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

The extreme f(x,y)=x^2+y^2+2 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 2 x^{3} - x^{2} - 4 x - 6

x^{2} + 2 y^{2} - 2 x - 12 y + 19

f (x) = 2 x^{3} - x^{2} - 4 x + 2

f (x) = 2 x^{3} - 12 x^{2} + 18 x + 5

f (x) = 2 x \cdot ln (x)

What is the extreme f(x,y)=x^2+y^2+2 ?
The extreme f(x,y)=x^2+y^2+2 is Minimum(0,0)