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Homework Help: G(x) and F(X) of following (I have work done!)

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x)=(2x-4)/(x+2)
    g(x) = 4x2-16

    find and simplyify:
    1/[g(1/x)]
    f(x)-g(x)


    2. Relevant equations



    3. The attempt at a solution

    For 1/[g(1/x)]:

    I got (-16x^2+4)/(x^2), but I cant figure out how to simplify this down further, this assignment had other problems and they all simplified down to a whole number so I assume this one would too.

    For f(x)-g(x)
    Plugged in variables, found common denominator, and ended with this mess:
    [(4x^3)+(8x^2)-14x+36]/(x+2)

    I am sure that can get simplified further, and tried and was really geting no where, any help on these two? Am I doing something wrong or is this the right answer ? Thanks
     
  2. jcsd
  3. Feb 3, 2009 #2
    How did you get that for 1/g(1/x)

    To work with f - g, first note that f = 2(x-2)/(x+2) and g = 4(x^2 - 4) = 4(x-2)(x+2)
     
  4. Feb 3, 2009 #3
    For 1/[g(1/x)] I did was plug in (1/x) in 1/g(x) which gave me -> 1/[4*((1/x)^2)] and then I just flipped it adding the ^-1

    For the fx-gx, I did see that but I didnt know what to do from there either, i still needed a common denominator and basically it sort of lead back to what I got.
     
  5. Feb 3, 2009 #4
    That sounds like you flipped twice, which brought you back to g(1/x). Do this stepwise: first substitute x by 1/x in g(x) to get g(1/x), and then take the inverse, i.e., 1/g(1/x).
     
  6. Feb 3, 2009 #5
    Did that and got :
    [(x^2)-1] / 16

    Is that correct?
     
  7. Feb 3, 2009 #6
    Correct - first plug in 1/x then flip it but I would be careful to use "inverse".

    Incorrect. If you want people to catch your mistake show your work! First show how you got g(1/x), SIMPLIFY FIRST. Then show what you do to get 1/g(1/x)
     
    Last edited: Feb 3, 2009
  8. Feb 3, 2009 #7
    ok so I start with:
    gx=4x^2-16

    inverse of that is:
    1/[4x^2]-16

    Then I plugin (1/x)
    and get:
    1/[4(1/x)^2]-16

    Is that correct so far? Then do I flip it back? making it:
    ([4(1/x)^2]-16)^-1
     
  9. Feb 3, 2009 #8
    No, if

    [tex] g(x) = 4x^2 - 16 [/tex]

    Then

    [tex] g(blah) = 4(blah)^2 - 16 [/tex]
     
  10. Feb 3, 2009 #9
    So :
    4(1/1/x)^2 - 16

    taking inverse:
    4(1/x)^2 - 16

    ?
     
  11. Feb 3, 2009 #10
    How are you getting 1/1/x? that notation doesn't even make sense since a/(b/c) <> (a/b)/c.

    Your 2nd line is correct, g(1/x) = 4(1/x)^2 - 16, what does simplify to?
     
  12. Feb 3, 2009 #11
    it should be:
    g(1/x) = 4(1/x)^2 - 16

    g(1/x) = 4(1^2/x^2) - 16

    g(1/x) = 4(1/x^2) - 16

    g(1/x) = (4/x^2) - 16

    g(1/x) = 4[(1/x^2) - 4]

    correct?
     
  13. Feb 3, 2009 #12
    By simplify I mean find common denominator
     
  14. Feb 3, 2009 #13
    ah so the common denominator being x^2 resulting in:
    [(-16x^2)+4]/x^2

    correct? (multiplied by x^2/x^2)
     
  15. Feb 3, 2009 #14
    Yes, so what's 1/that?
     
  16. Feb 3, 2009 #15
    x^2/[(-16x^2)+4] ?
     
  17. Feb 3, 2009 #16
    Yes, although I would write it as

    [tex] \frac{x^2}{4-16x^2}[/tex]
     
  18. Feb 3, 2009 #17
    Cool, and for the second one I think i figured it out:

    f = 2(x-2)/(x+2) and g = 4(x^2 - 4) = 4(x-2)(x+2)

    We can factor out the x-2 and x+2, which gives us:
    (x+2)(x-2)(-2)

    I am not sure if I should multiple this out or leave it as is?
     
  19. Feb 3, 2009 #18
    No... you need to find a common denominator again!
     
  20. Feb 3, 2009 #19
    f = 2(x-2)/(x+2) and g = 4(x-2)(x+2)

    the common denominator would be (x+2) so:
    2(x-2)-4(x-2)(x+2)(x+2)/(x+2)

    now do I multiple variables together, and subtract likes leaving it over denominator? (That gave me this mess:

    [(4x^3)+(8x^2)-14x+36]/(x+2)

    )
     
  21. Feb 3, 2009 #20
    You need to use parantheses better, you mean

    [2(x-2) - 4(x-2)(x+2)^2]/(x+2)

    At this point I would factor out the terms that are common to both parts in the numerator.
     
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