Provide an appropriate response.
-Find the absolute maximum and minimum values of $f ( x ) = \ln ( \sin x )$ on $\left[ \frac { \pi } { 6 } , \frac { 2 \pi } { 3 } \right]$.
A) Maximum $= 0$ at $x = 0$, minimum $= - \ln 2$ at $x = \frac { \pi } { 6 }$
B) Maximum $= 0$ at $x = \frac { \pi } { 2 }$, minimum $= \ln \frac { \sqrt { 3 } } { 2 }$ at $x = \frac { 2 \pi } { 3 }$
C) Maximum $= 0$ at $x = \frac { \pi } { 2 }$, minimum $= - \ln 2$ at $x = \frac { \pi } { 6 }$
D) Maximum $= 0$ at $x = \frac { \pi } { 2 }$, minimum $= - \ln 2$ at $x = \frac { 2 \pi } { 3 }$

Question

Quizplus · Accepted Answer

The maximum value of $f(x) = \ln(\sin x)$ occurs when $\sin x$ is at its maximum on the given interval, which is 1 at $x = \frac{\pi}{2}$, making $f(x) = \ln(1) = 0$. The minimum value occurs at the endpoints of the interval since $f(x)$ is continuous and differentiable on $\left[\frac{\pi}{6}, \frac{2\pi}{3}\right]$. At $x = \frac{\pi}{6}$, $\sin x = \frac{1}{2}$, so $f(x) = \ln\left(\frac{1}{2}\right) = -\ln(2)$. At $x = \frac{2\pi}{3}$, $\sin x = \frac{\sqrt{3}}{2}$, which gives a higher $f(x)$ value than at $x = \frac{\pi}{6}$. Therefore, the minimum is $-\ln(2)$ at $x = \frac{\pi}{6}$.

Provide an Appropriate Response f(x)=ln⁡(sin⁡x)f ( x ) = \ln ( \sin x )f(x)=ln(sinx)

Provide an Appropriate Response $f ( x ) = \ln ( \sin x )$