Implicit Function theorem and Bifurcation points So let us say we have a function $\dot{x} = f(x,r)$ that has some critical point at $(x_0,r_0)$ such that $f(x_0,r_0)=0$. The question now is: when is this a bifurcation point? I understand that $\frac{\partial{f(x_0,r_0)}}{\partial{x}} = 0$ works in practice, but I have two questions.
1.) Intuitively, why is this the case?
2.) The proof for this appealed to the implicit function theorem and said that if the derivative was non-zero, then there would be a (local) solution $x=X(r)$ such that $X(r_0)=x_0$, which cannot happen if the point is a bifurcation point. This doesn't click with me, why would this be not work if the point was a bifurcation? Also, how is it the case that the Jacobian is zero if only one entry is zero (namely the x-derivative entry)? Shouldn't it be the case that any 2 entries on opposite columns need to be zero?
Thanks
 A: Well, first of all do you assume $f:\mathbb{R}^n\rightarrow\mathbb{R}$ ? If not, the condition for the application of the theorem is slightly more complicated.
But let's say that $x,r\in\mathbb{R}$ and $f(x,r)=r-x^2$. We want to find the set $\mathbb{X}=\{(x,r)\in\mathbb{R^2}|f(x,r)=0\}$. Implicit function theorem says that most of the times this can be written in the form $x=X(r)$ in a neighbourhood of a point.
So in our case lets choose $(x_a,r_a)=(x_a,x_a^2)$ with $x_a>0$, then in a neighbourhood of this point $\mathbb{X}$ can be written as $x(r)=\sqrt{r}$. If we choose $(x_a,r_a)=(x_a,x_a^2)$ with $x_r<0$, then we get $x(r)=-\sqrt{r}$.
Do you see now why we cannot choose $(0,0)$? There is no neighbourhood of the point in which we can write $x$ as a function of $r$.
So, what's the big deal? We see that given $r_e$, the points for which $f(x,r_e)=0$ are the equilibria of the system. The implicit function theorem in this case says that the equilibria of the system can usually be written as a smooth (depending on f) function of $r$. This actually is very useful because we deduct that the topological properties of the equilibrium are constant and its behaviour changes smoothly with the parameter.
However if the implicit function theorem cannot be applied, we cannot say anything of the above and wild things can happen. We call these things bifurcations.
Also I didn't really understand the last part of your question.
