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I haven't done these in awhile. My analysis covered continuity but not differentiability. I have so far not revisited these in learning geometry or algebra. I am trying to help a calculus student, so any notion of 'interior' is intuitive.

First, please verify if these are correct. A rigorous yet intuitive summary of inflection and critical points for beginning calculus?

I am splitting this up to not cover a lot in one post.

Second, I use the above to rigorously (as rigorously as possible for calculus students) answer as follows. Please verify. If an answer or argument (such as if something in the first part above is wrong) is incorrect, then please give the corresponding correct answer or argument.:

  • We are given a function $h$, continuous on $[-2,2]$ and differentiable on a set which is initially stated to be $(-2,2)$ but then is corrected from $(-2,2)$ to $(-2,0) \cup (0,2)$. Then the graph of $h'$ is given below, and we are asked to give critical and inflection points in $(-2,2)$. I would also like to answer on $[-2,2]$.

enter image description here

  • My understanding:

    1. The correction of 'differentiable on $(-2,2)$' to 'differentiable on $(-2,0) \cup (0,2)$' should instead be to 'differentiable on $(-2,-0.5) \cup (0.5,2)$' because $h'(0)$ exists namely, $h'(0) = 1$.

    2. Whether or not $h$ is defined to have domain as $[-2,2]$, $\mathbb R$ or some proper $[-2,2] \subset D \subset \mathbb R$ is irrelevant, at least in asking for the critical and inflection points in $(-2,2)$.

    3. Besides $h$ being given as differentiable on $(-2,2)$ (except for some point in between), we are also given that $h$ is not differentiable at $x=-2$ and $x=2$ because the graph shows holes at $x=-2$ and $x=2$. (The problem never said differentiable only at $(-2,2)$). This too is irrelevant for in asking for the critical and inflection points in $(-2,2)$.

Critical points of $h$ in $(-2,2)$:

  • Based on Definition (1),

    1. $x=-1.5$ and $x=1$ are critical points of $h$ in $(-2,2)$ because they are interior points of $(-2,2)$ (because every point in $(-2,2)$ is interior. Hopefully this is intuitive) such that $h'(x)=0$.

    2. $x=-0.5$ is a critical point of $h$ because it is an interior point $(-2,2)$ such that

    3. $x=-2$ and $x=2$ are not critical points of $h$ in $(-2,2)$ because they are not in $(-2,2)$.

Critical points of $h$ in $[-2,2]$:

  1. This is the same as the critical points of $h$ in $(-2,2)$. In particular, the answer (3) does not change because $x=-2$ and $x=2$ are still not critical points of $h$ because they are not in the interior of $[-2,2]$, which is $(-2,2)$.

Inflection points of $h$ in $(-2,2)$:

  1. Based on Proposition (6), the inflection points in $(-2,2)$ are elements of $\{-1,0,1\}$

  2. $x=-1,0,1$ are all interior points so Definition (3) applies and then says they are all inflection points where why $x=1$ is not ruled out can be explained with Example (4.1).

Inflection points of $h$ in $[-2,2]$:

  1. Based on Proposition (6), the inflection points in $[-2,2]$ are elements of $\{-2,-1,0,1,2\}$

  2. Again, $x=-1,0,1$ are inflection points.

  3. Now $x=-2,2$ are not interior points, so Definition (3) does not apply. Applying Definition (2.1) gives us:

    • 9.1. If $h$ is defined on $[-2,2]$, then they are inflection points because (a) tangent lines exist at $x=-2$ and $x=2$ because $h$ is continuous at $x=-2$ and $x=2$ (b).

    • 9.2. If $h$ is defined on some $[-2,2] \subset D \subseteq \mathbb R$, then they may or may not be inflection points.

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    Derivatives defined on intervals satisfy the intermediate value property on that interval, and thus the graph shown CANNOT be the graph of a derivative. This is a common error I've seen teachers make (even a math professor with Ph.D. I was team-teaching a calculus course with), although it's probably less common in recent years (say, since mid 2000s or so) because "Darboux's theorem" has been so thoroughly discussed in internet math discussions in recent years. – Dave L. Renfro Feb 23 '19 at 09:02
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    Looking more carefully at the diagram, it might be OK after all, since the derivative is not defined at $-0.5.$ This "jump discontinuity" of the derivative can be achieved by having a left derivative of $0$ (levels out to $0$ in the limit from the left, to be more useful in obtaining the original graph) and a right derivative of about $0.75$ (again, levels out to ... from the right), by having a "corner point" (also "angular point", "point anguleux") there. Regarding such points, this construction may be of interest. – Dave L. Renfro Feb 23 '19 at 09:16
  • @DaveL.Renfro It says that Darboux's theorem's assumes differentiable on a closed interval $I=[a,b]$. Is it understood that differentiable on $I$ means differentiable on $(a,b)$, the interior of $I$? –  Feb 23 '19 at 11:59
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    As far as I can tell, the Darboux property result holds for any interval I of positive (or infinite) length, and when an interval has one or two endpoints, the meaning of differentiability at such an endpoint is that the appropriate one-sided derivative at the endpoint exists. I don't know why the Wikipedia article restricts itself to I being a closed interval, since it's the compactness of the subinterval $[a,b]$ that is actually used in the proof. This book is a good reference for a lot of these types of issues. – Dave L. Renfro Feb 23 '19 at 14:50
  • @DaveL.Renfro Thanks! Actually, which part is wrong exactly? It looks like the only problem is the no value at what $y_0$, the value that $h'(-2)$ would be if the hole was filled in, right? Also, $y_0$ can either be $h'(-2)$ or $h'(-0.5)$, right? –  Feb 24 '19 at 02:37
  • @DaveL.Renfro You said differentiability at the endpoint of an interval refers to one-sided derivative. Is that implicit in Definition (2.1) ? –  Feb 24 '19 at 02:58
  • I notice that I did not state something very well. When I said "when an interval has one or two endpoints ...", what I meant is that the interval HAS an endpoint and that endpoint is AN ELEMENT OF the interval. Thus, if $I = (a,b),$ then nothing about differentiability of the function at either $a$ or $b$ is assumed. Indeed, the function is not even assumed to be defined at $a$ or $b.$ Regarding your Definition (2.1), I note that if $A = {0, , 0.9,,0.99,,\ldots},$ then $x=1$ is in the boundary of $A$ but this set $A$ and its boundary point might be problematic. – Dave L. Renfro Feb 24 '19 at 07:03
  • @DaveL.Renfro I said $(x,f(x))$, so $x \in A$ ? Also if $f'(x)$ is defined, then $f(x)$ is defined? –  Feb 24 '19 at 16:32
  • @DaveL.Renfro Again, which part of the exercise is wrong such that the graph cannot be a derivative? –  Feb 24 '19 at 16:38
  • I think the graph is fine, although the meaning of "its derivative" is perhaps not fully up to the standards you're working at. Note that we're not told what the domain of the function is, only that it is continuous at each point of $[-2,2]$ (perhaps it is defined elsewhere, and at some of those elsewhere points it may or may not be continuous), and we're only told (when corrected) that it is differentiable at each point of $(-2,,-0.5) \cup (-0.5,,2),$ which does not imply non-differentiability at $-2$ or $2$ (or $-0.5,$ although we know that from Darboux's theorem). – Dave L. Renfro Feb 24 '19 at 17:57
  • @DaveL.Renfro I actually noted both of those in (2) and (3) of 'My understanding'. Am I wrong to think the both the domain and differentiability at $x=2$ or @ irrelevant since we are asked for the critical points in $(-2,2)$ ? –  Feb 28 '19 at 05:33
  • @DaveL.Renfro Wait so is it possible or impossible that $y_0 \notin im(h')$, where $y_0 = \lim_{x \to -2^{+}} h'(x) = \lim_{x \to -0.5^{+}} h'(x)$ ? Initially, your thought was that Darboux implies $y_0 \in im(h')$ ? –  Feb 28 '19 at 05:35

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