I have come to know that a bifurcation point must be non-hyperbolic. But I am not sure whether the converse is also true.
As I considered $f(x)=\mu-x^2$, I got $x^2=\mu$ are bifurcation points.
i) For $\mu<0$, there is no bifurcation point.
ii) For $μ=0$, there is one equilibrium point at $x=0$ which is non-hyperbolic and unstable.
iii) For $μ>0$, the system has two equilibrium points at $x=\pm\sqrtμ$, one of which is stable and the other is unstable.
So $\mu=0$ is the bifurcation point as for this point, the number of equilibrium points and their stabilities are changing.
Now to get my point, my classmate proposed me to choose a function $f(x)=\mu-x^5$. Doing the same analysis, I found that
i) for $\mu=0$, $x=0$ is the only critical point which is non-hyperbolic and so unstable
ii) for $\mu\ne0, x=\mu^{1/5}$ is the only critical point and hyperbolic and so stable.
My question arises here that though the number of critical points is not changing, their stability changes. So does there really any bifurcation occur at $\mu=0$?