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In a precalculus textbook, a sample problem might be something like "For the function $f(x)=x^2$, find the interval on which it is increasing". The expected answer is $[0,\infty)$. However, any subinterval of $[0,\infty)$ is also technically a valid answer, like, for example $[\pi,\infty)$, $(1,2)$, $[2,4]$, etc. But those answers would probably not be accepted as correct answers in the context of the question. So, then, my real question is, what is the formal and rigorous definition of the correct answers to those precalculus questions which ask of a function $f$ (which is usually a polynomial) those interval(s) where $f$ is increasing and where $f$ is decreasing? How does one formalize the question so as to exclude answers that are not the expected answer?

user107952
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    Replace "interval" with "maximal interval"? – Dermot Craddock Jun 07 '25 at 10:44
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    Another wrinkle: $f(x)=x^2$ isn't increasing at $x=0$. – Blue Jun 07 '25 at 12:26
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    @Blue sure, but $f\mid_{[0,a]}$ is an increasing function for any $a$. – Malady Jun 07 '25 at 13:28
  • Are you supposed to use calculus? $f^\prime>0$ means $f$ is increasing, etc.? – ultralegend5385 Jun 07 '25 at 16:01
  • Well, I'd argue about the mean "the". If you ask for "the" then I'd say the answer has to be distinct and unique so $(1,2)$ can't be correct. But then I'd argue it isn't clear what they do mean? Except it is. Then again what exactly are you asking and what answer do you want. You've found an example of a book that isn't rigorous in its questioning. Meh, that happens. I'd formalize it as "Describe where the function is increasing" and make it clear than I expect answers to be complete and thorough. It should be universal that partial answer are never complete. – fleablood Jun 07 '25 at 16:56
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    Alternate definition for increasing on an interval, $I$, for all $x,y\in I, x<y$ we have $f(x) < f(y)$ (or $f(x) \le f(y)$). So $f$ is increasing at $x=0$. – fleablood Jun 07 '25 at 16:59
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    @fleablood I'd say yours is the definition of increasing, not just an "alternate" definition. By such a definition, $f$ is increasing on the right and decreasing on the left at $x=0.$ I don't see that as a problem. – David K Jun 07 '25 at 18:56
  • But as $0 \in [0,\infty)$ then by definition it is increasing on the interval and increasing at $x=0$. (This was in response t Blue's comment). My comment was about increasing at $0$. – fleablood Jun 07 '25 at 20:47

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Consider $f:X \to\mathbb{R}$ where $X\subseteq \mathbb{R}$ is open and nonempty. For each of the binary relations $R\in\{<,=,>\}$:

  • Let $L_{R}$ be the poset of all connected subsets $S\subseteq X$, containing at least two elements, such that $$x<x'\Rightarrow f(x)Rf(x')$$ for all $x,x'\in S$.
  • Let $M_R$ be the set of maximal elements of $L_R$.

Then the question is:

For each $R\in\{<,=,>\}$, what are the elements of $M_R$?

To capture the naïve intuition that functions actually have components of their domain on which they are increasing, decreasing, or constant, we need further regularity assumptions:

  • If $f$ is continuously differentiable, then at least one of the three sets $X_R=\{x\in X \mid 0Rf'(x)\}$ will contain a nonempty open interval, so that the corresponding $L_R$ is nonempty (and therefore chain complete), so that the corresponding $M_R$ is nonempty.
  • We cannot relax this assumption to $f$ merely being differentiable everywhere: there exist such functions for which each of the three sets $X_R$ are dense (so that any open interval contained in one would have intersect the other two, a contradiction).
K B Dave
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  • These conditions seem too restrictive. $f(x)=|x|$ has clearly defined increasing and decreasing intervals. – David K Jun 07 '25 at 18:47
  • Can you give an example of a function for which each of the sets $X_R$ is dense? I'm having trouble visualizing this, since $X_R$ is supposed to be composed of connected subsets of $\mathbb R.$ – David K Jun 07 '25 at 18:52