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

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

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

extreme points

f (x = - 1, y = 1) = x^{2} + y^{2} + 2 x - 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) = 7 x + 2 y - 8

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

y = x^{2} e^{- z}

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

y = \frac{1}{3} x^{2} + 2 x + 5

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