Steiner's calculus problem

${\displaystyle g}$

${\displaystyle g'(x)={\frac {1-\ln x}{x^{2}}},}$

so ${\displaystyle g'(x)}$ is positive for ${\displaystyle 0<x<e}$ and negative for ${\displaystyle x>e}$, which implies that ${\displaystyle g(x)}$ – and therefore ${\displaystyle f(x)}$ – is

${\displaystyle 0<x<e}$

${\displaystyle x>e.}$

${\displaystyle x=e}$

${\displaystyle f(x).}$

References