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

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

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

extreme points

f (x, y) = y e^{x^{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}} = 4 e^{(x + y) (x - y)} y^{3} - 6 e^{(x + y) (x - y)} y

D (x, y) = 16 e^{2 x^{2} - 2 y^{2}} x^{2} y^{4} - 16 e^{2 (x^{2} - y^{2})} x^{2} y^{4} + 8 e^{2 x^{2} - 2 y^{2}} y^{4} - 8 e^{2 x^{2} - 2 y^{2}} x^{2} y^{2} - 12 e^{2 x^{2} - 2 y^{2}} y^{2} - 4 e^{2 (x^{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})

\frac{1}{2} (x^{3} - 3 x + 7)

f (x) = \frac{e ^{x}}{x ^{3}}

f (x) = \frac{( x ^{2} - 8 )}{x - 3}

q (x, y) = x^{2} - x y + 5 y^{2}

f (x, y) = 6 - x^{2} - 4 x - y^{2}