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})

f (x, y) = \frac{1}{3} x^{3} + \frac{4}{3} y^{3} - x^{2} - 3 x - 4 y - 3

f (x) = 3 x^{4} - 8^{3} - 6 x^{2} + 24 x - 2

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

f (x) = 9 x + 9 sin (x), 0 \leq x \leq 2 π

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