Identify the function's extreme values in the given domain, and say where they are assumed. Tell which of the extreme
values, if any, are absolute.
-$$f ( x ) = \sqrt { 64 - x ^ { 2 } } , - 8 \leq x

Question

Identify the function's extreme values in the given domain, and say where they are assumed. Tell which of the extreme
values, if any, are absolute.
-$$f ( x ) = \sqrt { 64 - x ^ { 2 } } , - 8 \leq x < 8$$
A) no local extrema; no absolute extrema
B) local and absolute minimum: 0 at x = -8; local and absolute maximum: 8 at x = 0
C) local and absolute minimum: 0 at x = -8 and x = 8; local and absolute maximum: 8 at x = 0
D) local and absolute maximum: 0 at x = -8; local and absolute minimum: 8 at x = 0

Quizplus · Accepted Answer

The function is defined on the interval [-8, 8), and its derivative is $$f'(x) = \frac{-x}{\sqrt{64-x^2}}$$which is negative for x < 0 and positive for x > 0. Therefore, the function has a local minimum at x = -8 and a local maximum at x = 0. Since the function is defined on a closed interval, these local extrema are also absolute extrema.

Identify the Function's Extreme Values in the Given Domain, and Say