What is the extreme f(x)=(ln(x))/(x^3) ?

The extreme f(x)=(ln(x))/(x^3) is Maximum(\sqrt[3]{e}, 1/(3e))

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extreme f(x)=(ln(x))/(x^3)

extreme points

f (x) = \frac{ln ( x )}{x ^{3}}

Maximum (3 e, \frac{1}{3 e})

Solution steps

f^{'} (x) = \frac{1 - 3 ln ( x )}{x ^{4}}

Find intervals:

Increasing

: 0 < x < 3 e,

Decreasing

: 3 e < x < \infty

Plug

x = 3 e

into

\frac{ln ( x )}{x ^{3}} : \frac{1}{3 e}

Maximum (3 e, \frac{1}{3 e})

f (x, y) = 2 x^{2} - 3 x y + 4

f (x) = x^{4} - 18 x^{2} + 11

f (x, y) = - 5 x^{2} y + 5 x y^{2}

f (x) = - x^{3} + 3 x^{2} + 9 x - 6

f (x, y) = \frac{x ^{2}}{8} + \frac{y ^{2}}{2} - 1

What is the extreme f(x)=(ln(x))/(x^3) ?
The extreme f(x)=(ln(x))/(x^3) is Maximum(\sqrt[3]{e}, 1/(3e))