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I was wondering whether a function can have a point in which its derivative is zero but the point is neither local min, local max or inflection point. I was considering the function $$f(x)=\begin{cases}x^2\sin\frac1x,\ x\neq 0;\\ 0,\ x=0.\end{cases}$$ The derivative at $x=0$ is zero (can be proven using limit definition of derivative). But for every $h>0$ there are positive and negative numbers for $f$ in $(-h,h)$, and therefore this isn't a local maximum or minimum. Is the point $x=0$ an inflection point or not? How can this be proven? $$$$ Thank you

ryang
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Omer Paz
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  • Are you sure that the derivative at x=0 exists? – A. Pesare Mar 14 '21 at 17:11
  • Yes, one proof is here https://math.stackexchange.com/questions/393602/differentiability-of-fx-x2-sin-frac1x-and-f – Omer Paz Mar 14 '21 at 17:44
  • $f'(0)$ is indeed $0$, but $f'$ is still discontinuous at $x=0$ and the second derivative isn't defined at that point. This makes it impossible to talk about the curvature of $f$ at $x=0$, and if one goes by the definition of an inflection point as a change of convexity, then $x=0$ can't be an inflection point for $f$. I would still consider it a 'critical' point, but it could be argued that this doesn't match the usual definition of critical points either. – user3733558 Mar 14 '21 at 21:28
  • @user3733558 A critical point is one whose first derivative is either zero or non-existent, isn't it? Going by this definition, here $x=0$ is certainly a stationary critical point. – ryang Mar 15 '21 at 06:26
  • @RyanG You're right. I thought I remembered an additional requirement for $f$ to be in $C^1$ if the derivative is $0$, but that doesn't seem to be the case. May have been specific to a study I was reading (many, many) years ago. So, I stand corrected, that's clearly a critical point. – user3733558 Mar 15 '21 at 07:33

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Good question!

  • The given function $f'(x)$ is zero but discontinuous† at $0.$

    So, for the same reason you cited as why it isn't a turning point, the stationary point $0$ isn't an inflection point either. (Also, neither the sign test nor second-derivative test are applicable.)

  • Another example of a non-inflection (curvature doesn't change sign), non-turning stationary point is the point $0$ of the function $g(x)=3.$

  • On the other hand, if a function has a continuous first derivative at, and is neither constant left of nor right of, its stationary point, then the stationary point is surely either a turning or inflection point.

  • Incidentally, the function $h(x)=|x|$ has a non-stationary turning point at $0.$

† $$f'(x)=\begin{cases}2x\sin\frac1x-\cos\frac1x,\ x\neq 0;\\ 0,\ x=0.\end{cases}$$ Let $\left\langle x_n:n\in\mathbb N\right\rangle=\left\langle \frac1{2n\pi}:n\in\mathbb N\right\rangle$ so that $x_n\rightarrow0$ and $n\neq0.$ But $f'(x_n)=-1\nrightarrow0=f'(0).$

ryang
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