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Math Help - Functional equation

  1. #1
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    Functional equation

    Suppose that f(x) is a function such that 3f(x) + 2f(1-x) = 2x + 9 for every real number x. What is the value of f(2)? Could someone explain how to do this?, i get how to do compostions but i dont get this.
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  2. #2
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    Quote Originally Posted by jarny View Post
    Suppose that f(x) is a function such that 3f(x) + 2f(1-x) = 2x + 9 for every real number x. What is the value of f(2)? Could someone explain how to do this?, i get how to do compostions but i dont get this.
    3f(2)+2f(-1)=2(2)+9=13
    3f(-1)+2f(2)=2(-1)+9=7
    Solve equations.
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  3. #3
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    ?

    I dont really get it and thats what i really want to know, not just the answer. I dont understand your logic because you did nothing with the coefficients of the functions, it would have been the same without them. Could you or someone else show me a step by step or give an good explanation? thanks
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  4. #4
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    Hello, jarny!

    ThePerfectHacker came up with an elegant solution . . .


    Suppose that f(x) is a function such that: 3f(x) + 2f(1-x) \:= \:2x + 9
    for every real number x. .What is the value of f(2) ?

    We are told that: . 3f(x) + 2f(1-x) \:=\:2x + 9 . for all values of x.


    . So it must be true for x = 2:\;\;3f(2) + 2f(\text{-}1)\;=\;13

    And it must be true for x = \text{-}1:\;\;3f(\text{-}1) + 2f(2)\;=\;7


    We have a system of equations: . \begin{array}{cc}3f(2) + 2f(\text{-}1) \;= \\ 2f(2) + 3f(\text{-}1) \;= \end{array}<br />
\begin{array}{cc} 13 \\ 7\end{array}

    Now solve for f(2).

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  5. #5
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    so is the answer 5? I solved by substitution
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