 Solution

A transcritical bifurcation has normal formIts 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 perturbationhas equilibria $x=(\mu \pm \sqrt{{\mu }^2+4\epsilon })/2$. For $\epsilon >0$ the two branches avoid the crossing; for $\epsilon <0$ 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.