2014 Paper 1 Question 1

$f^2(x)=f^{-1}(x)=\frac{x-1}{x}.$

$f^3(x)=x.$

$$\begin{array}{rl}{f}^2(x)& =f\left(\frac{1}{1-x}\right)\\ & =\frac{1}{1-\frac{1}{1-x}}\\ & =\frac{1}{-\frac{x}{1-x}}\\ & =-\frac{1-x}{x}\\ & =\frac{x-1}{x}\end{array}$$$$\begin{array}{c}\text{Let }y=\frac{1}{1-x}\\ y(1-x)=1\\ y-yx=1\\ yx=y-1\\ x=\frac{y-1}{y}\\ f^{-1}(x)=\frac{x-1}{x}\end{array}$$Hence $f^2(x)=f^{-1}(x)∎$

$$\begin{array}{rl}{f}^3(x)& =f^{2}f(x)\\ & =f^{-1}f(x)\text{since }{f}^2(x)=f^{-1}(x)\\ & =x∎\end{array}$$