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In this post, I am considering only functions of the form $f:\mathbb{R}\rightarrow\mathbb{R}$. It's common to classify stationary points into local maximum points, local minimum points, and saddle points / stationary inflection points.

Is this classification exhaustive? If so, what do we classify the point $x=0$ in the function $f$ given by $$ f(x) = \begin{cases} x^{3}\sin\left(\frac{1}{x}\right) &\text{ if }x\ne 0, \\ 0 &\text{ if }x = 0. \end{cases}. $$

This has $f'(x) = 0$, but the $x=0$ point seems to give neither a local maxima, local minima, nor a stationary inflection point.

Now this question is part semantics, but I must ask nonetheless: Is this point classified as a stationary point? An inflection point? A saddle point? A critical point?

MaximusIdeal
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1 Answers1

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The usual definition of inflection point (see e.g. Wikipedia) is "a point where the function changes from being concave (concave downward) to convex (concave upward), or vice versa." So this would not include a case where on at least one side of the point in question, there are points arbitrarily close to it where the function is strictly concave and others where it is strictly convex.
BTW, your example does not have a second derivative at $0$, but there are other examples that are smooth. For example, you could take $$f(x) = \cases{\exp(-1/x^2) \sin(1/x) & if $x \ne 0$\cr 0 & if $x=0$} $$

Robert Israel
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  • The function given by $f(x)=x^3\sin(1/x)$ when $x\ne0$ and $f(0)=0$ does have a second derivative (of $0$) at $0$. – Lucenaposition Aug 13 '24 at 22:45
  • @Lucenaposition If $f(x) = x^3 \sin(1/x)$, $f'(x) = 3 x^2 \sin(1/x) - x \cos(1/x)$. Thus there are points arbitrarily close to $0$ where $f'(x) = x$, and thus $\frac{f'(x) - f'(0)}{x} = 1$, and points arbitrarily close to $0$ where $f'(x) = -x$, and thus $\frac{f'(x) - f'(0)}{x} = -1$. So $f''(0) = \lim_{x \to 0} \frac{f'(x)-f'(0)}{x}$ does not exist. – Robert Israel Aug 14 '24 at 15:08
  • For some reason Desmos thinks it is defined at $0$. – Lucenaposition Aug 14 '24 at 22:39