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Instananeous rate of change

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the irc at X=2

    F(x) = [tex]\frac{4}{x-1}[/tex]



    2. Relevant equations

    [tex]\stackrel{lim}{h\rightarrow0}[/tex][tex]\frac{F(a+h)-F(a)}{h}[/tex]



    3. The attempt at a solution

    I end up with [tex]\frac{4}{h+h^{2}}[/tex] or some other form of an h on the bottom. Is there something else i can do to it??
     
  2. jcsd
  3. Oct 7, 2009 #2
    [tex]\frac{4}{(2+h)-1}[/tex] - [tex]\frac{4}{2-1}[/tex]

    all over h

    is how i set it up originally
     
  4. Oct 7, 2009 #3
    and then

    [tex]\frac{4(1)}{(1+h)1}[/tex] - [tex]\frac{4(1+4)}{1(1+h)}[/tex]

    still all over h
     
  5. Oct 7, 2009 #4
    which becomes


    [tex]\frac{h}{1+h}[/tex]

    all over h

    right??
     
  6. Oct 7, 2009 #5
    You made the mistake in your third post; that 4 in the very top right should be an h

    First factor out the four so it's easier to work with


    [tex]\frac{\frac{4}{h+1} - \frac{4}{1}*\frac{h + 1}{h + 1}}{h} = \frac{4\left(\frac{1}{h + 1} - \frac{h + 1}{h + 1}\right)}{h}[/tex]
    then see if you can cancel that h in the denominator.
     
  7. Oct 7, 2009 #6
    IRC = -4

    You're a genius! Thank you


    -Matt
     
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