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If 3f(x)+f(3-x)=x squared, what's f(x)?

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    I first tried replacing "x" when in parenthese with y. Therefor, 3y+3-y=x, yadda yadda, 2y=x-3, y=(x-3)/2, however, the real answer turned out to be some weird fractional thing. Please give me some kind of direction..

    EDIT: Ignore that...
     
    Last edited: Oct 17, 2007
  2. jcsd
  3. Oct 17, 2007 #2

    nrqed

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    OTHER EDIT!
    Sorry!! yes, my initial trick does work if you replace x by 3-x. I am obviously not thinking straight after having been teaching for 4 hours straight! just replace x by 3-x and then isolate your f(x)!


    Ignore the rest....



    EDIT: oops. I had read too quickly. My trick won't work right away. You will also have to consider replacing x by minus x to make it work

    Start from the initial equation. Substitute x -> x-3 everywhere you see an x (on both sides of the equation. This will give you a new equation containing f(x) and f(x-3). Now use those two equations to get rid of f(3-x) to leave you with an answer for f(x)
     
    Last edited: Oct 17, 2007
  4. Oct 17, 2007 #3
    3f(3-x)+f(x)=3-x is what I get if I plug in 3-x where x is...
     
  5. Oct 17, 2007 #4

    nrqed

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    you mean [itex](3-x)^2[/itex] on the right side??
     
  6. Oct 17, 2007 #5
    Oops haha
    So 3f(3-x)+f(x)=(x squared)+9-6x.

    So can you tell me where to go from here? I tried substituting y into x again, get (x squared - 6x)/-2, while the real answer is something like x/3+x/5-3 or something. Also, where did the 'plug 3-x for x everywhere' thing came from? And if this is a "topic" I can look up like for quadratic equations, or "f and g compositions"?
     
  7. Oct 17, 2007 #6

    nrqed

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    if you substitute again you will get back to the initial equation. No, just use the two equations you now have. get rid of f(3-x) (isolate it from one equation and plug in the other one) and you can isolate f(x) now.
     
  8. Oct 17, 2007 #7
    I am sorry, I was just reading through this thread: how can you substitute x for x-3? Doesn't this imply -3=0?
     
  9. Oct 17, 2007 #8

    Hurkyl

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    He said substitute for, not set equal to.

    If it makes you feel better, try the (equivalent) substitution x -> y - 3.
     
  10. Oct 17, 2007 #9
    3f(3-x)+f(x)=(x squared)+9-6x

    get rid of f(3-x) (isolate it from one equation and plug in the other one) and you can isolate f(x) now.
    ===
    If I want to isolate f(x).. it'd be f(x)=xsquared+9-6x-3f(3-x), right?
     
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