函数 $\mathrm{f}$ 定义为
$\mathrm{f} : x \mapsto 1 - \frac{4}{x^2}, \text{ 对于 } x \in \mathbb{R}, x < -1.$

1. Find the coordinates of the point where the curve crosses the axes.

Set $y = 0$ to find the $x$-intercept:
$1 - \frac{4}{x^2} = 0$
$x^2 = 4$
Since the domain is $x < -1$, we accept $x = -2$.
The curve crosses the axes at $(-2, 0)$.
(There is no $y$-intercept as $x=0$ is outside the domain).

2. Determine the equation of the asymptote.

Consider the behavior of $\mathrm{f}(x)$ as $x \to -\infty$.
$\frac{4}{x^2} \to 0$, so $y \to 1$.
The equation of the horizontal asymptote is $y = 1$.
(The vertical asymptote $x=0$ is not applicable as it lies outside the domain).

3. Identify the endpoint due to the restricted domain.

The domain is strictly $x < -1$. As $x \to -1^-$, $y \to 1 - \frac{4}{(-1)^2} = -3$.
The curve ends at an open (hollow) circle at $(-1, -3)$.

4. Sketch the graph.

Draw a set of axes. Plot the horizontal dotted line $y=1$, mark the $x$-intercept at $(-2,0)$, and plot an open circle at $(-1,-3)$. Connect these with a smooth, strictly decreasing curve.

1. Show that $\mathrm{f}$ is a one-to-one function to justify the existence of an inverse.

From the graph of $y = \mathrm{f}(x)$ sketched in (i), any horizontal line $y = k$ intersects the curve at most once. Hence, $\mathrm{f}$ is a one-to-one function, which implies its inverse $\mathrm{f}^{-1}$ exists.

(Alternatively, use an algebraic argument: Let $\mathrm{f}(a) = \mathrm{f}(b)$ for some $a, b < -1$. Then $1 - \frac{4}{a^2} = 1 - \frac{4}{b^2} \implies \frac{4}{a^2} = \frac{4}{b^2} \implies a^2 = b^2$. Since $a$ and $b$ are both strictly negative, we must have $a = b$. Thus $\mathrm{f}$ is one-to-one.)

2. Set up an equation and make $x$ the subject to find the inverse rule.

Let $y = 1 - \frac{4}{x^2}$.

Rearranging to solve for $x$:
$\frac{4}{x^2} = 1 - y$
$x^2 = \frac{4}{1 - y}$
$x = \pm \frac{2}{\sqrt{1 - y}}$

Since the domain of $\mathrm{f}$ is $x < -1$ (which means $x$ must be negative), we reject the positive root and take the negative root:
$x = -\frac{2}{\sqrt{1 - y}}$

Therefore, $\mathrm{f}^{-1}(x) = -\frac{2}{\sqrt{1 - x}}$.

3. Determine the domain of the inverse function.

The domain of $\mathrm{f}^{-1}$ is precisely the range of $\mathrm{f}$.

For the domain $x < -1$, we have:
$x^2 > 1$
$0 < \frac{1}{x^2} < 1$
Multiplying by $-4$ reverses the inequality signs:
$-4 < -\frac{4}{x^2} < 0$
Adding 1 to all sides:
$-3 < 1 - \frac{4}{x^2} < 1$

Thus, the range of $\mathrm{f}$ is $(-3, 1)$.
The domain of $\mathrm{f}^{-1}$ is therefore $-3 < x < 1$ (or $x \in (-3, 1)$).

1. Find the domain of $\mathrm{f}\mathrm{f}^{-1}(x)$.

The domain of $\mathrm{f}\mathrm{f}^{-1}(x)$ is the domain of $\mathrm{f}^{-1}(x)$, which is exactly the range of $\mathrm{f}$.
Given $\mathrm{f}(x) = 1 - \frac{4}{x^2}$ for $x < -1$:
As $x \to -1^-$, $\mathrm{f}(x) \to 1 - \frac{4}{1} = -3$.
As $x \to -\infty$, $\mathrm{f}(x) \to 1 - 0 = 1$.
Thus, the range of $\mathrm{f}$ is $-3 < \mathrm{f}(x) < 1$.
Therefore, the domain of $\mathrm{f}\mathrm{f}^{-1}(x)$ is $-3 < x < 1$.

2. State the domain of $\mathrm{f}^{-1}\mathrm{f}(x)$ and find the intersection of the two domains.

The domain of $\mathrm{f}^{-1}\mathrm{f}(x)$ is the domain of $\mathrm{f}(x)$, which is given as $x < -1$.
Since $\mathrm{f}\mathrm{f}^{-1}(x) = x$ and $\mathrm{f}^{-1}\mathrm{f}(x) = x$ on their respective domains, the equation $\mathrm{f}\mathrm{f}^{-1}(x) = \mathrm{f}^{-1}\mathrm{f}(x)$ requires $x$ to be valid in both domains simultaneously.
The intersection of the intervals $-3 < x < 1$ and $x < -1$ is:
$-3 < x < -1$

1. Determine the range of $\mathrm{g}$ to check the condition for existence

Since $\mathrm{e}^{2x} > 0$ for all $x \in \mathbb{R}$,
$-\mathrm{e}^{2x} < 0$
$-\mathrm{e}^{2x} - 2 < -2$
Thus, the range of $\mathrm{g}$ is $\mathrm{R}_{\mathrm{g}} = (-\infty, -2)$.

2. Explain why the composite function $\mathrm{fg}$ exists

The domain of $\mathrm{f}$ is given as $\mathrm{D}_{\mathrm{f}} = (-\infty, -1)$.
Since $\mathrm{R}_{\mathrm{g}} = (-\infty, -2) \subseteq (-\infty, -1) = \mathrm{D}_{\mathrm{f}}$, the composite function $\mathrm{fg}$ exists.

3. Find the rule and domain for the composite function $\mathrm{fg}$

$\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$
$= \mathrm{f}(-\mathrm{e}^{2x} - 2)$
$= 1 - \frac{4}{(-\mathrm{e}^{2x} - 2)^2}$
$= 1 - \frac{4}{(\mathrm{e}^{2x} + 2)^2}$

The domain of $\mathrm{fg}$ is the domain of $\mathrm{g}$, which is $x \in \mathbb{R}$.

4. Determine the range of the composite function $\mathrm{fg}$

Since $\mathrm{e}^{2x} > 0$:
$\mathrm{e}^{2x} + 2 > 2$
$(\mathrm{e}^{2x} + 2)^2 > 4$
$0 < \frac{4}{(\mathrm{e}^{2x} + 2)^2} < 1$
$-1 < -\frac{4}{(\mathrm{e}^{2x} + 2)^2} < 0$
$0 < 1 - \frac{4}{(\mathrm{e}^{2x} + 2)^2} < 1$

Thus, the range of $\mathrm{fg}$ is $(0, 1)$.