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

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

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

extreme points

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

Minimum (2, 2 e)

Solution steps

Find the critical points:

x = 2

Domain of

x e^{\frac{2}{x}} : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

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

Minimum (2, 2 e)

f (x) = - \frac{x ^{3}}{3} - 2 x^{2} + 5 x - 2

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

4 - 3 x - x^{3}

f (x) = 4 - x^{3}

- 2 x^{2} + 4 x + 8