Functional equation

**jarny** · November 22nd 2006, 07:15 PM

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.

**ThePerfectHacker** · November 22nd 2006, 07:37 PM

Originally Posted by jarny

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.

**jarny** · November 23rd 2006, 09:13 AM

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

**Soroban** · November 23rd 2006, 09:51 AM

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 $f(2).$

**jarny** · November 23rd 2006, 03:54 PM

so is the answer 5? I solved by substitution

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