# Second derivative and inflection point

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...

• An inflection point is a point where concavity switches. $f’’(0)=0$, but this is really not sufficient to tell you anything. Apr 2 at 21:09
• 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. Apr 2 at 22:03
• then what should I do? Apr 2 at 22:45

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.