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

1. Find the coordinates of the endpoint and the equation of the asymptote.

Substitute $x = -a$ into $\mathrm{f}(x)$:
$\mathrm{f}(-a) = \frac{a(-a)}{-a-a} = \frac{-a^2}{-2a} = \frac{a}{2}$
So the endpoint is $\left(-a, \frac{a}{2}\right)$.

The denominator is $0$ when $x = a$, so the vertical asymptote is $x = a$.

2. Sketch the curve for the given domain.

Notice that $\mathrm{f}(0) = 0$, so the curve passes through the origin $(0,0)$.
For $-a \le x < a$, the curve starts at a solid point at $\left(-a, \frac{a}{2}\right)$, decreases steadily through the origin, and tends towards $-\infty$ as $x$ approaches the vertical asymptote $x = a$ from the left.

3. Explain why the inverse function exists based on the properties of the function.

From the graph, any horizontal line $y = k$ cuts the curve at most once.
Therefore, $\mathrm{f}$ is a one-to-one function, which is the condition required for its inverse $\mathrm{f}^{-1}$ to exist.

1. Set $y = \mathrm{f}(x)$ and make $x$ the subject to find the inverse function.

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

2. Determine the range of $\mathrm{f}$ to state the domain of $\mathrm{f}^{-1}$.

The domain of $\mathrm{f}^{-1}$ is equal to the range of $\mathrm{f}$.
Evaluate $\mathrm{f}$ at the lower endpoint of its domain:
$\mathrm{f}(-a) = \frac{a(-a)}{-a-a} = \frac{-a^2}{-2a} = \frac{a}{2}$
As $x \to a^-$, the denominator $x-a \to 0^-$ and the numerator $ax \to a^2$ (which is positive since $a>0$). Thus, $\mathrm{f}(x) \to -\infty$.
Therefore, the range of $\mathrm{f}$ is $y \le \frac{a}{2}$.
The domain of $\mathrm{f}^{-1}$ is $x \le \frac{a}{2}$.

1. State the expression and domain of the inverse function (obtained from part b).

From part (b), we know that the inverse function is $\mathrm{f}^{-1}(x) = \frac{ax}{x-a}$ and its domain is $x \le \frac{a}{2}$.

2. Recognise that since the functions share the same algebraic expression, they are equal wherever both are defined.

Since $\mathrm{f}(x)$ and $\mathrm{f}^{-1}(x)$ have the identical algebraic expression $\frac{ax}{x-a}$, the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ will hold true for all values of $x$ that belong to the domains of both functions.

3. Find the intersection of the domains of $\mathrm{f}$ and $\mathrm{f}^{-1}$.

Domain of $\mathrm{f}$: $-a \le x < a$
Domain of $\mathrm{f}^{-1}$: $x \le \frac{a}{2}$

Finding the common range of values (intersection of the two domains) and noting that $\frac{a}{2} < a$ for a positive constant $a$:
$-a \le x \le \frac{a}{2}$

1. Deduce that applying the function an odd number of times is equivalent to applying it once.

From part (b), we know that $\mathrm{f}^{-1}(x) = \mathrm{f}(x)$, which means $\mathrm{f}$ is a self-inverse function and $\mathrm{f}^2(x) = x$.
For any positive integer $k$, applying $\mathrm{f}$ an even number of times returns the original input, so applying it an odd number of times gives:
$\mathrm{f}^{2k+1}(x) = \mathrm{f}(x)$

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

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

1. Set up the equation using the definition of the composite function $\mathrm{fg}(x)$.

Since $\mathrm{f}(x) = \frac{ax}{x-a}$, we have:
$\mathrm{f}(\mathrm{g}(x)) = \frac{a\mathrm{g}(x)}{\mathrm{g}(x) - a}$
Given that $\mathrm{fg}(x) = x + a$,
$\frac{a\mathrm{g}(x)}{\mathrm{g}(x) - a} = x + a$

2. Rearrange the equation to make $\mathrm{g}(x)$ the subject.

$a\mathrm{g}(x) = (x+a)(\mathrm{g}(x) - a)$
$a\mathrm{g}(x) = x\mathrm{g}(x) - ax + a\mathrm{g}(x) - a^2$
$x\mathrm{g}(x) = ax + a^2$
$\mathrm{g}(x) = \frac{a(x+a)}{x}$

3. State the domain of $\mathrm{g}(x)$.

The problem states the relationship holds for $x \le -a$. As $a$ is a positive constant, $x$ cannot be $0$, and no further restrictions are introduced. Thus, the domain is the given interval.
Domain of $\mathrm{g}$ is $x \le -a$.