The function $\text{g}$ is defined by $\text{g} : x \mapsto \ln(x - \sqrt{3})$, for $x \in \mathbb{R}$, $x > \sqrt{3}$.

1. Establish that the function is one-to-one by considering its behavior

For $x > \sqrt{3}$, as $x$ increases, $x - \sqrt{3}$ increases. Since the natural logarithm is a strictly increasing function, $\text{g}(x) = \ln(x - \sqrt{3})$ is also strictly increasing.

Alternatively, letting $\text{g}(x_1) = \text{g}(x_2)$ gives:
$\ln(x_1 - \sqrt{3}) = \ln(x_2 - \sqrt{3})$
$x_1 - \sqrt{3} = x_2 - \sqrt{3}$
$x_1 = x_2$
Thus, $\text{g}$ is a one-to-one function.

2. Conclude the existence of the inverse function

Because $\text{g}$ is a one-to-one function, it satisfies the condition for an inverse function to exist. Hence, $\text{g}^{-1}$ exists.

1. Set $y = \text{g}(x)$ and make $x$ the subject to find the rule for the inverse function.

$y = \ln(x - \sqrt{3})$
$e^y = x - \sqrt{3}$
$x = e^y + \sqrt{3}$
$\therefore \text{g}^{-1}(x) = e^x + \sqrt{3}$

2. Determine the domain of $\text{g}^{-1}$ by analyzing the range of $\text{g}$.

For the original function $\text{g}(x) = \ln(x - \sqrt{3})$ where $x > \sqrt{3}$, the term $(x - \sqrt{3})$ takes all positive real values $(0, \infty)$.
Thus, the range of $\text{g}$ is all real numbers, $\mathbb{R}$.
Since the domain of $\text{g}^{-1}$ is equal to the range of $\text{g}$, the domain of $\text{g}^{-1}$ is $x \in \mathbb{R}$.

1. Set the function to zero and convert the logarithmic equation to an exponential form

$\ln(x - \sqrt{3}) = 0$
$x - \sqrt{3} = e^0$

2. Solve for $x$ to find the exact value

$x - \sqrt{3} = 1$
$x = 1 + \sqrt{3}$

1. Identify critical values and apply the domain restriction.

From part (c), $\text{g}(x) = 0$ when $x = 1 + \sqrt{3}$.
The denominator is zero when $(x - 3)(x + 1) = 0$, giving $x = 3$ or $x = -1$.
Since the domain of $\text{g}$ is $x > \sqrt{3}$ (approx $1.732$), we reject $x = -1$.
The valid critical values to test are $x = 1 + \sqrt{3}$ and $x = 3$.

2. Determine the signs of the factors in the relevant intervals.

For $x > \sqrt{3}$, the factor $(x + 1)$ is always positive ($> \sqrt{3} + 1 > 0$). Thus, the sign of $\frac{\text{g}(x)}{(x - 3)(x + 1)}$ depends only on $\text{g}(x)$ and $(x - 3)$.
Consider the intervals defined by the critical values within the domain:

• For $\sqrt{3} < x < 1 + \sqrt{3}$: $\text{g}(x) < 0$ and $x - 3 < 0$, so the expression is positive.
• For $1 + \sqrt{3} < x < 3$: $\text{g}(x) > 0$ and $x - 3 < 0$, so the expression is negative.
• For $x > 3$: $\text{g}(x) > 0$ and $x - 3 > 0$, so the expression is positive.

3. State the final solution for the inequality, accounting for equality and undefined points.

The inequality is $\ge 0$, which means we include values where the numerator is zero ($x = 1 + \sqrt{3}$) but strictly exclude values where the denominator is zero ($x = 3$).
Combining the positive intervals and the zero point gives the final solution:
$\sqrt{3} < x \le 1 + \sqrt{3} \text{ or } x > 3$