Proof of Saddle Roof at (0,0) for f(x,y)=(y-3x^2)(y-x^2)

[Telemachus]

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

Theres is a critical point at (0,0). Let's see:

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

f_x=-8yx+12x^3
f_y=2y-4x^2
Its clear there that there is a critical point at (0,0)

The determinant of the partial derivatives of second order at (0,0)
f_{xx}=-8y+36x^2,
f_{xy}=-8x=f_{yx},
f_{yy}=2

f_{xx}(0,0)=0

\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 (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.

 

[LCKurtz]

In the xy plane draw the two curves y = x² and y = 3x². These two curves give the points in the domain where f(x,y) = 0. Everywhere else in the plane f(x,y) isn't zero and its sign depends on the signs of the two factors

(y -x²)(y-3x²)

So make a note in the plane where the product of those two factors is positive or negative. See if you can see a path through (0,0) where f achieves a max and another where it achieves a min.

 

[Telemachus]

Thank you very much :)