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.
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
Hello, jarny!
ThePerfectHacker came up with an elegant solution . . .
Suppose that $\displaystyle f(x)$ is a function such that: $\displaystyle 3f(x) + 2f(1-x) \:= \:2x + 9$
for every real number $\displaystyle x.$ .What is the value of $\displaystyle f(2)$ ?
We are told that: .$\displaystyle 3f(x) + 2f(1-x) \:=\:2x + 9$ . for all values of $\displaystyle x.$
. So it must be true for $\displaystyle x = 2:\;\;3f(2) + 2f(\text{-}1)\;=\;13$
And it must be true for $\displaystyle x = \text{-}1:\;\;3f(\text{-}1) + 2f(2)\;=\;7$
We have a system of equations: .$\displaystyle \begin{array}{cc}3f(2) + 2f(\text{-}1) \;= \\ 2f(2) + 3f(\text{-}1) \;= \end{array}
\begin{array}{cc} 13 \\ 7\end{array}$
Now solve for $\displaystyle f(2).$