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

The extreme f(x,y)=x^2+y^2-4 is Minimum(0,0)

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

extreme points

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

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) = 9 x^{2} + 2 y^{2} - 2 x y^{2}

f (x) = 2 x^{3} - 2 x^{2} - 16 x + 1

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

f (x) = e^{x + y}

f (x) = 6 x^{3} - 15 x^{2} + 12 x - 6

What is the extreme f(x,y)=x^2+y^2-4 ?
The extreme f(x,y)=x^2+y^2-4 is Minimum(0,0)