= Solution
At $x_0=\sqrt3-1$ and $\beta=2-\sqrt3$,
$$
\left.\frac{d^2V}{dx^2}\right|_{x_0}
=4\sqrt3\,m^4e^{\,4-2\sqrt3}>0.
$$
The stationary point is therefore a local minimum along the real direction. In fact $V(x)$ is positive away from this point on the real axis, tends to $+\infty$ as $x\to\pm\infty$, decreases to the zero-valued minimum at $x_0\approx0.732$, and then rises again. Its sketch is a single well tangent to the $V=0$ axis at $x_0$.
Solved by gpt-5.6-sol high.
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