Solving $f(f(x))=2x+1$ for $f$ 
let $f:\mathbb{R} \to \mathbb{R}$ be a  differentiable function  such that  $\forall x \in \mathbb{R} :  f'(x)>0 $ and $f(f(x))=2x+1$ then find the $f(x)$

My Try :
$$f(f(x))=2x+1 \\ f'(x) \cdot f'(f(x))=2 \\f'(f(x))=\dfrac{2}{f'(x)}$$
now what do i do ?
 A: Define $g(x)=f(x+1)-1$. Then the functional equation simplifies to
$$ g(g(x))=2x $$
Since $g(g(0))=0$ we must have $g(0)=0$; otherwise $g(g(x))=2x$ would not hold for $x=g(0)$.
Now define $h(x) = \log(g(e^x)/e^x)$. (The logarithm always exists because $f$ is assumed to be strictly increasing, so $g(e^x)$ is positive). Then the functional equation becomes
$$ h(x) + h(x+h(x)) = \log2$$
This equation implies that $h$ must be periodic with period $\log2$ -- namely, setting $y=x+h(x)$ we have
$ y+h(y) = x+\log2 $ and therefore $$ h(x) = \log2 - h(y) = h(y+h(y)) = h(x+\log2). $$
From the definition of $h$ we get
$$ g'(0) = \lim_{t\to 0}\frac{g(t)-g(0)}{t-0} = \lim_{t\to 0}\frac{g(t)}t = \lim_{x\to-\infty} e^{h(x)}$$
so in order for $g$ to be differentiable at $0$, the latter limit needs to exist.
Now, however, if $h(x)$ is not the constant function $h(x)=\frac12\log2$, then the periodicity together with the functional equation for $h$ means that $h(x)$ must take two different values $y$ and $\log(2)-y$ for arbitartily large negative $x$ -- so then $h(x)$ or $e^{h(x)}$ cannot have a limit for $x\to-\infty$.
Thus, $h$ is the constant function $\frac12\log2$, which means that $g(x)=\sqrt2 x$ for $x\ge 0$.  The same argument, mutatis mutandis, fixes $g(x)=\sqrt2 x$ for $x<0$.
A: Partial result. 
If we suppose that $f(x)=kx+n$, where $k>0$ since $f'>0$, then we get:
$$ k^2x+n(k+1)=2x+1\;\;\;\forall x\in \mathbb{R}$$ so $k=\sqrt{2}$ (and not $-\sqrt{2}$) and $n = {1\over \sqrt{2}+1} = \sqrt{2}-1$.

Any way we have from a given equation $f(2x+1) = 2f(x)+1$, I would strongly assume from here that the function I found is the only one.
