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Homework Help: Derivative physics question

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Each of the limits represent a derivative f'(a). Find f(x) and a.
    1) lim x-> 0 [(6^x) - 1]/x
    2) lim x-> 0 [((4+h)^(-2)) - 0.0625]/h

    2. The attempt at a solution

    i'm not sure what i'm suppose to be solving for, they give us a derivative, or a partial derivative thats not completely solved and they want us to go backward to find the original function and the number that gave us the derivative they have. How do we do this without integrating.
  2. jcsd
  3. Oct 10, 2007 #2
    How about using the definition of a derivative...
  4. Oct 10, 2007 #3
    Okay but i never even thought of that...
    lim x -> 0 [f(a+x)-f(a)]/x = lim x -> 0 [(6^x) - 1]/x

    so does that mean f(a+x) = 6^x, and f(a) = 1

    than to find f(x) you do f(a+x)-f(a) = 6^x - 1

    than a = 6^a = 2 than you take the log of that to find a?
  5. Oct 11, 2007 #4


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    You never even thought of that? The problem said each was a derivative? Don't you associate derivatives with limits?

    No, a is not 6^a and certainly not 2 (where did you get the 2 from?). What is 6^0?
    Now what do you think f(x) and a are?

    For the second problem you might want to calculate (1/4)2 as a decimal number.
  6. Oct 11, 2007 #5
    i was being sarcastic, of course that's the first thing i did...

    i think on the bottom of my last post i meant to put

    f(a+x) = 6^x, and f(a) = 1 Therefore, f(x) = 6^x - 1
    if you stuff a in for x, to get f(a), you get it f(a) = 6^x - 1
    Since f(a) = 1
    1 = 6^x -1
    6^x = 2
    do a log to find a

    OR... do i take the limit somewhere in there to find 6^0 = 1 and it'll all make sense...
  7. Oct 11, 2007 #6


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    What about f(0 + x) = 6^x (suggestive notation... hope that rings a bell)
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