The function $\text{f}$ is defined by $\text{f} : x \mapsto \frac{ax}{x - a}, \quad \text{for } x \in \mathbb{R}, \ -a \le x < a,$ where $a$ is a positive constant.

1. Find the coordinates of the endpoint and the axes intercepts.

For the domain $-a \le x < a$, the starting endpoint is at $x = -a$.
When $x = -a$, $y = \frac{a(-a)}{-a - a} = \frac{-a^2}{-2a} = \frac{a}{2}$.
Thus, the endpoint is $\left(-a, \frac{a}{2}\right)$.
To find the axes intercept, set $x = 0$:
$y = \frac{a(0)}{0 - a} = 0$. The curve passes through the origin $(0, 0)$.

2. Identify the relevant asymptote for the restricted domain.

The denominator is zero when $x = a$. As $x \to a^-$, $(x - a) \to 0^-$ and the numerator $ax > 0$, so $y \to -\infty$.
Thus, there is a vertical asymptote at $x = a$.
(Note: The horizontal asymptote $y=a$ of the full curve is not applicable here since the domain restricts $x < a$.)

3. Sketch the graph of the function.

Draw a set of axes. Mark the endpoint with a solid dot at $\left(-a, \frac{a}{2}\right)$. Draw a smooth curve passing through $(0,0)$ and curving downwards to approach the vertical dashed line $x = a$ asymptotically.

4. Explain why the function has an inverse.

Observe the sketched graph: the function is strictly decreasing over its domain. Any horizontal line $y = k$ will intersect the graph at most once. This confirms that $\text{f}$ is a one-to-one function, which is the required condition for an inverse function to exist.

1. Set $y = \text{f}(x)$ and rearrange the equation to make $x$ the subject.

$y = \frac{ax}{x-a}$
$y(x - a) = ax$
$xy - ay = ax$
$x(y - a) = ay$
$x = \frac{ay}{y - a}$

2. Write down the expression for $\text{f}^{-1}(x)$.

$\text{f}^{-1}(x) = \frac{ax}{x - a}$

3. Determine the domain of $\text{f}^{-1}$ by finding the range of $\text{f}$ for $-a \le x < a$.

The domain of $\text{f}^{-1}$ is the range of $\text{f}$.
$\text{f}(x)$ is strictly decreasing on the interval $-a \le x < a$.
At $x = -a$, $\text{f}(-a) = \frac{a(-a)}{-a - a} = \frac{-a^2}{-2a} = \frac{a}{2}$.
As $x \to a^-$, the denominator approaches $0^-$ while the numerator is positive ($a^2$), so $\text{f}(x) \to -\infty$.
Therefore, the range of $\text{f}$ is $y \le \frac{a}{2}$.
Hence, the domain of $\text{f}^{-1}$ is $x \le \frac{a}{2}$.

1. Equate $\text{f}(x)$ and $\text{f}^{-1}(x)$ to identify that they are algebraically identical

From part (b), we know that $\text{f}^{-1}(x) = \frac{ax}{x-a}$.
Thus, the equation $\text{f}(x) = \text{f}^{-1}(x)$ becomes:
$\frac{ax}{x-a} = \frac{ax}{x-a}$
This is an algebraic identity. Because a function is its own inverse, $\text{f}(x) = \text{f}^{-1}(x)$ holds true for all values of $x$ where both functions are simultaneously defined.

2. Determine the common domain where both $\text{f}$ and $\text{f}^{-1}$ are defined

The domain of $\text{f}$ is given in the question as $-a \le x < a$.

The domain of $\text{f}^{-1}$ corresponds to the range of $\text{f}$. Since $\text{f}(x)$ is a decreasing function on its domain, its maximum value occurs at $x = -a$:
$\text{f}(-a) = \frac{a(-a)}{-a-a} = \frac{1}{2}a$
As $x \to a^-$, $\text{f}(x) \to -\infty$. Thus, the domain of $\text{f}^{-1}$ is $x \le \frac{1}{2}a$.

The range of values of $x$ for which $\text{f}(x) = \text{f}^{-1}(x)$ is the intersection of their domains:
$-a \le x < a \quad \text{and} \quad x \le \frac{1}{2}a$

Since $a$ is a positive constant, $\frac{1}{2}a < a$. The intersection is therefore:
$-a \le x \le \frac{1}{2}a$

1. Deduce the simplification of the composite function $\text{f}^{2k+1}(x)$.

From part (b), we know that $\text{f}^{-1}(x) = \frac{ax}{x - a}$, which implies that $\text{f}(x) = \text{f}^{-1}(x)$.
Therefore, $\text{f}^2(x) = \text{f}(\text{f}^{-1}(x)) = x$.
For any positive integer $k$, applying $\text{f}$ an even number of times yields $x$, so $\text{f}^{2k}(x) = x$.
Thus, applying it an odd number of times gives $\text{f}^{2k+1}(x) = \text{f}(\text{f}^{2k}(x)) = \text{f}(x)$.

2. Evaluate $\text{f}\left(-\frac{1}{2}a\right)$.

$\text{f}\left(-\frac{1}{2}a\right) = \frac{a\left(-\frac{1}{2}a\right)}{-\frac{1}{2}a - a}$
$= \frac{-\frac{1}{2}a^2}{-\frac{3}{2}a}$
$= \frac{1}{3}a$

1. Find the inverse function $\text{f}^{-1}(x)$ (as derived in earlier part b).

Let $y = \text{f}(x) = \frac{ax}{x - a}$
$y(x - a) = ax$
$xy - ay = ax$
$x(y - a) = ay$
$x = \frac{ay}{y - a}$
Thus, $\text{f}^{-1}(x) = \frac{ax}{x - a}$

2. Use the property of inverse functions to isolate and find $\text{g}(x)$.

Given $\text{fg}(x) = x + a$, applying $\text{f}^{-1}$ to both sides gives:
$\text{g}(x) = \text{f}^{-1}(x + a)$
Substitute $x + a$ into the expression for $\text{f}^{-1}$:
$\text{g}(x) = \frac{a(x + a)}{(x + a) - a}$
$\text{g}(x) = \frac{a(x + a)}{x}$

3. State the domain of $\text{g}$.

The equation $\text{fg}(x) = x + a$ is explicitly given for $x \in \mathbb{R}, x \le -a$.
Therefore, the domain of $\text{g}$ is simply this restricted set of values.
Domain of $\text{g}$: $x \le -a$