Find critical points and inflection points given the derivative (also check for consistency with Darboux's theorem)?

Question

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.

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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 $[-2,2]$.

• 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 $[-2,2]$:

4. 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 $(-2,2)$:

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

6. $x=-1,0,1$ 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 $[-2,2]$:

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

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

9. 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.