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The following question was presented to me by a tutoring student in AP Calculus. It's supposedly from a practice test - not sure if it's official. Here's the issue.

Below I've reproduced the complete graph of some continuous function $f(x)$. The question asks us to identify what $x$ are inflection points and to offer an explanation as to why. My student's teacher provided her class with an answer that $x=2$ is an inflection point. I disagree.

$f'(x)$ is discontinuous at $x=2$ and $f''(x)$ does not exist! Furthermore, there is no tangent at $x=2$. I see no transition from concavity to convexity anywhere here. From the definitions of an inflection point provided in Stewart's and Thomas's textbooks, no inflection points exist. Can anyone chime in here? Is anyone aware of an alternate definition of inflection point - especially one that is taught in US high schools - that could have been intended here?

graph of f(t)

zahbaz
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    Yup, I don't see any inflection point in the graph, the question looks flawed to me. – learner Mar 10 '16 at 06:09
  • If $x=2$ is an inflection point, why not $-2$ also? I agree that there are none on this graph. – A. Thomas Yerger Mar 10 '16 at 06:31
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    $f|{[-2,2]}$ is concave. $f|{[2,3]}$ is affine-linear and hence convex (and also concave). So, one could say that $f$ transitions from being concave to being convex at $x=2$. – PhoemueX Mar 10 '16 at 06:55
  • @PhoemueX Hah! This is why I come here. We can fix this question! – zahbaz Mar 10 '16 at 07:19
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    @PhoemueX, if we allow concavity, rather than strict concavity, in the definition of "inflection point", then we should equally say that every point in $[-4,-2]$ and $[2,3)$ is an inflection point. This is rather silly. – vadim123 Mar 10 '16 at 14:29

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When I teach inflection points in calculus, I require that the second derivative change sign at a point. So if $f''$ changes from negative to positive at $x$, then I would say that $x$ is an inflection point.

Points where $f'' = 0$ but where $f''$ does not change sign are called "undulation points."

This graph has a very large number of undulation points, and no inflection points.

davidlowryduda
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