- #1
Telemachus
- 835
- 30
Hi there. I've got this function [tex]f(x,y)=(y-3x^2)(y-x^2)[/tex], and I have to analyze what happens at [tex](0,0)[/tex] 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 [tex](0,0)[/tex]. Let's see:
[tex]f(x,y)=(y-3x^2)(y-x^2)=y^2-4yx^2+3x^4[/tex]
[tex]f_x=-8yx+12x^3[/tex]
[tex]f_y=2y-4x^2[/tex]
Its clear there that there is a critical point at [tex](0,0)[/tex]
The determinant of the partial derivatives of second order at [tex](0,0)[/tex]
[tex]f_{xx}=-8y+36x^2[/tex],
[tex]f_{xy}=-8x=f_{yx}[/tex],
[tex]f_{yy}=2[/tex]
[tex]f_{xx}(0,0)=0[/tex]
[tex]\left| \begin{array}{ccc}0 \ 0 \\ 0 \ 2 \\ \end{array} \right| =0[/tex]
Then I can't say anything from there. And actually, if I try at any line that passes through [tex](0,0)[/tex] I would find a minimum. I know that it isn't a minimum, but I don't know how to prove it.
Bye there.
Theres is a critical point at [tex](0,0)[/tex]. Let's see:
[tex]f(x,y)=(y-3x^2)(y-x^2)=y^2-4yx^2+3x^4[/tex]
[tex]f_x=-8yx+12x^3[/tex]
[tex]f_y=2y-4x^2[/tex]
Its clear there that there is a critical point at [tex](0,0)[/tex]
The determinant of the partial derivatives of second order at [tex](0,0)[/tex]
[tex]f_{xx}=-8y+36x^2[/tex],
[tex]f_{xy}=-8x=f_{yx}[/tex],
[tex]f_{yy}=2[/tex]
[tex]f_{xx}(0,0)=0[/tex]
[tex]\left| \begin{array}{ccc}0 \ 0 \\ 0 \ 2 \\ \end{array} \right| =0[/tex]
Then I can't say anything from there. And actually, if I try at any line that passes through [tex](0,0)[/tex] I would find a minimum. I know that it isn't a minimum, but I don't know how to prove it.
Bye there.