What is the extreme f(x)=x^5*ln(x) ?

The extreme f(x)=x^5*ln(x) is Minimum(1/(\sqrt[5]{e)},-1/(5e))

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extreme f(x)=x^5*ln(x)

extreme points

f (x) = x^{5} \cdot ln (x)

Minimum (\frac{1}{5 e}, - \frac{1}{5 e})

Solution steps

f^{'} (x) = 5 x^{4} ln (x) + x^{4}

Find intervals:

Decreasing

: 0 < x < \frac{1}{5 e},

Increasing

: \frac{1}{5 e} < x < \infty

Plug

x = \frac{1}{5 e}

into

x^{5} \cdot ln (x) : - \frac{1}{5 e}

Minimum (\frac{1}{5 e}, - \frac{1}{5 e})

f (x) = ((42 - x) x - (2 x + 80))

f (x, y) = - 4 x + y - \frac{1}{x} + \frac{1}{y}

f (x) = \frac{x ^{2} + 4}{x ^{2} - 25}

f (x) = (x - 8) ln (x - 8)

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

What is the extreme f(x)=x^5*ln(x) ?
The extreme f(x)=x^5*ln(x) is Minimum(1/(\sqrt[5]{e)},-1/(5e))