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Math Help - Rolle's Theorem

  1. #1
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    Rolle's Theorem

    This is not exactly Rolle's Theorem but it's a problem that was given to me that is similar that is bugging me. It states "if f is differentiable on (a,b), and f(a) = f(b) = 0, then f is uniformly continuous on [a,b]" I understand since f is differentiable on (a,b) that it must also be continuous on (a,b).
    But continuity doesn't imply uniform continuity. So is there any way to show uniform continuity? Because if there is, I haven't been able to figure it out so far. Is this statement always false since our interval is open? Would providing a simple counter example suffice if that's the case?
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    Quote Originally Posted by ABigSmile View Post
    This is not exactly Rolle's Theorem but it's a problem that was given to me that is similar that is bugging me. It states "if f is differentiable on (a,b), and f(a) = f(b) = 0, then f is uniformly continuous on [a,b]" I understand since f is differentiable on (a,b) that it must also be continuous on (a,b).
    But continuity doesn't imply uniform continuity. So is there any way to show uniform continuity? Because if there is, I haven't been able to figure it out so far. Is this statement always false since our interval is open? Would providing a simple counter example suffice if that's the case?
    I don't understand the question? Does f have to be continuous on [a,b] otherwise this follows directly since any continuous function on a compact space is uniformly continuous.
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    Quote Originally Posted by ABigSmile View Post
    "if f is differentiable on (a,b), and f(a) = f(b) = 0, then f is uniformly continuous on [a,b]"
    The problem I am trying to solve is this. It doesn't say prove it so I am assuming it could be a true or false statement. Also I am pretty sure f does not have to be continuous on [a,b] .Does that clear anything up? Sorry for the confusion.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ABigSmile View Post
    The problem I am trying to solve is this. It doesn't say prove it so I am assuming it could be a true or false statement. Also I am pretty sure f does not have to be continuous on [a,b] .Does that clear anything up? Sorry for the confusion.
    Ok. So why doesn't the function f(x)=\begin{cases} 0 & \mbox{if} \quad x=0,1\\ 1 & \mbox{if} \quad 0<x<1\end{cases} serve as a counter example?. Clearly f(a)=f(b)=0 and f is differentiable on (0,1) but it isn't continuous, let alone uniformly continuous, on [0,1].
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