Question

If $f_1\left(x\right)=\frac{x}{x-1}$ and $f_n\left(x\right)=f_1\left(f_{n-1}\right(x\left)\right)$ for $n\ge 2$, then the integral value of $x$ that satisfies $f_{101}\left(x\right)=3x$ is

Answer 1

$f_1\left(x\right)=\frac{x}{x-1}$
$f_n\left(x\right)=f_1\left(f_{n-1}\right(x\left)\right),n\ge 2$
$\Rightarrow f_2\left(x\right)=f_1\left(f_1\right(x\left)\right)=f_1\left(\frac{x}{x-1}\right)=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}=x$
$\Rightarrow f_3\left(x\right)=f_1\left(f_2\right(x\left)\right)=\frac{x}{x-1}$
$\Rightarrow f_4\left(x\right)=f_1\left(f_3\right(x\left)\right)=f_1\left(\frac{x}{x-1}\right)=x$
$\Rightarrow f_5\left(x\right)=f_1\left(f_4\right(x\left)\right)=f_1\left(x\right)=\frac{x}{x-1}$

$\Rightarrow f_m\left(x\right)=\{\begin{array}{cc}\frac{x}{x-1},& \text{m is odd}\\ x,& \text{m is even}\end{array}$

$\Rightarrow f_{101}\left(x\right)=\frac{x}{x-1}=3x\Rightarrow 3x^2-3x-x=0\Rightarrow x(3x-4)=0$
$\Rightarrow x=0,\frac{4}{3}$
$x=0$