What is the extreme f(x,y)=x^2-2y^2+xy-3x+3y+4 ?

The extreme f(x,y)=x^2-2y^2+xy-3x+3y+4 is Saddle(1,1)

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extreme f(x,y)=x^2-2y^2+xy-3x+3y+4

extreme points

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

Saddle (1, 1)

Solution steps

Find the critical points:

(1, 1)

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

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Saddle (1, 1)

f (x) = \frac{x ^{3}}{3} + x^{2} - 8 x

- x^{4} y^{2} - x^{4} + 2 x^{3} y + 2 x^{2} y + x^{2} - 2 x - 2

f (x \cdot y) = x y + 8^{x y}

f (x) = x^{2} - 6 x + 12

f (t) = \frac{x}{x + 25} [0.9]

What is the extreme f(x,y)=x^2-2y^2+xy-3x+3y+4 ?
The extreme f(x,y)=x^2-2y^2+xy-3x+3y+4 is Saddle(1,1)