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Math Help - For which numbers a,b,c,d will the function f(x) satisfy...

  1. #1
    Member Mollier's Avatar
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    For which numbers a,b,c,d will the function f(x) satisfy...

    Problem:
    For which numbers a,b,c,d will the function
     f(x)=\frac{ax+b}{cx+d}
    satisfy f(f(x))=x for all x?

    ------------------------------------------------------------------------
    Attempt:
    <br />
f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\<br />
= \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\<br />
=\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}<br />

    By using polynomial division I get the following result:

    <br />
\left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd)  x+(bc+d^2)}<br />

    By doing some hopefully correct algebra I get:

    <br />
(a^2+bc)x+(ab+bd)=x<br />
    1.  (ab+bd)=0
    So, a=-d

    2.  (a^2+bc)=1
    So,  a=\sqrt{1-bc}

    If b and c have opposite signs, their product is negative, and they can be any number.

    If on the other hand, their signs are the same, their product is positive and:
     -1 \leq b \leq 1
     -1 \leq c \leq 1
    -----------------------------------------------------------------------

    Am I making this more complicated than it is?
    How can I better express my results regarding b, and c?

    Thanks!
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by Mollier View Post
    Problem:
    For which numbers a,b,c,d will the function
     f(x)=\frac{ax+b}{cx+d}
    satisfy f(f(x))=x for all x?
    if f[f(x)] = x , then f(x) is its own inverse, and as such, its graph is symmetrical to the line y = x

    \frac{a \cdot f(x) + b}{c \cdot f(x) + d} = x

    a \cdot f(x) + b = cx \cdot f(x) + dx

    a \cdot f(x) - cx \cdot f(x) = dx - b

    f(x)[a - cx] = dx - b

    f(x) = \frac{dx-b}{a-cx} = \frac{ax+b}{cx+d}

    \frac{-dx+b}{cx-a} = \frac{ax+b}{cx+d}


    the above equation only requires that a = -d.
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by Mollier View Post
    Problem:
    For which numbers a,b,c,d will the function
     f(x)=\frac{ax+b}{cx+d}
    satisfy f(f(x))=x for all x?

    ------------------------------------------------------------------------
    Attempt:
    <br />
f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\<br />
= \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\<br />
=\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}<br />
Correct up to here.

    By using polynomial division I get the following result:

    <br />
\left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd)  x+(bc+d^2)}<br />
The trouble with this is that you are about to find that a = d, so the denominator of the fraction in parentheses is zero.

    By doing some hopefully correct algebra I get:

    <br />
(a^2+bc)x+(ab+bd)=x<br />
    1.  (ab+bd)=0
    So, a=-d

    2.  (a^2+bc)=1
    So,  a=\sqrt{1-bc}

    If b and c have opposite signs, their product is negative, and they can be any number.

    If on the other hand, their signs are the same, their product is positive and:
     -1 \leq b \leq 1
     -1 \leq c \leq 1
    -----------------------------------------------------------------------

    Am I making this more complicated than it is?
    How can I better express my results regarding b, and c?
    The condition f(f(x)) tells you that \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} = x. Cross-multiply to get (a^2+bc)x+(ab+bd) = x\bigl((ac+cd)x+(bc+d^2)\bigr). If that quadratic equation holds for all x then you can equate each coefficient to zero. You should find that there are two sets of solutions:

    1. a+d=0 (b and c can be anything, so long as you don't have all four of a, b, c, d equal to 0);

    2. a = d \ne0,\;b=c=0.
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  4. #4
    Member Mollier's Avatar
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    Quote Originally Posted by skeeter View Post
    if [tex]
    the above equation only requires that a = -d.
    The way you come to that conclusion is very understandable, thank you.

    Quote Originally Posted by Opalg View Post


    \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd)  x+(bc+d^2)}<br />

    The trouble with this is that you are about to find that a = d, so the denominator of the fraction in parentheses is zero.
    Ah, I did not see that! Thank you!

    Merry Christmas
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