What is the extreme f(x,y)=x^2+y^2-2x+6y+10 ?

The extreme f(x,y)=x^2+y^2-2x+6y+10 is Minimum(1,-3)

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extreme f(x,y)=x^2+y^2-2x+6y+10

extreme points

f (x, y) = x^{2} + y^{2} - 2 x + 6 y + 10

Minimum (1, - 3)

Solution steps

Find the critical points:

(1, - 3)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, - 3) :

Minimum

Minimum (1, - 3)

f (x) = 18 x (x - 1)^{3}

f (x) = \frac{x ^{2} + 2 x - 7}{x ^{2} + 2 x - 3}

f (x) = 1 + x - 3 x^{2}

f (x) = \frac{x ^{2}}{2 x + 4}

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

What is the extreme f(x,y)=x^2+y^2-2x+6y+10 ?
The extreme f(x,y)=x^2+y^2-2x+6y+10 is Minimum(1,-3)