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.

there's 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 :)