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

The extreme f(x)=xe^{-x^2-y^2} is Minimum(-1/(sqrt(2)),0),Maximum(1/(sqrt(2)),0)

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extreme f(x)=xe^{-x^2-y^2}

extreme

Minimum (- \frac{1}{2}, 0), Maximum (\frac{1}{2}, 0)

Solution steps

Find the critical points:

(- \frac{1}{2}, 0), (\frac{1}{2}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 x (- 2 e^{- x^{2} - y^{2}} y^{2} + e^{- x^{2} - y^{2}})

D (x, y) = - 2 x (4 e^{- x^{2} - y^{2}} x^{3} - 6 e^{- x^{2} - y^{2}} x) (- 2 e^{- x^{2} - y^{2}} y^{2} + e^{- x^{2} - y^{2}}) - 4 e^{2 (- x^{2} - y^{2})} y^{2} (2 x + 1)^{2} (2 x - 1)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, 0) :

Minimum

Check critical point

(\frac{1}{2}, 0) :

Maximum

Minimum (- \frac{1}{2}, 0), Maximum (\frac{1}{2}, 0)

What is the extreme f(x)=xe^{-x^2-y^2} ?
The extreme f(x)=xe^{-x^2-y^2} is Minimum(-1/(sqrt(2)),0),Maximum(1/(sqrt(2)),0)