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

The extreme f(x,y)=ye^{x^2-2y^2} is Saddle(0,-1/2),Saddle(0, 1/2),Saddle(0,-1/2),Saddle(0, 1/2)

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 16 e^{(x + 2 y) (x - 2 y)} y^{3} - 12 e^{(x + 2 y) (x - 2 y)} y

D (x, y) = 64 e^{2 x^{2} - 4 y^{2}} x^{2} y^{4} - 64 e^{2 (x^{2} - 2 y^{2})} x^{2} y^{4} + 32 e^{2 x^{2} - 4 y^{2}} y^{4} - 16 e^{2 x^{2} - 4 y^{2}} x^{2} y^{2} - 24 e^{2 x^{2} - 4 y^{2}} y^{2} - 4 e^{2 (x^{2} - 2 y^{2})} x^{2}

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

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

Saddle

Check critical point

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

Saddle

Check critical point

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

Saddle

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

f (x) = x^{2} + x y + 12 y^{2} - 4 x + y

f (x, y) = 6 x + 6 x y + y

f (x) = (x t)

- (y - 3) (y + 5)

f (2) = x^{2} - 4 x - 4

What is the extreme f(x,y)=ye^{x^2-2y^2} ?
The extreme f(x,y)=ye^{x^2-2y^2} is Saddle(0,-1/2),Saddle(0, 1/2),Saddle(0,-1/2),Saddle(0, 1/2)