What is the extreme f(x)=xe^{1/(x^2)} ?

The extreme f(x)=xe^{1/(x^2)} is Maximum(-2^{1/2},-2^{1/2}e^{1/2}),Minimum(2^{1/2},2^{1/2}e^{1/2})

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extreme f(x)=xe^{1/(x^2)}

extreme points

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

Maximum (- 2^{\frac{1}{2}}, - 2^{\frac{1}{2}} e^{\frac{1}{2}}), Minimum (2^{\frac{1}{2}}, 2^{\frac{1}{2}} e^{\frac{1}{2}})

Solution steps

Find the critical points:

x = - 2^{\frac{1}{2}}, x = 2^{\frac{1}{2}}

Domain of

x e^{\frac{1}{x ^{2}}} : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 2^{\frac{1}{2}},

Decreasing

: - 2^{\frac{1}{2}} < x < 0,

Decreasing

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

Increasing

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

Plug the extreme point

x = - 2^{\frac{1}{2}}

into

x e^{\frac{1}{x ^{2}}} \Rightarrow y = - 2^{\frac{1}{2}} e^{\frac{1}{2}}

Maximum (- 2^{\frac{1}{2}}, - 2^{\frac{1}{2}} e^{\frac{1}{2}})

Plug the extreme point

x = 2^{\frac{1}{2}}

into

x e^{\frac{1}{x ^{2}}} \Rightarrow y = 2^{\frac{1}{2}} e^{\frac{1}{2}}

Minimum (2^{\frac{1}{2}}, 2^{\frac{1}{2}} e^{\frac{1}{2}})

Maximum (- 2^{\frac{1}{2}}, - 2^{\frac{1}{2}} e^{\frac{1}{2}}), Minimum (2^{\frac{1}{2}}, 2^{\frac{1}{2}} e^{\frac{1}{2}})

What is the extreme f(x)=xe^{1/(x^2)} ?
The extreme f(x)=xe^{1/(x^2)} is Maximum(-2^{1/2},-2^{1/2}e^{1/2}),Minimum(2^{1/2},2^{1/2}e^{1/2})