What is the extreme f(x)=x^2log_{5}(x) ?

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

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extreme f(x)=x^2log_{5}(x)

extreme points

f (x) = x^{2} lo g_{5} (x)

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

Solution steps

f^{'} (x) = \frac{x + 2 x ln ( x )}{ln ( 5 )}

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{1}{e}

into

x^{2} lo g_{5} (x) : - \frac{1}{2 e} lo g_{5} (e)

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

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

f (x) = (x - 1) (x - 6)^{3} + 6, 1 \leq x \leq 4

f (x) = x^{4} - 5 x^{3} + x^{2} + 21 x - 18

f (x, y) = x^{2} - 2 x y + 2 y^{2} - 2 x + 2 y + 1

g (x, y) = e^{x} - x + e^{y} - y

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