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

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

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

extreme points

f (x, y) = ln (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 ( - y ^{2} + x ^{2} - 1 )}{( x ^{2} + y ^{2} - 1 ) ^{2}}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} + 15 x - 9 y

f (x) = x^{3} ln (x)

\frac{1}{3} h d

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

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