Question

`f(x)=sin2x-x, -pi/2<=x<=pi/2`

Answer 1

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

\[ \Rightarrow f'\left( x \right) = 2 \cos 2x - 1\]

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

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

\[ \Rightarrow 2 \cos 2x - 1 = 0\]

\[ \Rightarrow \cos 2x = \frac{1}{2}\]

\[ \Rightarrow x = \frac{- \pi}{6} or \frac{\pi}{6}\]

Since  f'(x) changes from positive to negative when x increases through \[\frac{\pi}{6}\] x = \[\frac{\pi}{6}\] is the point of local maxima.

The local maximum value of  f (x) at x = \[\frac{\pi}{6}\] is given by \[\sin \left( \frac{\pi}{3} \right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6}\]

 

Sincef '(x) changes from negative to positive when x increases through \[- \frac{\pi}{6}\] x = \[- \frac{\pi}{6}\]  is the point of local minima.
The local minimum value of  f (x)  at x = \[- \frac{\pi}{6}\] is given by \[\sin \left( \frac{- \pi}{3} \right) + \frac{\pi}{6} = \frac{\pi}{6} - \frac{\sqrt{3}}{2}\]