Consider the discrete system $f\left ( x \right )=ax^{2}\left ( 1-x \right )$ on the interval $\left [ 0,1 \right ]$.
The fixed points are $x^{*}=0 \forall a $ and $x^{*}=\frac{a\pm \sqrt{a\left ( a-4 \right )}}{2a} \forall a>4 $
It is given that at a=4: $x^{*}=\frac{1}{2}$
and at $a= \frac{16}{3}: x^{*}=\frac{3}{4}$
I would like to know what kind of bifurcation occurs when a=4 and $a=\frac{16}{3}$.
Using the criteria for stability of a map:
$f'\left ( a=4, x^{*}=\frac{1}{2} \right ) =1$ so a fold bifurcation exists; saddle node, transcritical, pitchfork
$f'\left ( a=\frac{16}{3}, x^{*}=\frac{3}{4} \right )=-1$ so a flip bifurcation exists. Hence, it is expected at $x^{*}=\frac{3}{4}$ at $a=\frac{16}{3}$, there is a period doubling. The map $f\left ( x \right )$ has period 1 so it now has period 1.2; in general, a period-doubling cascade have the form $k\cdot 2^{n}$ where k is the k-cycle of a map and n is the number of times period-doubling occurs.
However, I would also like to know how the stability of these fixed point changes and whether new fixed points are created or annihilated. Any qualitative insight is greatly appreciated.
Edit: 