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Continuity and differentiability over a closed interval

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/69BmR.jpg

    2. Relevant equations



    3. The attempt at a solution
    a, c are right because f(c) is continuous.
    b, d are right because f'(c) is differentiable over the interval
    I am not sure about e. Can anyone explain to me?
     
    Last edited by a moderator: Oct 18, 2012
  2. jcsd
  3. Oct 18, 2012 #2
    Are you sure about that? What about a linear function? Then f'(x) = constant ≠ 0.

    What the claim is saying is that there exists a well-defined maximum value for f. How would the function f need to behave for this claim not to be true?
     
  4. Oct 18, 2012 #3

    HallsofIvy

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    You appear to have "cut off" a critical part: what is f(1)?
     
  5. Oct 18, 2012 #4
    Ah I see, b is the answer. Because f is not a constan, so it can only be linear. When it's differentiated it can't equal to zero under [-2,1].
     
    Last edited: Oct 18, 2012
  6. Oct 18, 2012 #5
    f(1)=4
     
  7. Oct 18, 2012 #6
    That makes no sense at all. Why won't you draw a picture to get some idea of what's happening, if you want to reason like this without any mathematical proofs.
     
  8. Oct 18, 2012 #7
    Thank you clamtrox. I am very rusty about the definition of continuity and differentiability, even I went back to my calc textbook, I still cannot figure out what I should do. Can you tell me what's wrong with my reasoning?
     
  9. Oct 19, 2012 #8
    I personally think drawing the picture is more informative

    I think the problem in your reasoning is that you don't fully understand the problem. So visually, these things you know about the function:
    -it starts from p=(-2,-5) and ends in q=(1,4)
    -there are no jumps
    -there are no sharp edges
    You are asked to check if there exists any such function for which the conditions A-E do not hold.

    A is easy: you cannot draw a curve from p to q without crossing f=0. This can be shown easily using intermediate value theorem.

    Now for B: here you need to check if all curves from p to q must have f'=0 at some point. In otherwords, they are parallel to the x-axis. Now I already gave you one counterexample for why this isn't true. I am sure you can draw several other curves that also are not parallel with x-axis at any points, but still satisfy the other conditions.

    Then D is the most interesting one: you should really think hard about this.
     
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