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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.

user264985
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  • Is there a derivative at a cusp? – Henry Nov 10 '15 at 23:55
  • derivative doesnt exist at a cusp – user264985 Nov 10 '15 at 23:55
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    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. – Henry Nov 11 '15 at 00:03

2 Answers2

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I would not consider $f(x)=x^{2/3}$ to be concave down on $(-\infty,\infty)$, but rather on $(-\infty, 0)\cup(0,\infty)$. If we use the definition of concave down to be "A function $f$ is concave down when $f'$ is decreasing", then we can see this is not the case with $f(x)=x^{2/3}$. Its derivative, $f'(x)=\frac{2}{3\sqrt[3]{x}}$, is decreasing on its domain, which is $(-\infty, 0)\cup(0,\infty)$. So by that definition, $f(x)$ is not concave down on $(-\infty,\infty)$.

While I agree that $x^{2/3}$ does not have an inflection point at its cusp, that doesn't mean we can't have inflection points at cusps. It really depends on your definition of inflection point. You can easily make these types of cusps appear by taking absolute values of functions. For example: $g(x)=\left| x^2-1\right|$ has cusps at $x=\pm 1$ and also changes concavity there.

A graph with a change in concavity at a cusp

Carly Vollet
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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.

abel
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