smc

Extend the definition of $f_n(x)$ to $n\le 0$: Let $f_{n-1}(x)$ be the function such that $f_1(f_{n-1}(x))=f_n(x)$. From the problem, $f_5(x)=f_{35}(x)$, so the functions $f_1(x),f_2(x),\dots$ must repeat in a cycle whose length is a cycle which is a divisor of $35-5=30$. Thus, if $f_a(x)=f_b(x)$ for integers $a$ and $b$, we know that $a\equiv b$ modulo $30$. Thus, because $28\equiv -2(\mathrm{mod}30)$, we know that $f_{-2}(x)=f_{28}(x)$. It is clear that $f_0(x)=x$, because $f_1(f_0(x))=f_1(x)$. Let $f_{-1}(x)=y$. Then, we know that $f_1(y)=f_0(x)=x$, so we have the following equation we can solve for $y$: $$\begin{array}{rl}\frac{2y-1}{y+1}& =x\\ 2y-1& =xy+x\\ y(2-x)& =x+1\\ y& =\frac{x+1}{2-x}\end{array}$$ Let $f_{-2}(x)=z$. Then, we know that $f_1(z)=f_{-1}(x)=\frac{x+1}{2-x}$, so we have the following equation we can solve for $z$: $$\begin{array}{rl}\frac{2z-1}{z+1}& =\frac{x+1}{2-x}\\ 4z-2-2xz+x& =xz+x+z+1\\ 3z-3xz& =3\\ z(1-x)& =1\\ z& =\frac{1}{1-x}\end{array}$$ We derived earlier the fact that $z=f_{-2}(x)=f_{28}(x)$, so $f_{28}(x)=\boxed{\frac{1}{1-x}}$.