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Limits of Average Rates of Change

  1. Dec 14, 2007 #1
    I'm not looking for an answer to a specific question, but I want to know in general how to evaluate the limit of average rates of change.

    1. The problem statement, all variables and given/known data

    lim[tex]_{}h \rightarrow0[/tex] f (x + h) - f (x) / h


    2. Relevant equations

    f(x) = x^2 , x = 1



    3. The attempt at a solution

    I really don't know what to do. Obviously we need the denominator not equal to 0. An example in my book showed them multiply by 1 by multiplying the numerator and denominator by the conjugate since the numerator had roots...but this has no roots.
     
  2. jcsd
  3. Dec 14, 2007 #2

    nicksauce

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    I don't want to do your homework, so I'll do f(x) = x^3.
    [tex]\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}[/tex]

    [tex]\lim_{h\rightarrow 0} \frac{(x+h)^3-x^3}{h}[/tex]

    [tex]\lim_{h\rightarrow 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}[/tex]

    [tex]\lim_{h\rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}[/tex]

    [tex]\lim_{h\rightarrow 0} 3x^2 + 3xh + h^2[/tex]

    [tex]=3x^2[/tex]
     
  4. Dec 14, 2007 #3
    If it says for example x = 1, all you do f (1) and evaluate?
     
  5. Dec 15, 2007 #4

    nicksauce

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    Right, so in the example I did, f ' (1) = 3, f ' (2) = 12 etc.
     
  6. Dec 15, 2007 #5

    Gib Z

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    Well Since you know that you must evaluate at x=1, you can do two things. Either do as nicksauce did, and sub in x=1 at the end, or simply evaluate

    [tex]\lim_{h\to 0} \frac{(1+h)^2 - 1^2}{h}[/tex] Directly.
     
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