2007 Paper 1 Question 2

$gf:x\mapsto \frac{1}{{(x-3)}^2},x\in ℝ,x\ne 3.$
$fg$ does not exist as $R_g\subseteq ̸D_f.$

$f^{-1}(x)=\frac{1+3x}{x}.$
$D_{f^{-1}}=(-\infty ,0)\cup (0,\infty ).$

$$\begin{array}{rl}gf(x)& =g\left(\frac{1}{x-3}\right)\\ & ={\left(\frac{1}{x-3}\right)}^2\\ & =\frac{1}{{(x-3)}^2}\end{array}$$ $$gf:x\mapsto \frac{1}{{(x-3)}^2},x\in ℝ,x\ne 3∎$$

$$\begin{array}{rl}{R}_g& =[0,\infty )\\ D_f& =(-\infty ,3)\cup (3,\infty )\end{array}$$

Since $R_g\subseteq ̸D_f,$ the composite function $fg$ does not exist $∎$

$$\begin{array}{c}\text{Let }y=\frac{1}{x-3}\\ y(x-3)=1\\ yx-3y=1\\ yx=1+3y\\ x=\frac{1+3y}{y}\end{array}$$

$$f^{-1}(x)=\frac{1+3x}{x}∎$$

$$\begin{array}{rl}{D}_{f^{-1}}& =R_f\\ & =(-\infty ,0)\cup (0,\infty )∎\end{array}$$