函数 $\mathrm{f}$ 定义为 $\mathrm{f} : x \mapsto \sqrt{5+2x}$，其中 $x \in \mathbb{R}$ 且 $x > -\frac{5}{2}$。

1. Use the definition of the composite function $\mathrm{f}^2(a)$ to find the value of $\mathrm{f}(a)$.

$\mathrm{f}^2(a) = 5$
$\mathrm{f}(\mathrm{f}(a)) = 5$
$\sqrt{5 + 2\mathrm{f}(a)} = 5$

Squaring both sides:
$5 + 2\mathrm{f}(a) = 25$
$2\mathrm{f}(a) = 20$
$\mathrm{f}(a) = 10$

2. Substitute the expression for $\mathrm{f}(a)$ and solve for $a$.

$\sqrt{5 + 2a} = 10$

Squaring both sides again:
$5 + 2a = 100$
$2a = 95$
$a = \frac{95}{2} = 47.5$

(Since $a = 47.5 > -\frac{5}{2}$, the answer is valid within the domain of $\mathrm{f}$.)

1. Identify the turning point of $\mathrm{g}(x)$ to find the condition for it to be one-to-one.

For $\mathrm{g}^{-1}$ to exist, $\mathrm{g}$ must be a one-to-one function.
Let $\mathrm{h}(x) = \frac{x}{1-2x}$. Since $\mathrm{h}'(x) = \frac{1}{(1-2x)^2} > 0$, $\mathrm{h}(x)$ is strictly increasing.
Taking the modulus reflects the negative parts of the graph in the $x$-axis, creating a turning point (minimum) where $\mathrm{h}(x) = 0$:
$\frac{x}{1-2x} = 0 \implies x = 0$

2. Determine the maximum value of $k$.

The function $\mathrm{g}(x)$ is decreasing for $x \le 0$ and increasing for $0 \le x < \frac{1}{2}$.
To ensure $\mathrm{g}(x)$ is one-to-one on the domain $x \le k$, the domain must not cross the turning point at $x = 0$.
Therefore, the largest value of $k$ is $0$.

1. Express $\mathrm{g}(x)$ without the modulus sign for the given domain.

For $x \le -1$, we have $x < 0$ and $1 - 2x \ge 3 > 0$.
Since the numerator is negative and the denominator is positive, $\frac{x}{1-2x} < 0$.
Thus, taking the absolute value flips the sign:
$\mathrm{g}(x) = -\left(\frac{x}{1-2x}\right) = \frac{-x}{1-2x} = \frac{x}{2x-1}$

2. Find the rule for $\mathrm{g}^{-1}(x)$.

Let $y = \frac{x}{2x-1}$.
$y(2x - 1) = x$
$2xy - y = x$
$2xy - x = y$
$x(2y - 1) = y$
$x = \frac{y}{2y-1}$
Thus, $\mathrm{g}^{-1}(x) = \frac{x}{2x-1}$.

3. Determine the domain of $\mathrm{g}^{-1}$ by finding the range of $\mathrm{g}$.

The domain of $\mathrm{g}^{-1}$ is the range of $\mathrm{g}$ for $x \le -1$.
Rewriting $\mathrm{g}(x)$ to make $x$ appear only once in the denominator:
$\mathrm{g}(x) = \frac{x}{2x-1} = \frac{\frac{1}{2}(2x-1) + \frac{1}{2}}{2x-1} = \frac{1}{2} + \frac{1}{2(2x-1)}$
Starting from the domain $x \le -1$:
$2x \le -2$
$2x - 1 \le -3$
Taking the reciprocal of both negative sides reverses the inequality (noting it remains strictly negative):
$-\frac{1}{3} \le \frac{1}{2x-1} < 0$
Multiply by $\frac{1}{2}$ and add $\frac{1}{2}$:
$\frac{1}{2} - \frac{1}{6} \le \frac{1}{2} + \frac{1}{2(2x-1)} < \frac{1}{2} + 0$
$\frac{1}{3} \le \mathrm{g}(x) < \frac{1}{2}$
Hence, the domain of $\mathrm{g}^{-1}$ is $\frac{1}{3} \le x < \frac{1}{2}$.

1. Find the range of the inner function $\mathrm{g}$ using the domain $x \le -1$.

From part (c), we are given $k = -1$, so the domain of $\mathrm{g}$ is $x \le -1$.
We find the range of $\mathrm{g}(x) = \left| \frac{x}{1-2x} \right|$ on this domain.
At $x = -1$:
$\mathrm{g}(-1) = \left| \frac{-1}{1-2(-1)} \right| = \left| -\frac{1}{3} \right| = \frac{1}{3}$
As $x \to -\infty$:
$\frac{x}{1-2x} = \frac{1}{\frac{1}{x} - 2} \to -\frac{1}{2}$
$\mathrm{g}(x) \to \left| -\frac{1}{2} \right| = \frac{1}{2}$
Since $\frac{x}{1-2x}$ is continuous and strictly monotonic for $x \le -1$, the range of $\mathrm{g}$ is $R_{\mathrm{g}} = \left[ \frac{1}{3}, \frac{1}{2} \right)$.

2. State the condition for the existence of $\mathrm{fg}$ and explain why it is met.

The domain of $\mathrm{f}$ is given as $D_{\mathrm{f}} = \left( -\frac{5}{2}, \infty \right)$.
For the composite function $\mathrm{fg}$ to exist, the range of $\mathrm{g}$ must be a subset of the domain of $\mathrm{f}$.
Since $R_{\mathrm{g}} = \left[ \frac{1}{3}, \frac{1}{2} \right)$ and $\left[ \frac{1}{3}, \frac{1}{2} \right) \subseteq \left( -\frac{5}{2}, \infty \right)$, the condition $R_{\mathrm{g}} \subseteq D_{\mathrm{f}}$ is satisfied. Therefore, $\mathrm{fg}$ exists.

3. Calculate the exact range of $\mathrm{fg}$.

The range of $\mathrm{fg}$ is the image of $R_{\mathrm{g}}$ under the function $\mathrm{f}(x) = \sqrt{5+2x}$.
Since $\mathrm{f}(x)$ is a strictly increasing function, we evaluate it at the boundaries of $R_{\mathrm{g}}$:
$\mathrm{f}\left(\frac{1}{3}\right) = \sqrt{5 + 2\left(\frac{1}{3}\right)} = \sqrt{\frac{17}{3}} = \frac{\sqrt{51}}{3}$
$\mathrm{f}\left(\frac{1}{2}\right) = \sqrt{5 + 2\left(\frac{1}{2}\right)} = \sqrt{5 + 1} = \sqrt{6}$
Thus, the exact range of $\mathrm{fg}$ is $\left[ \frac{\sqrt{51}}{3}, \sqrt{6} \right)$.