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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 12

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

8 x^{4} - 48 x^{2}

c

f (x) = 5 + (8 + 7 x)^{\frac{2}{7}}

F

y = \frac{6 x}{x ^{2} + 4}

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