Emily A.

asked • 12/08/19# If I have a cusp on a graph of f prime (at x=3), does that mean I will have an inflection point on the graph of f (at x=3)?

## 1 Expert Answer

Paul M. answered • 12/08/19

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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)x^{2} - 3x + 3 for x≥3

= -(1/2)x^{2} +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!

Arman G.

12/08/19

Arman G.

12/08/19

Stanton D.

Right, you don't deem a point of discontinuity to be a point of inflection of a function. However, a non-differentiable point on the first derivative might arise from a "semi-cusp" -- it has a non-zero or even infinite slope (2nd derivative) on one side of the point, but a zero slope (linear or constant value original function) on the other side. Don't think that that qualifies as a cusp, per se. To make that something in the real world -- what derivative(s) of position (of your body, along a line on the floor) with respect to time would you be capable of detecting, and what common names would you call each of those derivative(s)? How about measuring, using mechanical means (gauges, levers, etc.)? F(0) = position, F(1) = velocity, F(2) = acceleration, and so on. What derivatives would a motor vehicle be capable of supplying, using the engine only? Using any other part of the car? -- Cheers, -- Mr. d.12/12/19

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John M.

12/09/19