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})

f (x) = x^{3} - 4 x^{2} - 11 x + 30

f (x) = x^{2} - 8 x + 16

f (x) = \frac{1}{16 - x ^{2} - y ^{2}}

f (x) = - 3 x^{2} + 6 x

f (x, y) = x^{2} + (y - 2)^{2}

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