 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-4/3a/e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

V = (x^{2} - 4)^{2}

V^{'} = 4 x (x^{2} - 4)

. Thus $x = \pm 2$ are stable equilibria at the two potential minima, while $x = 0$ is unstable at the local maximum.

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