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

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

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extreme f(x,y)=3x^2-3y^2+2

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = \frac{- 6}{x ^{2} - 9}

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

f (x) = 2 x^{3} - 6 x^{2} - 48 x + 10

f (x) = e^{6 x} + e^{- x}

f (x) = \frac{x}{1 + x}

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