What is the extreme f(x,y)=x^2+y^2+6x-4y+13 ?

The extreme f(x,y)=x^2+y^2+6x-4y+13 is Minimum(-3,2)

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extreme f(x,y)=x^2+y^2+6x-4y+13

extreme points

f (x, y) = x^{2} + y^{2} + 6 x - 4 y + 13

Minimum (- 3, 2)

Solution steps

Find the critical points:

(- 3, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 3, 2) :

Minimum

Minimum (- 3, 2)

f (x) = (3 - 5 x) e^{2 x}

f (x) = 3 - 5 x^{2}

f (x) = x^{2} - 10 x - 2, 2 \leq x \leq 7

f (x, y) = x^{2} + x y - 5 x - 2 y^{2} + 2 y

\frac{x + 2}{x ^{2} - 7 x - 18}

What is the extreme f(x,y)=x^2+y^2+6x-4y+13 ?
The extreme f(x,y)=x^2+y^2+6x-4y+13 is Minimum(-3,2)