Solution
= Solution
For $V=(x^2-4)^2$, $V'=4x(x^2-4)$. Thus \b[$x=\pm2$ are stable equilibria] at the two potential minima, while \b[$x=0$ is unstable] at the local maximum.
= Solution
For $V=(x^2-4)^2$, $V'=4x(x^2-4)$. Thus \b[$x=\pm2$ are stable equilibria] at the two potential minima, while \b[$x=0$ is unstable] at the local maximum.