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))

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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}})

f (x, y) = - x^{3} + 2 y^{3} + 27 x - 24 y + 3

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

F (I, J) = (- 3 I + 2 J)

f (x) = x_{1} x

0.5 x^{2} - 130 x + 17555

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))