What is the extreme f(x,y)=3x^2+5xy-7y^2+1 ?

The extreme f(x,y)=3x^2+5xy-7y^2+1 is Saddle(0,0)

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extreme f(x,y)=3x^2+5xy-7y^2+1

extreme points

f (x, y) = 3 x^{2} + 5 x y - 7 y^{2} + 1

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 14

D (x, y) = - 109

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

What is the extreme f(x,y)=3x^2+5xy-7y^2+1 ?
The extreme f(x,y)=3x^2+5xy-7y^2+1 is Saddle(0,0)