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Problem with composite functions

  1. Feb 4, 2004 #1
    In a problem I was asked to find the domain of
    f of f, which is f(f(x))
    where f(x)=(ax+b)/(cx+d)

    so I get: f((ax+b)/(cx+d))
    (a)(ax+b)/(cx+d)
    (c)(ax+b)/(cx+d)

    after I multiply it out I get

    (a^2x+ab+bcx+bd)/(cx+d)
    (cax+bc+cdx+d^2)/(cx+d)

    now the "cx+d" cancels and you are left with

    (a^2x+ab+bcx+bd)/(cax+bc+cdx+d^2)

    therefore x can't be equal to (cax+bc+cdx+d^2)
    but when I try to reduce it into a form that works I get stuck with an x on both sides

    x=(-bc-cd-cdx)/(ca)


    Any help would be greatly appreciated. I have tried this problem several times and I continue to get the same answer but it dosent make sense to me. Also "x" obviously cant equal -d/c in the final answer


    Thanks
     
  2. jcsd
  3. Feb 4, 2004 #2

    NateTG

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    Homework Helper

    note that f(f(x)) should be:
    [tex]\frac{a \frac{ax+b}{cx+d} + b}{c \frac{ax+b}{cx+d}+d}[/tex]
    instead of what you had.

    For solving the equation:
    What you mean is that x cannot be a solution to:
    0=(cax+bc+cdx+d^2)
    If you group the x's together you get
    0=cax+cdx+bc+d^2
    then you can use the distributive propery
    0=x(ca+cd)+bc+d^2
    subtract from both sides
    -x(ca+cd)=bc+d^2
    and divide
    x=- (bc+d^2)/(ca+cd)

    In general, you might be better off finding the domain of f(x), first.
     
  4. Feb 4, 2004 #3
    Thanks,

    On paper i had everything you did with the extra +b on the top and bottom. I accidentally left it out when typing it out, not really used to typing out math on a computer.


    I thought it might be something easy like that but I just couldnt figure out the answer.

    Also, I found the domain of f(x) first because this was a four step problem where I had to find
    f(g(x))
    g(f(x))
    f(f(x))
    g(g(x))
    and I was also given g(x), I just failed to mention that I had already figured out the domain of f(x) till the end.

    But anyway thanks for the help.
     
  5. Feb 4, 2004 #4
    What program did you use to make your equation look so nice?
    It sure beats trying to figure out what parenthesies go where. And what part gets divided and such.
     
  6. Feb 4, 2004 #5

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    If you click on the image, it will show you exactly what he typed. :smile:


    Anyways, as was hinted, it's a lot easier to find out the domain of f, then find out what values of x aren't sent, by f, outside f's domain.
     
  7. Feb 5, 2004 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    [tex]f(x)=\frac{ax+b}{cx+d}[/tex]

    This is defined as long as the denominator is not 0 so the domain of f itself is "all x except -d/c".

    When we apply f (for f(f(x)) again, we cannot apply it to f(x)= -d/c
    so we need to determine what x gives that also.

    In general, to determine the domain of f(g(x)).

    1. Determine the domains of f(x) and g(x) separately. Typically, that will be "all numbers except ..."

    2. Determine what values of x make g(x) equal to numbers not in the domain of f.

    To find the domain of f(g(x)), remove the numbers found in (2) from the domain of g.
     
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