3

$$f(x)=\frac{1}{x-2}$$

number of points where $f$ is not differentiable?

I know that the domain of the function is $\mathbb{R}\setminus\{2\}$ and differentiability is checked only in the domain of the function so according to me the answer should be 0,

But my teacher is saying that as the function is not continuous at $x=2$, it must be non-differentiable also.

please help by solving this confusion.

Siminore
  • 35,136
Aryaman Bansal
  • 157
  • 1
  • 2
  • 10
  • 4
    He is wrong. Ask him if HE is odd or even. The same thing happens here. The concept of differentiability only makes sense on the domain of a function. Saying that $f$ is differentiable at $2$ or that is it false that $f$ is differentiable at $2$ is senseless. If he claims the answer is $1$, then you can tell him it's not differentiable at $i, \begin{bmatrix} 1 & 2\ 0 & 3\end{bmatrix}, (\mathbb R^*, \times)$ either. And the same goes for continuity. You can't meaningfully assert that it is false that $f$ is continuous at $2$. – Git Gud Jun 20 '14 at 08:49
  • He is posing me the question that is the function differentiable at x=2? According to me the answer to this should be that we can't say anything about differentiability. Am I correct? – Aryaman Bansal Jun 20 '14 at 08:53
  • @PraphullaKoushik I think it is not continuous at x=2 – Aryaman Bansal Jun 20 '14 at 08:54
  • 1
    is it defined at $x=2$?? there is no point in asking continuity at which the function is not defined.. you said the domain where it is not differentiable is $0$.. I believe you got the point but you should better sat empty set instead of what you have said.. –  Jun 20 '14 at 08:57
  • @PraphullaKoushik So are you saying that the function is continuous ? – Aryaman Bansal Jun 20 '14 at 09:03
  • No... i am saying function is not defined so i can not say anything... –  Jun 20 '14 at 09:04
  • @PraphullaKoushik Is the function continuous or non-continuous as it must be one of these? – Aryaman Bansal Jun 20 '14 at 09:06
  • It need not be... If you are not an Indian then you are neither north Indian nor south Indian (*Conditions apply) –  Jun 20 '14 at 09:08
  • 2
    A quick internet search shows that in 1-variable calculus classes, continuity of a function $f$ at $a$ is typically defined as satisfying all three of the following conditions: 1) $f(a)$ is defined. 2) Limit at $a$ exists. 3) Limit = $f(a)$. In this sense of this definition, $f(x) = 1/(x-2)$, with the usual domain "All Reals except 2", is not continuous at $2$ because it violates the first condition. – Braindead Jun 20 '14 at 09:21
  • 3
    I think poster and teacher agree on the facts but not on the vocabulary. I would say $f$ is differentiable everywhere in its domain, but I would also say it's not differentiable at $x=2$ since it's not even defined at $x=2$. – Gerry Myerson Jun 20 '14 at 09:23
  • 1
    @GerryMyerson The nature of the question itself makes me believe they do not agree on facts, because under your point of view, there are infinite objects are which the function is not differentiable, for example, it is not differentiable at $\sqrt{-1}$. – Git Gud Jun 20 '14 at 09:26
  • @GitGud What if the question was actually "How many points on $\mathbb{R}$ is $f$ not differentiable?" – Braindead Jun 20 '14 at 09:27
  • 1
    @Git, I thought about that, but I don't see why it isn't differentiable at $\sqrt{-1}$. – Gerry Myerson Jun 20 '14 at 09:27
  • @Braindead Under the definition of continuity you provided, I agree with you. But I (think I) reject that definition as being standard. – Git Gud Jun 20 '14 at 09:28
  • @Gerry I'm assuming $\text{dom}(f)=\mathbb R\setminus {2}$, but to make my point it suffices to say that under your point of view $f$ is not differentiable at $\begin{bmatrix} 1 & 2\ 0 & 3\end{bmatrix}$. – Git Gud Jun 20 '14 at 09:29
  • @GitGud If the question was "How many points in $\mathbb{R}$ is $f$ not differentiable?" then regardless of which definition is used, I would say that the answer is 1. – Braindead Jun 20 '14 at 09:31
  • 1
    @Git, I'd rather say $f$ is not differentiable at that matrix than say that it is. – Gerry Myerson Jun 20 '14 at 09:38
  • 1
    @Braindead And I would say that using a different definition it is the same as asking "How many points in $\mathbb R$, grgf gf rthtm45d?". Did it make sense to you? – Git Gud Jun 20 '14 at 09:43
  • @Braindead To make my point clearer, the question you propose is asking for the cardinality of the 'set' ${x\in \mathbb R\colon \neg($f$ \text{ is differentiable at }x)}$. So let's analyze what happens at $2$. If $2$ is in the set, then $\neg($f$ \text{ is differentiable at }2)$. So assuming this is something meaningful, $\neg \neg($f$ \text{ is differentiable at }x)$ also is meaningful, which implies (in classical logic) that $f \text{ is differentiable at }2$ is meaningful. – Git Gud Jun 20 '14 at 09:44
  • @Braindead Using a different definition of differentiability than the one you proposed, this is not meaningful. If $2$ isn't in the set, it is similar. – Git Gud Jun 20 '14 at 09:44
  • 1
    @GerryMyerson I'd rather say that as well, if I had only those two options. But there is a third option and that is where I stand: it doesn't make sense. Would you ask if $\frac 1 2$ is even? – Git Gud Jun 20 '14 at 09:46
  • @GitGud I get your point, but just I don't see how the statement "$f$ differentiable at $x$" requires $x \in dom(f)$ for it to be a meaningful statement. – Braindead Jun 20 '14 at 09:55
  • @GitGud Sorry for hijacking, but I would say $1/2$ is not an "even integer," since it is not an integer. – Braindead Jun 20 '14 at 09:56
  • Or better yet, $1/2 \not\in { 2n: n \in \mathbb{Z}}$. Therefore, $1/2$ is not an even integer. – Braindead Jun 20 '14 at 09:57
  • 1
    @Braindead Since the $\frac 1 2$ problem and the one in the question are similar in nature, I will talk about the $\frac 12$ only. If you define that an object in the universe of $\sf ZFC$ is even if it is an integer and it is of the form $2n$, then I agree that "$\frac 1 2$ is even" is a meaningful statement (and a false one). But that's not how the definition goes. The definition stars by restricting the universe to $\mathbb Z$ and then, in this restricted set, it says that a number is even if it is of the form $2n$. – Git Gud Jun 20 '14 at 10:01
  • So if you're given something outside of this restricted universe, how can you ever comment on whether is it even or not? Edit: I gotta go, can't continue this discussion for a few hours. – Git Gud Jun 20 '14 at 10:01
  • @GitGud If you start out by restricting the universe to $\mathbb{Z}$, yes, it doesn't make sense to ask "Is 1/2 even," since 1/2 does not belong to the universe, so any reference to "1/2" makes no sense. – Braindead Jun 20 '14 at 10:07
  • @GitGud But so what? In any universe where $1/2$ is included, $1/2$ is not going to be an even integer. Just as if in the original problem, if we define the universe to be $\mathbb{R}$, $2$ belongs to the universe, and would not belong to the set of real numbers where $f$ is differentiable. – Braindead Jun 20 '14 at 10:08
  • @Braindead I wasn't clear. You restrict the universe just for the definition, you still do the rest in the whole universe. So $\frac 1 2$ is in the universe. – Git Gud Jun 20 '14 at 13:15
  • @GitGud So...what wrong with the following? $E = { 2n : n \in \mathbb{Z}}$ and $1/2\not\in E$, so 1/2 is not an even integer. What's wrong with this statement? – Braindead Jun 22 '14 at 05:29
  • @Braindead Again, depends on the definition of even integer. Your set $E$ can be rewritten (in fact formally this is the 'true' form of $E$) as ${x\colon \exists n(n\in \mathbb Z\land x=2n)}$. So this is saying that element of the universe is an even integer if, and only if, it is an integer and it's a multiple of $2$. But if we define an even number like this "Given an integer $x$, it is said to be even if, and only if, $\exists n\in \mathbb Z(x=2n)$", the statement can be looked at (and should be looked at) as a conditional statement. – Git Gud Jun 22 '14 at 09:58
  • @Braindead The "given an integer part" translates to $\forall x\in \mathbb Z(\ldots)$ which in turn is short for $\forall x(x\in \mathbb Z\implies \ldots)$. As you very well know, if you're given an $x$ that doesn't satisfity the antecedent of the implication, that tells you nothing about the consequent of the implication. – Git Gud Jun 22 '14 at 09:59
  • @Braindead It's true that you can say that if $x$ doesn't satisfy the antecedent, then the whole statement is true. But ah!, the whole statement being true doesn't tell you anything about the consequent and the consequent is what we're interested in (depending on the definition, of course). With differentiality the same thing happens because the definition starts with "Given $x\in \text{dom}(f)\ldots$" (in fact it's $x\in \text{int}(\text{dom}(f))$, but that doesn't matter here). – Git Gud Jun 22 '14 at 09:59

3 Answers3

1

Calculus as a topic deals with real functions. Whenever we talk about the behaviour of ANY function with respect to ANY point or interval, it is a basic step to first find the DOMAIN. The question in this case is fundamentally wrong. You can not question the behaviour of the function for any point OUTSIDE its domain. Simple example, is tan(x) differentiable at $\frac{\pi}{2}$? Acc to your sources, answer is no! But $\frac{d \tan(x)}{dx}$ = $sec^2x$. Why?

Because, we consider the domain only!

EDIT: Your statement that a function is either continuous or discontinuous is also valid only in the domain.

Saurabh Raje
  • 1,007
  • So you're saying that my question itself is wrong? If yes, then what should have been the correct question? – Aryaman Bansal Jun 21 '14 at 12:46
  • "number of points..." is not wrong. "Is it differentiable at x=2" is wrong! – Saurabh Raje Jun 21 '14 at 12:47
  • I never asked that. I just said that as it is not continuous, it is not differentiable – Aryaman Bansal Jun 21 '14 at 12:55
  • answer to your question is zero – Saurabh Raje Jun 21 '14 at 13:01
  • You don't have to require that $f$ must be defined in $x=2$ since $2$ only has to be a limitpoint of it's domain. –  Jun 21 '14 at 15:08
  • @andre it is a necessity because continuity involves f(a), which is not defined – Saurabh Raje Jun 21 '14 at 16:28
  • I just don't understand your and GitGud's arguments. Let $E={ x\in dom(f): \text{$f$ is differentiable at $x$}}$. In the context of this problem, the universe is the set of real numbers, and the complement of $E$ is well-defined. For the function $f(x) = 1/(x-2)$, $E^c = {2}$. What is wrong with this argument? – Braindead Jun 22 '14 at 05:40
0

In my experience, teachers tend to to consider

  1. Non-continuity
  2. Discontinuity

as different ideas. In particular, many of us don't define discontinuity as the logical negation of continuity, since too many students would consider any point outside the domain of the given function as a point of discontinuity. The usual definition of discontinuity (at the point $x_0 \in E$ for the function $f \colon E \to \mathbb{R}$) is that either $\lim_{x \to x_0} f(x)$ does not exist in $\mathbb{R}$, or $f(x_0) \neq \lim_{x \to x_0} f(x)$.

But there is a second interpretation that can be rephrased as follows: a function $f \colon E \to \mathbb{R}$ is discontinuous at the point $x_0 \in \overline{E}$ if $f$ cannot be defined at $x_0$ in such a way that this extension is continuous.

For these teachers, $x_0=2$ is a discontinuity point of $f(x)=\frac{1}{x-2}$, since no definition of $f(2)$ will ensure continuity. I think that, at least in Italy, this alternative definition is very popular among high-school instructors and high-school textbooks.

This said, I have never read a book in which singular points (i.e. points where a function is not differentiable) may fall outside the domain.

Siminore
  • 35,136
  • But I am sure you will agree that conceptually, the question (is it differentiable at x=2) is wrong? – Saurabh Raje Jun 21 '14 at 12:53
  • So is there any difference between discontinuity and non continuity? And what according to you must be the answer to my question? – Aryaman Bansal Jun 21 '14 at 12:54
  • 1
    It depends on your definition of "point of discontinuity". I usually teach that a point of discontinuity must be a point of the domain. But there is a professor, in my department, who teaches that the function of your example has a discontinuity at $x=2$. – Siminore Jun 21 '14 at 14:53
  • Isn't there any specific definition given? :( – Aryaman Bansal Jun 22 '14 at 05:32
  • 2
    In my opinion, the whole concept of "point of discontinuity" is totally useless in higher mathematics. Continuity is useful in a positive sense, while its "negative" is used mainly for tests and exercises in calculus. – Siminore Jun 22 '14 at 08:15
  • Out of curiosity, what would your colleague say about about the discontinuity/non-continuity of a point $x_0\not \in \overline E$, for instance if $x_0$ was a matrix. – Git Gud Jun 22 '14 at 11:54
  • I am not sure I can understand your question @GitGud. I am sure that my colleague's definition applies ONLY to function of one real variable. But this is fairly typical: I've nvere read a topology book dealing with points of discontinuity... – Siminore Jun 23 '14 at 07:45
  • @Siminore My point is that if someone claims it makes sense to talk about points outside of the domain, I will take an extreme example that goes against intuition instead of taking something like $x=2$ as in the question. – Git Gud Jun 23 '14 at 07:50
  • @GitGud For many instructors, Calculus is a topic that should be taught in the same spirit as ancient mathematicians did: functions are regular graphs, definitions should go along with intuition, and so on. Anyway, I doubt that a student of calculus could appreciate en example with a function of a matrix, if he/she hardly understands what a function of a single real variable is... – Siminore Jun 23 '14 at 09:11
0

This is a very common kind of problem in calculus courses: there are multiple conventions for dealing with the basic notions of calculus.

This is particularly the case when we deal with "functions" that are defined by expressions that are undefined for some real numbers, e.g. $f(x) = 1/x$. If you ask different people this exact question, you will get several different answers: "If $f(x) = 1/x$ continuous everywhere?". Some will say yes, because the function is continuous everywhere in its domain. Some will say no, because the function is not defined at $x = 0$. The case of $g(x) = x/x$ is even more complicated.

Rather than trying to analyze the sources for the different conventions, I want to just emphasize practical advice. Because the conventions for these things do differ from textbook to textbook and instructor to instructor, you want to do two things. First, understand the mathematics of what is going on, apart from the conventions of terminology. Second, ask your instructor whenever you encounter one of these conventions, because only your instructor can tell you what answer is going to be required on an exam.

Carl Mummert
  • 81,604