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Core 3 help

  1. Sep 16, 2007 #1
    Hi.

    Could someone please help me with the following question? I would really appreciate it as I am really stuck.

    Express (x + 4)/(2x^2 + 3x + 1) - 2/(2x + 1) as a single fraction in its simplest form.

    I got that [(x + 4)(2x + 1) - 2(2x^2 + 3x + 1)]/(2x^2 + 3x + 1)(2x + 1)

    = (-2x^2 + 3x + 2)/(x^2 + 3x + 1)(2x + 1) but I am not sure where to go from here.

    Thank you.

    Cathy
     
  2. jcsd
  3. Sep 16, 2007 #2

    HallsofIvy

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    Wouldn't it make more sense to use the LEAST common denominator? Since x2+ 3x+ 1= (2x+1)(x+1) that IS a common denominator. You don't need to use the "2x+1" terrm twice. You have
    [tex]\frac{x+4}{x^2+ 3x+ 1}- \frac{2(x+1)}{x^2+ 3x+ 1}[/tex]
    That should be much simpler.

    (In your last formula, you have dropped the "2" in "2x2+ 3x+ 1".)
     
  4. Sep 16, 2007 #3
    Thanks so much for your help.

    I understand now.

    Cathy
     
  5. Sep 16, 2007 #4
    Hi.

    Could someone please help me with the following question?

    f(x) = 3 - x^2, x>= 0.

    (b) Sketch the graphs of y = f(x) and y = f^-1(x) on the same diagram.


    I got that the inverse function of f is (3 - x)^(1/2) but I do cannot figure out how to draw it in comparison to f(x). Could someone please show me.

    g(x) = 8/(3 - x), x cannot = 3.

    (e) Solve the equation f^-1(x) = g(x).


    I have no idea how to even start this last part.

    Any help would be really appreciated.

    Thank you.

    Cathy
     
  6. Sep 16, 2007 #5

    arildno

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    (e).
    So, you have:

    [tex]\sqrt{3-x}=\frac{8}{(3-x)}\to\sqrt{3-x}=\frac{8}{(\sqrt{3-x})^{2}}[/tex]

    Does that help?
     
  7. Sep 16, 2007 #6
    Thanks for your help.

    I have the answer now.

    Cathy
     
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