What is the extreme f(x,y)=4x^2-2y^2+2xy ?

The extreme f(x,y)=4x^2-2y^2+2xy is Saddle(0,0)

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extreme f(x,y)=4x^2-2y^2+2xy

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = a e^{2 x} - e^{3 x}

f (x) = x^{4} + 8 x^{3} + 18 x^{2} + 4

f (x) = x^{4} + 8 x^{3} + 18 x^{2} - 8

f (x, y) = 4 x y - x^{3} - 2 y^{2}

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

What is the extreme f(x,y)=4x^2-2y^2+2xy ?
The extreme f(x,y)=4x^2-2y^2+2xy is Saddle(0,0)