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Given some function $f: I \subseteq\mathbb R \rightarrow \mathbb R$, Which is differentiable twice at some point $a\in I$. Can one say that there is a region around the point where the function is differentiable twice, without any other information?

so I assume that its true because if we look at the first derivative which is:

$lim_{h\rightarrow0} \frac {f(a+h)-f(a)}h $ then obliviously we can "take" h to be smaller as we want and then I can assume that if the limit exists then it exists at some region of that point a.

so I guess the same goes for the second derivative.

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  • $\begingroup$ How do you even define twice differentiable in a single point? If we have a function that is only differentiable in a single point, than the first derivative is only defined in this single point, and taking the derivative of that again is kind of useless. $\endgroup$ – Dirk Aug 14 at 11:29
  • $\begingroup$ @Dirk I didn't defined the function to be differentiable only at a single point, I defined some function which can be differentiate twice in some point a. $\endgroup$ – Serlok Aug 14 at 11:32
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Let $f\colon\mathbb R\longrightarrow\mathbb R$ a continuous function which is differentable nowhere, let $F$ be a primitive of $f$ and let $g(x)=x^2F(x)$. Then $g$ is differentiable: $g'(x)=2xF(x)+x^2f(x)$. And $g'$ is differentiable at $0$. But that's the only point of $\mathbb R$ at which $g'$ is differentiable.

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  • $\begingroup$ But because differentiable is defined by a limit, then if we can say the limit exists for 0, cant we say the limit exists in a region around zero? $\endgroup$ – Serlok Aug 14 at 11:45
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    $\begingroup$ I am puzzled by your comment. Did I not prove that that does not occur? Short answer: no, you cannot say that. $\endgroup$ – José Carlos Santos Aug 14 at 11:48
  • $\begingroup$ +1. Nice construction. $\endgroup$ – Kavi Rama Murthy Aug 14 at 11:48
  • $\begingroup$ @KaviRamaMurthy Thank you. $\endgroup$ – José Carlos Santos Aug 14 at 11:50

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