What is the extreme f(x)=x^4ln(x) ?

The extreme f(x)=x^4ln(x) is Minimum(1/(\sqrt[4]{e)},-1/(4e))

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extreme f(x)=x^4ln(x)

extreme points

f (x) = x^{4} ln (x)

Solution steps

f^{'} (x) = 4 x^{3} ln (x) + x^{3}

Find intervals:

Decreasing

Increasing

Plug

into

x^{4} ln (x) : - \frac{1}{4 e}

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

f (x) = \frac{x + 1}{x - 1}

f (x) = e^{x} (24 - x^{2})

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

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 3 y^{2} - 9 x

What is the extreme f(x)=x^4ln(x) ?
The extreme f(x)=x^4ln(x) is Minimum(1/(\sqrt[4]{e)},-1/(4e))