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

x^{2} + 4 x y + y^{2}

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

x + 2

f (x, y) = 80 x + 80 y - x^{2} - y^{2}

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)