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Another confusing problem

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    For which numbers a, b, c, and d will the function

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

    satisfy f(f(x)) = x for all x (for which this equation makes sense.

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure if the answer will consist of specific values, or just a general solution, but here is my attempt so far:

    [tex]
    \begin{align*}
    f(f(x)) &= \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d} + d}\\
    x &= \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}\\
    0 &= cx^2 + (d-a)x - b
    \end{align*}
    [/tex]

    I skipped a bunch of steps in between, but that was the final equation I ended up with after simplification. Which was really weird, because I don't know what to do with that. If I had to guess, from the last equation, a, b, c, d must be values that satisfy the equation

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

    but again, I'm really confused and am not sure how to proceed.
     
  2. jcsd
  3. Oct 10, 2009 #2
    uh...

    i can see that it will work when a=d and b=c=0, but that seems a bit trivial.
     
  4. Oct 10, 2009 #3

    Dick

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    Homework Helper

    No, you won't get specific values for a,b,c and d. You'll get a condition from them to satisfy. But you are equating the wrong things. Your expression for f(f(x)) isn't equal to x. It's equal to f(x)=(ax+b)/(cx+d). Equate those. I.e.
    [tex]\frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}=\frac{ax+b}{cx+d}[/tex].
    If you clear out the denominators and move everything to one side, you'll get a quadratic in x^2 that must be 0 for all x. So the coefficients of x^2, x and the constant must all be equal to zero. See if you can figure out a condition on a,b,c and d that makes that true.
     
  5. Oct 10, 2009 #4
    but the question says "for which a, b, c ,d will the f(f(x)) = x for all x"

    i thought that means that f(f(x)) = x?
     
  6. Oct 10, 2009 #5

    Dick

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    Indeed you are right. I was reading the problem wrong. Let me work it out again. But you still should turn the expression into a quadratic in x equalling 0 and then set the coefficients to zero.
     
  7. Oct 10, 2009 #6
    [tex]

    \begin{align*}
    f(f(x)) &= \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d} + d}\\
    x &= \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}\\
    0 &= cx^2 + (d-a)x - b
    \end{align*}

    [/tex]

    that's the quadratic i got
     
  8. Oct 10, 2009 #7

    Dick

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    That's great. You are way ahead of me. Now the only way that can be zero for all x is if each coefficient equals zero. What do you conclude?
     
  9. Oct 10, 2009 #8
    d=a, c=0, b=0

    i think...

    thanks for your help today dick.
     
  10. Oct 10, 2009 #9

    Dick

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    Homework Helper

    And that corresponds to f(x)=x, the trivial solution. I'm noticing you dropped an overall factor in the quadratic. Shouldn't it be multiplied by (a+d)? That gives you another solution possibility. Careful!
     
  11. Oct 10, 2009 #10
    oh...yeah, there was an a+d in there. does that mean that if a+d = 0, it is also true?
     
  12. Oct 10, 2009 #11

    Dick

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    It sure does! Put d=(-a) into your expression for f(f(x)).
     
  13. Oct 10, 2009 #12
    Thanks very much!
     
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