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

The extreme f(x,y)=(x^2-y^2)/(sqrt(9-2x^2-y^2)) is Saddle(0,0)

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{4 x ^{2} y ^{2} - 9 y ^{2} - 10 x ^{4} + 81 x ^{2} - 162}{( 9 - 2 x ^{2} - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = \frac{- 180 x ^{6} + 162 x ^{4} y ^{2} - 162 x ^{4} + 54 x ^{2} y ^{4} - 2592 x ^{2} y ^{2} + 10206 x ^{2} - 36 y ^{6} - 162 y ^{4} + 7290 y ^{2} - 26244}{( - y ^{2} + 9 - 2 x ^{2} ) ^{5}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

What is the extreme f(x,y)=(x^2-y^2)/(sqrt(9-2x^2-y^2)) ?
The extreme f(x,y)=(x^2-y^2)/(sqrt(9-2x^2-y^2)) is Saddle(0,0)