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) = \frac{x ^{2} + 1}{x ^{2} - 4}

f (x) = \frac{1}{3} x^{3} - 9 x + 2

f (x) = e^{x^{2} - 7 x - 1}

f (x) = 25 - x^{2} - y^{2}

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

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