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

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

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

extreme points

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

Minimum (- 2, - 2)

Solution steps

Find the critical points:

(- 2, - 2)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 2, - 2) :

Minimum

Minimum (- 2, - 2)

x + 3, 0 < x < 2

f (x) = x + 4 - x

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} - 7

f (x, y) = e^{y} (4 x^{2} + 2 x y + y^{2} - 6 x - 15)

2 (x - 1)^{2} + 5

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