What is the extreme F(x,y)=3y^2-2y^2-3x^2+6xy ?

The extreme F(x,y)=3y^2-2y^2-3x^2+6xy is Saddle(0,0)

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extreme F(x,y)=3y^2-2y^2-3x^2+6xy

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 48

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

\frac{( x - 1 )}{e ^{x + 2}}

14 + 14 x^{- \frac{3}{17}}

f (x, y) = x^{4} + 2 y^{2} - 8 x y

f (x) = x^{4} + y^{3} - 3 \cdot x^{2} + y^{2} + x - 2 \cdot y + 1

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

What is the extreme F(x,y)=3y^2-2y^2-3x^2+6xy ?
The extreme F(x,y)=3y^2-2y^2-3x^2+6xy is Saddle(0,0)