Can a function be increasing or decreasing at a point?

Question

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?

Answer 1

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.

Answer 2

As monotonicity is based on comparisons, you obviously need more than a point. The usual definitions work on an interval.

Answer 3

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.

Answer 4

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 $e.$

Answer 5

"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?