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F(x)+2*f((x+2000)/(x-1))=4011-x -> f(x)=?

  1. Aug 21, 2004 #1
    f(x)+2*f((x+2000)/(x-1))=4011-x ---> f(x)=?

    f(x)+2*f((x+2000)/(x-1))=4011-x ---> f(x)=?
  2. jcsd
  3. Aug 21, 2004 #2
    [tex] f(x) = 4011 - x - 2f((x+2000)/(x-1)) [/tex] lol

    you cant just ask a quastion and expec to get a straight up answer. Its not done that way here. Tell me what you have so far, and where you are stuck.
  4. Aug 21, 2004 #3
    If i could solve it, i wouldn't have written here! How can we annihilate f((x+2000)/(x-1)), i guess we should equalize it to f(x), but how?
  5. Aug 21, 2004 #4
    Oops! Maybe, i should write this question in ''homework'' or something like that part. Sorry for that!
  6. Aug 22, 2004 #5
    Even a little "help, I don't even know how to start this problem" in the first post goes a long way towards getting people to not feel like you're just using them.

    Anywho, suppose that you could find an x so that f((x+2000)/(x-1)) could be simplified to just f(x), i.e. a solution y to (y+2000)/(y-1) = x... Maybe that hint will help.
  7. Aug 22, 2004 #6

    :tongue: Thats the difficulty of learning English! Again, sorry, if i was rude.

    By the way i finally solved the problem, but one more thing..... Another function.... The question is; f ((x-3)/(x+1))+f((3+x)/(1-x))=x ---> f(x)=?

    I showed that (x-3)/(x+1)=a and -(x+3)/(1-x)=a' ---> f'(a)=b, f(b)=a,

    f(a)+f(b)=x can i use linear function to continue the solution?
  8. Aug 22, 2004 #7


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    Dearly Missed

    First, make a trial solution f(x)=ax+b.
    If you can find constants a,b such that the equation is valid for every x, then you have found a solution of your functional equation.
    (It is by no means certain that you have found every solution)

    If the above approach doesn't work, try another trial solution f(x)=ax^2+bx+c
    Good luck!
  9. Aug 22, 2004 #8
    SEE THE Attach Files,


    Attached Files:

    Last edited: Aug 22, 2004
  10. Aug 22, 2004 #9


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    Question: What kind of function is this? Is it geometric, quadratic, 5th degree, rational? (What are you currently studying in math? That information can narrow down the possibilities)

    Also, don't you need two equations if you're going to solve for x AND f(x) ? Couldn't you just say "f(x) = 0 and x = 4011" and save yourself some time?
  11. Aug 22, 2004 #10
    do not post answers like that. Hints are fine, answers are not.
  12. Aug 22, 2004 #11


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    From that picture:
    x^2-x = x+2000
    x^2-2x-2000 = 0
    x = (2 +- sqr(4 - 4*1*-2000))/2 = 1 +- sqr(-7996) = 1 +- 89.4

    He assigned a value to x to solve the problem. Although half of the work there doesn't make any sense since it seems to go in circles.

    f(91.4) + 2*f(91.4) = 4011 - 91.4
    After that it's sortof... very... easy.
  13. Aug 22, 2004 #12
    ''Question: What kind of function is this? Is it geometric, quadratic, 5th degree, rational?'' --> It wasn't clarified in the question. When it comes to me, i just learnt second degree. These are not homeworks of course, i just want to devolop myself. And again sorry for that, i don't know the rules here.
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