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Yet again, I find myself confused about something that seems basic: critical points and regions of increasing / decreasing.

Previously, I thought that to identify regions of increasing / decreasing, one simply had to:

  1. identify the critical points of the function;
  2. determine whether the derivative was positive or negative on either side of those critical points.

However, I recently discovered that I was using an imprecise definition of a critical point, my definition was simply:

$x_o$ is a critical point of the function $f(x)$ if $f'(x_o) = 0$ or $f'(x_o)$ is undefined.

And a correct definition is:

$x_o \in D\,(f)$ is a critical point of the function $f(x)$ if $f'(x_o) = 0$ or $f'(x_o)$ is undefined.

Which makes sense, if one thinks of critical points as "critical points to consider when looking for a max or min."

However, this kind of ruins my algorithm for finding regions of increasing / decreasing, as the sign of the derivative may change at points that are not critical points (eg singular points).

The only reason this bothers me is that I have consulted 3 major calculus textbooks (the standard references at the top universities in my country) and not one of them addresses this point: when they discuss regions of increasing / decreasing, they inevitably focus on trivial examples that have no points of discontinuity (eg cubic polynomials), and when they discuss local / global extrema, they don't discuss the fact that the process will not indicate ALL regions of increasing / decreasing.

Am I correct in concluding then, that in order to identify all regions of increasing / decreasing, one has to find all points $\mathbf{x_o}$ at which $\mathbf{f'(x_o) = 0}$ or $\mathbf{f'(x_o)}$ DNE, even if $\mathbf{x_o}$ is not in the domain of $\mathbf{f(x)}$? I realize the formulation is imperfect for functions with idiosyncratic domains, but assuming one is dealing with functions whose domain is given by the domain convention (eg $f = \frac{1}{x^2}$), would this be appropriate?

N.B. I am looking for an explanation that is appropriate for an introductory applied maths type of course, not a course in analysis.

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    $\begingroup$ I don't understand exactly what you what mean when you say we have to consider points where $f'(x_0)$ does not exist, even if $x_0$ is not in the domain of $f$. Take $f(x)=\sqrt{x}.$ What does $f'(-1)$ mean? What does it have to do with the problem? $\endgroup$ – saulspatz Apr 21 at 0:47
  • $\begingroup$ For example, if $f(x) = \frac{1}{x^2}$, $x = 0$ is not in the domain, but the fact that the derivative is not defined there makes it an important "dividing point" in the domain: the function is increasing to the left of that point and decreasing to the right of it. Even though this point on the number line is not in the domain of the function, it demarcates a change in the function's behaviour. For me, $x=0$ in this example is somehow different from $x = -1$ in $f = \sqrt{x}$, but I know of no good language to distinguish the two. $\endgroup$ – Rax Adaam Apr 21 at 14:52
  • $\begingroup$ @saulspatz does the above clarify? or rather, do you think my proposed answer addresses the issues raised by the ambiguity you underlined? Really appreciate your comment, as it prompted me to think more clearly about the issue :) $\endgroup$ – Rax Adaam Apr 23 at 23:28
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    $\begingroup$ Yes, that sounds correct to me. $\endgroup$ – saulspatz Apr 23 at 23:31
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After further reflection, thanks to saulspatz's comment, a better formulation occurred to me, that seems to avoid the imprecision (and especially the overly inclusive language) of the formulation given in the OP:

  1. Identify critical points;
  2. Identify singular points;
  3. Identify points of discontinuity.
  4. Verify if $f'(x)$ is positive or negative on each subinterval demarcated by the above points.

In this way, the change from increasing to decreasing that happens in a function like $f(x) = \frac{1}{x^2}$ would be identified (as the change happens across a singular point), whereas irrelevant points like $x = -1$ for $f = \sqrt{x}$ would not be considered.

I'm proposing this as the answer to my question, but will hold off on marking it as correct, in case someone is able to provide an even better formulation (more concise or precise etc).

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