Q. Difference between the greatest and the least values of the function $f(x)=x(\ln x-2) \text { on }\left[1, e^{2}\right] \text { is }$
Application of Derivatives
Solution:
$ \mathrm{y}=\mathrm{x}(\ln \mathrm{x}-2)$
$y^{\prime}=x\left(\frac{1}{x}\right)+(\ln x-2)=\ln x-1$
$\frac{d y}{d x}=\ln x-1=0 \Rightarrow x=e$
now $\mathrm{f}(1)=-2$
$\mathrm{f}(\mathrm{e})=-\mathrm{e} $ (least)
$\mathrm{f}\left(\mathrm{e}^{2}\right)=0 $ (greatest)
$\therefore $ difference $=0-(-\mathrm{e})=\mathrm{e}$
