What is the extreme f(x,y)= 1/(1-x^2-y^2) ?

The extreme f(x,y)= 1/(1-x^2-y^2) is Minimum(0,0)

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extreme f(x,y)= 1/(1-x^2-y^2)

extreme points

f (x, y) = \frac{1}{1 - x ^{2} - y ^{2}}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( 3 y ^{2} - x ^{2} + 1 )}{( 1 - x ^{2} - y ^{2} ) ^{3}}

D (x, y) = \frac{- 12 x ^{4} - 24 x ^{2} y ^{2} + 8 x ^{2} - 12 y ^{4} + 8 y ^{2} + 4}{( - x ^{2} - y ^{2} + 1 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

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

x^{2} y + y^{3} - 12 y

f (x, y) = 2 x^{2} + 3 x y + y^{2} + a x + 5

f (x, y) = (x - y) (1 - x y)

What is the extreme f(x,y)= 1/(1-x^2-y^2) ?
The extreme f(x,y)= 1/(1-x^2-y^2) is Minimum(0,0)