函数 $\mathrm{f}$ 和 $\mathrm{g}$ 的定义为
$\mathrm{f} : x \to \frac{4}{(x-4)^2}, \quad x \in \mathbb{R}, x \ne 4,$
$\mathrm{g} : x \to \ln\left(1 + \frac{1}{x}\right), \quad x \in \mathbb{R}, x > 0.$

1. Find the equations of the asymptotes.

The function is undefined when the denominator is zero:
$(x - 4)^2 = 0 \implies x = 4$
Thus, the vertical asymptote is $x = 4$.

As $x \to \pm\infty$, $(x - 4)^2 \to \infty$, so $y \to 0$.
Thus, the horizontal asymptote is $y = 0$.

2. Determine the coordinates of the points where the graph crosses the axes.

To find the $y$-intercept, set $x = 0$:
$y = \mathrm{f}(0) = \frac{4}{(0 - 4)^2} = \frac{4}{16} = \frac{1}{4}$
The graph crosses the $y$-axis at $\left(0, \frac{1}{4}\right)$.

To find the $x$-intercept, set $y = 0$:
$\frac{4}{(x - 4)^2} = 0$
This has no real solution, so the graph does not cross the $x$-axis.

3. Determine if there are any turning points by examining the first derivative.

$\mathrm{f}'(x) = \frac{d}{dx} \left[ 4(x - 4)^{-2} \right]$
$\mathrm{f}'(x) = -8(x - 4)^{-3} = -\frac{8}{(x - 4)^3}$

For turning points, $\mathrm{f}'(x) = 0$.
Since $-\frac{8}{(x - 4)^3} \ne 0$ for all $x \in \mathbb{R}, x \ne 4$, there are no turning points.

4. Sketch the graph using the identified features.

The graph of $y = \mathrm{f}(x)$ consists of two branches:

• For $x < 4$, the curve passes through $\left(0, \frac{1}{4}\right)$ and approaches $+\infty$ as $x \to 4^{-}$.
• For $x > 4$, the curve approaches $+\infty$ as $x \to 4^{+}$ and approaches $0$ as $x \to \infty$.
  The graph is strictly positive for all $x \ne 4$.

1. Show that the composite function exists by checking the range of $f$ against the domain of $g$.

From the definition of $\mathrm{f}$, $(x-4)^2 > 0$ for all $x \ne 4$, so the range of $\mathrm{f}$ is $\mathrm{R}_{\mathrm{f}} = (0, \infty)$.
The domain of $\mathrm{g}$ is given as $\mathrm{D}_{\mathrm{g}} = (0, \infty)$.
Since $\mathrm{R}_{\mathrm{f}} \subseteq \mathrm{D}_{\mathrm{g}}$, the composite function $\mathrm{g}\mathrm{f}$ exists.

2. Find the rule of the composite function.

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

3. Determine the domain of the composite function.

The domain of $\mathrm{g}\mathrm{f}$ is simply the domain of the inner function $\mathrm{f}$.
$\mathrm{D}_{\mathrm{g}\mathrm{f}} = \mathrm{D}_{\mathrm{f}} = \mathbb{R} \setminus \{4\}$ (or $x \in \mathbb{R}, x \ne 4$)

4. Determine the range of the composite function.

To find the range of $\mathrm{g}\mathrm{f}$, we pass the range of $\mathrm{f}$ into $\mathrm{g}$.
For $y \in \mathrm{R}_{\mathrm{f}} = (0, \infty)$, we have $\frac{1}{y} \in (0, \infty)$.
Therefore, $1 + \frac{1}{y} \in (1, \infty)$.
Taking the natural logarithm, $\ln\left(1 + \frac{1}{y}\right) \in (0, \infty)$.
$\mathrm{R}_{\mathrm{g}\mathrm{f}} = (0, \infty)$

1. State the largest value of $k$ for $\mathrm{f}$ to be one-to-one

For $\mathrm{f}^{-1}$ to exist, the function $\mathrm{f}$ must be a one-to-one function.
Since $\mathrm{f}(x) = \frac{4}{(x-4)^2}$ has a vertical asymptote and is symmetric about the line $x = 4$, it is strictly increasing for $x < 4$ and strictly decreasing for $x > 4$.
Therefore, the largest restricted domain of the form $x < k$ that prevents the function from repeating values is $x < 4$.
$k = 4$

1. Set $y = \mathrm{f}(x)$ and rearrange to solve for $x$ in terms of $y$.

Let $y = \frac{4}{(x-4)^2}$.
$(x-4)^2 = \frac{4}{y}$
$x - 4 = \pm \sqrt{\frac{4}{y}} = \pm \frac{2}{\sqrt{y}}$

2. Apply the domain restriction to select the correct root.

From part (c), the largest value of $k$ for a one-to-one function is $4$, so the restricted domain is $x < 4$.
Since $x < 4$, the expression $(x - 4)$ must be negative.
Therefore, we reject the positive root:
$x - 4 = -\frac{2}{\sqrt{y}}$
$x = 4 - \frac{2}{\sqrt{y}}$

3. Determine the domain of $\mathrm{f}^{-1}$ and express the inverse function in the required notation.

For $x < 4$, the range of the restricted function $\mathrm{f}$ is $(0, \infty)$, meaning $y > 0$. This becomes the domain of $\mathrm{f}^{-1}$.
Expressing the inverse function in the "similar form" given in the question stem:
$\mathrm{f}^{-1} : x \mapsto 4 - \frac{2}{\sqrt{x}}, \quad x \in \mathbb{R}, x > 0$