如果对于 $\text{h}$ 定义域中的所有 $x$，都有 $\text{h}(x) = \text{h}^{-1}(x)$，则称函数 $\text{h}$ 为自反函数。 函数 $\text{f}$ 和 $\text{g}$ 定义为
$\begin{aligned} \text{f}(x) &= 2 + \frac{1}{x-2}, \quad \text{for } x \in \mathbb{R}, x > 2 \\ \text{g}(x) &= \begin{cases} 3x - 1 & \text{for } x \le 1, \\ (x-1)^2 + 2 & \text{for } x > 1. \end{cases} \end{aligned}$

1. Sketch the graph of $y = \text{f}(x)$ for the restricted domain $x > 2$

For $x > 2$, $y = 2 + \frac{1}{x-2}$.
The graph is a translation of $y = \frac{1}{x}$ by $2$ units to the right and $2$ units upwards.
The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 2$.
The domain is strictly $x > 2$, so only the upper-right branch of the hyperbola is drawn.
The curve approaches $x=2$ as $y \to \infty$, and approaches $y=2$ as $x \to \infty$.

(Student produces a sketch showing the axes, dashed lines for asymptotes $x=2$ and $y=2$, and a decreasing curve bounded by these asymptotes in the first quadrant.)

2. Use the graph to explain why the inverse function exists

From the sketch, it is visually apparent that any horizontal line $y = c$ will intersect the curve at most once.
Because every output $y$ is mapped from exactly one input $x$, $\text{f}$ is a one-to-one function.
A function must be one-to-one for its inverse to exist.

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

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

2. State the inverse function and conclude that $\text{f}$ is self-inverse.

$\text{f}^{-1}(x) = 2 + \frac{1}{x-2}$
Since $\text{f}^{-1}(x) = \text{f}(x)$ for all $x > 2$, $\text{f}$ is self-inverse (shown).

3. Find $\text{f}^2(x)$ using direct algebraic substitution.

$\text{f}^2(x) = \text{f}(\text{f}(x))$
$= 2 + \frac{1}{\text{f}(x) - 2}$
$= 2 + \frac{1}{\left(2 + \frac{1}{x-2}\right) - 2}$

4. Simplify the expression to obtain the final composite function.

$= 2 + \frac{1}{\left(\frac{1}{x-2}\right)}$
$= 2 + (x - 2)$
$= x$
(Alternatively, since $\text{f}$ is self-inverse, $\text{f}^2(x) = \text{f}(\text{f}^{-1}(x)) = x$)

1. Deduce that $\text{f}^{2025}(4) = \text{f}(4)$ using the self-inverse property.

From part (b), we know $\text{f}^2(x) = x$.
This means that for any even power $2k$, $\text{f}^{2k}(x) = x$.
Since $2024$ is even, $\text{f}^{2024}(x) = x$.
Therefore, $\text{f}^{2025}(4) = \text{f}(\text{f}^{2024}(4)) = \text{f}(4)$.

2. Evaluate $\text{f}(4)$.

$\text{f}(4) = 2 + \frac{1}{4-2}$
$= 2 + \frac{1}{2}$
$= 2.5$

1. Determine the range of $\text{f}(x)$ to select the correct part of the piecewise function $\text{g}(x)$.

For the domain $x > 2$, $x - 2 > 0$, so $\frac{1}{x-2} > 0$.
Therefore, $\text{f}(x) = 2 + \frac{1}{x-2} > 2$.
Since $\text{f}(x) > 2$, it satisfies the condition $\text{f}(x) > 1$.

2. Substitute $\text{f}(x)$ into the expression for $\text{g}(x)$ corresponding to $x > 1$.

Using the second piece of the function $\text{g}$, $\text{g}(u) = (u-1)^2 + 2$:
$\text{gf}(x) = \text{g}(\text{f}(x)) = (\text{f}(x) - 1)^2 + 2$
$\text{gf}(x) = \left(\left(2 + \frac{1}{x-2}\right) - 1\right)^2 + 2$
$\text{gf}(x) = \left(1 + \frac{1}{x-2}\right)^2 + 2$
$\text{gf}(x) = \left(\frac{x-2+1}{x-2}\right)^2 + 2$
$\text{gf}(x) = \left(\frac{x-1}{x-2}\right)^2 + 2$

3. State the domain of $\text{gf}$.

The domain of a composite function $\text{gf}$ consists of all $x$ in the domain of $\text{f}$ such that $\text{f}(x)$ is in the domain of $\text{g}$.
Since the domain of $\text{g}$ is $x \in \mathbb{R}$, this imposes no further restrictions.
The domain of $\text{gf}$ is simply the domain of $\text{f}$.
$\text{Domain: } x \in \mathbb{R}, x > 2$

1. Determine the range of the inner function $\text{f}(x)$.

Given the domain $x > 2$, we have $x - 2 > 0$, so $\frac{1}{x-2} > 0$.
Therefore, $\text{f}(x) = 2 + \frac{1}{x-2} > 2$.
The range of $\text{f}$ is $(2, \infty)$.

2. Use the range of $\text{f}$ as the domain for $\text{g}$ to find the range of the composite function $\text{gf}$.

The range of $\text{f}$ forms the domain for $\text{g}$ in the composite function. Let $y = \text{f}(x)$.
Since $y > 2$, we use the corresponding piece of $\text{g}$: $\text{g}(y) = (y-1)^2 + 2$.
For $y > 2$, $y - 1 > 1 \implies (y - 1)^2 > 1$.
Thus, $\text{g}(y) > 1 + 2 = 3$.
The range of $\text{gf}$ is $(3, \infty)$.