What is the extreme f(x)=x^3ln(x),x>0 ?

The extreme f(x)=x^3ln(x),x>0 is Minimum(1/(\sqrt[3]{e)},-1/(3e))

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extreme f(x)=x^3ln(x),x>0

extreme points

f (x) = x^{3} ln (x), x > 0

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

Solution steps

Find the critical points:

x = \frac{1}{3 e}

Domain of

x^{3} ln (x), x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{3 e},

Increasing

: \frac{1}{3 e} < x < \infty

Plug the extreme point

x = \frac{1}{3 e}

into

x^{3} ln (x) \Rightarrow y = - \frac{1}{3 e}

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

f (x) = e^{x^{3} - x}

f (x) = x - cos (x)

f (x, y) = 25 - x^{2} - y^{2}

\frac{1}{ln ( 3 )} (x + 3 ln (x))

f (x, y) = 3 x^{2} + 2 x y - 8 y^{2}

What is the extreme f(x)=x^3ln(x),x>0 ?
The extreme f(x)=x^3ln(x),x>0 is Minimum(1/(\sqrt[3]{e)},-1/(3e))