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Limit/Derivative Question

  1. Apr 25, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img159.imageshack.us/img159/8161/64625334ex7.png

    2. Relevant equations



    3. The attempt at a solution

    (f(a)+h)-f(a)
    ------------
    h

    f(a)=x^3 and f(a)=8
    f(a)=2

    I'm pretty sure that the answer of f(a)=2 is right but is there a better way to show it and I'm also unsure if I answered it fully.
     
  2. jcsd
  3. Apr 25, 2007 #2
    f(a) is not 2. The limit actually refers to the derivaive of the function at the point 2. Therefore, a=2.
     
  4. Apr 25, 2007 #3
    Okay; a=2

    How do I show this though because if there's a test, the teacher isn't going to give me marks because I didn't show my work.
     
  5. Apr 25, 2007 #4
    I would write:
    f(a+h) = (2 + h)^3, hence a+h=2+h, hence a=2. Therefore f(x)=x^3. Assume it is correct and use the other part of the equation to confirm the assumption.

    f(a)=8. We use previously found formula -> f(a)=a^3=8. We solve for a.
    a=2 what holds according to the data provided.
     
  6. Apr 25, 2007 #5

    mjsd

    User Avatar
    Homework Helper

    i guess it is a matter of comparing with the standard formula:
    [tex]\lim_{h\rightarrow 0}\; \frac{f(x+h)-f(x)}{h}[/tex]

    so all I think you need to do is the following:
    [tex]\lim_{h\rightarrow 0}\; \frac{(2+h)^3-8}{h} =
    \lim_{h\rightarrow 0}\; \frac{(2+h)^3-2^3}{h}
    = \left.\lim_{h\rightarrow 0}\; \frac{(x+h)^3-x^3}{h}\right|_{\text{at}\; x=2}
    [/tex]
    then you can identify what f(x) and x=a are.
     
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