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Can we write Rolle's Theorem this way?

  1. Jan 30, 2014 #1
    Rolle's theorem:

    Statements:
    If y =f(x) is a real valued function of a real variable such that:

    1) f(x) is continuous on [a,b]
    2) f(x) is differentiable on (a,b)
    3) f(a) = f(b)

    then there exists a real number c[itex]\in[/itex](a,b) such that f'(c)=0

    what if the the f(x) is like the following graph:
    [​IMG]

    here there is a point 'c' for which f'(c) =0 but f(a) [itex]\neq[/itex] f(b)

    So to take such cases in consideration can we make a change to the last statement of Rolle's theorem as:
    3)f(c) > [f(a),f(b)] Or f(c)<[f(a),f(b)]

    are there any exceptions to the above statement?
     

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    Last edited: Jan 30, 2014
  2. jcsd
  3. Jan 30, 2014 #2

    Office_Shredder

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    Yes, that statement works. The easiest way to prove this is to notice that if f(c.) > f(b) > f(a) (for example), then there exists some d such that a<d<c and f(d) = f(b) by the intermediate value theorem. Then applying Rolle's theorem to the interval [d,c] completes the proof.

    This computer seems to insist on writing f(c.) without the period as f©, hence the strange notation.
     
  4. Jan 30, 2014 #3
    Do you mean interval[d,b]
     
  5. Jan 30, 2014 #4

    Office_Shredder

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    Yeah good catch
     
  6. Jan 30, 2014 #5
    Thanks :smile:
     
    Last edited: Jan 30, 2014
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