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Differentiation question

  1. Nov 16, 2007 #1
    Problem statement

    Show that if F:(a,b)->R is differentiable at every point of (a,b) and increasing then f'(x) >= 0 for all x belonging to (a,b)

    Problem Solution

    Definition of differentiation:
    Assume sequence {an} -> a then
    f(an) - f(x) / (an -x) converges to some value

    Using this definition I am saying since it is known that f is differentiable at a,b then the following is true

    Assume sequence an->a and bn->b
    then f(bn) -f(an) / (bn-an) --- (eq1)

    Since this is an increasing function both numerator and denominator will be always greater than 0 and thus the derivative by eq1 is always positive.

    Am I on the right track?

    If not plz advise.


    Thanks

    Asif
     
  2. jcsd
  3. Nov 17, 2007 #2

    EnumaElish

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    I think you meant [f(b) -f(bn)] / (b-bn) and alternatively [f(an) -f(a)] / (an-a). Even then this only addresses f'(a) > 0 and f'(b) > 0.
     
  4. Nov 17, 2007 #3
    No I meant what I had written - however I did realise after posting that it was wrong. I worked through a specific f(x) and figured it was wrong.

    If I use the definition of differentiability as you have done (f(b) - f(bn))/(b-bn), can I safely say that it is f'(b) >= 0;
     
  5. Nov 17, 2007 #4

    HallsofIvy

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    Is that really the definition of "differentiation" your text gives?

    It should be "f is differentiable at a if and only if for every sequence {an} that converges to a, (f(an)- f(a))/(an-a) converges to the same limit. If f is differentiable at a, then its derivative is that common limit."
    (NOT "f(an) - f(x) / (an -x)" that makes no sense. You must mean (f(an)- f(a))/(an-a) with "a", not "x". )

    The way you've written it, it sounds like just one such sequence would be sufficient to determine that a function is differentiable. Of course, here, since you are given that f is differentiable, then it really doesn't matter.

    I see no reason for looking at two sequences {an} and {bn}. Take {an} to be a sequence converging to a with an> a for all n.
     
    Last edited: Nov 17, 2007
  6. Nov 17, 2007 #5
    Just use mean-value theorem.
     
  7. Nov 17, 2007 #6

    HallsofIvy

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    Well, if you want to do things the easy way!! :)
     
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