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प्रश्न
`f(x)=2sinx-x, -pi/2<=x<=pi/2`
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उत्तर
\[\text { Given: } \hspace{0.167em} f\left( x \right) = 2 \sin x - x\]
\[ \Rightarrow f'\left( x \right) = 2 \cos x - 1\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 2 \cos x - 1 = 0\]
\[ \Rightarrow \cos x = \frac{1}{2}\]
\[ \Rightarrow x = \frac{\pi}{3} or \frac{- \pi}{3}\]
Sincef '(x) changes from positive to negative when x increases through \[\frac{\pi}{3}\] x = \[\frac{\pi}{3}\] is the point of local maxima.
The local maximum value of f (x) at x = \[\frac{\pi}{3}\] is given by \[2 \sin \left( \frac{\pi}{3} \right) - \frac{\pi}{3} = \sqrt{3} - \frac{\pi}{3}\]
Since f '(x) changes from negative to positive when x increases through \[- \frac{\pi}{3}\] x = \[- \frac{\pi}{3}\] is the point of local minima.
The local minimum value of f (x) at x = \[- \frac{\pi}{3}\] is given by \[2 \sin \left( \frac{- \pi}{3} \right) + \frac{\pi}{3} = \frac{\pi}{3} - \sqrt{3}\]
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