Answer
Absolute minimum value is at $f(1/2)=-0.5$ and absolute maximum value is at $f(e/2)=0$
Work Step by Step
Let us consider $y=f(x)=x \ln 2x-x$
Now, $f'(x)=\dfrac{2x}{2x}+\ln 2x -1= \ln 2x$
Now, $f'(1/2)= \ln 2(1/2)=\ln 1=0$
Thus, at $x=\dfrac{1}{2}; f'(x)=0$
This shows a critical point.
Now, $f(1/2)= (1/2) \ln 2(1/2)-(1/2)=\dfrac{-1}{2}=-0.5$
So, absolute minimum value is at $f(1/2)=-0.5$
and
$f(e/2)= (e/2) \ln 2(e/2)-(e/2)=0$
So, absolute maximum value is at $f(e/2)=0$
