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(C) Equilibrium points occur where the force $F = -\frac{dV}{dx} = 0$, which corresponds to the points where the slope of the $V(x)$ versus $x$ graph

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Three equilibrium points, one stable two unstable

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(C) Equilibrium points occur where the force $F = -\frac{dV}{dx} = 0$, which corresponds to the points where the slope of the $V(x)$ versus $x$ graph is zero (i.e.,the peaks and valleys of the curve).Looking at the graph, there are three such points: two local maxima (peaks) and one local minimum (valley).$1$. At a local minimum, the potential energy is at a minimum, so $\frac{d^2V}{dx^2} > 0$. This represents a stable equilibrium point.$2$. At a local maximum, the potential energy is at a maximum, so $\frac{d^2V}{dx^2} Since there are two peaks (unstable) and one valley (stable), there are three equilibrium points in total: one stable and two unstable.

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

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