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

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

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

extreme

Saddle (0, - \frac{3}{2})

Solution steps

Find the critical points:

(0, - \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, - \frac{3}{2}) :

Saddle

Saddle (0, - \frac{3}{2})

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