Equation $2f(x+2)+3f(x-2)=x+5$ Find $f(1)$ Here is equation :
$$2f(x+2)+3f(x-2)=x+5$$
How to find $f(1)$ ?
 A: If $x=y-2$ then $2f(y+4)+3f(y)=y+3$, or $$f(y+4)=\frac{y+3-3f(y)}{2}$$
So you can choose any set of values for $f$ on $[0,4)$ and complete the function to the real line.
This means that $f(1)$ is not determined by this equation alone.
Perhaps there was another condition on $f$? Was it supposed to be linear?
A: You could make some hypothesis. If you suppose that $f(x)$ is a polynomial, then you could write $f(x)=ax+b$ and find $a$ and $b$ and therefore find $f(1)$. But you need start somewhere...
A: If we assume that $f(x)$ is linear, that is to say that $f(x)=a x+b$, we have $$2f(x+2)+3f(x-2)=2[a(x+2)+b]+3[a(x-2)+b]=5ax-2a+5b$$ If this has to be equal to $x+5$, we have two equations and two unknowns $$5a=1$$ $$5b-2a=5$$ from which $a=\frac{1}{5}$ and $b=\frac{27}{25}$. Then, $f(1)=\frac {28} {25}$.
If you assume that $f(x)$ is quadratic, that is to say that $f(x)=a x^2+b x+c$, expanding $2f(x+2)+3f(x-2)$ will lead to the fact that $a=0$; then, if polynomial, $f(x)$ is linear.
In fact, you could easily prove that, if $f(x)$ is a polynomial, it must be of the same degree as the rhs.
