= Solution
{c}
Linearizing at the positive <steady state> gives $n'=f'(N^*)n$, so a small displacement decays as $n(t)=n(0)e^{-t/\tau}$. The <recovery time of a stable equilibrium> is therefore
$$
\boxed{\tau=-\frac1{f'(N^*)}=\frac r{2\delta(r-\delta)}.}
$$
As $\delta\uparrow r$ with $r,b$ fixed, $N^*\to0$ 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 $N'\sim-rbN^3$, which gives algebraic decay $N(t)\sim(2rb t)^{-1/2}$ instead of exponential recovery.
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