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Find f(x)

  1. Feb 12, 2008 #1
    I am not getting the answer that my calculator is giving me for the following:

    Find the following values for x^2-1 (all over) x+4

    f(2a-1)

    I am getting 4a^2 (all over) 2a+3 for my final answer and my calculator is getting (2a-1)^2-1 (all over) 2a+3.

    What did I do wrong? Also, how do you work the math latex references?
    these:
    [x[2]^{}[/tex]-1]\frac{}{}[/x+4]
     
  2. jcsd
  3. Feb 12, 2008 #2
    funny. I got x-4+(15)/(x+4)
     
  4. Feb 12, 2008 #3

    HallsofIvy

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    Since the answer must have "a" rather that "x" that's not particularly funny!

    Okay, if x= 2a- 1, then x2= (2a-1)2= 4a2- 4a+ 1 so x2- 1= 4a2- 4a. How did you get "4a2"? Also x+ 4= (2a-1)+ 4= 2a+ 3. You have
    [tex]\frac{4a^2- 4a+ 1}{2a+ 3}[/ite]

    The [/tex] only ends a latex code. You need [tex] in front. Did you know that you can see the code for a latex formula by clicking on it?
     
  5. Feb 12, 2008 #4
    Where does the+1 come from? I had this before I got my answer: [tex]4a^{2}-2a-2a+1-1...the ones would cancel out right?
     
  6. Feb 14, 2008 #5
    [tex] f(x) = \frac{x^2-1}{x+4}[/tex]
    If x = 2a - 1 then we have:
    [tex] f(2a-1) = \frac{(2a-1)^2-1}{(2a-1) + 4} = \frac{(4a^2 - 4a +1) -1}{2a +3} = \frac{4a^2 - 4a}{2a +3}[/tex]

    You could end here or factor out the 4a on the numerator if you really wanted to.
    [tex] \frac{4a(a - 1)}{2a+3}[/tex]
     
  7. Feb 14, 2008 #6

    HallsofIvy

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    Yes, and I showed that it did. I said (2a-1)2= 4a2- 4a+ 1 so (2a-1)2+ 1= 4a2- 4a just as you have now and just as I had before. But 4a2- 4a is NOT the 4a2 you had before!
     
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