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

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

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

extreme points

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