Find all function $f:\mathbb Q^+ \rightarrow \mathbb Q^+$ that satisfy $$f(x)+f\bigg(\dfrac{1}{x}\bigg)=1$$ and $$f(2x+1)=\dfrac{f(x)}{2}$$ for all $x$ in domain of $f$
My Approach:
Put $x=1$ in $f(x)+f\bigg(\dfrac{1}{x}\bigg)=1$
$\implies$
$f(1)+f(1)=1$
$\implies$ $f(1)=\dfrac{1}{2}$
Now put $x=2$ in $f(x)+f\bigg(\dfrac{1}{x}\bigg)$
we obtain $f(2)+f(\dfrac{1}{2})=1 \cdots \cdots (1)$
Now using $f(2x+1)=\dfrac{f(x)}{2}\cdots \cdots (2)$
Put$x=\dfrac{1}{2}$ in eqation $(2)$
$\implies$ $f\bigg(\dfrac{1}{2}\bigg)=2f(2)$
Using above result in Equation $(1)$ we obtain $f(2)=\dfrac{1}{3}$
In similar fashion we finally obtain $f(x)=\dfrac{1}{1+x}$ For Integers
I want to know Some other method to solve this
I also tried using $g(x)=f(x-1)$ like this Help in solving a simple functional equation: $3f(2x+1)=f(x) + 5x$. But couldn't get desired result
Also this question is similar to Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$