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

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

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extreme f(x,y)=1+x^2-y^2

extreme points

f (x, y) = 1 + x^{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) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2

f (x) = x e^{- \frac{x ^{2}}{32}} [- 3.8]

U (X, Y) = 2 X + 5 Y

f (x, y) = 5 x^{2} - 5 y^{2}

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

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