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

1. Show that the function evaluates to the same output only if the inputs are identical.

Let $g(a) = g(b)$ for some $a, b > \sqrt{3}$.
$\ln(a - \sqrt{3}) = \ln(b - \sqrt{3})$
$a - \sqrt{3} = b - \sqrt{3}$
$a = b$

2. Conclude that the function is one-to-one, satisfying the condition for an inverse to exist.

Since $g(a) = g(b) \implies a = b$, $g$ is a one-to-one function.
Therefore, $g^{-1}$ exists.

1. Set $y = g(x)$ and make $x$ the subject.

Let $y = \ln(x - \sqrt{3})$.
$e^y = x - \sqrt{3}$
$x = e^y + \sqrt{3}$

2. State the inverse function $g^{-1}(x)$.

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

3. Determine the domain of $g^{-1}$ by considering the range of $g$.

The domain of $g^{-1}$ is exactly the range of $g$.
For $x > \sqrt{3}$, the expression $x - \sqrt{3}$ takes all values in $(0, \infty)$.
Therefore, $\ln(x - \sqrt{3})$ takes all real values.
$\text{Domain of } g^{-1} \text{ is } x \in \mathbb{R}$

1. Set up the equation to solve for $x$.

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

2. Convert the logarithmic equation to an exponential form to find the exact value of $x$.

$x - \sqrt{3} = e^0$
$x - \sqrt{3} = 1$
$x = 1 + \sqrt{3}$

1. Establish the domain of the inequality and the root from the numerator.

From the definition of $g(x)$, the domain is $x > \sqrt{3}$.
To find the critical point from the numerator, we solve $g(x) = 0$ (as calculated in part c):
$\ln(x - \sqrt{3}) = 0 \implies x - \sqrt{3} = 1 \implies x = 1 + \sqrt{3}$

2. Simplify the inequality by identifying any strictly positive factors within the domain.

For the domain $x > \sqrt{3}$, the factor $(x+1)$ in the denominator is strictly positive, since $x + 1 > \sqrt{3} + 1 > 0$.
Thus, we can multiply both sides of the inequality by $x+1$ without reversing the inequality sign, simplifying it to:
$\frac{g(x)}{x-3} \ge 0$

3. Identify the remaining critical value and determine their relative order without a calculator.

We identify the remaining critical value from the denominator as $x = 3$.
To construct a sign test without a calculator, we determine the relative order of the domain boundary $\sqrt{3}$ and the critical values $1+\sqrt{3}$ and $3$.
Since $1 < \sqrt{3} < \sqrt{4} = 2$, it logically follows that:
$1 + \sqrt{3} < 1 + 2 = 3$
So the critical points in ascending order are $\sqrt{3}$, $1+\sqrt{3}$, and $3$.

4. Perform a sign test on the valid intervals and deduce the final solution.

We perform a sign test on the intervals bounded by the domain and critical values:

• For $\sqrt{3} < x < 1+\sqrt{3}$: $g(x) < 0$ and $x-3 < 0 \implies \frac{g(x)}{x-3} > 0$
• For $1+\sqrt{3} < x < 3$: $g(x) > 0$ and $x-3 < 0 \implies \frac{g(x)}{x-3} < 0$
• For $x > 3$: $g(x) > 0$ and $x-3 > 0 \implies \frac{g(x)}{x-3} > 0$

The inequality includes equality to $0$, which occurs at the numerator's zero, $x = 1+\sqrt{3}$. At $x = 3$, the denominator is zero so the expression is undefined.
Combining these observations gives the final valid intervals.