Find the absolute extreme values of the function on the interval.
-$$F ( x ) = \sqrt [ 3 ] { x } , - 2 \leq x \leq 64$$
A) absolute maximum is 0 at $x = 0$; absolute minimum is 4 at $x = 64$
B) absolute maximum is 4 at $x = 64$; absolute minimum is $- 4$ at $x = - 64$
C) absolute maximum is 4 at $x = - 64$; absolute minimum is 0 at $x = 0$
D) absolute maximum is 4 at $x = 64$; absolute minimum is 0 at $x = 0$

Question

Quizplus · Accepted Answer

To find the absolute extreme values, we need to find the critical values where either the function attains its maximum or minimum.

First, we check the endpoints of the interval:

$$F(-2) = \sqrt[3]{-2} \approx -1.259$$ 
$$F(64) = \sqrt[3]{64} = 4$$

Next, we find where the derivative of the function is zero or undefined:

$$F'(x) = \frac{1}{3\sqrt[3]{x^2}}$$

This derivative is undefined at x = 0, but since 0 is not in the interval [-2, 64], we don't consider it.

Setting the derivative to zero:

$$\frac{1}{3\sqrt[3]{x^2}} = 0$$

This equation has no real solutions.

Therefore, the absolute maximum of the function occurs at x = 64, where F(x) = 4, and the absolute minimum occurs at x = -2, where F(x) ≈ -1.259.

Therefore, the answer is choice D.

Find the Absolute Extreme Values of the Function on the Interval