Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 32A / iii / Solution

For $\lambda >1$, the new fixed point in $[0,1]$ isThe origin has multiplier $F_x(0,\lambda )=\lambda >1$ and is therefore unstable. At the new fixed point,HenceThe point $x^{\ast }$ is stable in that interval and unstable for $\lambda >2$. At $\lambda =2$ its multiplier is $-1$, the threshold for a period-doubling bifurcation.