What is the extreme f(x,y)=ln(x^2+y^2-9) ?

The extreme f(x,y)=ln(x^2+y^2-9) is Maximum(0,0)

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extreme f(x,y)=ln(x^2+y^2-9)

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( - y ^{2} + x ^{2} - 9 )}{( x ^{2} + y ^{2} - 9 ) ^{2}}

D (x, y) = - \frac{4 ( x ^{2} + y ^{2} + 9 )}{( x ^{2} + y ^{2} - 9 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

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

4 x^{2} y + 2 x y^{2} - 12 x y - 5

K (r, s) = 4 r - s

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 2

What is the extreme f(x,y)=ln(x^2+y^2-9) ?
The extreme f(x,y)=ln(x^2+y^2-9) is Maximum(0,0)