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Can cusps be considered points of inflection?

I'm getting conflicting information but my thought process is that cusps cannot be points of inflection?

Can points of inflection exist when there is a vertical tangent to the graph? Assume there is change in concavity and the function is continuous.

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  • $\begingroup$ Is there a derivative at a cusp? $\endgroup$ – Henry Nov 10 '15 at 23:55
  • $\begingroup$ derivative doesnt exist at a cusp $\endgroup$ – user264985 Nov 10 '15 at 23:55
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    $\begingroup$ So it may depend on whether smoothness is part of your definition of a point of inflection or whether you use a definition that does not require smoothness (e.g. any change between convexity and concavity). My personal inclination would be to say not. $\endgroup$ – Henry Nov 11 '15 at 0:03
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This posting does not answer the original question. The original question was whether a cusp point should be considered as an inflection point if the concavity (sign of second derivative) changes across this point.

In most Calculus textbooks, authors define inflection point "loosely" so that cusp point can be an inflection point. ( Typical Definition: A continuous function f has inflection at c if the sign of f'' changes across c.) Authors of Calculus books usually don't require the function to be differentiable at the inflection point.

But, in the end, none of these authors exhibit examples of cusp type inflection points. For all practical purpose, therefore, cusps should not be allowed as inflection point.

Some other authors (minority) require that f should be differentiable at an inflection point (so that a cusp cannot be an inflection point.) But, this definition is now too restrictive; with this definition, cube-root-of x function would not have an inflection point at x=0.

Of course, there is a way to define inflection point so that vertical tangent points are included but cusps are excluded. And, that is the exact meaning of inflection in geometry.

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no. look at the graph of $y = x^{2/3}.$ this has a cusp at $(0,0)$ but concave down on $(-\infty, \infty)$ and $(0,0)$ is certainly not a point of inflection.

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