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Let $f\in C^{\infty}(\mathbb{R})$. A point of inflection of $f$ is a point where the function changes from concave to convex or vice-versa. Equivalently, it is a point at which $f''$ changes sign.

Is it correct to say that points of inflection of $f$ are local minima/maxima of $f'$?

Can points of inflection of $f'$ also be points of inflection of $f$? Or are these points of undulation of $f$?

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I think that once you read my answer to your second question, you may begin to doubt your definition of inflection point, or at least realize that you need to make very clear the notion of "changes sign".

For the second question, you're looking for a function $f$ where both $f''$ and $f'''$ change sign at some point $a$, which we might as well assume (by shifting coordinates) is zero. Now look at $g = f''$. That's a continuous function such that $g$ and $g'$ both change sign at zero and are continuous.

Now we have to make sense of "change sign"; I'm going to assume that "$h$ changes sign at $0$" means that for some small interval around $0$, we have $h(x) < 0$ for $x < 0$, and $h(x) > 0$ for $x > 0$, or vice versa. An alternative notion would be "$h$ changes sign at $0$ if, for every number $c>0$, there are numbers $p$ and $q$ with $-c < p < 0 < q < c$ and $h(p) \cdot h(q) < 0$." I'll discuss that second case in a moment.

For the first definition:

By continuity, $g(0) = 0$. By negating, if necessary, we can assume that for $0 < x < c$, for some small $c$, we have $g'(x) > 0$, and for $-c < x < 0$, we have $g'(x) < 0$. Use these facts, and the fundamental theorem of calculus, we can conclude that for $-c < x < c$, we have $f(x) \ge 0$, hence $f$ does not change sign.

For the second definition:

Let $$ h(x) = \begin{cases} exp(\frac{-1}{x^2}) & x \ne 0 \\ 0 & x = 0 \end{cases} $$ Then $h$ is infinitely differentiable, and all derivatives of $h$ at $0$ are $0$. It's "extremely flat" at zero. It's also even (i.e., $h(-x) = h(x)$), so its derivative is odd, i.e., $h'(-x) = -h'(x)$. Let $k = h'$.

Let's define $$ k(x) = \begin{cases} h(x) \sin\frac{1}{x} & x \ne 0 \\ 0 & x = 0 \end{cases} $$ Clearly $k$ is nice away from $0$. What about derivatives? Well, \begin{align} k'(0) &= \lim_{s \to 0} \frac{k(s) - k(0)}{s} \\ &= \lim_{s \to 0} \frac{h(s) \sin\frac{1}{s} - 0}{s} &= \lim_{s \to 0} \frac{h(s) \sin\frac{1}{s}}{s} \end{align} Now $$ \frac{-h(s)}{s} \le \frac{h(s) \sin\frac{1}{s}}{s} \le \frac{h(s)}{s} $$ so by the squeeze lemma, the limit is between $-h'(0)$ and $h'(0)$, which are both $0$, so the limit exists and is zero. I believe (but have not checked!) that a similar computation shows that all other derivatives also exist at zero.

But clearly both $k$ and its derivative oscillate wildly (but with very tiny amplitude!) near $0$, so they change sign infinitely often near zero, hence satisfy the second definition of "change sign at zero".

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A point of inflection $a$ of $f$ may occur when $f'$ does not exist in a whole neighborhood of $a$. For example $f'$ may fail to exist on a countable dense set.

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  • $\begingroup$ How about if $f$ is smooth? Is it true then? $\endgroup$ – A. Goodier Dec 8 '18 at 13:58

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