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प्रश्न
f(x) = sin x \[-\] cos x, 0 < x < 2\[\pi\] .
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उत्तर
\[\text { Given: } \hspace{0.167em} f\left( x \right) = \sin x - \cos x\]
\[ \Rightarrow f'\left( x \right) = \cos x + \sin x\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \cos x + \sin x = 0\]
\[ \Rightarrow \cos x = - \sin x\]
\[ \Rightarrow \tan x = - 1\]
\[ \Rightarrow x = \frac{3\pi}{4} or \frac{7\pi}{4}\]
Sincef '(x) changes from positive to negative when x increases through \[\frac{3\pi}{4}\], x = \[\frac{3\pi}{4}\] is the point of local maxima.
The local maximum value of f (x) at x = \[\frac{3\pi}{4}\] is given by \[\sin\left( \frac{3\pi}{4} \right) - \cos\left(\frac{3\pi}{4} \right) = \sqrt{2}\]
Since f '(x) changes from negative to positive when x increases through \[\frac{7\pi}{4}\],x= \[\frac{7\pi}{4}\] is the point of local minima.
The local minimum value of f (x) at x = \[\frac{7\pi}{4}\] is given by \[\sin\left( \frac{7\pi}{4} \right) - \cos\left(\frac{7\pi}{4} \right) = - \sqrt{2}\]
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