Find the local maximum and local minimum values for the function $f(x) = x\sqrt{1 - x}$ where $0 < x < 1$.

(N/A) Given $f(x) = x\sqrt{1 - x}$ for $0 < x < 1$.
Using the product rule,$f'(x) = (1)\sqrt{1 - x} + x \cdot \frac{1}{2\sqrt{1 - x}}(-1) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}$.
Simplifying,$f'(x) = \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}}$.
Setting $f'(x) = 0$,we get $2 - 3x = 0$,which implies $x = \frac{2}{3}$.
Now,find the second derivative $f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2(1 - x)^{1/2}} \right)$.
Using the quotient rule,$f''(x) = \frac{1}{2} \left[ \frac{-3(1 - x)^{1/2} - (2 - 3x) \cdot \frac{1}{2}(1 - x)^{-1/2}(-1)}{1 - x} \right] = \frac{-6(1 - x) + (2 - 3x)}{4(1 - x)^{3/2}} = \frac{3x - 4}{4(1 - x)^{3/2}}$.
At $x = \frac{2}{3}$,$f''\left(\frac{2}{3}\right) = \frac{3(2/3) - 4}{4(1 - 2/3)^{3/2}} = \frac{2 - 4}{4(1/3)^{3/2}} = \frac{-2}{4(1/3)^{3/2}} < 0$.
Since $f''\left(\frac{2}{3}\right) < 0$,$x = \frac{2}{3}$ is a point of local maxima.
The local maximum value is $f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.
There is no local minimum value in the interval $(0, 1)$.