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

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

The denominator cannot be zero. As $x \to a^-$, the denominator $x - a \to 0^-$, causing $y \to -\infty$.
Thus, the vertical asymptote is $x = a$.

The domain starts at $x = -a$.
Substitute $x = -a$ into $f(x)$:
$f(-a) = \frac{a(-a)}{-a - a} = \frac{-a^2}{-2a} = \frac{a}{2}$
The coordinates of the endpoint are $\left(-a, \frac{a}{2}\right)$.

2. Find the axes intercepts and sketch the graph.

Substitute $x = 0$:
$f(0) = \frac{a(0)}{0 - a} = 0$
The curve passes through the origin $(0,0)$.

Sketching Details:

• Draw a set of axes and mark the vertical dashed line $x = a$.
• Plot a solid dot at $\left(-a, \frac{a}{2}\right)$.
• Draw a smooth decreasing curve from $\left(-a, \frac{a}{2}\right)$ that passes through $(0,0)$ and curves downwards to approach the asymptote $x = a$ as $y \to -\infty$.

3. Explain why the function has an inverse.

From the sketch, the curve is strictly decreasing for its entire domain $-a \le x < a$.
Any horizontal line $y = k$ will intersect the graph of $y = f(x)$ at most once (Horizontal Line Test).
Therefore, $f$ is a one-to-one function, which means $f$ has an inverse.

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

Let $y = \dfrac{ax}{x-a}$

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

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

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

3. Determine the domain of $f^{-1}$ by finding the range of $f$.

The domain of $f^{-1}$ is exactly the range of $f$.
Given the domain of $f$ is $-a \le x < a$, we evaluate $f(x)$ at the endpoint:
$f(-a) = \dfrac{a(-a)}{-a-a} = \dfrac{-a^2}{-2a} = \dfrac{a}{2}$

From the graph of $f(x)$ (or by observing the horizontal asymptote and vertical asymptote $x=a$), the function is strictly decreasing for $x < a$. As $x \to a^-$, $f(x) \to -\infty$.
Thus, the range of $f$ is $y \le \dfrac{a}{2}$.

Therefore, the domain of $f^{-1}$ is $x \le \dfrac{a}{2}$.

1. Find the algebraic expression for the inverse function $f^{-1}(x)$ (required from part b).

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

2. Determine the domain of $f^{-1}(x)$ by finding the range of $f(x)$ (required from part b).

The domain of $f^{-1}$ corresponds to the range of $f$.
Express $f(x)$ as:
$f(x) = \frac{a(x-a)+a^2}{x-a} = a + \frac{a^2}{x-a}$
Given the domain $-a \le x < a$:
$-2a \le x - a < 0$
$\frac{1}{x-a} \le -\frac{1}{2a}$
Since $a > 0$, multiply by $a^2$:
$\frac{a^2}{x-a} \le -\frac{a^2}{2a} = -\frac{a}{2}$
$a + \frac{a^2}{x-a} \le a - \frac{a}{2} = \frac{a}{2}$
As $x \to a^-$, $x-a \to 0^-$, so $f(x) \to -\infty$.
Thus, the range of $f(x)$ is $(-\infty, \frac{a}{2}]$, meaning the domain of $f^{-1}(x)$ is $x \le \frac{a}{2}$.

3. Find the range of $x$ for which $f(x) = f^{-1}(x)$ by intersecting their domains.

Since $f(x)$ and $f^{-1}(x)$ share the exact same algebraic expression $\frac{ax}{x-a}$, the equation $f(x) = f^{-1}(x)$ holds for all $x$ where both functions are defined.
We find the intersection of their valid domains:
Domain of $f$: $-a \le x < a$
Domain of $f^{-1}$: $x \le \frac{a}{2}$
Taking the intersection of these two sets yields $-a \le x \le \frac{a}{2}$.

1. Determine the composite function $f^2(x)$ to establish the pattern for $f^{2k+1}(x)$.

Evaluate $f^2(x) = f(f(x))$:
$f^2(x) = f\left(\dfrac{ax}{x-a}\right) = \dfrac{a\left(\dfrac{ax}{x-a}\right)}{\dfrac{ax}{x-a} - a}$
$= \dfrac{\dfrac{a^2x}{x-a}}{\dfrac{ax - a(x - a)}{x-a}} = \dfrac{a^2x}{a^2} = x$
Since $f^2(x) = x$, applying the function an even number of times returns the original input $x$. Thus, an odd number of applications simplifies to a single application: $f^{2k+1}(x) = f(x)$ for any positive integer $k$.

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

Substitute $x = -\frac{1}{2}a$ into $f(x)$:
$f^{2k+1}\left(-\frac{1}{2}a\right) = f\left(-\frac{1}{2}a\right) = \dfrac{a\left(-\frac{1}{2}a\right)}{-\frac{1}{2}a - a}$
$= \dfrac{-\frac{1}{2}a^2}{-\frac{3}{2}a} = \dfrac{1}{3}a$

1. Apply the inverse function $f^{-1}$ to find an expression for $g(x)$.

$f(g(x)) = x+a \implies g(x) = f^{-1}(x+a)$

2. Substitute $x+a$ into $f^{-1}(x)$ to obtain $g(x)$.

Using the result from part (b) where $f^{-1}(x) = \frac{ax}{x-a}$, substitute $x+a$ into $f^{-1}$:
$g(x) = \frac{a(x+a)}{(x+a)-a}$
$g(x) = \frac{a(x+a)}{x}$

3. State the domain of $g(x)$.

The relationship $fg(x) = x+a$ is explicitly given for $x \le -a$. This defines the valid input values for $g$.
Domain of $g$: $x \le -a$