= Solution
Put $k=\beta/\gamma$. The <nullclines> of the space-clamped <FitzHugh-Nagumo model> are
$$
\dot v=0:\quad w=f(v)+I,\qquad
\dot w=0:\quad w=kv.
$$
The first <nullcline> is a downward-leading <cubic polynomial>, shifted vertically by $I$; the second <nullcline> is a line through the origin with slope $k>0$. Their intersections satisfy
$$
I=h(v),\qquad h(v)=kv-f(v)=v^3-(1+\alpha)v^2+(\alpha+k)v.
$$
Writing $M=(1-\alpha+\alpha^2)/3$, differentiation gives
$$
f'(v)=-3\left(v-\frac{1+\alpha}{3}\right)^2+M,\qquad h'(v)=k-f'(v).
$$
If $k>M$, $h$ increases strictly, so there is exactly one <equilibrium>. If $k=M$, $h$ is still strictly increasing, although its <derivative> vanishes at one point, so the <equilibrium> is again unique. If $0<k<M$, define
$$
u_\pm=\frac{1+\alpha\pm\sqrt{1-\alpha+\alpha^2-3k}}3,\qquad
I_{\rm low}=h(u_+),\quad I_{\rm high}=h(u_-).
$$
Here $h$ has a local maximum at $u_-$ and a local minimum at $u_+$; therefore
$$
\boxed{\begin{array}{ll}
I_{\rm low}<I<I_{\rm high}:&\text{three distinct equilibria},\\
I<I_{\rm low}\ \text{or}\ I>I_{\rm high}:&\text{one equilibrium}.
\end{array}}
$$
At either endpoint there are two distinct <equilibria>, one a double root. These are the <saddle-node bifurcation> boundaries omitted by the generic one-or-three wording. At $k=M$ and $I=h((1+\alpha)/3)$ there is one triple root. This is the <Cubic FitzHugh-Nagumo nullcline criterion>.
\Image[/past-exam-of-the-further-mathematical-biology-course-of-the-university-of-oxford/oxford-2023-further-mathematical-biology-nullclines.png]
{title=One and three intersections of the <FitzHugh-Nagumo model> <nullclines>}
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