Find the absolute extreme values of the function on the interval.
-$$F ( x ) = - \frac { 3 } { x ^ { 2 } } , 0.5 \leq x \leq 3$$
A) absolute maximum is $- \frac { 1 } { 3 }$ at $x = \frac { 1 } { 2 }$; absolute minimum is $- 12$ at $x = - 3$
B) absolute maximum is $- \frac { 1 } { 3 }$ at $x = 3$; absolute minimum is $- 12$ at $x = \frac { 1 } { 2 }$
C) absolute maximum is $\frac { 1 } { 3 }$ at $x = \frac { 1 } { 2 }$; absolute minimum is $- 12$ at $x = 3$
D) absolute maximum is $- \frac { 1 } { 3 }$ at $x = 3$; absolute minimum is $- 12$ at $x = - \frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

To find the absolute extreme values, we first take the derivative of the function.
$$F'(x)=\frac{6}{x^3}$$
The critical points are where the derivative is zero or undefined. Setting the derivative to zero, we get $x=\sqrt[3]{6}$ which is not in the interval. Therefore, there are no critical points inside the interval. Checking the endpoints, we get:

$$F(0.5)=-12$$

$$F(3)=-\frac{1}{3}$$

Therefore, the absolute minimum is $-12$ at $x=\frac{1}{2}$ and the absolute maximum is $-\frac{1}{3}$ at $x=3$.

Thus, the best choice is B.

Find the Absolute Extreme Values of the Function on the Interval