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

The extreme f(x,y)=x^2+2x+y^2-2y+2 is Minimum(-1,1)

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

extreme points

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

Minimum (- 1, 1)

Solution steps

Find the critical points:

(- 1, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Minimum

Minimum (- 1, 1)

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

y = \frac{x ^{2} - 9 x + 39}{x - 7}

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

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

f (x, y) = (x + y^{2}) \cdot e^{2 x}

What is the extreme f(x,y)=x^2+2x+y^2-2y+2 ?
The extreme f(x,y)=x^2+2x+y^2-2y+2 is Minimum(-1,1)