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

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

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

extreme points

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

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)

y = \frac{x ^{2} - 9 x + 39}{x - 7}

\frac{1}{( x + 1 ) ^{2}}

f (x) = x (x^{2} - 12)

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

f (x, y) = 8 e^{(\frac{( - x ^{2} - y ^{2} )}{4})}

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