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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 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

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

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

h (t) = - 16 t^{2} + 32 t + 8

4 x + 15

f (x) = 2 n + 1

y = 5 x - 4

8

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