Find the absolute extreme values of the function on the interval.
-$$f ( x ) = \tan x , - \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }$$
A) absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$ and $- \frac { \pi } { 6 }$; absolute minimum does not exist
B) absolute maximum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $\frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$
C) absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { 2 \pi } { 18 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 12 }$
D) absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$

Question

Quizplus · Accepted Answer

To find the absolute extreme values of the function, we need to find the critical points and the endpoints of the interval.

Critical points occur where the derivative is either zero or undefined. Since $f(x)=	an x$, we have $f'(x)=\sec^2 x$, which is never zero. The derivative is also undefined at $x=\frac{\pi}{2}$ and $x=-\frac{\pi}{2}$, but these are outside the given interval.

Therefore, the only candidates for the absolute extreme values occur at the endpoints, which are $-\frac{\pi}{6}$ and $\frac{\pi}{6}$. We can evaluate the function at these points to see which one gives the maximum and which one gives the minimum:

$f(-\frac{\pi}{6})=	an (-\frac{\pi}{6})=-\frac{\sqrt{3}}{3}$

$f(\frac{\pi}{6})=	an (\frac{\pi}{6})=\frac{\sqrt{3}}{3}$

Thus, the absolute maximum value of the function is $\frac{\sqrt{3}}{3}$ at $x=\frac{\pi}{6}$, and the absolute minimum value does not exist. Therefore, the answer is D.

Find the Absolute Extreme Values of the Function on the Interval