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Math Help - Differentiable Func.

  1. #1
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    Differentiable Func.

    Given that f is function that's differentiable on the interval [a,b] where f'(x) \neq 1\,\,\, \forall x \in [a,b].

    1.) Prove f has at most 1 fixed point on the interval [a,b]. (ie., show there's at most 1 point x \in [a,b] where f(x) = x

    2.) By giving specific examples, show it's possible that f either has no fixed points or that f has 1 fixed point on [a,b].
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    2.) By giving specific examples, show it's possible that f either has no fixed points or that f has 1 fixed point on [a,b].
    what are the restrictions here? it seems to me you can come up with a lot of trivial examples for this. example, f(x) = 2 on the interval [0,1]
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    Suppose that there are two point in [a,b] such that f\left( {x_1 } \right) = x_1 \,\& \,f\left( {x_2 } \right) = x_2 .

    We can assume that x_1  \le x_2 so by the mean value theorem
    \left( {\exists c \in \left( {x_1 ,x_2 } \right)} \right)\left[ {f'(c) = \frac{{f\left( {x_2 } \right) - f\left( {x_1 } \right)}}{{x_2  - x_1 }} = 1} \right].
    Do you see the contradiction?
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    I thought f'(x) =! 1, but you have that f'(c) = 1..

    I could see a contradiction if x_1 was strictly less than x_2, but it can also be equal, so perhaps I'm overlooking it. Then, if x_1 were equal to x_2, there'd be a contradiction. But you have the stipulation that it can be equal to it.

    And to Jhevon: that was the exact question. I'm not sure if I can use a trivial example such as that, as this was a big question and I doubt it'd be that easy.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    I thought f'(x) =! 1, but you have that f'(c) = 1..

    I could see a contradiction if x_1 was strictly less than x_2, but it can also be equal, so perhaps I'm overlooking it. Then, if x_1 were equal to x_2, there'd be a contradiction. But you have the stipulation that it can be equal to it.
    yes, that's the contradiction. if we have more than 1 fixed point, we end up having something they told us we couldn't have in the first place, thus we must have at most one fixed point, as the theorem says
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