Functions f and g are defined by $f : x \mapsto \dfrac{1}{x-3}$ for $x \in \mathbb{R}$, $x \ne 3$, $g : x \mapsto x^{2}$ for $x \in \mathbb{R}$. (i) Only one of the composite functions fg and gf exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist. (ii) Find $f^{-1}(x)$ and state the domain of $f^{-1}$.

1. Determine that gf exists, evaluate its definition, and state its domain.

For $\text{gf}$ to exist, we require the range of f ($R_f$) to be a subset of the domain of g ($D_g$).
Since $D_g = \mathbb{R}$, $R_f \subseteq D_g$ is satisfied, so $\text{gf}$ exists.
$\text{gf}(x) = \text{g}(\text{f}(x)) = \text{g}\left(\dfrac{1}{x-3}\right) = \left(\dfrac{1}{x-3}\right)^2 = \dfrac{1}{(x-3)^2}$
The domain of $\text{gf}$ is the same as the domain of the inner function $\text{f}$, so the domain is $x \in \mathbb{R}, x \ne 3$.

2. Explain why the composite function fg does not exist.

For $\text{fg}$ to exist, we require the range of g to be a subset of the domain of f ($R_g \subseteq D_f$).
Domain of f: $D_f = \{x \in \mathbb{R} : x \ne 3\}$
Range of g: $R_g = [0, \infty)$
Since $3 \in R_g$ but $3 \notin D_f$, $R_g \not\subseteq D_f$. Therefore, the composite function $\text{fg}$ does not exist.

1. Set $y = f(x)$ and rearrange to make $x$ the subject.

Let $y = \dfrac{1}{x-3}$.
$y(x-3) = 1$
$xy - 3y = 1$
$xy = 3y + 1$
$x = \dfrac{3y + 1}{y}$

2. Express the inverse function in terms of $x$.

$f^{-1}(x) = \dfrac{3x+1}{x} \quad \text{or} \quad \dfrac{1}{x} + 3$

3. Determine the domain of the inverse function by finding the range of $f$.

The domain of $f^{-1}$ is the range of $f$.
For $f(x) = \dfrac{1}{x-3}$, the numerator is a non-zero constant ($1$), so $f(x)$ can never be $0$.
Therefore, the range of $f$ is $f(x) \ne 0$.
Hence, the domain of $f^{-1}$ is $x \in \mathbb{R}, x \ne 0$.