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

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

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

extreme points

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

f (x) = cos (2 x)

f (x) = \frac{1}{x} + 9

(- 4, - 2), (3, 3)

f (x) = \frac{e ^{x}}{1 - x}

\frac{x ^{2} - 16}{7 x ^{2}}

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