The functions $\text{f}$ and $\text{g}$ are defined as follows
$\text{f} : x \mapsto \frac{\ln(3x - 2)}{x}, x > \frac{2}{3}$
$\text{g} : x \mapsto 3x^2 - 12x + 13, x \in \mathbb{R}$

1. Determine the equations of the vertical and horizontal asymptotes.

As $x \to \left(\frac{2}{3}\right)^+$, $(3x - 2) \to 0^+$ and $\ln(3x - 2) \to -\infty$.
Thus, $\text{f}(x) \to -\infty$, yielding a vertical asymptote at $x = \frac{2}{3}$.

As $x \to \infty$, $\frac{\ln(3x - 2)}{x} \to 0$.
Thus, there is a horizontal asymptote at $y = 0$.

2. Find the coordinates of the intersection with the axes.

Since the domain is $x > \frac{2}{3}$, $x$ cannot be $0$. Hence, there is no $y$-intercept.

To find the $x$-intercept, set $y = 0$:
$\frac{\ln(3x - 2)}{x} = 0$
$\ln(3x - 2) = 0$
$3x - 2 = e^0 = 1$
$3x = 3 \implies x = 1$

The intersection with the axes is at $(1, 0)$.

3. Use a graphing calculator to find the coordinates of the turning point.

Using the maximum point feature on a graphing calculator for the function $y = \frac{\ln(3x - 2)}{x}$:
$x \approx 2.106$
$y \approx 0.6946$

Correcting to 3 significant figures, the maximum turning point is $(2.11, 0.695)$.

1. Find the range of g.

By completing the square,
$g(x) = 3(x^2 - 4x) + 13 = 3(x-2)^2 - 12 + 13 = 3(x-2)^2 + 1$
Since $(x-2)^2 \ge 0$ for all real $x$, the minimum value of $g(x)$ is $1$.
Therefore, the range of $\text{g}$ is $\text{R}_\text{g} = [1, \infty)$.

2. Show that the composite function fg exists.

The domain of $\text{f}$ is $\text{D}_\text{f} = \left(\frac{2}{3}, \infty\right)$.
Since $\text{R}_\text{g} = [1, \infty)$ and $[1, \infty) \subseteq \left(\frac{2}{3}, \infty\right)$, we have $\text{R}_\text{g} \subseteq \text{D}_\text{f}$.
Hence, the composite function $\text{fg}$ exists.

3. Find the range of fg.

The range of $\text{fg}$ corresponds to the range of $\text{f}$ restricted to the domain $x \in \text{R}_\text{g} = [1, \infty)$.
At the lower bound $x = 1$, $\text{f}(1) = \frac{\ln(3(1)-2)}{1} = 0$.
As $x \to \infty$, $\text{f}(x) \to 0$.
Using a graphing calculator to find the maximum point of $y = \text{f}(x)$ for $x \ge 1$, we obtain a maximum value of $y \approx 0.69468$ at $x \approx 2.11$.
Therefore, the range of $\text{fg}$ is $[0, 0.695]$ (to 3 significant figures).

1. Complete the square for $\text{g}(x)$ to find its turning point.

$\text{g}(x) = 3x^2 - 12x + 13$
$\text{g}(x) = 3(x^2 - 4x) + 13$
$\text{g}(x) = 3[(x - 2)^2 - 4] + 13$
$\text{g}(x) = 3(x - 2)^2 - 12 + 13$
$\text{g}(x) = 3(x - 2)^2 + 1$

The turning point is a minimum at $x = 2$.

2. Determine the largest value of $k$ for $\text{h}$ to have an inverse.

For $\text{h}^{-1}$ to exist, $\text{h}$ must be a one-to-one function.
Since the minimum point is at $x = 2$, restricting the domain to $x \le k$ limits the curve to its decreasing portion, which is one-to-one as long as it does not pass the vertex.
Therefore, the largest value of $k$ is $2$.

3. Express $x$ in terms of $y$ to find the inverse function formula.

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

Since the domain of $\text{h}$ is $x \le 2$, we must have $x - 2 \le 0$, so we reject the positive root:
$x - 2 = -\sqrt{\frac{y - 1}{3}}$
$x = 2 - \sqrt{\frac{y - 1}{3}}$

4. Determine the domain of the inverse and state it in a similar mapping form.

The domain of $\text{h}^{-1}$ is the range of $\text{h}$.
For $x \le 2$, the minimum value of $\text{h}(x)$ is $1$, so the range of $\text{h}$ is $y \ge 1$.
Thus, the domain of $\text{h}^{-1}$ is $x \ge 1$.

In similar form:
$\text{h}^{-1} : x \mapsto 2 - \sqrt{\frac{x - 1}{3}}, x \ge 1$