Answer
$g(x)$ has the absolute maximum value of $\frac{2\sqrt3}{3}$ at $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ and the absolute minimum value of $1$ at $x=\frac{\pi}{2}$ in the defined domain.
Work Step by Step
$$g(x)=\csc x\hspace{1cm}\frac{\pi}{3}\le\theta\le\frac{2\pi}{3}$$
1) Find the derivative $g'(x)$: $$g'(x)=-\csc x\cot x=-\frac{\cot x}{\sin x}$$
- $g'(x)=0$ when $\cot x=0$, which is when $x=\pi/2+k\pi$ for $k\in Z$.
- $g'(x)$ is undefined when $\sin x=0$, which is when $x=k\pi$ $(k\in Z)$.
All these values of $x$ are the critical points of $g(x)$.
However, among all these values, only $x=\pi/2$ lying in the defined domain $[\pi/3,2\pi/3]$. So only the critical point $x=\pi/2$ would be taken into consideration here.
2) Evaluate $g'(x)$ at the critical points and endpoints:
- For $x=\pi/3$: $$g\Big(\frac{\pi}{3}\Big)=\csc\Big(\frac{\pi}{3}\Big)=\frac{1}{\sin\frac{\pi}{3}}=\frac{1}{\frac{\sqrt3}{2}}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}$$
- For $x=\pi/2$: $$g\Big(\frac{\pi}{2}\Big)=\csc\Big(\frac{\pi}{2}\Big)=\frac{1}{\sin\frac{\pi}{2}}=\frac{1}{1}=1$$
- For $x=2\pi/3$: $$g\Big(\frac{2\pi}{3}\Big)=\csc\Big(\frac{2\pi}{3}\Big)=\frac{1}{\sin\frac{2\pi}{3}}=\frac{1}{\frac{\sqrt3}{2}}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}$$
This means $g(x)$ has the absolute maximum value of $\frac{2\sqrt3}{3}$ at $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ and the absolute minimum value of $1$ at $x=\frac{\pi}{2}$ in the defined domain.