Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 4 / 3A / e

For $V=(x^2-4{)}^2$, $V^{\prime }=4x(x^2-4)$. Thus $x=\pm 2$ are stable equilibria at the two potential minima, while $x=0$ is unstable at the local maximum.

For $x\ne 0$, $V^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$ never vanishes. At $x=0$, the smooth flat extension has $V^{\prime }(0)=0$, and $V(x)>V(0)=0$ for every $x\ne 0$. Hence $x=0$ is the unique equilibrium and is stable.