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

The extreme f(x)=2x^2(ln(x)) is Minimum(1/(sqrt(e)),-1/e)

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

extreme points

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

Minimum (\frac{1}{e}, - \frac{1}{e})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

: \frac{1}{e} < x < \infty

Plug

x = \frac{1}{e}

into

2 x^{2} ln (x) : - \frac{1}{e}

Minimum (\frac{1}{e}, - \frac{1}{e})

f (x, y) = ln (y - x)

f (x) = x^{\frac{2}{3}} (x + 2)

y

f (x) = 2 x y - 5 x^{2} - 2 y^{2} + 4 x - 4 y + 4

f (x) = \frac{2 x + 1}{x ^{2} + x + 1}

What is the extreme f(x)=2x^2(ln(x)) ?
The extreme f(x)=2x^2(ln(x)) is Minimum(1/(sqrt(e)),-1/e)