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

The extreme f(x,y)=-e^{x-y^2+xy} is Saddle(-2,-1)

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

extreme points

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

Saddle (- 2, - 1)

Solution steps

Find the critical points:

(- 2, - 1)

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

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

Check the sign of at each critical point

Check critical point

(- 2, - 1) :

Saddle

Saddle (- 2, - 1)

(15, 3), (2, - 14)

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

f (x) = \frac{x}{x - 2}

2 e^{2 x}

y = x^{3} ln (x)

What is the extreme f(x,y)=-e^{x-y^2+xy} ?
The extreme f(x,y)=-e^{x-y^2+xy} is Saddle(-2,-1)