What is the extreme f(x,y)=x^2+y^2+4x+2y+5 ?

The extreme f(x,y)=x^2+y^2+4x+2y+5 is Minimum(-2,-1)

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extreme f(x,y)=x^2+y^2+4x+2y+5

extreme points

f (x, y) = x^{2} + y^{2} + 4 x + 2 y + 5

Minimum (- 2, - 1)

Solution steps

Find the critical points:

(- 2, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 2, - 1) :

Minimum

Minimum (- 2, - 1)

f (x) = 4 x^{2} - 6 x

y = 2 x

- 6 x^{5} + 15 x^{4} + 20 x^{3} + 6

f (x) = 3 - ∣ t - 3 ∣

f (x) = 6 x^{2} + 48 x + 72

What is the extreme f(x,y)=x^2+y^2+4x+2y+5 ?
The extreme f(x,y)=x^2+y^2+4x+2y+5 is Minimum(-2,-1)