Answer
(a) The maximum value is approximately $0.30$
The minimum value is $0$
(b) The maximum value is $0.32$
The minimum value is $0$
Work Step by Step
(a) $f(x) = x~\sqrt{x-x^2}$
We can use the zoom function on a graphing calculator to estimate the maximum and minimum values of the function in the interval $[0,1]$ where the function is defined.
The maximum value is approximately $0.30$
The minimum value is $0$
(b) $f(x) = x~\sqrt{x-x^2}$
$f'(x) = \sqrt{x-x^2}+(x)~(\frac{1-2x}{2\sqrt{x-x^2}})$
$f'(x) = \frac{2x-2x^2}{2\sqrt{x-x^2}}+\frac{x-2x^2}{2\sqrt{x-x^2}}$
$f'(x) = \frac{3x-4x^2}{2\sqrt{x-x^2}} = 0$
$3x-4x^2 = 0$
$x(3-4x) = 0$
$x = 0~~$ or $~~x = \frac{3}{4}$
When $x = 0$:
$f(0) = (0)~\sqrt{(0)-(0)^2} = 0$
The minimum value is $0$
When $x = \frac{3}{4}$:
$f(x) = (\frac{3}{4})~\sqrt{(\frac{3}{4})-(\frac{3}{4})^2} = 0.32$
The maximum value is $0.32$