Find the absolute extreme values of the function on the interval.
-$$F ( x ) = \sqrt [ 3 ] { x } , - 3 \leq x \leq 8$$
A) absolute maximum is 2 at $x = 8$; absolute minimum is $- 2$ at $x = - 8$
B) absolute maximum is 2 at $x = - 8$; absolute minimum is 0 at $x = 0$
C) absolute maximum is 0 at $x = 0$; absolute minimum is 2 at $x = 8$
D) absolute maximum is 2 at $x = 8$; absolute minimum is 0 at $x = 0$

Question

Quizplus · Accepted Answer

To find the absolute extreme values of the function, we need to find the critical points within the given interval and evaluate the function at these points as well as at the endpoints of the interval.
Taking the first derivative, we get
$$F'(x)=\frac{1}{3\sqrt[3]{x^2}}$$
The critical point is at $x=0$ where $F'(x)$ is not defined. Evaluating the function at the critical point and endpoints of the interval, we get
$$F(-3)=\sqrt[3]{-3} = -\sqrt[3]{3}\approx -1.442,$$
$$F(0)\text{ is undefined,}$$
$$F(8)=\sqrt[3]{8}=2.$$
Therefore, the absolute maximum value is 2 at $x=8$ and the absolute minimum value is 0 at $x=0$. The correct choice is D.

Find the Absolute Extreme Values of the Function on the Interval