The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $\mathrm{f} : x \mapsto \left| 4 + 2x - x^2 \right|, \ x \in \mathbb{R}, \ x \ge 3.5,$ $\mathrm{g} : x \mapsto 4 + \mathrm{e}^{ax}, \ x \in \mathbb{R}, \ x \ge -1,$ where $a > 0$.

1. Remove the modulus sign by determining the sign of the quadratic expression for the given domain.

By completing the square, we have $4 + 2x - x^2 = 5 - (x-1)^2$.
For $x \ge 3.5$, $x-1 \ge 2.5$, so $(x-1)^2 \ge 6.25$.
Thus, $5 - (x-1)^2 \le 5 - 6.25 = -1.25 < 0$.
Since the expression inside the modulus is strictly negative, we drop the modulus by negating the expression:
$\mathrm{f}(x) = (x-1)^2 - 5$

2. Let $y = \mathrm{f}(x)$ and rearrange the equation to make $x$ the subject.

Let $y = (x-1)^2 - 5$.
$(x-1)^2 = y + 5$
$x - 1 = \pm\sqrt{y + 5}$

3. Choose the correct sign for the square root based on the domain to write the inverse function.

Since the domain restricts $x \ge 3.5$, it implies $x - 1 \ge 2.5 > 0$. We must therefore reject the negative root.
$x = 1 + \sqrt{y + 5}$
Therefore, swapping the variables gives the inverse function:
$\mathrm{f}^{-1}(x) = 1 + \sqrt{x + 5}$

4. Determine the domain of the inverse function by finding the range of $\mathrm{f}$.

The domain of $\mathrm{f}^{-1}$ is equivalent to the range of $\mathrm{f}$.
Since $x \ge 3.5$, we have $(x-1)^2 \ge 6.25$.
$\mathrm{f}(x) = (x-1)^2 - 5 \ge 6.25 - 5 = 1.25$
The range of $\mathrm{f}$ is $[1.25, \infty)$.
Thus, the domain of $\mathrm{f}^{-1}$ is $x \ge 1.25$.

1. Use the property that intersections of a function and its inverse lie on the line $y = x$ to form an equation.

$\mathrm{f}^{-1}(x) = \mathrm{f}(x) \implies \mathrm{f}(x) = x$
For $x \ge 3.5$, $4 + 2x - x^2 < 0$, so the modulus resolves to $\mathrm{f}(x) = x^2 - 2x - 4$.
$x^2 - 2x - 4 = x$
$x^2 - 3x - 4 = 0$

2. Solve the resulting quadratic equation and reject any invalid roots based on the domain constraint.

$(x - 4)(x + 1) = 0$
$x = 4 \quad \text{or} \quad x = -1$
Since the domain of $\mathrm{f}$ is $x \ge 3.5$, we reject $x = -1$.
$x = 4$

1. Find the range of $\mathrm{g}(x)$ to establish the condition for the existence of $\mathrm{fg}$.

For the domain $x \ge -1$ and constant $a > 0$, the exponential function $\mathrm{g}(x) = 4 + \mathrm{e}^{ax}$ is strictly increasing.
The minimum value occurs at $x = -1$:
$\mathrm{g}(-1) = 4 + \mathrm{e}^{-a}$
As $x \to \infty$, $\mathrm{e}^{ax} \to \infty$.
Therefore, the range of $\mathrm{g}$ is $\mathrm{R}_{\mathrm{g}} = [4 + \mathrm{e}^{-a}, \infty)$.

2. Show that the range of $\mathrm{g}$ is a subset of the domain of $\mathrm{f}$.

The domain of $\mathrm{f}$ is given as $\mathrm{D}_{\mathrm{f}} = [3.5, \infty)$.
Since $a > 0$, we know that $\mathrm{e}^{-a} > 0$. Thus,
$4 + \mathrm{e}^{-a} > 4 > 3.5$
This establishes that $\mathrm{R}_{\mathrm{g}} \subseteq \mathrm{D}_{\mathrm{f}}$.
Because the range of $\mathrm{g}$ lies entirely within the domain of $\mathrm{f}$, the composite function $\mathrm{fg}$ exists.

3. Evaluate the composite function $\mathrm{fg}(x)$ using the simplified form of $\mathrm{f}(x)$ for its domain.

For $x \ge 3.5$, we have $x - 1 \ge 2.5$, so $(x - 1)^2 \ge 6.25 > 5$.
Therefore, the expression inside the absolute value of $\mathrm{f}(x) = |5 - (x - 1)^2|$ is strictly negative. Dropping the absolute value flips the signs:
$\mathrm{f}(x) = (x - 1)^2 - 5$
Substituting $\mathrm{g}(x)$ into $\mathrm{f}$:
$\mathrm{fg}(x) = \mathrm{f}(4 + \mathrm{e}^{ax})$
$= (4 + \mathrm{e}^{ax} - 1)^2 - 5$
$= (3 + \mathrm{e}^{ax})^2 - 5$
$= 9 + 6\mathrm{e}^{ax} + \mathrm{e}^{2ax} - 5$
$= 4 + 6\mathrm{e}^{ax} + \mathrm{e}^{2ax}$

4. Determine the exact range of $\mathrm{fg}$.

For $x \ge -1$ and $a > 0$, the minimum possible value of $\mathrm{e}^{ax}$ is $\mathrm{e}^{-a}$.
Since $\mathrm{fg}(x) = 4 + 6\mathrm{e}^{ax} + \mathrm{e}^{2ax}$ is a strictly increasing function with respect to $\mathrm{e}^{ax}$ (all terms are positive and increasing), its absolute minimum occurs when $x = -1$:
$\text{Minimum } \mathrm{fg}(x) = 4 + 6\mathrm{e}^{-a} + (\mathrm{e}^{-a})^2 = 4 + 6\mathrm{e}^{-a} + \mathrm{e}^{-2a}$
As $x \to \infty$, $\mathrm{e}^{ax} \to \infty$, meaning $\mathrm{fg}(x) \to \infty$.
Therefore, the exact range of $\mathrm{fg}$ is $[4 + 6\mathrm{e}^{-a} + \mathrm{e}^{-2a}, \infty)$.

1. Establish that the numerator is strictly positive, hence the denominator must be positive.

Since $\mathrm{e}^{ax} > 0$ for all real $x$, the numerator $\mathrm{g}(x) = 4 + \mathrm{e}^{ax} > 0$.

For the rational expression to be non-negative ($\ge 0$), the denominator cannot be zero and must have the same sign as the numerator:
$x^2 - 2x - 2 > 0$

2. Solve the quadratic inequality for the denominator.

Find the roots of $x^2 - 2x - 2 = 0$ using the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2}$
$x = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$

Solving $x^2 - 2x - 2 > 0$ gives:
$x < 1 - \sqrt{3} \quad \text{or} \quad x > 1 + \sqrt{3}$

3. Apply the domain restriction of $\mathrm{g}(x)$ to find the final solution.

The function $\mathrm{g}(x)$ is defined only for $x \ge -1$.
Since $1 - \sqrt{3} \approx -0.732$, we have $-1 < 1 - \sqrt{3}$.

Intersecting the domain $x \ge -1$ with the solution to the quadratic inequality gives the final valid ranges for $x$:
$-1 \le x < 1 - \sqrt{3} \quad \text{or} \quad x > 1 + \sqrt{3}$