 Solution

Write $h={\mu }^2+\epsilon$. The entire unfolded diagram is obtained from $g(x)=h$, with the same dynamical stability sign $-x(x+2)$ as before. A saddle-node bifurcation at $x=0$ requires $h=0$, while a saddle-node bifurcation at $x=-2$ requires $h=4/3$.

For $0<\epsilon <4/3$, the crossing separates into a positive stable branch and a negative unstable branch, with no saddle-node bifurcation near $\mu =0$. Three equilibria exist for $|\mu |<\sqrt{4/3-\epsilon }$, and the two negative branches annihilate at the outer saddle-node bifurcations $\mu =\pm \sqrt{4/3-\epsilon }$. Outside that interval only the positive stable equilibrium remains. If $\epsilon >4/3$, the diagram has just that one branch everywhere; equality is the limiting saddle-node bifurcation contact at $\mu =0,x=-2$.

For $\epsilon <0$, let $d=-\epsilon >0$. Two inner saddle-node bifurcations occur at $\mu =\pm \sqrt{d},x=0$, and two outer saddle-node bifurcations at $\mu =\pm \sqrt{4/3+d},x=-2$. For $|\mu |<\sqrt{d}$ there is a single stable equilibrium below $-3$; for $\sqrt{d}<|\mu |<\sqrt{4/3+d}$ there are three equilibria with stable–unstable–stable ordering; outside the outer saddle-node bifurcations only the positive stable equilibrium remains. The preceding figure shows representative small unfoldings. In particular this unrestricted perturbation destroys the exact transcritical bifurcation crossing.