What is the extreme f(x,y)=x^2+y^2+1 ?

The extreme f(x,y)=x^2+y^2+1 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2+1

extreme points

f (x, y) = x^{2} + y^{2} + 1

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

f (x, y) = x y^{2} - 16 x^{2} - 2 y^{2}

f (x) = \frac{5}{3} x^{3} + 18 x^{2} - 32 x - 30

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

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

What is the extreme f(x,y)=x^2+y^2+1 ?
The extreme f(x,y)=x^2+y^2+1 is Minimum(0,0)