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

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

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

extreme points

f (x, y) = 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)

C (t) = 6.5 (e^{- a t} - e^{- b t})

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

f (x) = 9 x^{3} - 7 x^{2} + 3 x + 10 [- 5.6]

f (x) = x^{3} - 5 x^{2} + 8 x

f (x, y) = ln (2 x + 2 y)

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