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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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) = \frac{x}{( 1 - x )}

4 - x - x^{2} - 1

f (x) = x^{3} - 27 x + 7

f (x) = 8^{- x}

y - 9 = \frac{3}{4} (x - 4)

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