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

The extreme f(x,y)=yxe^{-(x^2+y^2)} is

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = x (4 e^{- x^{2} - y^{2}} y^{3} - 6 e^{- x^{2} - y^{2}} y)

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

Check critical point

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

Minimum

Check critical point

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

Minimum

Check critical point

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

Maximum

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

f (x, y) = - 2 x^{4} + y^{2} + x^{2} - 2 y

f (x) = \frac{x}{x ^{2} - x + 1}, 0 \leq x \leq 3

f (x, y) = x^{2} + x y + \frac{y ^{2}}{2} + 2 x

F (I, J) = 25 I N + 12 J N

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

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