What is the extreme f(x)=2x^5ln(x) ?

The extreme f(x)=2x^5ln(x) is Minimum(1/(\sqrt[5]{e)},-2/(5e))

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extreme f(x)=2x^5ln(x)

extreme points

f (x) = 2 x^{5} ln (x)

Minimum (\frac{1}{5 e}, - \frac{2}{5 e})

Solution steps

f^{'} (x) = 2 (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

2 x^{5} ln (x) : - \frac{2}{5 e}

Minimum (\frac{1}{5 e}, - \frac{2}{5 e})

f (x) = 48 x^{3} + 57 x^{2} - 54 x + 18

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

f (x) = \frac{x ^{2} + 3}{x ^{2} - 16}

3 x^{4} + 3 y^{4} - x y

f (x) = \frac{x ^{2} - 1}{x - 1}

What is the extreme f(x)=2x^5ln(x) ?
The extreme f(x)=2x^5ln(x) is Minimum(1/(\sqrt[5]{e)},-2/(5e))