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Math Help - How to prove $f(a)\leq f(b)$?

  1. #1
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    How to prove $f(a)\leq f(b)$?

    Dear all, suppose f is a continous function on [TEX[a,b][/TEX]. E\subset[a,b] is countable. Assume f is differentiable on [a,b]\backslash E, the the derivative >=0. Show that f(a)\leq f(b).

    Note: We assume only that f is differentiable on [a,b]\backslash E.

    I do not know how to prove. Do you know? Help me, thank you...
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  2. #2
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    Re: How to prove $f(a)\leq f(b)$?

    Since E is a countable subset of (a,b), let E = {ei} with a < e1 < e2< ... < b.

    Then we have that f is continuos on [a, e1], and differentiable on (a, e1). The mean value theorem applies and we find that f(a) \leq f(e1). We can prove this by contradiction. Assuming f(a) > f(e1), we would have by the MVT a c1, s.t. f'(c1) = (f(e1​)-f(a))/(e1-a) < 0, a contradiction to f'(x) \geq 0.

    Similiarly, we can find that f(e1) \leq f(e2). Continuing in this way we can argue that for any ei < ej, f(ei) \leq f(ej).

    I think this may be a step toward proving the desired, f(a) \leq f(b), but I haven't yet worked it completely. Will finish and post the rest soon, unless you or someone else can finish it.
    Last edited by RBowman; November 20th 2012 at 01:26 PM.
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  3. #3
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    Re: How to prove $f(a)\leq f(b)$?

    Do anyone knows how to prove it? Thank you ...
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