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Let $f(x) = \begin{cases} \sin(\frac{1}{x})\cdot e^{-\frac{1}{x^2}}, & \text{if $x\neq0$} \\[2ex] 0, & \text{if $x=0$} \end{cases}$

Does $f''(0)$ exist? Is $x_0=0$ inflection point?

Regarding the first part, do I have to check continuity of $f'$ in $x_0=0$ and if $\lim_{x\rightarrow0^+}f''(x)=\lim_{x\rightarrow0^-}f''(x)$? And I don't even know how to start the second part - should I find $f^{(4)}$? I don't think so...

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  • $\begingroup$ An inflection point is a point where concavity switches. $f’’(0)=0$, but this is really not sufficient to tell you anything. $\endgroup$ Apr 2 at 21:09
  • $\begingroup$ Since $\exp(-1/x^2)$ just completely bulldozes the other factors for $x \rightarrow 0$, $f^{(n)}(0)=0$ so that's not really helping. $\endgroup$ Apr 2 at 22:03
  • $\begingroup$ then what should I do? $\endgroup$
    – zaba12
    Apr 2 at 22:45

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To answer the first part, it's pretty easy to check that $\lim \limits_{x \to 0-}f(x) = f(0) = \lim \limits_{x \to 0+}f(x)$ and because the localized area around $0$ acts linear you can treat the subsequent derivatives like they are $0$. Because of that, we can say that $f''(x)=0$, but because the surrounding points are also $0$, we can't say that it is an inflection point just like how the second derivative of a linear function or a constant also can't have an inflection point.

As for checking continuity, it's always best practice to at least try to if you can. And because you are checking a differentiable function around one point, the limits around that point will always equal each other: $$\lim \limits_{x \to 0-}f(x) = \lim \limits_{x \to 0+}f(x) \mid \lim \limits_{x \to 0-}f'(x) = \lim \limits_{x \to 0+}f'(x)\mid \lim \limits_{x \to 0-}f''(x)=\lim \limits_{x \to 0+}f''(x)\mid... $$ so it isn't necessary to hand check $\lim \limits_{x \to 0}f^{(4)}(x)$.If you want more information, you can check this page for the rules about integrating piecewise functions.

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