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

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

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

extreme points

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

Maximum (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 )}{( x ^{2} + y ^{2} - 1 ) ^{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) :

Maximum

Maximum (0, 0)

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