 Solution

Linearizing at the positive steady state gives $n^{\prime }=f^{\prime }(N^{\ast })n$, so a small displacement decays as $n(t)=n(0)e^{-t/\tau }$. The recovery time of a stable equilibrium is thereforeAs $\delta \uparrow r$ with $r,b$ fixed, $N^{\ast }\to 0$ and $\tau \to \infty$. Thus the equilibrium population approaches extinction and recovers increasingly slowly after small perturbations. This is critical slowing down. Exactly at the threshold the leading equation is $N^{\prime }\sim -rb{N}^3$, which gives algebraic decay $N(t)\sim (2rbt{)}^{-1/2}$ instead of exponential recovery.