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

The extreme f(x)=x^2+y^2-2x-2y+3 is Minimum(1,1)

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

extreme points

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

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^{2} + 3 x - 1

f (x) = ln (6 - ln (x))

7 x^{4} - 4 x^{2} - 8 x - 5

1 - (y + 1) 1 - y^{2} + y

f (x, y) = e^{x} - e^{x y}

What is the extreme f(x)=x^2+y^2-2x-2y+3 ?
The extreme f(x)=x^2+y^2-2x-2y+3 is Minimum(1,1)