Is this function continuous and differentiable?

In summary: So we have 2 values for x = 5. So function is discontinuous.I'd say the function is discontinuous at x = 5 because the one-sided limits don't agree. (I'm assuming you meant "f(x) = 2x" in the first part of your post.)Yes, I did. And yes, that makes sense. Thanks for explanation.In summary, the conversation discusses the application of limits and the continuity and differentiability of functions. A specific function is given as an example, and it is shown that the function
  • #1
jaus tail
615
48

Homework Statement


upload_2018-5-14_10-10-57.png


Homework Equations


Solve using limits. Function is continuous if it's graph is continuous throughout. here the (x-1) term gets canceled in numerator and denominator. So we have a continuous graph of (x+1).

The Attempt at a Solution


The (x-1) term gets canceled from numerator and denominator and we get a continuous function (x+1) which is defined everywhere.
It is also differentiable with derivative of (x+1) = 1.
So answer is D.
Book answer is C. How?
 

Attachments

  • upload_2018-5-14_10-8-31.png
    upload_2018-5-14_10-8-31.png
    51.8 KB · Views: 448
  • upload_2018-5-14_10-8-56.png
    upload_2018-5-14_10-8-56.png
    18.6 KB · Views: 382
  • upload_2018-5-14_10-10-57.png
    upload_2018-5-14_10-10-57.png
    8.3 KB · Views: 755
Physics news on Phys.org
  • #2
jaus tail said:

Homework Statement


View attachment 225698

Homework Equations


Solve using limits. Function is continuous if it's graph is continuous throughout. here the (x-1) term gets canceled in numerator and denominator. So we have a continuous graph of (x+1).

The Attempt at a Solution


The (x-1) term gets canceled from numerator and denominator and we get a continuous function (x+1) which is defined everywhere.
It is also differentiable with derivative of (x+1) = 1.
So answer is D.
Book answer is C. How?
What do you get when you evaluate the function at x=0?
 
  • Like
Likes jaus tail
  • #3
jaus tail said:

Homework Statement


View attachment 225698

Homework Equations


Solve using limits. Function is continuous if it's graph is continuous throughout. here the (x-1) term gets canceled in numerator and denominator. So we have a continuous graph of (x+1).

The Attempt at a Solution


The (x-1) term gets canceled from numerator and denominator and we get a continuous function (x+1) which is defined everywhere.
Cancellation is applicable only if the factors you cancel aren't zero. Your function is not defined at x = 1, so how could it be continuous there?
jaus tail said:
It is also differentiable with derivative of (x+1) = 1.
If it's not continuous at x = 1, how could it be differentiable there?
jaus tail said:
So answer is D.
Book answer is C. How?
 
  • Like
Likes jaus tail
  • #4
The book answer looks wrong. As well as the way you do it, we can apply l'Hopital's rule, which gives the answer as ##\lim_{x\to 1}\frac{2x}1=2##.

Part of the problem is that the problem statement fails to state what the function's domain is. Hence, strictly speaking, the function is not properly specified and hence the question is unanswerable.
 
  • Like
Likes jaus tail
  • #5
tnich said:
What do you get when you evaluate the function at x=0?
At x = 0, I get (-1)/(-1) = 1
Mark44 said:
Cancellation is applicable only if the factors you cancel aren't zero. Your function is not defined at x = 1, so how could it be continuous there?
If it's not continuous at x = 1, how could it be differentiable there?
I didn't understand this 'cancellation is not application only if factors aren't 0'
Could you please explain?
 
  • #6
jaus tail said:
I didn't understand this 'cancellation is not application only if factors aren't 0'
Could you please explain?
It's basically "you can't divide by zero."
##\frac 0 0## is undefined.
For example, ##\frac{x - 1}{x - 1} = 1##, as long as x isn't 1.
The graph of ##y = \frac{x - 1}{x - 1}## is two disconnected horizontal line segments, which a "hole" at (1, 1), where the graph is discontinuous. You totally ignored the fact that your function isn't continuous.

C is indeed the correct answer.
 
  • Like
Likes jaus tail
  • #7
Ok. Got it. So a function is continuous if there is no place where the function is not defined.
 
  • #8
andrewkirk said:
The book answer looks wrong.
Not to me.

andrewkirk said:
Part of the problem is that the problem statement fails to state what the function's domain is.
Clearly the domain is ##\{x \in \mathbb R | x \neq 1\}##. The function is neither continuous nor differentiable at x = 1.
 
  • Like
Likes jaus tail
  • #9
jaus tail said:
Ok. Got it. So a function is continuous if there is no place where the function is not defined.
No, that isn't it.
A function can be defined at each real number, but still not be continuous.
 
  • #10
But if a function has multiple values then it's not continous. Like there must not be any ambiguity.
Example: f(x) = 2x for x = 0 to 5
and f(x) = 0 for x < 0
and f(x) = 12 for x more than equal to 5
is discontinuous at x = 5
 
  • #11
Mark44 said:
Clearly the domain is ##\{x \in \mathbb R | x \neq 1\}##. The function is neither continuous nor differentiable at x = 1.
It can't be very 'clearly' since I disagree with the claim.
 
  • #12
andrewkirk said:
It can't be very 'clearly' since I disagree with the claim.

As shown in post #1:
The function f(x) = ##\frac {x^2 - 1}{x - 1}## at x = 1 ...

It seems clear enough to me that the domain is exactly as I wrote in post #8. It should be obvious that the domain, whatever it is, doesn't include 1. A very common practice in textbooks at the precalculus and calculus level is to not explicitly state the domain for each problem, but in a short phrase at the beginning of a group of problems, or implied by the context of the section the problem is in.
 
Last edited:
  • #13
jaus tail said:
But if a function has multiple values then it's not continous. Like there must not be any ambiguity.
Example: f(x) = 2x for x = 0 to 5
and f(x) = 0 for x < 0
and f(x) = 12 for x more than equal to 5
is discontinuous at x = 5
Yes, and this is what I was thinking of in my reply in post #9.
Here you have a function that is piecewise defined (not "has multiple values").
 
  • Like
Likes jaus tail
  • #14
Mark44 said:
Yes, and this is what I was thinking of in my reply in post #9.
Here you have a function that is piecewise defined (not "has multiple values").
But this function has multiple values at x = 5.
 
  • #15
jaus tail said:
But this function has multiple values at x = 5.

I'd be interested to know what you think the multiple values are?

##f(5) = 6## and what else?
 
  • #16
PeroK said:
I'd be interested to know what you think the multiple values are?

##f(5) = 6## and what else?
One is 10 and the other is 12.

This is function:
f(x) = 2x for x = 0 to 5 so for x = 5, f(x) = 10
and f(x) = 0 for x < 0
and f(x) = 12 for x more than equal to 5, so f(x) = 12 for x = 5.

So we have 2 values for x = 5. So function is discontinuous.
 
  • #17
jaus tail said:
One is 10 and the other is 12.

This is function:
f(x) = 2x for x = 0 to 5 so for x = 5, f(x) = 10
and f(x) = 0 for x < 0
and f(x) = 12 for x more than equal to 5, so f(x) = 12 for x = 5.

So we have 2 values for x = 5. So function is discontinuous.

That's not a function in the first place.
 
  • #18
But these were the kind of functions in the examples to give us examples of discontinuous functions.
 
  • #19
jaus tail said:
But these were the kind of functions in the examples to give us examples of discontinuous functions.

Ah, but it's important that you don't define the function more than once at the same point. There's an important difference between:

A)

##f(x) = 0 \ (x < 0); \ \ f(x) = 1 \ (x \ge 0)##

That is a well-defined function that is defined on all of ##\mathbb{R}##. It is discontinuous at ##x = 0##.

B)

##g(x) = 0 \ (x < 0); \ \ g(x) = 1 \ (x > 0)##

This is a well-defined function that is defined on ##\mathbb{R} - \{0\}##. Strictly speaking ##g## is continuous (on its domain), although it can be said (informally) to have a jump discontinuity at ##x = 0##. But, you should really specify a value for ##g(0)## before you talk about it being discontinuous at ##x = 0##.

C)

##h(x) = 0 \ (x \le 0); \ \ h(x) = 1 \ (x \ge0)##

This is not a well-defined function, because it has an ambiguous value at ##x = 0##. So, you can't start talking about what other properties it has without sorting out the definition.
 
Last edited:
  • Like
Likes jaus tail
  • #20
jaus tail said:
Example: f(x) = 2x for x = 0 to 5
and f(x) = 0 for x < 0
and f(x) = 12 for x more than equal to 5
To add to what PeroK said, your example is unclear.
"for x = 0 to 5" -- Which of the following does this mean?
##0 < x < 5##
##0 < x \le 5##
##0 \le x < 5##
##0 \le x \le 5##
 
  • Like
Likes jaus tail
  • #21
jaus tail said:

Homework Statement


View attachment 225698

Homework Equations


Solve using limits. Function is continuous if it's graph is continuous throughout. here the (x-1) term gets canceled in numerator and denominator. So we have a continuous graph of (x+1).

The Attempt at a Solution


The (x-1) term gets canceled from numerator and denominator and we get a continuous function (x+1) which is defined everywhere.
It is also differentiable with derivative of (x+1) = 1.
So answer is D.
Book answer is C. How?

I think the correct answer is (E) Undefined.
 
  • Like
Likes jaus tail and StoneTemplePython
  • #22
andrewkirk said:
The book answer looks wrong. As well as the way you do it, we can apply l'Hopital's rule, which gives the answer as ##\lim_{x\to 1}\frac{2x}1=2##.

Part of the problem is that the problem statement fails to state what the function's domain is. Hence, strictly speaking, the function is not properly specified and hence the question is unanswerable.
l'Hopital's rule is nice, but it does not work where the definition of continuous can not be applied. Since f(1)=0/0 is undefined, it is not valid to say that the definition of continuous is satisfied at x=1.
 
  • Like
Likes jaus tail
  • #23
Mark44 said:
It seems clear enough to me that the domain is exactly as I wrote in post #8. It should be obvious...
As a general principle in mathematics, if the only support one can offer for a claim one makes is 'it should be obvious' or 'clearly', one needs to reconsider the claim.

If we allow the use of words like 'obviously' or 'clearly' in mathematics, it rapidly descends into a hand-waving chaos. If a claim is obvious, it is possible to say why it is true - if not at first then at least when challenged.

I maintain that the function is inadequately specified, as specification of a function requires specification of domain, range and mapping and in this case two of the three are not specified. If you want to maintain a different view, that's fine. We can agree to disagree.
 
  • Like
Likes jaus tail
  • #24
jaus tail said:

Homework Statement


View attachment 225698

Homework Equations


Solve using limits. Function is continuous if it's graph is continuous throughout. here the (x-1) term gets canceled in numerator and denominator. So we have a continuous graph of (x+1).

The Attempt at a Solution


The (x-1) term gets canceled from numerator and denominator and we get a continuous function (x+1) which is defined everywhere.
It is also differentiable with derivative of (x+1) = 1.
So answer is D.
Book answer is C. How?
Keep in mind that f(1) = 0/0 is undefined. Canceling factors in the numerator and denominator makes a new function that is defined at x=1 but does not change the undefined nature of the original function. And the new function is not the same as the original function. Look closely at your book definitions of "continuous" and "differentiable". If they require a value for f(1), then those definitions can not be satisfied.

PS. If one wanted to make this continuous, f(1) would have to be specifically defined as = 2.
 
Last edited:
  • Like
Likes jaus tail
  • #25
andrewkirk said:
As a general principle in mathematics, if the only support one can offer for a claim one makes is 'it should be obvious' or 'clearly', one needs to reconsider the claim.

If we allow the use of words like 'obviously' or 'clearly' in mathematics, it rapidly descends into a hand-waving chaos. If a claim is obvious, it is possible to say why it is true - if not at first then at least when challenged.

I maintain that the function is inadequately specified, as specification of a function requires specification of domain, range and mapping and in this case two of the three are not specified. If you want to maintain a different view, that's fine. We can agree to disagree.
in whatever way you define your domain it is obvious that $$f(x) = \frac{x^2 - 1}{x - 1}$$ is not defined at ##x = 1##. Don't know what book the question comes from, however, I would take the position without having any further information that the function is defined everywhere where it can be evaluated.

So the answer to the initial question that the function is not defined at ##x = 1## and so its neither continuous nor differentiable.
 
  • Like
Likes jaus tail
  • #26
Ray Vickson said:
I think the correct answer is (E) Undefined.
Not one of the available choices, but since f(1) is not defined, nor a fortiori is f differentiable at 1, answer C fits quite well.

andrewkirk said:
As a general principle in mathematics, if the only support one can offer for a claim one makes is 'it should be obvious' or 'clearly', one needs to reconsider the claim.
I don't buy this. Sometimes things really are obvious, as in the case of this problem, where f(x) is defined to be equal to ##\frac{x^2 - 1}{x - 1}##.
Is it not obvious to you that whatever the domain happens to be, it does not include x = 1?
andrewkirk said:
If we allow the use of words like 'obviously' or 'clearly' in mathematics, it rapidly descends into a hand-waving chaos. If a claim is obvious, it is possible to say why it is true - if not at first then at least when challenged.
If the "when challenged" bit is aimed at me, I gave a reason why the domaiin doesn't include 1. If you think there are any other real numbers not included in the domain, with no information in the problem suggesting this, then you need to say why you think this is the case.
andrewkirk said:
I maintain that the function is inadequately specified, as specification of a function requires specification of domain, range and mapping and in this case two of the three are not specified.
Have you never seen an intro calculus textbook? It is rare to the point of nonexistence for a calculus textbook to specify both the domain and range for a given function for every example and problem in the textbook. For a simple rational function such as the one in this thread, it is a reasonable expectation that a reader will be able to figure out what the domain is.
andrewkirk said:
If you want to maintain a different view, that's fine. We can agree to disagree.
It should be obvious that I disagree.
 
  • Like
Likes jaus tail
  • #27
Marc Rindermann said:
in whatever way you define your domain it is obvious that $$f(x) = \frac{x^2 - 1}{x - 1}$$ is not defined at ##x = 1##.
Well, it seems obvious to me, but some might fear that such a statement could lead to "handwaving chaos." :oldbiggrin:

Marc Rindermann said:
So the answer to the initial question that the function is not defined at x=1x=1x = 1 and so its neither continuous nor differentiable.
I.e., answer C.
 
  • Like
Likes jaus tail
  • #28
Mark44 said:
Not one of the available choices, but since f(1) is not defined, nor a fortiori is f differentiable at 1, answer C fits quite well.I don't buy this. Sometimes things really are obvious, as in the case of this problem, where f(x) is defined to be equal to ##\frac{x^2 - 1}{x - 1}##.
Is it not obvious to you that whatever the domain happens to be, it does not include x = 1?
If the "when challenged" bit is aimed at me, I gave a reason why the domaiin doesn't include 1. If you think there are any other real numbers not included in the domain, with no information in the problem suggesting this, then you need to say why you think this is the case.Have you never seen an intro calculus textbook? It is rare to the point of nonexistence for a calculus textbook to specify both the domain and range for a given function for every example and problem in the textbook. For a simple rational function such as the one in this thread, it is a reasonable expectation that a reader will be able to figure out what the domain is.
It should be obvious that I disagree.

I know very well that "(E) Undefined" was not one of the available choices, but I think it should have been. I agree that (C) is the best of the available choices, and I guess you could argue that if a function is not defined somewhere it cannot be continuous or differentiable there.
 
  • Like
Likes jaus tail
  • #29
Mark44 said:
For a simple rational function such as the one in this thread, it is a reasonable expectation that a reader will be able to figure out what the domain is.
If the domain doesn't include 1 - which is what you seem to be implying here - then the question is meaningless because continuity of a function at a point is a concept that is only applicable to points in the function's domain (unless we adopt an extension of the notion of continuity to cover domains obtained by removing isolated points from a larger set - in which case the function is continuous and differentiable at the missing point because the left and right limits match).

Alternatively, if the domain does include 1 the question is still meaningless because the function value is not defined on the entire domain.

That's the mathematician's approach to the question. Alternatively we can take a physics approach to the question, which would be that removing a point from a curve, or indeed removing any set of points with measure zero, has no physical impact, so the curve is continuous and differentiable as long as the set of points by which it differs from a curve that is continuous and differentiable has measure zero.

Having said that, I would suggest to @jaus tail that, if the lecturer's approach is that (1) functions may be specified by a formula alone without indicating domain and (2) with such a specification, the domain is taken to be ##\mathbb R## with the removal of all points where the formula's value is undefined, and (3) students should say the function is continuous and nondifferentiable at points that are not in the domain, then in the interests of maximising marks, it is advisable to take that approach in answers to quiz questions for as long as one has that particular lecturer.
 
  • Like
Likes jaus tail
  • #30
andrewkirk said:
If the domain doesn't include 1 - which is what you seem to be implying here
No, I am saying this.

andrewkirk said:
then the question is meaningless
How is it meaningless? The goal is to complete the sentence (it's not a question) with one of the four given choices, base on the given information.
andrewkirk said:
because continuity of a function at a point is a concept that is only applicable to points in the function's domain
Well, of course.
andrewkirk said:
(unless we adopt an extension of the notion of continuity to cover domains obtained by removing isolated points from a larger set - in which case the function is continuous and differentiable at the missing point because the left and right limits match).
That's quite a stretch, since there was no indication whatsoever that the student should extend the notion of continuity or fill in isolated point discontinuities or whatever. If you look at the problem, as given, the only reasonable choice is option C.

andrewkirk said:
Alternatively, if the domain does include 1 the question is still meaningless because the function value is not defined on the entire domain.
Pray tell, how could the domain of this function possibly include 1? In what world is division by 0 meaningful?
 
  • Like
Likes jaus tail
  • #31
Closed temporarily for moderation
 
  • Like
Likes jaus tail
  • #32
Mes chers messieurs, there is no need for such a sophisticated debate, which most likely confuse the OP.

One commonly speaks of discontinuities if a point in a graph is missing, regardless of the domain. This is necessary to investigate the various types of singularities. Whether this is correct in a strict logical sense, as a non defined location might as well be considered outside of consideration, is in the end completely meaningless and a matter for philosophers and maybe logicians, although I doubt the latter will be interested. It is the type of discussion which never ends and which won't lead anywhere, and it is the reason we canceled philosophy from the board.

So whoever feels right, let it be so. Thirty posts for such a simple question are definitely twenty-five too many.

Thread will remain closed.
 
  • Like
Likes jaus tail and andrewkirk

1. What does it mean for a function to be continuous and differentiable?

A continuous function is one that has no sudden jumps or breaks in its graph, meaning that it can be drawn without lifting the pen from the paper. A differentiable function is one that has a well-defined derivative at every point, meaning that it has a smooth and continuous slope throughout its graph.

2. How can I determine if a function is continuous and differentiable?

To determine if a function is continuous, you can graph it and check for any breaks or jumps. To determine if it is differentiable, you can take its derivative and check if it exists at every point. Alternatively, you can use the definition of continuity and differentiability to analyze the function mathematically.

3. What are the conditions for a function to be continuous and differentiable?

A function is continuous if it satisfies the three conditions of continuity: it is defined at the point in question, it has a limit at that point, and the limit equals the function's value at that point. A function is differentiable if it has a well-defined derivative at every point within its domain.

4. Can a function be continuous but not differentiable?

Yes, a function can be continuous but not differentiable. This means that the function has a smooth and continuous graph, but it does not have a well-defined derivative at every point. An example of this is the absolute value function, which is continuous but not differentiable at the point where the graph changes direction.

5. What are the practical applications of knowing if a function is continuous and differentiable?

Knowing if a function is continuous and differentiable is important in many areas of science, such as physics, engineering, and economics. It allows us to model and analyze real-world phenomena using mathematical functions, and make predictions and decisions based on the behavior of these functions. Additionally, the concepts of continuity and differentiability are fundamental in calculus, the branch of mathematics that is essential for understanding many scientific and technological fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
26
Views
804
  • Calculus and Beyond Homework Help
Replies
8
Views
851
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
128
  • Calculus and Beyond Homework Help
Replies
2
Views
454
  • Calculus and Beyond Homework Help
Replies
1
Views
130
  • Calculus and Beyond Homework Help
Replies
3
Views
796
Back
Top