What is the extreme f(x,y)=x^2+y^2-3 ?

The extreme f(x,y)=x^2+y^2-3 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2-3

extreme points

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

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)

x^{2} - x - ln (x)

x^{2} + \frac{8}{x}

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

f (x) = x^{3} - 9 x^{2} + 7 x - 6

f (x, y) = x + y - 4

What is the extreme f(x,y)=x^2+y^2-3 ?
The extreme f(x,y)=x^2+y^2-3 is Minimum(0,0)