# What are the absolute extrema of  f(x)= x-ln(3x) in [1,e]?

Find the roots of first derivative you get

$f ' \left(x\right) = 0 \implies 1 - \frac{1}{x} = 0 \implies x = 1$

But $f ' ' \left(x\right) = \frac{1}{x} ^ 2 > 0$

Hence $f \left(1\right) = 1 - \ln 3$ is a minimum.

Hence $f \left(e\right) > f \left(x\right)$ for every $x$ in $\left[1 , e\right]$ its a maximum

where $f \left(e\right) = e - \ln \left(3 e\right)$