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Math Help - Constants for Composition of functions

  1. #1
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    Constants for Composition of functions

    Let f(x) = ax+b and g(x) =cx+d where a,b,c,d are constants. Determine for which constants a,b,c,d it is true that f o g = g o f.

    I am not very sure how to approach this problem. I simplifed the f o g and g o f but after that I am not sure how to go from there.

    Here is what I did first

    f(g(x)) = g(f(x))
    a(cx+d) +b = c(ax+b) +d
    acx+ad+b = cax + cb +d

    then after that I got stuck. I tried switching around the varibles on either side and tried to get them to equal to eachother, however that didn't work. Am I'm going in the right direction or am I totally off? Thank you for your help!!!

    acx-acx = cb +d
    ad+b = cb +d
    ad -d = cb -b
    d(a-1) = b(c-1)

    that is where I got stuck




    Thank you for your help!
    Last edited by MathGuru; February 10th 2006 at 10:03 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hotmail590
    Let f(x) = ax+b and g(x) =cx+d where a,b,c,d are constants. Determine for which constants a,b,c,d it is true that f o g = g o f.

    I am not very sure how to approach this problem. I simplifed the f o g and g o f but after that I am not sure how to go from there.

    Here is what I did first

    f(g(x)) = g(f(x))
    a(cx+d) +b = c(ax+b) +d
    acx+ad+b = cax + cb +d

    then after that I got stuck. I tried switching around the varibles on either side and tried to get them to equal to eachother, however that didn't work. Am I'm going in the right direction or am I totally off? Thank you for your help!!!

    acx-acx = cb +d
    ad+b = cb +d
    ad -d = cb -b
    d(a-1) = b(c-1)

    that is where I got stuck




    Thank you for your help!
    You are almost there, but may be expecting something cleverer than is
    actualy the case.

    First f \circ g = g \circ f gives as you observe:

    acx+ad+b = cax + cb +d,

    which rearranges to:

    (ac-ca)x + ad+b-cb+d=0,

    Which holds when:

    ad+b-cb+d=0.

    Which is all there is to it; other than rearranging this last equation to as
    neat a form as possible. I like:

    \frac {a-1}{b}=\frac{c-1}{d}

    RonL
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  3. #3
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    Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?

    Thank you very much for your quick reply CaptainBlack!!!
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  4. #4
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    Quote Originally Posted by hotmail590
    Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?
    Yes. When a=c and b=d, the equation becomes an identity.

    f(x) = ax +b
    g(x) = cx +d

    For f o g = g o f,
    a(cx +d) +b = c(ax +b) +d
    acx +ad +b = acx +bc +d
    The acx cancels out,
    ad +b = bc +d
    ad -d = bc -b
    d(a-1) = b(c-1) ---------***

    So if a=c and b=d,
    b(c-1) = b(c-1)
    Or,
    d(a-1) = d(a-1)
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by hotmail590
    Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?

    Thank you very much for your quick reply CaptainBlack!!!
    It is true that a, b, c, and d may be any real numbers which satisfy:

    <br />
\frac {a-1}{b}=\frac{c-1}{d}<br />

    So if a=2 and c=4 then any b and d
    will do as long as d=3b.

    RonL
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