 Solution

For an autonomous scalar differential equation, linear stability of a simple equilibrium point $N_e$ is determined by the derivative $f_N(N_e,a)=P^{\prime }(N_e)$: a negative derivative attracts nearby solutions and a positive derivative gives repulsion in a dynamical system of nearby solutions.

Here $f(0,a)=1+a\ge 1/2>0$ throughout the prescribed mathematical model parameter range, while $f(N,a)\to -\infty$ as $N\to \infty$. If there is only one equilibrium point, the function graph must cross the axis on one of its decreasing branches, outside the interval between its two extrema. The crossing is simple and $P^{\prime }(N_e)<0$. Thus the unique equilibrium point is linearly asymptotically stable. At a saddle-node bifurcation there are two distinct equilibrium points, one of them double, so that case does not fall under the one-equilibrium hypothesis.