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) = a + 2 b = 53

F (x, y) = x^{3} - 4 x^{2} - x y - y^{2}

f (x) = x^{2} - 4 x - 7

f (x, y) = 4 x^{2} - 8 y^{2}

f (x) = 5 x + 4 y

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)