= Solution
A <transcritical bifurcation> has normal form
$$
\dot x=\mu x-x^2.
$$
Its equilibrium branches are $x=0$ and $x=\mu$. Linearization gives <derivative> $\mu$ on the first and $-\mu$ on the second, so the branches cross at $(\mu,x)=(0,0)$ and exchange stability: for $\mu<0$, $x=0$ is stable and $x=\mu$ unstable; for $\mu>0$ the labels reverse.
A small constant perturbation
$$
\dot x=\mu x-x^2+\varepsilon
$$
has equilibria $x=(\mu\pm\sqrt{\mu^2+4\varepsilon})/2$. For $\varepsilon>0$ the two branches avoid the crossing; for $\varepsilon<0$ there is a parameter interval with no equilibrium, bounded by two saddle-node points. Thus arbitrarily small perturbations change the bifurcation diagram, so the crossing is not structurally stable.
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