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Mean value theorem variation proof

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f[/itex] is differentiable function on [itex][0,1][/itex] and [itex]f^{'}(0)=1,f^{'}(1)=0[/itex]. Prove that [itex]\exists c\in(0,1) : f^{'}(c)=f(c)[/itex].

    2. Relevant equations
    -Mean Value Theorem

    3. The attempt at a solution

    The given statement is not true. Counter-example is [itex]f(x)=\frac{2}{\pi}\sin\frac{\pi}{2}x+10[/itex].
    Does this mean that the statement can't be proved?
     
  2. jcsd
  3. Jan 23, 2016 #2

    PeroK

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    I suspect you should also have ##f(0) = 0## and ##f(1) = 1##.

    Otherwise, your counterexample is fine
     
  4. Jan 23, 2016 #3

    HallsofIvy

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    And, yes, if a statement is not true, it cannot be proved!
     
  5. Jan 24, 2016 #4

    Ray Vickson

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    There is a difference between "can't be proved" and "is false". If a statement has a counterexample, it is false; that is stronger than the claim that it 'cannot be proved' (but, of course, it cannot be proved as well).
     
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