What is the extreme f(x,y)=x^2+y^2-9 ?

The extreme f(x,y)=x^2+y^2-9 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2-9

extreme points

f (x, y) = x^{2} + y^{2} - 9

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) = - 11 y^{2} + (x + 16)^{2} + 1

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

f (x) = 4 x^{3} - 6 x^{2} - 72 x + 30

f (x) = 5 - 5 x^{2}

\frac{x}{3 x ^{2} - 1}

What is the extreme f(x,y)=x^2+y^2-9 ?
The extreme f(x,y)=x^2+y^2-9 is Minimum(0,0)