I did not realize that Arman G. had already supplied an answer to this question.
However, I would like to suggest a different possibility.
Consider the following function:
F(x) = (1/2)x2 - 3x + 3 for x≥3
= -(1/2)x2 +3x for x<3
F'(x)=f(x)= x-3 for x≥3
= -(x-3) for x<3
f clearly has a "cusp" at x=3, but F(x) has a discontinuity at x = 3, not a point of inflection.
In order for the conclusion to be true, I think you need to impose continuity of the
antiderivative at the point where f has the "cusp"...but I am not totally sure of that!