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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{4 ( 3 y ^{2} - x ^{2} + 1 )}{( 1 - x ^{2} - y ^{2} ) ^{3}}

D (x, y) = \frac{- 48 x ^{4} - 96 x ^{2} y ^{2} + 32 x ^{2} - 48 y ^{4} + 32 y ^{2} + 16}{( - 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) = - 2 x^{2} - 3 x + 5

\frac{1}{x} + x y - \frac{8}{y}

f (x, y) = y \cdot ln (x)

f (x) = x^{2} + x y + y^{2} - 37 y + 456

f (x) = 4 - 2 (x - 1)^{\frac{2}{3}}

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