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

The extreme y=((ln(x))^2)/x is Minimum(1,0),Maximum(e^2, 4/(e^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme points

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

Minimum (1, 0), Maximum (e^{2}, \frac{4}{e ^{2}})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < e^{2},

Decreasing

: e^{2} < x < \infty

Plug

x = 1

into

\frac{ln ^{2} ( x )}{x} : 0

Minimum (1, 0)

Plug

x = e^{2}

into

\frac{ln ^{2} ( x )}{x} : \frac{4}{e ^{2}}

Maximum (e^{2}, \frac{4}{e ^{2}})

Minimum (1, 0), Maximum (e^{2}, \frac{4}{e ^{2}})

What is the extreme y=((ln(x))^2)/x ?
The extreme y=((ln(x))^2)/x is Minimum(1,0),Maximum(e^2, 4/(e^2))