What is the extreme f(x)=x^2log_{10}(2x) ?

The extreme f(x)=x^2log_{10}(2x) is Minimum(1/(2sqrt(e)),-1/(8e)log_{10}(e))

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extreme f(x)=x^2log_{10}(2x)

extreme points

f (x) = x^{2} lo g_{10} (2 x)

Minimum (\frac{1}{2 e}, - \frac{1}{8 e} lo g_{10} (e))

Solution steps

f^{'} (x) = \frac{x + 2 x ln ( 2 x )}{ln ( 10 )}

Find intervals:

Decreasing

: 0 < x < \frac{1}{2 e},

Increasing

: \frac{1}{2 e} < x < \infty

Plug

x = \frac{1}{2 e}

into

x^{2} lo g_{10} (2 x) : - \frac{1}{8 e} lo g_{10} (e)

Minimum (\frac{1}{2 e}, - \frac{1}{8 e} lo g_{10} (e))

f (x) = - [x + 2]

f (x) = 2 x^{3} + 9 x^{2} - 24 x, - 5 \leq x \leq 2

22 x^{2}

f (x) = (x^{3} - 1)^{4}, - 1 \leq x \leq 2

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

What is the extreme f(x)=x^2log_{10}(2x) ?
The extreme f(x)=x^2log_{10}(2x) is Minimum(1/(2sqrt(e)),-1/(8e)log_{10}(e))