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प्रश्न
f(x) =\[x\sqrt{1 - x} , x > 0\].
उत्तर
\[\text { Given }: f\left( x \right) = x\sqrt{1 - x}\]
\[ \Rightarrow f'\left( x \right) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}}\]
\[\text { For the local maxima or minima, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{2 - 3x}{2\sqrt{1 - x}} = 0\]
\[ \Rightarrow x = \frac{2}{3}\]
Since, f '(x) changes from positive to negative when x increases through \[\frac{2}{3}\], x = \[\frac{2}{3}\] is a point of maxima.
The local maximum value of f (x) at x = \[\frac{2}{3}\] is given by \[\frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}\]
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