The functions $\text{f}$ and $\text{g}$ are defined by
$\begin{aligned} \text{f} &: x \mapsto x^2 - 2x + 5 \quad \text{for } x \in \mathbb{R}, \\ \text{g} &: x \mapsto \frac{1}{x - 1} \quad \text{for } x \in \mathbb{R}, \ x \ne 1. \end{aligned}$

1. State the range of $\text{g}$ and the domain of $\text{f}$.

The domain of $\text{f}$ is given as all real numbers:
$\text{D}_{\text{f}} = \mathbb{R}$

For $\text{g}(x) = \frac{1}{x-1}$, the output can be any real number except $0$. Therefore, the range of $\text{g}$ is:
$\text{R}_{\text{g}} = \mathbb{R} \setminus \{0\}$

2. Verify the condition for the existence of the composite function $\text{fg}$.

For the composite function $\text{fg}$ to exist, the range of $\text{g}$ must be a subset of the domain of $\text{f}$ (i.e., $\text{R}_{\text{g}} \subseteq \text{D}_{\text{f}}$).

Since $\mathbb{R} \setminus \{0\} \subseteq \mathbb{R}$, the condition is satisfied.

Therefore, the composite function $\text{fg}$ exists.

1. Substitute the expression for $\text{g}(x)$ into $\text{f}(x)$ to find $\text{fg}(x)$.

$\text{fg}(x) = \text{f}(\text{g}(x))$
$= \text{f}\left(\frac{1}{x-1}\right)$
$= \left(\frac{1}{x-1}\right)^2 - 2\left(\frac{1}{x-1}\right) + 5$
$= \frac{1}{(x-1)^2} - \frac{2}{x-1} + 5$

2. State the domain of the composite function $\text{fg}$.

The domain of $\text{fg}$ is the set of all $x$ in the domain of $\text{g}$ such that $\text{g}(x)$ is in the domain of $\text{f}$.
Since the domain of $\text{f}$ is $\mathbb{R}$, there are no additional restrictions on the output of $\text{g}$.
Thus, the domain of $\text{fg}$ is simply the domain of $\text{g}$.
$\text{D}_{\text{fg}} = \text{D}_{\text{g}} = \{x \in \mathbb{R} : x \ne 1\}$

1. Determine the range of the inner function $\text{g}(x)$.

Let $y = \text{g}(x)$. For $x \in \mathbb{R}, x \ne 1$:
$y = \frac{1}{x - 1}$
The range of $\text{g}$ is all real numbers except $0$, so $y \in \mathbb{R}, y \ne 0$.

2. Find the range of $\text{f}(y)$ over the domain $y \in \mathbb{R}, y \ne 0$.

We find the range of values taken by $\text{f}(y)$ for $y \ne 0$. By completing the square:
$\text{f}(y) = y^2 - 2y + 5 = (y - 1)^2 + 4$
The minimum value occurs at $y = 1$, which is within our domain, giving a minimum value of $4$.

Although $y = 0$ is excluded from the domain (which would give $\text{f}(0) = 5$), the value $5$ is still achieved by the function at $y = 2$, since $\text{f}(2) = (2 - 1)^2 + 4 = 5$ and $y=2$ is a valid output of $\text{g}(x)$.

As $y \to \pm\infty$, $\text{f}(y) \to \infty$.
Thus, the range of $\text{fg}$ is $[4, \infty)$.

1. Express $x$ in terms of $y$ by completing the square for $\text{f}(x)$.

Let $y = x^2 - 2x + 5$.
$y = (x - 1)^2 - 1^2 + 5$
$y = (x - 1)^2 + 4$
$(x - 1)^2 = y - 4$

2. Take the square root and choose the correct sign based on the restricted domain.

$x - 1 = \pm\sqrt{y - 4}$
Since the domain of $\text{f}$ is restricted to $(-\infty, 1]$, we have $x \le 1$, which means $x - 1 \le 0$. Thus, we reject the positive root.
$x - 1 = -\sqrt{y - 4}$
$x = 1 - \sqrt{y - 4}$
Therefore, $\text{f}^{-1}(x) = 1 - \sqrt{x - 4}$.

3. Determine the domain of $\text{f}^{-1}$ from the range of the restricted $\text{f}$.

The domain of $\text{f}^{-1}$ is equal to the range of $\text{f}$.
For $x \le 1$, $(x - 1)^2 \ge 0$, so $\text{f}(x) = (x - 1)^2 + 4 \ge 4$.
The range of $\text{f}$ is $[4, \infty)$.
Therefore, the domain of $\text{f}^{-1}$ is $[4, \infty)$ (or $x \ge 4$).

1. Determine the domain of $\text{f}^{-1}$ by finding the range of the restricted function $\text{f}$

By completing the square, $\text{f}(x) = x^2 - 2x + 5 = (x - 1)^2 + 4$.
For the restricted domain $x \le 1$, we have $(x - 1)^2 \ge 0$, which means $\text{f}(x) \ge 4$.
The domain of $\text{f}^{-1}$ is equal to the range of $\text{f}$, so $\text{D}_{\text{f}^{-1}} = [4, \infty)$.
(Note: This would typically have been found in part (iv).)

2. Use the identity property of inverse functions to determine the valid values of $x$

The identity $\text{ff}^{-1}(x) = x$ is satisfied for all values of $x$ in the domain of $\text{f}^{-1}$.
Therefore, the range of values of $x$ is $x \ge 4$.

1. Rewrite the inverse function equation in terms of the original function $\text{g}$.

By the definition of inverse functions, if $\text{g}^{-1}(a) = b$, then $\text{g}(b) = a$.

2. Substitute $b$ into $\text{g}(x)$ to find $a$.

Substituting $x = b$ into $\text{g}(x) = \frac{1}{x - 1}$, we get:
$a = \frac{1}{b - 1}$