Question

1. find the absolute maximum and minimum values of ( f(x)=csc x ), ( left\frac{pi}{3}, \frac{2 pi}{3}

ight ).

Explanation:

Step1: Find the derivative of \(y = \csc x\)

The derivative of \(y=\csc x= \frac{1}{\sin x}\). Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) where \(u = 1\), \(u^\prime=0\), \(v=\sin x\), \(v^\prime=\cos x\). So \(y^\prime=-\frac{\cos x}{\sin^{2}x}\)

Step2: Find the critical points

Set \(y^\prime = 0\), then \(-\frac{\cos x}{\sin^{2}x}=0\). Since \(\sin^{2}x
eq0\) for \(x\in[\frac{\pi}{3},\frac{2\pi}{3}]\), we solve \(\cos x = 0\). In the interval \([\frac{\pi}{3},\frac{2\pi}{3}]\), \(x = \frac{\pi}{2}\)

Step3: Evaluate the function at the critical point and endpoints

• When \(x=\frac{\pi}{3}\), \(f(\frac{\pi}{3})=\csc(\frac{\pi}{3})=\frac{2\sqrt{3}}{3}\)
• When \(x=\frac{\pi}{2}\), \(f(\frac{\pi}{2})=\csc(\frac{\pi}{2}) = 1\)
• When \(x=\frac{2\pi}{3}\), \(f(\frac{2\pi}{3})=\csc(\frac{2\pi}{3})=\frac{2\sqrt{3}}{3}\)

Answer:

The absolute maximum value is \(\frac{2\sqrt{3}}{3}\) and the absolute minimum value is \(1\)