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

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

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

extreme points

f (x, y) = 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) = 2 x^{3} - 6 x^{2} - 18 x + 7

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

f (x) = 3 x^{4} - 4 x^{3} - 8

f (x, y) = - 250 - 2 x^{2} - 3 y^{2}

f (x) = x - ln (1 + x)

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