The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by
$\mathrm{f} : x \mapsto x^2 + 2x + 4, \quad x \le -1 \quad \text{and} \quad \mathrm{g} : x \mapsto \ln(2x - 1), \quad x > \frac{1}{2}.$

1. Set $y = \mathrm{f}(x)$ and complete the square to make $x$ the subject.

$y = x^2 + 2x + 4$
$y = (x + 1)^2 - 1^2 + 4$
$y = (x + 1)^2 + 3$
$(x + 1)^2 = y - 3$
$x + 1 = \pm\sqrt{y - 3}$

2. Apply the domain restriction to choose the correct root and state the inverse function.

Since the domain of $\mathrm{f}$ is $x \le -1$, we have $x + 1 \le 0$.
Therefore, we reject the positive root.
$x + 1 = -\sqrt{y - 3}$
$x = -1 - \sqrt{y - 3}$
$\therefore \mathrm{f}^{-1}(x) = -1 - \sqrt{x - 3}$

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

The domain of $\mathrm{f}^{-1}$ is equal to the range of $\mathrm{f}$.
For $x \le -1$, the minimum value of $(x + 1)^2$ is $0$.
So, $\mathrm{f}(x) = (x + 1)^2 + 3 \ge 3$.
Therefore, the range of $\mathrm{f}$ is $[3, \infty)$.
Thus, the domain of $\mathrm{f}^{-1}$ is $x \ge 3$.

1. Sketch the graph of $y = \mathrm{f}(x)$.

Completing the square gives $\mathrm{f}(x) = (x + 1)^2 + 3$.
Since the domain is restricted to $x \le -1$, the graph is a half-parabola with its starting (and minimum) point at the vertex $(-1, 3)$, curving upwards and moving towards the left (second quadrant).

2. Sketch the graph of $y = \mathrm{f}^{-1}(x)$.

The graph of $y = \mathrm{f}^{-1}(x)$ is the reflection of the graph of $y = \mathrm{f}(x)$ across the line $y = x$.
Reflecting the starting point $(-1, 3)$ gives $(3, -1)$. The reflected curve is a half-parabola starting at $(3, -1)$, curving downwards and moving towards the right (fourth quadrant).

3. Deduce the nature of the solutions for $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ and state the reason.

Solutions to the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ correspond to the points of intersection between their graphs.
From the sketch, the graph of $y = \mathrm{f}(x)$ lies entirely in the region $x \le -1, y \ge 3$, while the graph of $y = \mathrm{f}^{-1}(x)$ lies entirely in the region $x \ge 3, y \le -1$.
Since the two graphs are completely separated and do not intersect, there are no solutions to the equation.

1. Find the range of the inner function $\mathrm{f}$.

$\begin{aligned} \mathrm{f}(x) &= x^2 + 2x + 4 \\ &= (x+1)^2 - 1 + 4 \\ &= (x+1)^2 + 3 \end{aligned}$
For the domain $x \le -1$, $x+1 \le 0$, which implies $(x+1)^2 \ge 0$.
Thus, $\mathrm{f}(x) \ge 3$, so the range of $\mathrm{f}$ is $R_{\mathrm{f}} = [3, \infty)$.

2. Check if the condition for the existence of $\mathrm{gf}$ is satisfied.

The domain of $\mathrm{g}$ is given as $D_{\mathrm{g}} = (\frac{1}{2}, \infty)$.
Since $R_{\mathrm{f}} = [3, \infty) \subseteq (\frac{1}{2}, \infty) = D_{\mathrm{g}}$, the composite function $\mathrm{gf}$ exists.

3. Find the expression for $\mathrm{gf}(x)$.

$\begin{aligned} \mathrm{gf}(x) &= \mathrm{g}(\mathrm{f}(x)) \\ &= \mathrm{g}(x^2 + 2x + 4) \\ &= \ln(2(x^2 + 2x + 4) - 1) \\ &= \ln(2x^2 + 4x + 8 - 1) \\ &= \ln(2x^2 + 4x + 7) \end{aligned}$

4. Find the exact range of $\mathrm{gf}$.

To find the range of $\mathrm{gf}$, we evaluate the outer function $\mathrm{g}$ using the range of the inner function $R_{\mathrm{f}} = [3, \infty)$ as the input domain.
Since $\mathrm{g}(u) = \ln(2u - 1)$ is a strictly increasing function, its minimum value over this interval occurs at $u=3$:
$\mathrm{g}(3) = \ln(2(3) - 1) = \ln 5$.
As $u \to \infty$, $\mathrm{g}(u) \to \infty$.
Therefore, the exact range of $\mathrm{gf}$ is $[\ln 5, \infty)$.