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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = ln^{2} (x) + 2 ln (x)

Find intervals:

Increasing

: 0 < x < \frac{1}{e ^{2}},

Decreasing

: \frac{1}{e ^{2}} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = \frac{1}{e ^{2}}

into

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

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

Plug

x = 1

into

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

Minimum (1, 0)

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

f (x) = x [36 - 2 (x)]^{2}

f (x) = x + 12 - 8

f (x, y) = 4 x y + 120 y - 20 y^{2} - \frac{1}{10} x^{2} y - 90

f (x, y) = 2 x^{2} + 4 y^{2} - 2 x y - 10 x - 2 y + 2

f (x) = \frac{7 x}{x ^{2} + 9}, 0 \leq x \leq 9

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