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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{1}{e}

into

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

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

f (x) = x (21 - 46 + 2 x) (\frac{46}{2} - x)

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

f (x) = x^{2} - 4 x, - \infty < x \leq 4

f (x) = (x - 2) (x + 2) (x + 4)

f (x) = \frac{x + 1}{x ^{2} - 6 x - 7}

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