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An Impossible Solution

  1. Oct 28, 2011 #1
    So, yesterday in class we were asked to try and solve the following problem:

    Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.

    Now in class we arrived at the conclusion that there exists no such solution, however I beg to disagree.

    Let's suppose that a = -1 and b = 1 which satisfies the inequality. Now for f(a) to = f(b) one function comes to mind, the associated power series:

    f(x) = 1 + 2x + 4x2 + 8x3 + ... + 2nxn = [itex]\frac{1}{1-2x}[/itex]

    Such that, f2(1) = 1 and f2(-1) = 1 therefore, f(a) = f(b).

    = [itex]\frac{d}{dx}[/itex]([itex]\frac{1}{1-2x}[/itex])2

    = ([itex]\frac{-4}{2x - 1}[/itex])3

    Therefore, there does not exist f'(c) = 0

    If we take the limit as x [itex]\rightarrow[/itex] [itex]\infty[/itex] then f'(x) = 0 however there is no particular element c of (a, b) s.t. f'(c) = 0.

    My question is am I right? If not where did I go horribly wrong.
    Last edited: Oct 28, 2011
  2. jcsd
  3. Oct 28, 2011 #2
    Going too fast ?
    Differentiation that power series and see that at x=0
    f ' (x) = 0
  4. Oct 28, 2011 #3
    I accidently posted before I finished lol.
  5. Oct 28, 2011 #4


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    This statement is indeed true.

    I'm not sure why you're taking a limit as x goes to infinity. In any case, if you think the statement is wrong, you need to show there exists a c in (a,b) such that f'(c) = 0 for every f(x). Unfortunately, showing one example doesn't prove anything.

    Consider the example f(x) = |x| and a = -1, b = 1. Does this satisfy the hypothesis? Is the conclusion satisfied?
  6. Oct 28, 2011 #5
    Which statement is true? That there is no solution? Or that there is a solution?
  7. Oct 28, 2011 #6


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    This statement:

    It is possible to construct an example of a continuous function f such that there does not exist a c [itex]\in[/itex] (a,b) such that f'(c) = 0. Consider f(x) = |x|.
  8. Oct 28, 2011 #7
    f(x) is discontinuous at 0.
  9. Oct 28, 2011 #8


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    f(x) = |x| is continuous at x=0. However, it isn't differentiable at x=0.
  10. Oct 28, 2011 #9
    Sorry, that's what I meant.
  11. Oct 28, 2011 #10


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    In your OP, f(x) = 1/(1-2x) is not continuous (in fact it is not even defined) when x = 1/2.
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