Finding a suitable function in a specified domain Find a fuction $f$ $:$  $[0, \infty) \rightarrow \mathcal{R}$  so that :-
$f(2x+1)=3f(x)$  $\forall  x \in [0, \infty)  $ .
Here ,$\mathcal{R}$ denotes the set of all real numbers .
 A: I'd try an experimental approach here.  Let's insert a linear function $f(x)=mx+b$ to see if that will work:
$$f(2x+1)=m(2x+1)+b=3(mx+b)$$
$$2mx+m+b=3mx+b$$
which yields $m=0$,$b=0$, so no non-trivial solution there.
Let's try quadratic: $f(x)=ax^2+bx+c$.  Then
$$a(2x+1)^2+b(2x+1)+c=3(ax^2+bx+c)$$
$$2ax^2+(4a+2b)x+a+b+c=3ax^2+3bx+3c$$
which again leads to no non-trivial solutions.  The pattern is clear here: the leading order always leads to a zero coefficients, so there is no non-trivial polynomial.  
Let's try exponential next: $f(x)=ae^{bx}$.  Then:
$$ae^{b(2x+1)}=3ae^{bx}$$
or
$$ae^{2b}e^{2bx}=3ae^{bx}$$
which again leads to no-nontrivial solutions.  
Nice continuous functions are not cooperating.  So let's try something else.  The pattern suggests perhaps a piecewise definition.  In particular, from $f(0)$ we know $f(1)$, from which we know $f(3)$, from which we know $f(7)$, etc.  So we can use these as the breaks of the piecewise definition.  You can let $f(x)$ equal anything on $0 \leq x <1$, and f(x) can be defined recursively on the other intervals.  So for example let $f(x)=1$ on $0 \leq x <1$, $f(x)=3$ 
on $1 \leq x <3$, etc.  In general, you can write $$f(x)=3^{a-1}$$, $$2^{a-1}-1 \leq x \leq 2^a-1, a \in \mathbb{N}$$ will give you a simple, but non-trivial solution.
