Identifying Errors in Calculus: Understanding Continuity

[cvc121]

Hello,
I recently wrote a test for an introductory calculus class and was confused by the grading of one of the questions regarding continuity. I have attached a screenshot of the test question and my work below. There was a comment reading "Continuity for all x ≠ 1?" and I was deducted marks as a result. I am wondering if anyone could shed some light on where and what the error that I made was, because I am having difficulty identifying it. Thank you! All help is much appreciated.

 

[Orodruin]

I would assume points were deducted because you just stated that the function is continuous for $x \neq 1$, you did not actually show it.

 

[bhobba]

Had a look at it - its basically correct but somewhat overly complex in explaining it and a subtle point not made clear. It's really very simple, if its continuous at 1 limit c → 1+ (limit from the roght) gives c^2 =4 or c = +-2. Also limit c → 1- (limit from the ;left) gives 3c -2 =4 or c =2 thus the answer is c=2. But your answer made it more complex than that and didn't mention one sided limits were involved. I personally would have deducted a little bit and asked you to see me where I would explain why - try and be more concise in your answers. Have a chat to your teacher about it and remember continuity involves both limit c → 1+ and c → 1- are true and if the function is different on either side of the limit you can't really use limit → c. Its a small point, most don't worry about it, but the question did ask you to careful and fully justify your answer. Strictly speaking you would also need to show both functions above and below 1 are continuous for all reals above and below 1 respectfully, but I personally would not have deducted marks for that - just mention it to the student - the reason is its obvious looking at the functions they are - but that's just me - your teacher may not look at it that way.

I am personally of the belief when studying analysis to try and not 'scare' students. Most students hate it (I was weird and loved it) because all that does stop people taking it - which is a pity.

Thanks
Bill

 

[mathman]

I would assume points were deducted because you just stated that the function is .continuous for $x \neq 1$, you did not actually show it.

I agree. The Grader is nitpicking, although the criticism is valid.

 

[member 587159]

I agree. The Grader is nitpicking, although the criticism is valid.

Same here. This is what the grader meant.

To the OP: rather mark this as a basic question. Nothing intermediate about this.

 

[PeroK]

Hello,
I recently wrote a test for an introductory calculus class and was confused by the grading of one of the questions regarding continuity. I have attached a screenshot of the test question and my work below. There was a comment reading "Continuity for all x ≠ 1?" and I was deducted marks as a result. I am wondering if anyone could shed some light on where and what the error that I made was, because I am having difficulty identifying it. Thank you! All help is much appreciated.

Your answer looks good to me. Yes, you could have avoided the quadratic, but that's not wrong. Also, I found the way you set out the proof very clear and logical.

If the question asked "find all values of $c$ for which the function $c^2x$ or $3cx-2$ is contnuous", then what answer would be expected? And what knowledge of continuity would be assumed?

For me, given the question and the implied level, I would not expect to have to prove that $c^2$ and $x$ are continuous functions; or that the product or sum of two continuous functions is itself continuous. If the question expected proofs of these things then it should have stated "from first principles", IMHO.

 

[Mark44]

Had a look at it - its basically correct but somewhat overly complex in explaining it and a subtle point not made clear. It's really very simple, if its continuous at 1 limit c → 1- gives c^2 =4 or c = +-2. Also limit c → 1+ gives 3c -2 =4 or c =2 thus the answer is c=2. But your answer made it more complex than that and didn't mention one sided limits were involved.

I don't think it was necessary to look at the limits as c approaches 1 or as c approaches 2. The only limits needed were the one-sided limits as x approaches 1.

To the OP: rather mark this as a basic question. Nothing intermediate about this.

"I" level seems appropriate to me, indicating undergrad level.

Your answer looks good to me. Yes, you could have avoided the quadratic, but that's not wrong. Also, I found the way you set out the proof very clear and logical.
<snip>
For me, given the question and the implied level, I would not expect to have to prove that $c^2$ and $x$ are continuous functions; or that the product or sum of two continuous functions is itself continuous. If the question expected proofs of these things then it should have stated "from first principles", IMHO.

I agree completely.

The only objection I have to the proof given, and this is really a nit, is the statement near the top about f(x) being continuous "for x < 1 and x > 1" instead of "for x < 1 or x > 1." I wouldn't have deducted points for this, but I would have written a comment about "x < 1 and x > 1" being an empty set.

 

[cvc121]

Thank you all for your input. How would I go about showing that the function is continuous for all x cannot equal 1? I know that f(x) is defined as a polynomial when x < 1 and a linear function when x > 1 and therefore are continuous everywhere on their domain, but is there a clear way of showing this?

 

[Orodruin]

How about using the same type of argumentation that you used for x=1?

 

[student34]

Wait a minute, the marker is totally wrong. The question says "for all real numbers". Not allowing f(1) is not allowing for all real numbers. I would argue this.

 

[verty]

I think the marker might have meant that writing "c = 1 or c = 2" means you think that is the answer. But that only shows continuity for $x \neq 1$. It would need something like "c = 1 and c = 2 are candidate solutions but need to be checked" so that you only write the answer when it is the answer. I can't see what else the marker might have meant.

 

[Mark44]

I think the marker might have meant that writing "c = 1 or c = 2" means you think that is the answer. But that only shows continuity for $x \neq 1$. It would need something like "c = 1 and c = 2 are candidate solutions but need to be checked" so that you only write the answer when it is the answer. I can't see what else the marker might have meant.

In the image the OP attached, he stated that f(x) is not continuous if c = 1, but is continuous if c = 2.

 

[student34]

Right but he/she then put, "not continuous if c=1". It's almost like the marker quit reading past the 2nd step. This was meant for verty

 

[verty]

In the image the OP attached, he stated that f(x) is not continuous if c = 1, but is continuous if c = 2.

Actually I don't think what I was thinking made sense. I was trying to justify what the teacher wrote but actually I don't see a better way to solve the problem without at some point concluding c = 1 or c = 2. It seems perfectly fair to do that as a step toward the answer.

 

[Mark44]

Right but he/she then put, "not continuous if c=1". It's almost like the marker quit reading past the 2nd step.

Yes. If I were the OP, I would definitely ask the instructor why points were deducted.

 

[Orodruin]

Yes. If I were the OP, I would definitely ask the instructor why points were deducted.

Agreed. I still think what I said in #2, but there is no way of knowing for sure without asking whoever graded it.

 

[vela]

Thank you all for your input. How would I go about showing that the function is continuous for all x cannot equal 1? I know that f(x) is defined as a polynomial when x < 1 and a linear function when x > 1 and therefore are continuous everywhere on their domain, but is there a clear way of showing this?

It could be that the grader simply missed the first line of what you wrote, or perhaps the grader wanted you to say explicitly f is continuous for $x\ne 1$ because f is a polynomial when x<1 and when x>1.

 

[cvc121]

Thank you all! I will follow your suggestion and reach out to my instructor for definitive clarification.