What is the extreme f(x,y)= 3/(x^2+y^2-9) ?

The extreme f(x,y)= 3/(x^2+y^2-9) is Maximum(0,0)

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extreme f(x,y)= 3/(x^2+y^2-9)

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{6 ( - 3 y ^{2} + x ^{2} - 9 )}{( x ^{2} + y ^{2} - 9 ) ^{3}}

D (x, y) = \frac{- 108 x ^{4} - 216 x ^{2} y ^{2} + 648 x ^{2} - 108 y ^{4} + 648 y ^{2} + 2916}{( x ^{2} + y ^{2} - 9 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

6 x + 12

- 2 x^{2} - x + 3

y = 5 x^{2} - 3 x - 1

f (x, y) = (3 x^{2} - y^{2}) e^{5 x}

f (x) = 2 x^{3} - 24 x - 1

What is the extreme f(x,y)= 3/(x^2+y^2-9) ?
The extreme f(x,y)= 3/(x^2+y^2-9) is Maximum(0,0)