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

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Answer 1 · 2016-05-28T16:25:56

Find the roots of first derivative you get

$f'(x)=0=>1-1/x=0=>x=1$

But $f''(x)=1/x^2>0$

Hence $f(1)=1-ln3$ is a minimum.

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

where $f(e)=e-ln(3e)$

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