Question

f(x) = sin 2x, 0 < x < \[\pi\] .

Answer 1

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \sin 2x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos 2x\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 2 \cos 2x = 0\]

\[ \Rightarrow \cos 2x = 0\]

\[ \Rightarrow x = \frac{\pi}{4} or \frac{3\pi}{4}\] 

Sincef '(x) changes from positive to negative when x increases through \[\frac{\pi}{4}\], x = \[\frac{\pi}{4}\] is the point of maxima.
The local maximum value of  f (x) at x = \[\frac{\pi}{4}\] is given by \[\sin\left( \frac{\pi}{2} \right) = 1\] 

Sincef '(x) changes from negative to positive when x increases through

\[\frac{3\pi}{4}\] x = \[\frac{3\pi}{4}\] is the point of minima.
The local minimum value of  f (x) at x = \[\frac{3\pi}{4}\] is given by \[\sin\left( \frac{3\pi}{2} \right) = - 1\]