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

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

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extreme f(x,y)=x^2+y^2+2x-6y-3

extreme points

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

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 y) = \frac{3 y + 5 x}{x y + 3}

f (x, y) = - 4 x^{2} + 8 x y - 2 y^{4} - 4

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

x^{4} + 2 x^{3} + 6 x^{2} - 9 x + 12

f (x) = ∣ 2 x - 4 ∣

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