Question

Find the local maxima and local minima for the given function and also find the local maximum and local minimum values$f\left(x\right)=x\sqrt{1-x},0<x<1$

Answer 1

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

$f^{{}^{\prime }}\left(x\right)=\frac{-x}{2\sqrt{1-x}}+\sqrt{1-x}$

Putting this equal to zero we get

$f^{{}^{\prime }}\left(x\right)=\frac{-x}{2\sqrt{1-x}}+\sqrt{1-x}=0$

$\Rightarrow -3x+2=0$

$\Rightarrow x=\frac{2}{3}$.

Now let's see the double derivative of this function.
$f^{{}^{\prime \prime }}\left(x\right)=\frac{-2\sqrt{1-x}-\frac{2x}{\sqrt{1-x}}}{4(1-x)}-\frac{1}{\sqrt{1-x}}$

At $x=\frac{2}{3}$

$f^{{}^{\prime \prime }}\left(\frac{2}{3}\right)=\frac{-2\sqrt{1-\frac{2}{3}}-\frac{2\times \frac{2}{3}}{\sqrt{1-\frac{2}{3}}}}{4(1-\frac{2}{3})}-\frac{1}{\sqrt{1-\frac{2}{3}}}$
$\Rightarrow f^{{}^{\prime \prime }}\left(\frac{2}{3}\right)=-4\sqrt{3}$

This is negative so function will take maximum value at $x=\frac{2}{3}$

Maximum value of the function is

$f\left(\frac{2}{3}\right)=\frac{2}{3}\times \sqrt{1-\frac{2}{3}}$

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