Question

Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:

f(x) ${\displaystyle =x\sqrt{1-x},0<x<1}$

Answer 1

Given function f(x) ${\displaystyle =x\sqrt{1-x},0<x<1}$ ....(1)

${\displaystyle \therefore f\prime \left(x\right)=1.\sqrt{1-x}+\frac{1}{2\sqrt{1-x}}(-1)\cdot x}$

${\displaystyle =\frac{2(1-x)-x}{2\sqrt{1-x}}}$

${\displaystyle =\frac{2-3x}{2\sqrt{1-x}}}$

If f'(x) = 0 then ${\displaystyle \frac{2-3x}{2\sqrt{1-x}}=0,}$

${\displaystyle \therefore x=\frac{2}{3}}$

At ${\displaystyle x=\frac{2}{3}}$, the sign changes from positive to negative when x passes through x ${\displaystyle =\frac{2}{3}}$

${\displaystyle \therefore }$ There is a local maximum at the point f

Thus, the local maximum value is f(x) = f ${\displaystyle \left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3}\sqrt{\frac{1}{\sqrt{3}}}}$            ... [Substituting equation (1), x = ${\displaystyle \frac{2}{3},}$ in (1)]

${\displaystyle =\frac{2}{2\sqrt{3}}=\frac{2\sqrt{3}}{9}}$