What is the extreme f(x)=(3x^2-4xy+2y^2)/(4x+9y^2) ?

The extreme f(x)=(3x^2-4xy+2y^2)/(4x+9y^2) is Saddle(0,-2/9)

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extreme f(x)=(3x^2-4xy+2y^2)/(4x+9y^2)

extreme

Saddle (0, - \frac{2}{9})

Solution steps

Find the critical points:

(0, - \frac{2}{9})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 x ( - 108 x ^{2} + 729 x y ^{2} + 432 x y + 32 x - 324 y ^{3} - 216 y ^{2} )}{( 4 x + 9 y ^{2} ) ^{3}}

D (x, y) = \frac{- 104976 y ^{8} + 314928 x y ^{7} - 93312 y ^{7} - 236196 x ^{2} y ^{6} + 163296 x y ^{6} - 20736 y ^{6} - 104976 x ^{3} y ^{4} + 62208 x ^{2} y ^{4} - 9216 x y ^{4} - 62208 x ^{3} y ^{3} + 18432 x ^{2} y ^{3} - 13824 x ^{3} y ^{2}}{( 9 y ^{2} + 4 x ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, - \frac{2}{9}) :

Saddle

Saddle (0, - \frac{2}{9})

What is the extreme f(x)=(3x^2-4xy+2y^2)/(4x+9y^2) ?
The extreme f(x)=(3x^2-4xy+2y^2)/(4x+9y^2) is Saddle(0,-2/9)