Determine the location and value of the absolute extreme value of f on the given interval, if they exist. f(x)=x\ \ln \frac{x}{5}\ on\ \left[0.1, 5\right]

Step 1
Given Data
The function is ${\displaystyle f\left(x\right)=x\frac{\ln \left(x\right)}{5}}$.
The interval is ${\displaystyle [0.1,5]}$.
Differentiate the function ${\displaystyle f\left(x\right)=x\frac{\ln \left(x\right)}{5}}$ with respect to x and equate to zero to evaluate the location of the absolute extreme,
${\displaystyle \frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(x\frac{\ln \left(x\right)}{5}\right)}$
${\displaystyle f^{prime}\left(x\right)=\frac{\ln \left(x\right)}{5}\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\frac{\ln \left(x\right)}{5}\right)}$
${\displaystyle 0=\frac{\ln \left(x\right)}{5}\cdot 1+x\left(\frac{1}{x}\right)\left(\frac{1}{5}\right)}$
${\displaystyle \frac{\ln \left(x\right)}{5}+\frac{1}{5}=0}$
${\displaystyle \frac{\ln \left(x\right)+1}{5}=0}$
${\displaystyle \ln \left(x\right)=-1}$
${\displaystyle x=-e^{-1}}$
${\displaystyle x=\frac{1}{e}}$
The value of absolute extreme ${\displaystyle x=\frac{1}{e}}$ lie in given interval ${\displaystyle [0.1,5]}$.
Hence the location of the absolute extreme value of f on the given interval is ${\displaystyle x=\frac{1}{e}}$.
Step 2
Evaluate the function ${\displaystyle f\left(x\right)=x\frac{\ln \left(x\right)}{5}}$ by substituting the value ${\displaystyle x=\frac{1}{e}}$,
${\displaystyle f\left(x\right)=x\frac{\ln \left(x\right)}{5}}$
${\displaystyle f\left(\frac{1}{e}\right)=\frac{1}{e}\frac{\ln \left(\frac{1}{e}\right)}{5}}$
${\displaystyle =-0.0735}$
${\displaystyle <0}$ Hence the minimum value of the absolute extreme value of f on the given interval is -0.0735, which is less than zero.