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Fixed Point Problems

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all real values x that are fixed by the function y=4-x^2
    f(x)=4-x^2

    2. Relevant equations
    x=y


    3. The attempt at a solution
    x=4-x62
    0=-x^2-x+4
    0=-(x^2+x+(1/4))+(17/4)

    This is where i get stuck.

    I also have two other problems which iIdo not understand how to work with.
    f(x)=7+sqrt(x-1)
    f(x)=sqrt(10+3x) -4

    The main problem keeping me from doing the the two above is not knowing what to do with the square root of the expression underneath. Thanks in advance.
     
  2. jcsd
  3. Oct 11, 2011 #2

    verty

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    Homework Helper

    Remember that you want to isolate x. Focus on doing that. If you are completing the square, be sure you know that method.
     
  4. Oct 11, 2011 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A "fixed point" for a function f is a value of x such that f(x)= x.

    1) [itex]4- x^2= x[/itex] gives [itex]x^2+ x- 4= 0[/itex].
    Solve that by completing the square or using the quadratic formula.

    2) [itex]7+ \sqrt{x- 1}= x[/itex] is the same as [itex]\sqrt{x- 1}= x- 7[/itex].
    Square both sides to get [itex]x- 1= (x- 7)^2= x^2- 14x+ 49[/itex].
    That is also a quadratic equation- but this time is easily factorable. Be sure to check your answers in the original equation. "Squaring both sides" of an equation can introduce spurious solutions.

    3). [itex]\sqrt{10+ 3x}- 4= x[/itex] is the same as [itex]\sqrt{10+ 3x}= x+ 4[/itex]. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.
     
  5. Oct 11, 2011 #4
    So for the 1st problem would the answer be [itex](\sqrt{17}/2)-(1/2)[/itex] or [itex](-\sqrt{17}/2)-(1/2)[/itex] Thanks for all of your help by the way.
     
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