Answer
relative minimum: $( \displaystyle \frac{1}{2}, 1)$
Work Step by Step
$y=2x-\ln(2x)$
Logarithmic functions have a restriction on the domain, the argument must be positive.
Domain: $x>0 $
$y^{\prime}=2-(\displaystyle \frac{1}{2x}\cdot 2)=2-\frac{1}{x}=\frac{2x-1}{x}$
Critical numbers:
Since 0 is not in the domain,
we have one critical number, $x=\displaystyle \frac{1}{2}$ .
Second derivative test:
$y^{\prime\prime}=(2-\displaystyle \frac{1}{x})^{\prime}=2-(-x^{-2})=2+\frac{1}{x^{2}},$
which is always positive.
So, at x=0.5
$y=2(0.5)-\ln 1=1,$
we have a
relative minimum: $( \displaystyle \frac{1}{2}, 1)$
