Question

`f(x)=xsqrt(1-x),  x<=1` .

Answer 1

\[\text { Given }: f\left( x \right) = x\sqrt{1 - x}\]

\[ \Rightarrow f'\left( x \right) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}\]

\[\text { For the local maxima or minima, we must have }\]

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

\[ \Rightarrow \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = 0\]

\[ \Rightarrow \sqrt{1 - x} = \frac{x}{2\sqrt{1 - x}}\]

\[ \Rightarrow 2 - 2x = x\]

\[ \Rightarrow 3x = 2\]

\[ \Rightarrow x = \frac{2}{3} \]

\[\text { Thus, x } = \frac{2}{3} \text { is the possible point of local maxima or local minima }. \]

\[\text { Now }, \]

\[ f''\left( x \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left( \frac{\sqrt{1 - x} + \frac{x}{2\sqrt{1 - x}}}{\left( 1 - x \right)} \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left[ \frac{2 - x}{\left( 1 - x \right)\sqrt{1 - x}} \right]\]

\[\text { At  }x = \frac{2}{3}: \]

\[ f''\left( \frac{2}{3} \right) = \frac{- 1}{\sqrt{1 - \frac{2}{3}}} - \frac{1}{2}\left[ \frac{2 - \frac{2}{3}}{\left( 1 - \frac{2}{3} \right)\sqrt{1 - \frac{2}{3}}} \right] = - \sqrt{3} - \frac{\frac{4}{3}}{\frac{1}{3 \times \sqrt{3}}} = - \sqrt{3} - 4\sqrt{3} < 0\]

\[\text { So,} x = \frac{2}{3}\text {  is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( \frac{2}{3} \right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}}\]