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Where to start I need to multiply and divide by somthing! anyone c it?

  1. Nov 13, 2005 #1
    Hello everyone, i'm really lost on this problem, am i proving like the chain rule or composition of functions? Here is the problem:
    or here:

    THe professor told me to multiply and dvide by the right thing which didn't help much. I started to multiply
    (h(p)-h(a))/(p-a) through and it just got really ugly, I don't see how this is gonna work at all! Any tips on how I can get this started? :bugeye:
  2. jcsd
  3. Nov 13, 2005 #2


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    Am I missing something? You are asked to show that
    [tex]\frac{f(p)-f(a)}{p-a}= \frac{f(p)-f(a)}{h(p)-h(a)}\frac{h(p)-h(a)}{p-a}[/tex]
    Isn't that obvious? You multiply and divide the left hand side by h(p)- h(a)!

    Or are you missing the fact that f(p)-f(a)= g(h(p))- g(h(a)) by definition of f?
  4. Nov 13, 2005 #3
    wait you are right...what the heck, i didn't even look at the left hand side, i figured I should try and make the right hand side equal to the left but that is so easy to prove the left is equal to the right by doing waht you said. How is this going to prove anything? it seems so easy to just do that, is that it?
  5. Nov 15, 2005 #4
    can someone tell me what i'm trying to prove? is it the chain rule? this will help me find out where i'm suppose to go with this.
  6. Nov 15, 2005 #5
    It kinda looks like some kind of application of the mean value theorem to me.

    Why not use the mean value theorem:

    (f(p) - f(a))/(p-a) = f`(some value c between p and a)

    for it to work, you only have to take the derivative of g(h(c)).
    Last edited: Nov 15, 2005
  7. Nov 15, 2005 #6
    Will that show what he is asking? I think he wants me to keep it in the forum of the given information of the problem, but I like your idea.
  8. Nov 16, 2005 #7


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    I can't answer "Will that show what he is asking?" since I have no idea who "he" is! It looks to me as if you are doing part of the proof of the mean value theorem.
  9. Nov 16, 2005 #8
    THanks for the responce, all he wants is what hte problem wants, is to just show that one side is equal to the other, it seems like you already told me how to do that. It just seems so simplistic.
  10. Nov 16, 2005 #9
    Thanks for the help everyone, Hall was right on the proof. It also wasn't proving the mean value, it was proving the chain rule if anyone cares.
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