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

The extreme f(x)=x^2ln(x/2) is Minimum(2/(sqrt(e)),-2/e)

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{2}{e}

into

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

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

y = 2 x^{2} + 2 x - 4

f (x) = - \frac{5}{x} + 2

- 2 cos (\frac{x}{4}) - 2

4 y = 8 x + 9

\frac{1}{1 - e ^{- x}}

What is the extreme f(x)=x^2ln(x/2) ?
The extreme f(x)=x^2ln(x/2) is Minimum(2/(sqrt(e)),-2/e)