A 辑：纯粹数学 [40 分] 函数 $\text{f}$ 的定义为 $\text{f} : x \mapsto \frac{2x}{x-2}$，其中 $x \in \mathbb{R}$，$x \ne 2$。

1. Find the asymptotes and axis intercepts of the graph.

Rewrite $\text{f}(x)$ to clearly identify the horizontal asymptote:
$\text{f}(x) = \frac{2x}{x-2} = \frac{2(x-2) + 4}{x-2} = 2 + \frac{4}{x-2}$

• The vertical asymptote is given by the undefined value in the denominator: $x = 2$
• The horizontal asymptote is given by the constant term: $y = 2$
  To find the axis intercepts, substitute $x = 0$:
  $y = \frac{2(0)}{0-2} = 0$
  The graph intersects both axes at the origin $(0, 0)$.

2. Sketch the graph of the function.

Draw the Cartesian axes and include the following key features:

• A dashed vertical line at $x = 2$, labelled.
• A dashed horizontal line at $y = 2$, labelled.
• A rational curve (hyperbola) with two separate branches: one in the top-right quadrant formed by the asymptotes ($x>2, y>2$), and one in the bottom-left quadrant passing through the origin $(0,0)$.

3. Determine the range of the function.

From the equation $y = 2 + \frac{4}{x-2}$ and the sketched graph, the curve approaches the horizontal asymptote $y = 2$ but never touches it. It extends infinitely upwards and downwards for all other values.
Therefore, the function takes all real values except $2$.
$\text{Range of } \text{f} = \mathbb{R} \setminus \{2\}$

1. Find the range of the inner function $\text{g}$.

Since the absolute value is always non-negative, $|x + 2| \ge 0$ for all $x \in \mathbb{R}$.
$\text{g}(x) = 3 + |x + 2| \ge 3$
Thus, the range of $\text{g}$ is $R_\text{g} = [3, \infty)$.

2. Compare $R_\text{g}$ with the domain of $\text{f}$ to prove the existence of $\text{fg}$.

The domain of $\text{f}$ is $D_\text{f} = \mathbb{R} \setminus \{2\}$.
Since $R_\text{g} = [3, \infty)$ and $2 \notin [3, \infty)$, we have $R_\text{g} \subseteq D_\text{f}$.
Because the range of $\text{g}$ is a subset of the domain of $\text{f}$, the composite function $\text{fg}$ exists.

3. Find the expression for $\text{fg}(x)$.

$\text{fg}(x) = \text{f}(\text{g}(x))$
$= \text{f}(3 + |x + 2|)$
$= \frac{2(3 + |x + 2|)}{3 + |x + 2| - 2}$
$= \frac{2|x + 2| + 6}{|x + 2| + 1}$

4. State the domain of $\text{fg}$.

The domain of the composite function $\text{fg}$ is the domain of the inner function $\text{g}$.
$D_\text{fg} = \mathbb{R}$

5. Find the range of $\text{fg}$.

The range of $\text{fg}$ is the set of values of $\text{f}(y)$ for $y \in R_\text{g}$.
Let $y = \text{g}(x)$. We know $y \ge 3$.
$\text{f}(y) = \frac{2y}{y-2} = \frac{2(y-2) + 4}{y-2} = 2 + \frac{4}{y-2}$
As $y \ge 3$, we have $y - 2 \ge 1$.
$0 < \frac{1}{y-2} \le 1$
$0 < \frac{4}{y-2} \le 4$
$2 < 2 + \frac{4}{y-2} \le 6$
Hence, the range of $\text{fg}$ is $(2, 6]$.