Question

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

$f\left(x\right)=x\sqrt{1-x},x>0$

Answer 1

Given function is, $f\left(x\right)=x\sqrt{1-x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x(-1)}{2\sqrt{1-x}}+\sqrt{1-x}=\frac{-x+2(1-x)}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}f"\left(x\right)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3x)\left(\frac{-1}{2\sqrt{1-x}}\right)}{(1-x)}\right]=\frac{\sqrt{1-x}(-3)-(2-3x)\left(\frac{-1}{2\sqrt{1-x}}\right)}{2(1-x)}=\frac{-6(1-x)+(2-3x)}{4(1-x{)}^{\frac{3}{2}}}=\frac{3x-4}{4(1-x{)}^{\frac{3}{2}}}$
For maxima or minima put f'(x)=0
$\frac{2-3x}{2\sqrt{1-x}}=0\Rightarrow 2-3x=0\Rightarrow x=\frac{2}{3}Atx=\frac{2}{3},f"\left(\frac{2}{3}\right)=\frac{3\left(\frac{2}{3}\right)-4}{4{(1-\frac{2}{3})}^{\frac{3}{2}}}=\frac{2-4}{4{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}=\frac{-1}{2{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}<0$
$\therefore x=\frac{2}{3}$ is a point of maxima.
Maximum value $=f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3}\times \sqrt{\frac{1}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}$