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

1. Show algebraically that the function is one-to-one.

Let $g(x_1) = g(x_2)$ for $x_1, x_2 > \sqrt{3}$.
$\ln(x_1 - \sqrt{3}) = \ln(x_2 - \sqrt{3})$
$x_1 - \sqrt{3} = x_2 - \sqrt{3}$
$x_1 = x_2$
Since $g(x_1) = g(x_2)$ strictly implies $x_1 = x_2$, the function $g$ is a one-to-one function.

2. State the conclusion linking the one-to-one property to the existence of an inverse.

An inverse function exists if and only if the function is one-to-one. Therefore, $g^{-1}$ exists.

1. Set up an equation and make $x$ the subject to find the inverse function.

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

2. State the inverse function in terms of $x$.

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

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

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

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

$\ln(x - \sqrt{3}) = 0$
$x - \sqrt{3} = e^0$
$x - \sqrt{3} = 1$

2. Solve for $x$ and state the exact value.

$x = 1 + \sqrt{3}$

1. Find the critical values from the numerator and denominator.

First, we concisely reproduce the result of part (c) by solving $g(x) = 0$:
$\ln(x - \sqrt{3}) = 0 \implies x - \sqrt{3} = e^0 = 1 \implies x = 1 + \sqrt{3}$
Next, we find the critical values of the denominator by setting it to zero:
$(x-3)(x+1) = 0 \implies x = 3 \quad \text{or} \quad x = -1$

2. Analyse the sign of the rational expression within the domain of the function.

The domain of $g(x)$ is given as $x > \sqrt{3}$. Therefore, $x = -1$ is rejected since it falls outside the domain, and we note that $x + 1 > 0$ for all valid $x$.
We test the signs of the expression $\dfrac{g(x)}{(x-3)(x+1)}$ across the intervals defined by the valid critical values $x = 1 + \sqrt{3}$ and $x = 3$:

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

3. State the final solution for the inequality.

The inequality requires the rational expression to be greater than or equal to $0$.
At $x = 1 + \sqrt{3}$, the numerator is $0$, making the expression exactly $0$, which satisfies the inequality. At $x = 3$, the expression is undefined.
Combining this with the sign analysis, we obtain the final solution:
$\sqrt{3} < x \le 1 + \sqrt{3} \quad \text{or} \quad x > 3$