Question

f(x) =\[x\sqrt{1 - x} , x > 0\].

Answer 1

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

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

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

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

\[ \Rightarrow \frac{2 - 3x}{2\sqrt{1 - x}} = 0\]

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

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 \[\frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}\]