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Why is F=dU/dx=0 either side of inflexion point?
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[QUOTE="walking, post: 6174385, member: 656726"] 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##. [/QUOTE]
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Why is F=dU/dx=0 either side of inflexion point?
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