What is the extreme f(x,y)=3x^2-3y^2+8 ?

The extreme f(x,y)=3x^2-3y^2+8 is Saddle(0,0)

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extreme f(x,y)=3x^2-3y^2+8

extreme points

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

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)

y = x e^{- \frac{x ^{2}}{18}}

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

f (x) = \frac{40}{r} + 4 π r^{2}

f (x) = x \cdot x + 2

What is the extreme f(x,y)=3x^2-3y^2+8 ?
The extreme f(x,y)=3x^2-3y^2+8 is Saddle(0,0)