What is the extreme f(x)=(ln(x))/(x^2) ?

The extreme f(x)=(ln(x))/(x^2) is Maximum(sqrt(e), 1/(2e))

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extreme f(x)=(ln(x))/(x^2)

extreme points

f (x) = \frac{ln ( x )}{x ^{2}}

Maximum (e, \frac{1}{2 e})

Solution steps

f^{'} (x) = \frac{1 - 2 ln ( x )}{x ^{3}}

Find intervals:

Increasing

: 0 < x < e,

Decreasing

: e < x < \infty

Plug

x = e

into

\frac{ln ( x )}{x ^{2}} : \frac{1}{2 e}

Maximum (e, \frac{1}{2 e})

What is the extreme f(x)=(ln(x))/(x^2) ?
The extreme f(x)=(ln(x))/(x^2) is Maximum(sqrt(e), 1/(2e))