What is the extreme F(X,Y)=1-X^2+2Y^2 ?

The extreme F(X,Y)=1-X^2+2Y^2 is Saddle(0,0)

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extreme F(X,Y)=1-X^2+2Y^2

extreme points

F (X, Y) = 1 - X^{2} + 2 Y^{2}

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) = 8 + 12 x^{2} - 8 x^{3}

f (x) = x^{3} - 15 x^{2} + 78 x + 5

x^{2} - 2 x y + 2 y^{2} - 2 x + 4 y - 1

f (x) = - 2 sin (x) cos (x)

f (x, y) = - 2 x^{3} + 6 x y + 3 y^{3}

What is the extreme F(X,Y)=1-X^2+2Y^2 ?
The extreme F(X,Y)=1-X^2+2Y^2 is Saddle(0,0)