函数 $\mathrm{f}$ 定义为 $\mathrm{f} : x \mapsto \ln(x+1)^2 + 2, \quad x \in \mathbb{R}, \ x \ge 0.$

1. Set up the condition for a one-to-one function by assuming $\mathrm{f}(a) = \mathrm{f}(b)$ and simplify the equation.

Let $a$ and $b$ be two values in the domain ($a \ge 0, b \ge 0$). Assume $\mathrm{f}(a) = \mathrm{f}(b)$.
$\ln(a+1)^2 + 2 = \ln(b+1)^2 + 2$
$\ln(a+1)^2 = \ln(b+1)^2$
$(a+1)^2 = (b+1)^2$

2. Use the domain restriction to deduce that $a = b$ and conclude.

Taking the square root on both sides:
$a+1 = \pm (b+1)$
Since $a \ge 0$ and $b \ge 0$, both $(a+1)$ and $(b+1)$ are strictly positive. We must reject the negative root.
$a+1 = b+1$
$a = b$
Since $\mathrm{f}(a) = \mathrm{f}(b) \implies a = b$, the function $\mathrm{f}$ is a one-to-one function.
Therefore, $\mathrm{f}$ has an inverse.

1. Set $y = \mathrm{f}(x)$ and manipulate the equation to express $x$ in terms of $y$.

$y = \ln(x+1)^2 + 2$
Since $x \ge 0$, $x+1 > 0$, so we can use the power rule for logarithms:
$y - 2 = 2\ln(x+1)$
$\frac{y-2}{2} = \ln(x+1)$
$\mathrm{e}^{\frac{y-2}{2}} = x + 1$
$x = \mathrm{e}^{\frac{y-2}{2}} - 1$

2. State the inverse function $\mathrm{f}^{-1}(x)$ in terms of $x$.

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

3. Find the domain of $\mathrm{f}^{-1}$ by evaluating the range of $\mathrm{f}$.

The domain of $\mathrm{f}^{-1}$ is exactly the range of $\mathrm{f}$.
Given the domain of $\mathrm{f}$ is $x \ge 0$:
$x + 1 \ge 1$
$\ln(x+1)^2 \ge \ln(1) = 0$
$\ln(x+1)^2 + 2 \ge 2$
Therefore, the range of $\mathrm{f}$ is $\mathrm{f}(x) \ge 2$.
The domain of $\mathrm{f}^{-1}$ is $x \in \mathbb{R}, x \ge 2$.

1. Determine the range of $\mathrm{f}^{-1}$ and the domain of $\mathrm{g}^{-1}$.

$\text{Range of } \mathrm{f}^{-1} = \text{Domain of } \mathrm{f} = [0, \infty)$
To find the domain of $\mathrm{g}^{-1}$, we find the range of $\mathrm{g}$. Let $y = \frac{1-x}{x+1}$.
$y(x+1) = 1-x$
$xy + y = 1 - x$
$x(y+1) = 1 - y$
$x = \frac{1-y}{y+1}$
For $x$ to be real and $x \ne -1$, we must have $y \ne -1$.
Hence, $\text{Domain of } \mathrm{g}^{-1} = \text{Range of } \mathrm{g} = \mathbb{R} \setminus \{-1\}$.

2. Show that the composite function $\mathrm{g}^{-1}\mathrm{f}^{-1}$ exists.

For the composite function $\mathrm{g}^{-1}\mathrm{f}^{-1}$ to exist, we require the range of the inner function to be a subset of the domain of the outer function:
$\text{Range of } \mathrm{f}^{-1} \subseteq \text{Domain of } \mathrm{g}^{-1}$
Since $[0, \infty) \subseteq \mathbb{R} \setminus \{-1\}$ (as $-1$ is not in the interval $[0, \infty)$), the condition is satisfied.
Therefore, $\mathrm{g}^{-1}\mathrm{f}^{-1}$ exists.

3. Find the range of $\mathrm{g}^{-1}\mathrm{f}^{-1}$.

The range of $\mathrm{g}^{-1}\mathrm{f}^{-1}$ is precisely the image of the range of $\mathrm{f}^{-1}$ under $\mathrm{g}^{-1}$.
From step 1, $\mathrm{g}^{-1}(x) = \frac{1-x}{x+1}$ for $x \ne -1$. Rewriting this:
$\mathrm{g}^{-1}(x) = \frac{2 - (x+1)}{x+1} = \frac{2}{x+1} - 1$
We evaluate $\mathrm{g}^{-1}(x)$ over the interval $x \in [0, \infty)$:
$x \ge 0 \implies x + 1 \ge 1$
$0 < \frac{1}{x+1} \le 1$
$0 < \frac{2}{x+1} \le 2$
$-1 < \frac{2}{x+1} - 1 \le 1$
Hence, the range of $\mathrm{g}^{-1}\mathrm{f}^{-1}$ is $(-1, 1]$.

1. Apply the function $\mathrm{g}$ to both sides to isolate $\mathrm{f}^{-1}(x)$.

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

2. Evaluate $\mathrm{g}\left(\frac{1}{2}\right)$.

$\mathrm{g}\left(\frac{1}{2}\right) = \frac{1 - \frac{1}{2}}{\frac{1}{2} + 1}$
$\mathrm{g}\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$

3. Apply the function $\mathrm{f}$ to both sides and evaluate to find exact value of $x$.

$\mathrm{f}^{-1}(x) = \frac{1}{3}$
$x = \mathrm{f}\left(\frac{1}{3}\right)$
$x = \ln\left(\frac{1}{3} + 1\right)^2 + 2$
$x = \ln\left(\frac{4}{3}\right)^2 + 2 = \ln\left(\frac{16}{9}\right) + 2$