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Intro to Analysis (Differentiation)

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove or disprove:
    Suppose f:[a,b]->R is continuous. If f is diff on interval (a,b) and f'(x) has a limit at b, then f is diff at b.


    2. Relevant equations

    We say that f is differentiable at x0 to mean that there exists a number A such that:
    f(x)=f(x0)+A(x-x0)+REM
    where,
    lim(x->x0) REM(x)/(x-x0) = 0


    3. The attempt at a solution

    We will prove f is diff at b by showing that that there exists a number A so that f(x)=f(b)+A(x-b)+REM
    so,
    lim(x->x0) REM(x)/(x-b) = 0

    I have gotten good at normal proofs in this course and am very confused on how to build proofs with this diff theorem. I know that this is true, but am confused how to fashion the plan and how to start the proof. How do I establish the existence of an A that satisfies this?

    Thanks!
     
  2. jcsd
  3. Apr 24, 2012 #2

    Dick

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    Write a difference quotient for f'(b) and use the Mean Value Theorem. That works, doesn't it?
     
  4. Apr 24, 2012 #3
    Hmm. How would I go about doing that? I know by the mean value theorem there is a c in (a,b) that is satisfied by f'(c)=f(b)-f(a)/b-a but how can I show that since f'(c) exists f'(b) exists since it is not for every c in the interval?
     
  5. Apr 24, 2012 #4

    Dick

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    f'(b)=limit x->b (f(b)-f(x))/(b-x). There is a c such that f'(c) equals that quotient in (x,b).
     
  6. Apr 24, 2012 #5
    So I can say that because f'(x) has a limit at b the function limit x->b (f(b)-f(x))/(b-x)=f'(b)?

    So a proof would go something like:

    We will prove f is diff at b by showing that f'(b)=f'(b)=limit x->b (f(b)-f(x))/(b-x).

    Since f'(x) has a limit at b, we know limit x->b (f(b)-f(x))/(b-x)=f'(b).

    Thus, we know that the function is diff. at b.
     
  7. Apr 25, 2012 #6

    Dick

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    You need to mention the mean value theorem and show where you used it.
     
  8. Apr 25, 2012 #7
    Alright I have been working on this proof all dayish trying to get everything work out right.

    I have for my plan:

    Suppose x<c<b

    We will prove f'(c) has a limit m at b by showing that for every ε>0 there exists an δ>0 so that for every x in (a,b) 0<abs(x-b)<δ where abs(f'(c)-m)<ε.

    Consider ε>0 arb.
    Since by the mean value theorem and the def of derivative we know that limit x->b (f(b)-f(x))/(b-x)=lim c->b (f'(c)) and (f(b)-f(a))/(b-a)=f'(c) as a<x<c<b. We know that there is a δ that satisfies this. Choose such a δ.

    Consider x in (a,b) arb.

    for such an x,

    abs(f'(c)-m))....

    Then I don't know how to build that final inequality to make it work.


    Any suggestions?
     
  9. Apr 25, 2012 #8

    Dick

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    You don't need epsilons or deltas. You know f'(b)=limit x->b (f(b)-f(x))/(b-x). The MVT tells you that there is a c(x) in (x,b) such that f'(c(x))=(f(b)-f(x))/(b-x). As x->b it must be that c(x)->b. Since you know f'(c(x)) approaches a limit as c(x)->b doesn't that tell you the difference quotient must have a limit?
     
  10. Apr 26, 2012 #9
    May be a dumb question but we know there is a limit at b, but trying to say if it is differentiable. How can we make the leap that there is a limit at b to that it is differentiable at b?

    I am following all of your logic (hopefully!) but that is the part I am hung up about. Thanks!
     
  11. Apr 27, 2012 #10

    Dick

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    If the limit of the difference quotient exists then f(x) is differentiable at b. At least, it has a one sided derivative. There's not much you can say about the other side because f is only defined on [a,b].
     
  12. Apr 27, 2012 #11
    I feel like disproving it. Take [itex]\displaystyle f\left(x\right)=\sqrt[3]{x}[/itex]. Take a = -1, b = 0. Is it differentiable on (-1,0)? Yes. Note that it's open. Is it continuous? Yes. f'(a) as a->0 does have a limit, -∞. (Yes, this is considered a limit.) But ... it's not differentiable at 0.
     
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