Question

Find the greatest and the least values of the following function:
$f\left(x\right)=\sqrt{\frac{x^2}{2x-1}}$ in the interval $[\frac{3}{4},2].$

Answer 1

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

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

$\Rightarrow \frac{x-1}{2x-1}=0$

$\Rightarrow x=1$ is the critical point

$f^{\prime \prime }\left(x\right)=\frac{1}{(2x-1{)}^2}$, at $x=1,f^{\prime \prime }\left(x\right)>0$

Therefore, $x=1$ is the point of minima.

Now, $f\left(2\right)=2\sqrt{\frac{1}{3}}=\frac{2}{\sqrt{3}}$

$f\left(\frac{3}{4}\right)=\frac{3\sqrt{2}}{5}$

$f\left(1\right)=1$

Hence, at $x=2,f\left(x\right)=\frac{2}{\sqrt{3}}$ is the greatest value

at $x=\frac{3}{4}.f\left(x\right)=\frac{3\sqrt{2}}{5}$ has the least value.