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 points

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

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)

5 cos (x), - \frac{3 π}{2} \leq x \leq \frac{3 π}{2}

(7 x (1 + 12 x^{2})^{- 2}), - 7 \leq x \leq 12

f (x) = \frac{x - 7}{x ^{2} - 25}, 0 \leq x < 5

f (x) = x 600 - x

f (x, y) = x (y + 1) + (x + y + 1)

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)