# Why is force 0 either side of an inflexion point in neutral equilibrium?

In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where $$\frac{dU}{dx}=0$$ and also $$\frac{dU}{dx}=0$$ for a small displacement either side of the point. However I do not understand why $$\frac{dU}{dx}$$ remains $$0$$ either side of the inflexion point. Surely the gradient of an inflexion point is just $$0$$ at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that $$\frac{dU}{dx}\ne 0$$.

The stable or unstable equilibrium is the point at which $$dU/dx=0$$. Let's use:

$$F = -\frac{dU}{dx} \tag{1}$$

to write it as $$F=0$$. Now suppose we move a small distance $$\delta x$$ from this position, then to first order the change in the force $$F$$ is:

$$\delta F = \frac{dF}{dx} \delta x$$

and substituting using equation (1) we get:

$$\delta F = \frac{d^2U}{dx^2} \delta x \tag{2}$$

For a neutral equilibrium we require $$\delta F = 0$$ and hence from (2) we get:

$$\frac{d^2U}{dx^2} = 0$$

which is of course the condition for a point of inflexion. The condition given in Tipler and Mosca:

$$\frac{dU}{dx}=0$$ for a small displacement either side of the point

is just the condition that $$d^2U/dx^2 = 0$$.