Functions f and g are defined by $f : x \mapsto \dfrac{4}{(x-4)^2},\quad x \in \mathbb{R},\ x \ne 4,$ $g : x \mapsto \ln\!\left(1 + \dfrac{1}{x}\right),\quad x \in \mathbb{R},\ x > 0.$

1. Determine the equations of the asymptotes

The function is undefined when the denominator is zero:
$x - 4 = 0 \implies x = 4$
This gives a vertical asymptote at $x = 4$.
As $x \to \pm\infty$, $(x-4)^2 \to \infty$, so $y \to 0$.
This gives a horizontal asymptote at $y = 0$.

2. Find coordinates where the graph crosses the axes

Substitute $x = 0$ into the equation to find the $y$-intercept:
$y = \frac{4}{(0-4)^2} = \frac{4}{16} = \frac{1}{4}$
The $y$-intercept is $(0, \frac{1}{4})$.
Since $y = \frac{4}{(x-4)^2} > 0$ for all $x \neq 4$, $y$ is never zero, so the graph never crosses the $x$-axis.

3. Determine the existence of turning points

Find the first derivative $f'(x)$:
$f(x) = 4(x-4)^{-2}$
$f'(x) = -8(x-4)^{-3} = \frac{-8}{(x-4)^3}$
For a turning point to exist, $f'(x) = 0$.
Since the numerator is $-8$, $f'(x) \neq 0$ for all valid $x$, therefore there are no turning points.

1. Determine the range of $f$ and use it to show that the composite function $gf$ exists.

From the definition $f(x) = \dfrac{4}{(x-4)^2}$ for $x \ne 4$, we know that $(x-4)^2 > 0$, which implies $f(x) > 0$.
Since $f(x) \to \infty$ as $x \to 4$ and $f(x) \to 0$ as $x \to \pm\infty$, the range of $f$ is:
$R_f = (0, \infty)$

The domain of $g$ is given as $D_g = (0, \infty)$.
Because $R_f \subseteq D_g$, the composite function $gf$ exists.

2. Find the rule for the composite function $gf$.

$gf(x) = g(f(x)) = g\left( \frac{4}{(x-4)^2} \right)$
$gf(x) = \ln\left(1 + \frac{1}{\frac{4}{(x-4)^2}}\right)$
$gf(x) = \ln\left(1 + \frac{(x-4)^2}{4}\right)$

3. State the domain of $gf$.

The domain of a composite function $gf$ is the domain of the initial inner function $f$.
$D_{gf} = D_f = \{x \in \mathbb{R} : x \ne 4\}$.

4. Find the range of $gf$.

To find the range of $gf$, we observe the behaviour of $g(y)$ for all inputs $y \in R_f = (0, \infty)$.
$g(y) = \ln\left(1 + \frac{1}{y}\right)$
As $y \to \infty$, $\frac{1}{y} \to 0$, so $g(y) \to \ln(1) = 0$.
As $y \to 0^+$, $\frac{1}{y} \to \infty$, so $g(y) \to \infty$.
Since $g$ is continuous and strictly decreasing on $(0, \infty)$, the range of $gf$ is:
$R_{gf} = (0, \infty)$

1. State the condition for the inverse function to exist and find the largest $k$

For $f^{-1}$ to exist, the function $f$ must be one-to-one.
From the graph of $f(x) = \dfrac{4}{(x-4)^2}$, there is a vertical asymptote at $x = 4$ and the graph is symmetric about $x = 4$.
To maintain a one-to-one relationship for a domain of $x < k$, the interval cannot cross the vertical asymptote $x = 4$.
Therefore, the largest possible value for $k$ is $4$.

1. Express $x$ in terms of $y$, selecting the correct square root based on the restricted domain.

Let $y = f(x) = \dfrac{4}{(x-4)^2}$.
$(x-4)^2 = \dfrac{4}{y}$
$x - 4 = \pm\dfrac{2}{\sqrt{y}}$
From part (c), the maximum restricted domain for $f^{-1}$ to exist is $x < 4$, which means $x - 4 < 0$. Thus, we must reject the positive root and take the negative root:
$x - 4 = -\dfrac{2}{\sqrt{y}}$
$x = 4 - \dfrac{2}{\sqrt{y}}$

2. Determine the domain of $f^{-1}$ by finding the range of the restricted function $f$.

The domain of $f^{-1}$ is the range of $f$ on the restricted domain $x < 4$.
For $x < 4$, $(x-4)^2 > 0 \implies \dfrac{4}{(x-4)^2} > 0$.
Since the function can take any positive value as $x \to 4^-$, the range of $f$ is $y > 0$. This range becomes the domain of $f^{-1}$.

3. Write the inverse function in the required mapping form.

The question asks for the answer in a "similar form" to the given definitions, so we use mapping notation and include the domain:
$f^{-1} : x \mapsto 4 - \dfrac{2}{\sqrt{x}}, \quad x \in \mathbb{R}, \ x > 0$