Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 32B / b

At zero, $F^{\prime }(0)=-(1+\mu )$, so it is stable for $-2<\mu <0$, loses stability through multiplier $+1$ at $-2$ and $-1$ at $0$. Nonzero fixed points areborn in a saddle-node at $\mu =-2-q^2/(4p)$. Their stability follows from $F^{\prime }(x_{\ast })=3+\mu +p{x}_{\ast }^2$; $|F^{\prime }|<1$ marks solid branches. Thus the three bifurcations are the saddle-node, the exchange at $-2$, and the flip at $0$.

With $q=0$, expand $F(F(x))$ for $x=O(\sqrt{\mu })$ through $O({\mu }^{3/2})$. Besides zero, $F^2(x)=x$ gives $x^2=\mu /p+O({\mu }^2)$, and the two points are exchanged by $F$. The derivative product is $1-4\mu +O({\mu }^2)$, so the period-two orbit $|x|=\sqrt{\mu /p}+O({\mu }^{3/2})$ is stable.