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Mean Value Theorem/Rolle's Theorem and differentiability

  1. Jul 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Let f(x) = 1 - x2/3. Show that f(-1) = f(1) but there is no number c in (-1,1) such that f'(c) = 0. Why does this not contradict Rolle's Theorem?

    2. Relevant equations


    3. The attempt at a solution
    f(x) = 1 - x2/3.
    f(-1) = 1 - 1 = 0
    f(1) = 1 - 1 = 0

    f' = 2/3 x -1/3.

    I don't understand why this doesn't have a number c in f'(c), or why Rolle's theorem excludes nondifferentiable points?
     
  2. jcsd
  3. Jul 9, 2015 #2

    LCKurtz

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    For a function to violate Rolle's theorem, it would need to do two things:
    1. Satisfy all the hypotheses of Rolle's theorem.
    2. Fail to satisfy the conclusion of Rolle's theorem.

    Does this function do those two things?
     
  4. Jul 9, 2015 #3
    Hm... I'm not sure it satisfies all the hypotheses--from what the back of the book says, it isn't differentiate on the interval of (-1, 1) but I'm not sure how? Did I do a bad job differentiating it?
     
  5. Jul 9, 2015 #4

    LCKurtz

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    What do you get for f'(0)?
     
  6. Jul 9, 2015 #5
    Is it not 0? Or is that not a valid answer?
     
  7. Jul 9, 2015 #6

    LCKurtz

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    If you think ##\frac 1 0 = 0## then you must also think ##0\cdot 0 = 1##?
     
  8. Jul 9, 2015 #7
    ...This is honestly the most embarrassing moment of my life. Thank you though!
     
  9. Jul 10, 2015 #8

    HallsofIvy

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    Just wait!
     
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