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Exercise with lagrange and derivatives

  1. Dec 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Being a>0 and f:[a,b]--->R continuos and differentiable in (a,b), show that there exists a t ##\in## (a,b) such that:
    ## \frac{bf(a)-af(b)}{b-a}=f(t)-tf'(f)##

    3. The attempt at a solution
    For lagrange's theorem, we have:
    ## \frac{f(a)-f(b)}{b-a}= -f'(t) ##

    thought i could find f(t) from the equation ##f(x) = f(t) + f'(t)(x-t)+R_1(x-t)## ignoring R.
    but then a "x" would appear and I don't know how to deal with it.
     
    Last edited by a moderator: Dec 22, 2012
  2. jcsd
  3. Dec 22, 2012 #2

    mfb

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    Staff: Mentor

    To simplify the equation, consider g:[a,b]->R with
    $$g(x)=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}$$
    This is chosen to get g(a)=g(b)=0 and ##g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}##

    Is there some t ∈ (a,b) such that ##0=g(t)-tg'(t)##?
    Working backwards, you should be able to get the original equation.
     
  4. Dec 23, 2012 #3
    sounds good, but I'm explicitly asked to use Lagrange's theorem
     
  5. Dec 23, 2012 #4

    mfb

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    I think you can use that theorem to solve the reduced problem ;).
     
  6. Dec 23, 2012 #5
    another hint? i don't get it, i always end up in a loop :(
     
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