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Fixed points

  1. May 12, 2009 #1
    ive got a question on how to get a fixed point. on the equation for.

    [tex]\frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}[/tex]

    do you find the two factors to get the fixed points. or run the equation though a quadratic formula to get the fixed points.

    i have an = which is x^2-3X-4=0 but i dont know how the fraction 11/8X =3X in the equation.

    1. The problem statement, all variables and given/known data


    2. Relevant equations



    3. The attempt at a solution
    A fixed point for a function, f, is a value of x such that f(x)= x.
    Here
    [tex]f(x)= \frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}= x[/tex]
    The easiest way to handle the fractions is to multiply the entire equation by 8:
    [tex]x^2+ 11x+ 1= 8x[/tex]
    or
    [tex]x^2+ 3x+ 1= 0[/tex]
     
    Last edited by a moderator: May 12, 2009
  2. jcsd
  3. May 12, 2009 #2

    Tom Mattson

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    Yes, that's how you do it. But I wouldn't use both [itex]X[/itex] and [itex]x[/itex] for the variable. Use the same symbol in each instance.
     
  4. May 13, 2009 #3
    thank you for your help.
     
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