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Prove this result

  1. Jun 7, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let f be continuous in [a,b] and differentiable in (a,b). If f(a)=a and f(b)=b, then prove that f'(x1)+f'(x2)=2 for some x1,x2 ε (a,b)

    2. Relevant equations

    3. The attempt at a solution
    Using Lagrange's Mean Value Theorem
    f'(x)=1 for some x in (a,b). But the question asks the sum of derivatives at 2 points. How do I figure it out?
     
  2. jcsd
  3. Jun 7, 2013 #2
    Do x1 and x2 have to be different? Because otherwise it seems pretty trivial...
     
  4. Jun 7, 2013 #3

    utkarshakash

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    Yes of course.
     
  5. Jun 7, 2013 #4

    haruspex

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    Not sure why you say "of course". That condition is not in the OP. Have you quoted the problem exactly? If not, I wonder whether it is actually asking for a much stronger result: given x1 ε (a,b) show there exists x2 ε (a,b) (not necessarily different from x1) such that f'(x1)+f'(x2) = 2.
    If it really is as you say, consider splitting the interval in half.
     
  6. Jun 7, 2013 #5

    utkarshakash

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    This is what I mean to say.

    But if I split the interval in half how will I be able to apply LMVT separately to those two intervals?
     
    Last edited: Jun 7, 2013
  7. Jun 7, 2013 #6
    You get two intervals, (a,(a+b)/2) and ((a+b)/2,b). Simply apply LMVT for both of them.
     
  8. Jun 8, 2013 #7

    haruspex

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    Are you sure? Having thought about it some more, I don't think my suggested version is true. Please post the problem word for word.
     
  9. Jun 8, 2013 #8
    Why isn't that true? Your suggestion gives the right answer. :confused:
     
  10. Jun 8, 2013 #9

    haruspex

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    You're confusing two things. Let me recap:
    The question as posed is trivial since it allows x1=x2, so all you need is to use LMVT in the usual way.
    DimReg asked whether x1 and x2 had to be different, and utkarshakash said yes.
    If that is the correct version of the question, then my range halving idea works, but I'm suspicious that we still don't have the right question. I thought perhaps that instead of finding an x1 and an x2, the task was to find an x2 for some arbitrary given x1. That would be much more interesting. Indeed utkarshakash replied that I was right about that, but then I realised that what I'd suggested as the question is probably not even true. It still may be that the correct version of the question is something like that - we need to wait and see what utkarshakash says.
     
  11. Jun 8, 2013 #10

    HallsofIvy

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    So when DimReg asked "Do x1 and x2 have to be different? Because otherwise it seems pretty trivial... " and you answered "Yes, of course", you were wrong?
     
  12. Jun 8, 2013 #11

    utkarshakash

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    if x1 and x2 were same then it would not have been a problem to prove the result. but the question does not mention anywhere that they are same. It simply says "for some x1 and x2". This leads me to conclude that they must be different. What's your opinion on this?
     
  13. Jun 8, 2013 #12
    Be careful, sometimes the wording of easy problems is designed to make it seem more complicated than it is. The case of x1 = x2 is an example of the theorem, as written. If this is for class, you should ask the grader for clarification, not the internet.
     
  14. Jun 8, 2013 #13

    utkarshakash

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    OK I will let you know what my teacher says tomorrow.
     
  15. Jun 9, 2013 #14

    haruspex

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    I think you mean, it doesn't mention that they could be the same. But if it does not clearly indicate that you must not choose them so then you are within your rights to assume that you can.
     
  16. Jun 9, 2013 #15

    utkarshakash

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    But assuming x1 and x2 to be different is not a problem either. Your method gives the right answer very easily. I only mean that they are not same because then the problem would have been too easy to solve and I don't expect my teacher to give such easy question. But anyways, I'll discuss this with my teacher.
     
  17. Jun 9, 2013 #16

    Office_Shredder

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    It's definitely not true. Let f(x) = xn on the interval [0,1], for a really big choice of n, and let x1 be a number really close to 1. Then f'(x1) > 2 and there are no negative derivatives to use
     
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