What is the extreme f(x)=3x^2ln(x) ?

The extreme f(x)=3x^2ln(x) is Minimum(1/(sqrt(e)),-3/(2e))

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extreme f(x)=3x^2ln(x)

extreme points

f (x) = 3 x^{2} ln (x)

Minimum (\frac{1}{e}, - \frac{3}{2 e})

Solution steps

f^{'} (x) = 3 (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

3 x^{2} ln (x) : - \frac{3}{2 e}

Minimum (\frac{1}{e}, - \frac{3}{2 e})

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

P (y, x) = - y^{3} - 2 y^{2} - x y

x 2 - x

y = \frac{x}{x ^{2} + 1}

f (x) = - e^{- x}

What is the extreme f(x)=3x^2ln(x) ?
The extreme f(x)=3x^2ln(x) is Minimum(1/(sqrt(e)),-3/(2e))