函数 f 和 g 的定义为 $f : x \mapsto x^2 + 3x + 1,\ x \in \mathbb{R},\ x \le -\frac{3}{2},$ $g : x \mapsto 3 + e^{-x},\ x \in \mathbb{R}.$

1. Express $f(x)$ in completed square form to identify the vertex.

$f(x) = x^2 + 3x + 1$
$f(x) = \left(x + \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 1$
$f(x) = \left(x + \frac{3}{2}\right)^2 - \frac{5}{4}$

2. Use the restricted domain to show that $f$ is a one-to-one function.

The graph of $y = f(x)$ is a parabola with its vertex at $x = -\frac{3}{2}$.
Given the domain restriction $x \le -\frac{3}{2}$, the graph consists only of the left branch of the parabola, where $f(x)$ is strictly decreasing.
Therefore, any horizontal line $y=k$ intersects the graph at most once, which means $f$ is a one-to-one function. Hence, $f$ has an inverse.

1. Find the range of $f$ to determine the domain of $f^{-1}$.

$f(x) = x^2 + 3x + 1 = \left(x + \frac{3}{2}\right)^2 - \frac{5}{4}$
For $x \le -\frac{3}{2}$, $\left(x + \frac{3}{2}\right)^2 \ge 0$.
So $f(x) \ge -\frac{5}{4}$.
The range of $f$ is $R_f = \left[-\frac{5}{4}, \infty\right)$.
The domain of $f^{-1}$ is $D_{f^{-1}} = R_f = \left[-\frac{5}{4}, \infty\right)$.

2. Determine the range of $g$ and check the condition for the existence of $f^{-1}g$.

For all $x \in \mathbb{R}$, $e^{-x} > 0$.
So $g(x) = 3 + e^{-x} > 3$.
The range of $g$ is $R_g = (3, \infty)$.
Since $R_g = (3, \infty) \subseteq \left[-\frac{5}{4}, \infty\right) = D_{f^{-1}}$, the composite function $f^{-1}g$ exists.

3. Find the rule for the inverse function $f^{-1}(x)$.

Let $y = f(x)$.
$y = \left(x + \frac{3}{2}\right)^2 - \frac{5}{4}$
$\left(x + \frac{3}{2}\right)^2 = y + \frac{5}{4}$
$x + \frac{3}{2} = \pm\sqrt{y + \frac{5}{4}}$
Since the domain of $f$ is $x \le -\frac{3}{2}$, we have $x + \frac{3}{2} \le 0$, so we reject the positive root.
$x = -\frac{3}{2} - \sqrt{y + \frac{5}{4}}$
Hence, $f^{-1}(x) = -\frac{3}{2} - \sqrt{x + \frac{5}{4}}$.

4. Find the rule for the composite function $f^{-1}g(x)$.

$f^{-1}g(x) = f^{-1}(3 + e^{-x})$
$= -\frac{3}{2} - \sqrt{(3 + e^{-x}) + \frac{5}{4}}$
$= -\frac{3}{2} - \sqrt{e^{-x} + \frac{17}{4}}$

5. State the domain of the composite function $f^{-1}g$.

The domain of a composite function $f^{-1}g$ is equal to the domain of the inner function $g$.
Hence, $D_{f^{-1}g} = D_g = \mathbb{R}$.