The function $\text{f}$ is defined by
$\text{f} : x \mapsto \frac{-x - 3}{x + 1}, \text{ for } x \in \mathbb{R}, x \ne k.$

1. State the value of $k$ by identifying where the denominator is zero

For the rational expression $\frac{-x - 3}{x + 1}$, the denominator is $x + 1$.
Set $x + 1 = 0 \implies x = -1$.
Since the domain excludes $k$, we have $k = -1$.

2. Explain why $k$ is excluded from the domain

When $x = -1$, the denominator evaluates to zero. Division by zero is undefined, so $\text{f}(-1)$ is not a valid real number. Hence, $x = -1$ must be excluded to ensure the function is well-defined.

1. Set up an equation to find the inverse function.

Let $y = \text{f}(x)$.
$y = \frac{-x - 3}{x + 1}$

2. Make $x$ the subject of the formula.

$y(x + 1) = -x - 3$
$xy + y = -x - 3$
$xy + x = -y - 3$
$x(y + 1) = -(y + 3)$
$x = \frac{-y - 3}{y + 1}$

3. State the inverse function in terms of $x$.

$\text{f}^{-1}(x) = \frac{-x - 3}{x + 1}$

4. Use the result to deduce $\text{f}^2(x)$.

From the result above, we can observe that $\text{f}^{-1}(x) = \text{f}(x)$.
Since $\text{f}(\text{f}^{-1}(x)) = x$, replacing $\text{f}^{-1}(x)$ with $\text{f}(x)$ gives:
$\text{f}(\text{f}(x)) = x$
Therefore, $\text{f}^2(x) = x$.

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

For the domain $x > -1$, we have $x + 1 > 0$.
Since $\text{g}(x) = \sqrt{x+1}$, the values of $\text{g}(x)$ will be all positive real numbers.
Thus, the range of $\text{g}$ is $(0, \infty)$, or $\text{g}(x) > 0$.

2. Determine the range of the outer function $\text{f}$ over the interval from the range of $\text{g}$.

The range of $\text{fg}$ is the set of values of $\text{f}(x)$ when its domain is restricted to the range of $\text{g}$ (i.e., $x > 0$).
Rewrite $\text{f}(x)$ to easily find its bounds:
$\text{f}(x) = \frac{-x-3}{x+1} = \frac{-(x+1)-2}{x+1} = -1 - \frac{2}{x+1}$
For $x > 0$:
$x + 1 > 1$
$0 < \frac{1}{x+1} < 1$
Multiply by $-2$ (reversing the inequality):
$-2 < -\frac{2}{x+1} < 0$
Subtract $1$:
$-3 < -1 - \frac{2}{x+1} < -1$

3. State the final range of $\text{fg}$.

Based on the inequality, the range of the composite function is between $-3$ and $-1$.
Hence, the range of $\text{fg}$ is $-3 < \text{fg}(x) < -1$.