Find $f(3)$ if $f(f(x))=3+2x$ A function $f\colon \mathbb{R} \to \mathbb{R}$ is defined as $f(f(x))=3+2x$
Find $f(3)$ if $f(0)=3$
My try:
Method $1.$ Put $x=0$ we get $f(f(0))=3$ $\implies$ $f(3)=3$
Method $2.$ Replace $x$ with $f(x)$ we get
$$f(f(f(x)))=3+2f(x)$$ $\implies$
$$f(3+2x)=3+2f(x)$$
Put $x=0$
$$f(3)=9$$
I feel Method $2.$ is Correct since $f(f(x))=3+2x$ is Injective which means $f(x)$ should be Injective.
 A: It looks like you have found a contradiction. Apparently, $f(f(x)) = 3 + 2x$ and $f(0) = 3$ are not reconcilable. There is no such function.
A: Not only is $f(f(x))=3+2x$ injective, it is bijective, and $f(f(x))\not=x$ except when $x=-3$
so you must have $f(x)\not =x$ and thus must have have $f(f(x)) \not =f(x)$, except  for the case $x=-3$ in which case you must have $f(-3)=-3$
in particular you cannot have $f(0)=f(f(0))=3$
In fact you can have $f(0)$ with any values apart from $-3$ or those in $\left\{\ldots, -\frac{45}{16},-\frac{21}{8},-\frac{9}{4},-\frac{3}{2},0,3,9,21,45,\ldots\right\}$, and you will then have $f(3)=3+2f(0)$ 
For a general solution to the functional equation, 


*

*you partition the reals into equivalence classes with the equivalence relation $y = 3+2x \implies y R x$ 

*then pair these equivalence classes up, apart from $\{-3\}$ as a special case paired with itself, (there are some easy ways of doing this, and many more difficult ones) 

*so $f(x)=z$ takes you from one equivalence class to its pair 

*and $f(z)=f(f(x))=3+2x$ takes you back but one step up  


One example solution to the functional equation is $f(x)=-6-x$ when $x\ge -3$ and $f(x)=-9-2x$ when $x\le -3$, which has $f(0)=-6$ and $f(3)=-9$ 
