What is the extreme f(x,y)=ln(4-x^2-y^2)-1/(1-x^2) ?

The extreme f(x,y)=ln(4-x^2-y^2)-1/(1-x^2) is Maximum(0,0)

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extreme f(x,y)=ln(4-x^2-y^2)-1/(1-x^2)

extreme

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = \frac{- 4 x ^{10} - 8 x ^{8} y ^{2} + 24 x ^{8} - 4 x ^{6} y ^{4} + 36 x ^{6} y ^{2} - 88 x ^{6} - 116 x ^{4} y ^{2} + 340 x ^{4} - 12 x ^{2} y ^{6} + 32 x ^{2} y ^{4} + 168 x ^{2} y ^{2} - 384 x ^{2} - 4 y ^{6} + 20 y ^{4} + 64 y ^{2} - 320}{( x + 1 ) ^{3} ( x - 1 ) ^{3} ( 4 - x ^{2} - y ^{2} ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

What is the extreme f(x,y)=ln(4-x^2-y^2)-1/(1-x^2) ?
The extreme f(x,y)=ln(4-x^2-y^2)-1/(1-x^2) is Maximum(0,0)