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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - 2 x^{2} + 8 x - 6

\frac{1}{x ^{2} + 2 x + 9}

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

f (x) = - \frac{7}{x ^{2} + 1}

f (x) = \frac{x}{25 - x + x ^{2}}

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