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

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

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extreme f(x)=xe^{(x/2)}

extreme points

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

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

Solution steps

Find the critical points:

x = - 2

Domain of

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \infty

Plug the extreme point

x = - 2

into

x e^{\frac{x}{2}} \Rightarrow y = - \frac{2}{e}

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

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

f (x, y) = - x^{3} + 4 x y - 2 y^{2} + 1

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

\frac{d}{d B} (a (B - h))

f (x, y) = 2 x + y

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