What is the extreme f(x,y)=x^2+2xy-xy^2-3 ?

The extreme f(x,y)=x^2+2xy-xy^2-3 is Saddle(0,0),Saddle(0,2),Minimum(-1/2 ,1)

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extreme f(x,y)=x^2+2xy-xy^2-3

extreme points

f (x, y) = x^{2} + 2 x y - x y^{2} - 3

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 x

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, 2) :

Saddle

Check critical point

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

Minimum

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

f (x) = \frac{1}{4} x^{4} - 2 x^{2}

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

f (x) = 1 + 3 x^{2} - 2 x^{3}

f (x) = \frac{1}{4} x + 3 + \frac{400}{x}

f (x) = cos (x) - x

What is the extreme f(x,y)=x^2+2xy-xy^2-3 ?
The extreme f(x,y)=x^2+2xy-xy^2-3 is Saddle(0,0),Saddle(0,2),Minimum(-1/2 ,1)