= Solution
Set $g(x)=x^2+x^3/3$. <Equilibria> satisfy $\mu^2=g(x)$, and their scalar linear <eigenvalue> is
$$
F_x=-x(x+2).
$$
It vanishes only at $x=0$ or $x=-2$. Substitution in the <equilibrium> relation gives the three nonhyperbolic parameter-state points
$$
\boxed{(\mu,x)=(0,0),\quad(2/\sqrt3,-2),\quad(-2/\sqrt3,-2).}
$$
The branches with $x<-2$ and $x>0$ are stable, while $-2<x<0$ is unstable. For $0<|\mu|<2/\sqrt3$ there are three <equilibria>; beyond the two outer <saddle-node bifurcations> there is just the positive stable one. At $\mu=0$ there is also the distinct stable <equilibrium> $x=-3$.
Near the crossing, use the smooth coordinate $y=x\sqrt{1+x/3}$, whose derivative is positive near zero. The equation becomes $\dot y=p(y)(\mu^2-y^2)$ with $p>0$. A positive time rescaling removes $p$, and the parameter-dependent state shift $u=y-\mu$ gives
$$
\frac{du}{d\tau}=-2\mu u-u^2.
$$
Thus \b[the crossing at $\mu=0$ is a <transcritical bifurcation>], with parameter $-2\mu$. The two smooth branches $y=\pm\mu$ exchange <dynamical stability>; labeling the stable branch by $|\mu|$ instead would obscure that exchange. The <parameter-dependent coordinate shift in a transcritical bifurcation> explains why the original coordinate need not display a branch identically equal to zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-60-scalar-unfoldings.png]
{title=Scalar equilibrium diagrams before and after positive and negative unfolding, with stable branches solid and unstable branches dashed}
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