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Math Help - Rolle's Theorem

  1. #1
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    Rolle's Theorem

    I'm really stuck on this problem...

    If I let the function:

    f : II be continuous on I and differentiable on the open set I
    for I := [0,1]

    I know that theres at least 1 point t [0, 1] such that f(t) = t.

    Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t

    I've tried 3 different proofs for this, but none of them are giving me uniqueness of t. Please help!
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  2. #2
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    Quote Originally Posted by MatthewD View Post
    I'm really stuck on this problem...



    If I let the function:

    f : II be continuous on I and differentiable on the open set I
    for I := [0,1]

    I know that theres at least 1 point t [0, 1] such that f(t) = t.

    Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t


    This cannot be correct since f(x)=x is a counterexample...Perhaps it should be that f'(x)\neq 0 in (0,1) ? Because then it is easy to prove the point t must be unique...
    Check this.

    Tonio




    I've tried 3 different proofs for this, but none of them are giving me uniqueness of t. Please help!
    .
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  3. #3
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    It may be a typo on the worksheet...

    How would I go about proving it if it was 0 instead? Either way, my proof methods weren't getting me anywhere.
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  4. #4
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    Quote Originally Posted by MatthewD View Post
    It may be a typo on the worksheet...

    How would I go about proving it if it was 0 instead? Either way, my proof methods weren't getting me anywhere.

    Assume t_1\,,\,t_2\in[0,1]\,,\,\,t_1\neq t_2 are s.t. f(t_i)=t_i\,,\,\,i=1,2. and apply directly Rolle's theorem to f in the interval [t_1,t_2] ...

    Tonio
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  5. #5
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    I'm a little confused as to why f(x) = x would be a counterexample (if there isn't a mistake)... because f'(x) = 1, so we don't have a problem since we are looking at f'(x) \neq 1...? or am I missing something here
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  6. #6
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    Quote Originally Posted by MatthewD View Post
    I'm a little confused as to why f(x) = x would be a counterexample (if there isn't a mistake)... because f'(x) = 1, so we don't have a problem since we are looking at f'(x) \neq 1...? or am I missing something here

    My bad: I misread your first post and thought it said "if f'(x) IS EQUAL to 1 then there is a unique..." .

    Tonio
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  7. #7
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    Right. If f'(x) is NOT 1 in [0, 1] f(x)= x is NOT a counter example and the statement as given is true.

    Apply Rolle's theorem to the function g(x)= f(x)- x. If there are two points, x1 and x2, such that f(x1)= x1 and f(x2)= x2, then g(x1)= g(x2)= 0. Therefore by Rolle's theorem...
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