1. anonymous

@Spacelimbus

2. anonymous

hmm I can't see the error myself yet, my only guess could be with the fixated values.

3. anonymous

i know that the 3 is wrong but idk why

4. anonymous

The Jacobian Matrix looks right to me too, have you tried with different fixated values? other critical points I mean, I see you have found three of them.

5. anonymous

yea and they all seem to be wrong somehow one of the values in the jacobian end up being wrong

6. anonymous

so it seems like the critical points are the problem then, not your setup.

7. anonymous

$\Large x'=y \\ \Large y'=3x-3x^3$ yes?

8. anonymous

yea

9. anonymous

but my critical points are correct because in a previous question it asked to solve for all the critical points and this was a follow up question

10. anonymous

yes I see.

11. anonymous

$\Large J(x,y)=\left[\begin{matrix}0 & 1 \\ -9x^2+3 & 0\end{matrix}\right]$

12. anonymous

I caught my mistake!!!

13. anonymous

oh? Mind to share?

14. anonymous

I haven't seen it yet.

15. anonymous

when i plugged in the critical poiints into the jacobian there was an algebraic error so the jacobian wouldve been 0 1 -6 0

16. anonymous

point (1,0) huh?

17. anonymous

yea

18. anonymous

hmm I understand it for that point, but I am not sure why point (0,0) shouldn't work, if it's in fact a critical point.

19. anonymous

because this question is asking for the first critical point we found, which we had to put in order from lowest value ot highest, and so we had to use (-1,0)

20. anonymous

Oh I understand, I wasn't aware of that. (-: Well, well done!

21. anonymous

Thank you!!!