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Homework Help: Math homework challenge

  1. Jun 5, 2008 #1
    1. The problem statement, all variables and given/known data
    suppose f(x-1/x+1)+f(-1/x)+f(1+x/1-x)=x then find f(x)

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 5, 2008 #2
    What's your attempt at finding a solution?
    Last edited: Jun 5, 2008
  4. Jun 5, 2008 #3
    change x to tan(x)
  5. Jun 5, 2008 #4
    Are the brackets in the right place? It seems to me that you're missing some. Furthermore is this the entire question?
  6. Jun 5, 2008 #5
    would you like to know the answer
  7. Jun 5, 2008 #6
    Yes please!
  8. Jun 5, 2008 #7
    f(x)=2xxxx+3xx-1/3x-3xxx it was a question of the iranian mathematical olympiad note xx means x power two
  9. Jun 5, 2008 #8
    So you mean: [tex]
    f(x)=2x^4+3x^2- \frac{1}{3} x-3x^3
    This isn't possible the function is defined for 1, -1 and 0 and the exercise says that that isn't possible.

    What method did they use to get the solution other than just trying different order of polynomials? Can you give me the website where you got this?
    Last edited: Jun 5, 2008
  10. Jun 5, 2008 #9
    t=x-1/x+1 ------ > tx +t=x-1 -------- >x=1+t/1-t , -1/x=t-1/t+1 ,1+x/1-x=-1/t

    ------->f(t)+f(t-1/t+1)+f(-1/t)=1+t/-t -------->f(x)+f(x-1/x+1)+f(-1/x)=1+x/1-x

    t=-1/x --------->x=-1/t , x-1/x+1=t+1/1-t , 1+x/1-x=t-1/t+1
    ------->f(t+1/1-t)+f(t)+f(t-1/t+1)=-1/t >f(x+1/1-x)+f(x)+f(x-1/x+1)=-1/x

    t=1+x/1-x----------->t-tx=1+x ----->x=t-1/1+t------ > -1/x =t+1/1-t , x-1/x+1=-1/t
    ------>f(-1/t)+f(t+1/1-t)+f(t)=t-1/t+1------->f(-1/x)+f(x+1/1-x)+f(x)= x-1/x+1

    two more step remains that i think you can do them yourself
    what about this one
    prove that: Arctan(1)+Arctan(2)+Arctan(3)=Pi
  11. Jun 5, 2008 #10
    Is there a website or so where I can see the question and the answers of the previous question?
  12. Jun 5, 2008 #11


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    Homework Helper

    I can actually do this one...haha

    just use tan(A+B+C) where tanA=1,tanB=2,tanC=3

  13. Jun 5, 2008 #12
    a geometric proof
  14. Jun 5, 2008 #13
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