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For example, the I.O.I of y=x^2 is (0,infinite), with the round brackets meaning that the value is excluded. Are there any scenarios where a square bracket would be used when stating the intervals of increase/ decrease for a function? If it narrows it down, the only functions I deal with are: linear, exponential, quadratic, root, reciprocal, sinusoidal, and absolute

Martin Sleziak
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counterhash
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    Can you define what an "interval of increase" is? Is it just the interval on which the function is increasing? Also, using abbreviations in questions (especially the first time you use the term" can make questions harder to read for people who don't know that (non-standard) terminology. – Stella Biderman Sep 09 '17 at 23:01
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    The function $x^2$ is strictly increasing on $[0,\infty)$ and strictly decreasing on $(-\infty,0]$, actually. –  Sep 09 '17 at 23:11
  • Wow! It's amazing to know how complicated the world is out there. – trying Sep 09 '17 at 23:48
  • Interval of increase are the intervals (of x) when the y value increases. So my above example would mean that y increases between but not including those two x values (0 and infinity). Sorry about that – counterhash Sep 10 '17 at 00:09
  • See also: Notation for intervals – Martin Sleziak Jun 13 '20 at 07:17

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For $a,b\in\mathbb R,a<b$, real intervals are defined as follows: $$(a,b):=\{x\in\mathbb R\mid a<x<b\}$$ $$(a,b]:=\{x\in\mathbb R\mid a<x\leq b\}$$ $$[a,b):=\{x\in\mathbb R\mid a\leq x<b\}$$ $$[a,b]:=\{x\in\mathbb R\mid a\leq x\leq b\}$$ Each function is defined on domain. If the domain is a subset of $\mathbb R$ that contains intervals, you can ask which behavior the function has on these intervals.

For example, $f(x)=x^2,x\in\mathbb R$

  • is increasing on any interval $(a,b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(a,\infty)$, $[a,\infty)$ with $a,b\in\mathbb R$, $0\leq a<b$ (these are all intervals on which $f$ increases) and
  • decreasing on any interval $(a,b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(\infty,b)$, $(\infty,b]$ with $a,b\in\mathbb R$, $a<b\leq 0$ (these are all intervals an which $f$ decreases).
Xaver
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  • Right you are: so, the squaring function is increasing on $[0.\infty)$ and decreasing on $(-\infty,0]$. At first, paradoxical, but not when you recognize, as you have here, that increasingness happens on a set, not at a point. Plus one. – Lubin Jul 18 '20 at 02:10