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This is supposed to be for bullet 5.2 in this question: A rigorous yet intuitive summary of inflection and critical points for beginning calculus?

I'm missing something very obvious, but what's an example where $f''(x)$ is undefined, but $(x,f(x))$ is not an inflection point of $f$?


After some thought, I have come up with some examples. Are these correct?

First example: This example is one where $f$ is defined but not continuous at $x$.

  • $f: \mathbb R \to \mathbb R, f(t)=\text{sign}(t), x=0$. $f'(0)$ is undefined and thus so is $f''(0)$. However $f''(t)=0$ for $t \ne 0$.

  • In case someone asks further 'Okay, what if we assumed $f$ is continuous? Then $x$ would have to be an inflection point right?', then we proceed:

Second example: This example is one where $f$ is continuous but not differentiable at $x$.

  • A sharp turn like $g(t)=|t|$ but always increasing at an increasing rate but at different rates $f: \mathbb R \to \mathbb R, f(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$. This gives us $f': \mathbb R \setminus \{0\} \to \mathbb R, f'(t) = \text{sign}(t)+2$. $f'(0)$ is undefined and thus so is $f''(0)$. However $f''(t)>0$ for $t \ne 0$.

  • In case someone asks further 'Okay, what if we assumed $f$ is differentiable? Then $x$ would have to be an inflection point right?', then we proceed:

Third example: This example is one where $f$ is differentiable at $x$.

  • $f': \mathbb R \to \mathbb R, f'(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$.
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  • $\begingroup$ Case 3 seems like the best answer here, since $f’’(0)$ isn’t defined in the first two cases. I would consider case 3 a good answer to this. $\endgroup$ – David M. Feb 24 at 13:11
  • $\begingroup$ @DavidM. $f''(0)$ is supposed to be undefined in all cases. What do you mean? $\endgroup$ – Mitjackson Feb 24 at 16:30
  • $\begingroup$ Oops I completely misread. What exactly is your question? $\endgroup$ – David M. Feb 24 at 16:40
  • $\begingroup$ @DavidM. Edited: This is supposed to be for bullet 5.2 in this question: A rigorous yet intuitive summary of inflection and critical points for beginning calculus? $\endgroup$ – Mitjackson Feb 28 at 5:38
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I can’t tell what you're asking, so I’ll answer the question in the title. Or, more precisely, I’ll give an example of an $f$, differentiable on the whole real line, with the property you describe in the title. I can’t tell what you’re doing by listing all those cases.

Take $f$ to be $x^2$ when $x$ is negative and $x^4$ when $x$ is non-negative.

Then $f’’(0)$ is undefined, and moreover $f$ does not have an inflection point at the origin.

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  • $\begingroup$ Thanks! I was giving several answers with different conditions. One might say after the first answer $f(t) = sign(t)$, something like 'Okay, what if we assumed $f$ is continuous? Then $x$ would have to be an inflection point right?' Then comes the second answer. Does that clarify things? $\endgroup$ – Mitjackson Feb 28 at 6:06
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    $\begingroup$ So what is your question??? $\endgroup$ – symplectomorphic Feb 28 at 6:07
  • $\begingroup$ The first question is give an example. If a user gives an answer of a non-continuous function, the next question is give an example of continuous. If a user gives an answer of a non-differentiable function, the next question is give an example of a differentiable function, which you did. Thanks! (I will still edit my post to clarify) $\endgroup$ – Mitjackson Feb 28 at 6:08
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    $\begingroup$ No, you aren’t understanding. What is your question? It seems like you aren’t asking a question at all. $\endgroup$ – symplectomorphic Feb 28 at 6:09
  • $\begingroup$ Okay I edited my question now. Anyway, thanks! $\endgroup$ – Mitjackson Feb 28 at 6:12

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