Hi there. I've got this function $\displaystyle f(x,y)=(y-3x^2)(y-x^2)$, and I have to analyze what happens at $\displaystyle (0,0)$ in terms of maxims and minims. But what I actually have to proof is that theres a saddle roof at that point.

Theres is a critical point at $\displaystyle (0,0)$. Lets see:

$\displaystyle f(x,y)=(y-3x^2)(y-x^2)=y^2-4yx^2+3x^4$

$\displaystyle f_x=-8yx+12x^3$

$\displaystyle f_y=2y-4x^2$

Its clear there that there is a critical point at $\displaystyle (0,0)$

The determinant of the partial derivatives of second order at $\displaystyle (0,0)$

$\displaystyle f_{xx}=-8y+36x^2$,

$\displaystyle f_{xy}=-8x=f_{yx}$,

$\displaystyle f_{yy}=2$

$\displaystyle f_{xx}(0,0)=0$

$\displaystyle \left| \begin{array}{ccc}0 \ 0 \\ 0 \ 2 \\ \end{array} \right| =0$

Then I can't say anything from there. And actually, if I try at any line that passes through $\displaystyle (0,0)$ I would find a minimum. I know that it isn't a minimum, but I don't know how to prove it.

Bye there.