I was solving: Determine the intervals of increase and decrease for $f(x) = \frac {2x}{ ln x}$ and I stumbled upon the fact that f(x) is decreasing on (0,e] and increasing on [e, $\infty$). This would otherwise suggest that the function is both increasing and decreasing at x=e. Is that true? Or a function can be sure to be increasing or decreasing only in the vicinity of a point but at a point?
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You need a precise definition of increasing, decreasing to answer this. Usual definition involoves $f(x)$ at two points. – coffeemath Dec 13 '17 at 18:39
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Do you consider a constant function $f(x)=c$ to be increasing and decreasing at the same time? If so then yes, if not then no. – CyclotomicField Dec 13 '17 at 18:42
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increase or decrease is a difference between two values we cannot use one value to determine it. – Vasili Dec 13 '17 at 18:44
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For increasing, if $x1>x2$ then, $f(x1)>f(x2)$ and vice-versa for decreasing. You clearly need $2$ points to know whether its increasing or decreasing. Also, here $x=e$ would just be a point of local minima or the point where sign of the slope changes w.r.t to $e+\epsilon$ and $e-\epsilon$ for $\epsilon>0$. – Your IDE Dec 13 '17 at 18:49
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I think you meant that $f$ is decreasing on $(1, e]$. – mucciolo Dec 20 '17 at 15:21
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A function can't be increasing or decreasing unless you can compare it to another point.
So it depends on definition.
I believe there are 3 (or 4... or 5...) incompatible options.
1) Increasing at a point $x$ means that there is an $\epsilon > 0$ so that for every $x-\epsilon < y < x < z < x+\epsilon$ such that $f(y) \le f(x) \le f(z)$. (strictly increasing would mean strict inequalities.) and Increasing on an an interval would mean increasing on every point of an interval.neither at $e$.
However I have NEVER seen anyone or any text use this definition.
In fact, I just made it up.
2) Increasing at a point is a logical inconsistancy and makes no sense. Increasing on an interval (whether open or closed or mixed) means for any two points $x,y$ in the interval so that $x<y$ then it must follow that $f(x) \le f(y)$.
This seems to be the most accepted definition. So $f$ is decreasing on $(0,e]$ and decreasing on $[e,\infty)$.
And increasing at $x = e$ simply is not a meaningful concept.
2a) Same as above but allowing "increasing at a point" to mean the point is within and interval where the function is increasing.
In this case $f$ is both increasing and decreasing "at" $e$. I've seen people say this but it's really semantics and not mathematics.
2b) Same as above but allowing $\{e\}$ to be a "single point interval". THus every function is vacuuously both increasing and decreasing at every point because there are no $x < y$ in the "interval" than for all (all zeor of them) $x < y$ we have $f(x) \le f(y)$. (We also have $f(x)$ is a blue dragon eating colorless yellow thoughts.)
Again... semantics; not mathematics.
3) $f$ is increasing on a set of points $S$ so that for any $x,y \in S$ and $x< y$ then $f(x) < f(y)$.
This would mean vacuously that every function is increasing and decreasing on a set with a single point.
However I have never seen anyone use this definition and I just made this up. It is probably useless as I can say something line $f(x) = x$ if $x \in \mathbb Q$ and $f(x) = 0$ if $x$ is irrational, would be increasing on the rationals. WHich I think avoids the issues.
Any way...
I think most would use definition 2. (But there are always exceptions.) But practically, I don't think claiming $f$ is increasing at a single point makes much sense or is useful unless you are claiming the point is in an interval on which the function is increasing.
fleablood
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1Your first definition is not too uncommon. I read it in Hardy's Pure Mathematics. And the definition of increasing on an interval is pretty common (your second definition) and is not based on "increasing at a point". It is then a theorem that if a function is increasing at every point of an interval then it is increasing in that interval. The proof involves completeness of real numbers. – Paramanand Singh Dec 20 '17 at 16:20
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My own opinion and taste are quite different from those of most of the commenters. I think it makes sense only to speak of increase or decrease on an interval, whether open or closed. I would say that $f$ is increasing on an interval $I$ when $\forall x,x'\in I$, $x<x'\Rightarrow f(x)<f(x')$. In particular, the squaring function is increasing on $[0,\infty\rangle$ and decreasing on $\langle-\infty,0]$.
And to say that $f$ was increasing at a point I would require that there be an open interval containing the point on which $f$ was increasing.
Lubin
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I think most of us are okay with increasing on a closed interval. But I think most of us think saying it is increasing at a point doesn't make much sense. – fleablood Dec 13 '17 at 20:34
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3The notion of strictly increasing at a point is widely used in real analysis, and it means that left of the point you're lower and right of the point you're higher. This is a weaker notion that that of strictly increasing in some interval of the point, a notion that has less use in mathematics. I don't have time to say more now, but googling will easily turn up lots of stuff, and the notion is even in Spivak's well known book Calculus. – Dave L. Renfro Dec 13 '17 at 22:14
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"A function can't be increasing or decreasing unless you can compare it to another point." "increase or decrease is a difference between two values we cannot use one value to determine it."
I agree with this, BUT if this is the case why does the first derivative test use ONE point to establish that a function is increasing decreasing on the interval in question?
MMM
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As monotonicity is based on comparisons, you obviously need more than a point. The usual definitions work on an interval.
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In set theory a function $f:A\to B$ is a certain type of subset of $A\times B.$ So the function $f_1:(0,e]\to \Bbb R$ and the function $f_2:[e,\infty)\to \Bbb R$ are two different objects, even though $f_1\cup f_2$ is also a function because $f_1(e)=f_2(e).$ The function $f_1$ is decreasing on its domain. And $f_2$ is increasing on its domain.
Usually a function $f:A\to \Bbb R,$ where $A\subset \Bbb R,$ is said to be increasing at some $x\in A$ when there exists some $r>0$ such that $f(y)\leq f(x)$ for all $y\in (-r,x)\cap A$ and $f(z)\geq x$ for all $z\in (x,x+r)\cap A.$ (...If we write $<$ and $>$ rather than $\leq$ and $\geq,$ we say "strictly increasing at $x$"...). This terminology is usually only used when $x$ is not an isolated point of the domain of $f.$
In your Q, $f_1\cup f_2$ is neither increasing nor decreasing at $e.$
DanielWainfleet
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