (0, 1) has no absolute maximum; the interval is not closed.
The values of f get arbitrarily close to 1 but never reach it, because 1 is not in the domain. Other examples: f(x) = 1/x on (0, 1], which is unbounded, or a function with a jump discontinuity on a closed interval.
Question 4 Multiple Choice
What are the candidates for absolute extrema of a continuous function f on [a, b]?
AOnly the points where f'(x) = 0
BOnly the endpoints a and b
CThe endpoints and the critical numbers of f in (a, b)
DThe inflection points of f
By Fermat's theorem, an interior extremum occurs only at a critical number, a point where f' is zero or undefined. The endpoints must be checked separately because Fermat's theorem does not apply there.