Answer
The absolute maximum is $f(\frac{2}{3}) = \frac{2\sqrt{3}}{9}$
The absolute minimum is $f(-1) = -\sqrt{2}$
$f(\frac{2}{3}) = \frac{2\sqrt{3}}{9}$ is a local maximum.
There is no local minimum.
Work Step by Step
$f(x) = x~\sqrt{1-x}$
Note that this function is continuous on the interval $[-1,1]$ and differentiable on the interval $(-1,1)$.
We can find the points where $f'(x) = 0$:
$f'(x) = \sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=0$
$\frac{2(1-x)}{2\sqrt{1-x}}-\frac{x}{2\sqrt{1-x}}=0$
$\frac{2-3x}{2\sqrt{1-x}}=0$
$2-3x=0$
$x = \frac{2}{3}$
We can verify the value of the function when $f'(x) = 0$ and at the endpoints of the interval $[-1,1]$:
$f(-1) = (-1)~\sqrt{1-(-1)} = -\sqrt{2}$
$f(\frac{2}{3}) = (\frac{2}{3})~\sqrt{1-(\frac{2}{3})} = \frac{2\sqrt{3}}{9}$
$f(1) = (1)~\sqrt{1-(1)} = 0$
The absolute maximum is $f(\frac{2}{3}) = \frac{2\sqrt{3}}{9}$
The absolute minimum is $f(-1) = -\sqrt{2}$
$f(\frac{2}{3}) = \frac{2\sqrt{3}}{9}$ is a local maximum.
There is no local minimum.
