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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 5

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - x^{3} - 3 y^{2} + 12 x y - 21 x + 8

f (x) = \frac{x - 3}{x ^{2} - 9}

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

f (x) = x (ln (x))^{5}

f (x, y) = \frac{e ^{(- x^{2} + y^{2})} ( x + y ^{2} ) + 1}{2}

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