 Solution

A constant orbit is a fixed point of $F(u)=u(1+u)/2$. Solving $F(u)=u$ gives $u(u-1)=0$, henceFor fixed point stability for an iteration, perturb a fixed point $u_{\ast }$ by $v_n$. Its local evolution is $v_{n+1}=F^{\prime }(u_{\ast })v_n+O(v_n^2)$, where $F^{\prime }(u)=u+1/2$. At zero, $F^{\prime }(0)=1/2$, so perturbations contract: zero is locally asymptotically stable. For example, if $|u|\le r<1$, then $|F(u)|\le (1+r)|u|/2$; this gives an invariant neighborhood and geometric convergence to zero.

At one, $F^{\prime }(1)=3/2$, so perturbations expand: one is unstable. More explicitly, putting $u_n=1+v_n$ gives $v_{n+1}=3v_n/2+v_n^2/2$, whose magnitude increases while a nonzero perturbation remains sufficiently small. The derivative test concerns local stability; it does not assert attraction from every real initial value.