The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\begin{aligned} \mathrm{f} &: x \mapsto \frac{2x+14}{3x-2}, \quad x \in \mathbb{R}, x > \frac{2}{3}, \\ \mathrm{g} &: x \mapsto \frac{\sin x}{x}, \quad x \in \mathbb{R}, x > 0. \end{aligned}$$

1. Set up an equation with $y = \mathrm{f}(x)$ and make $x$ the subject of the formula.

Let $y = \frac{2x+14}{3x-2}$.
$y(3x-2) = 2x+14$
$3xy - 2y = 2x+14$
$3xy - 2x = 2y+14$
$x(3y - 2) = 2y+14$
$x = \frac{2y+14}{3y-2}$

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

$\mathrm{f}^{-1}(x) = \frac{2x+14}{3x-2}$

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

The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$.
Rewrite $\mathrm{f}(x)$ to analyse its range:
$\mathrm{f}(x) = \frac{\frac{2}{3}(3x-2) + \frac{46}{3}}{3x-2} = \frac{2}{3} + \frac{46}{3(3x-2)}$
Given the domain of $\mathrm{f}$ is $x > \frac{2}{3}$, it follows that $3x-2 > 0$.
Thus, $\frac{46}{3(3x-2)} > 0$, giving $\mathrm{f}(x) > \frac{2}{3}$.
Therefore, the range of $\mathrm{f}$ is $( \frac{2}{3}, \infty)$, which means the domain of $\mathrm{f}^{-1}$ is $x > \frac{2}{3}$.

1. Show that $\mathrm{f}$ is a self-inverse function and deduce $\mathrm{f}^2(x) = x$.

From part (a), finding the inverse of $y = \frac{2x+14}{3x-2}$ gives $3xy - 2y = 2x + 14 \implies x(3y - 2) = 2y + 14 \implies x = \frac{2y+14}{3y-2}$.
Thus, $\mathrm{f}^{-1}(x) = \frac{2x+14}{3x-2}$.
Since $\mathrm{f}^{-1}(x) = \mathrm{f}(x)$, the function is self-inverse.
$\mathrm{f}^2(x) = \mathrm{f}(\mathrm{f}(x)) = \mathrm{f}(\mathrm{f}^{-1}(x)) = x$

2. Extend the result to the 2024th power.

Since $\mathrm{f}^2(x) = x$, applying the function $\mathrm{f}$ repeatedly will yield $x$ for any even number of iterations.
As $2024$ is an even number, $\mathrm{f}^{2024}(x) = (\mathrm{f}^2)^{1012}(x) = x$.

1. Identify the equation of the composite function to be sketched.

From part (b), we established that $\mathrm{f}^{2024}(x) = x$ for $x > \frac{2}{3}$.
Thus, the graph of $y = \mathrm{f}^{2024}(x)$ is simply the straight line $y = x$ restricted to the domain $x > \frac{2}{3}$.

2. Find the equations of the asymptotes for $y = \mathrm{f}(x)$.

For the rational function $y = \frac{2x+14}{3x-2}$, the vertical asymptote occurs where the denominator is zero:
$3x - 2 = 0 \implies x = \frac{2}{3}$
The horizontal asymptote is determined by the limit as $x \to \infty$:
$y = \lim_{x \to \infty} \frac{2x+14}{3x-2} = \frac{2}{3}$
Hence, the equations of the asymptotes are $x = \frac{2}{3}$ and $y = \frac{2}{3}$.

3. Describe the features of the combined sketch.

When sketching the graphs together for $x > \frac{2}{3}$:

• $y = \mathrm{f}^{2024}(x)$: Draw the line $y = x$ starting from an empty circle at $(\frac{2}{3}, \frac{2}{3})$ and extending diagonally upwards.
• $y = \mathrm{f}(x)$: Draw a single hyperbola branch. As $x \to \frac{2}{3}^+$, $y \to \infty$. As $x \to \infty$, the curve decreases strictly and approaches $y = \frac{2}{3}^+$. The curve intersects the line $y = x$.

1. State the range of f and the domain of g

From part (a), the domain of $\mathrm{f}^{-1}$ gives the range of $\mathrm{f}$. Thus, the range of $\mathrm{f}$ is $\mathrm{R}_{\mathrm{f}} = \left(\frac{2}{3}, \infty\right)$.
From the definition of $\mathrm{g}$, the domain of $\mathrm{g}$ is $\mathrm{D}_{\mathrm{g}} = (0, \infty)$.

2. Compare the range of f with the domain of g to conclude

Since all values in $\left(\frac{2}{3}, \infty\right)$ are strictly greater than $0$, $\mathrm{R}_{\mathrm{f}} \subseteq \mathrm{D}_{\mathrm{g}}$.
Because the range of $\mathrm{f}$ is a subset of the domain of $\mathrm{g}$, the composite function $\mathrm{g}\mathrm{f}$ exists.

1. Identify the relevant domain for the outer function $\mathrm{g}$ when evaluating the composite function $\mathrm{g}\mathrm{f}$.

The range of the composite function $\mathrm{g}\mathrm{f}$ is given by $\mathrm{g}(\mathrm{R}_\mathrm{f})$.
From part (a), the domain of $\mathrm{f}^{-1}$ corresponds to the range of $\mathrm{f}$. For $x > \frac{2}{3}$, $\mathrm{f}(x) = \frac{2x+14}{3x-2}$ strictly decreases with a horizontal asymptote of $y = \frac{2}{3}$.
Thus, $\mathrm{R}_\mathrm{f} = \left(\frac{2}{3}, \infty\right)$.
Consequently, we need to find the range of $\mathrm{g}(x) = \frac{\sin x}{x}$ for $x \in \left(\frac{2}{3}, \infty\right)$.

2. Find the minimum and supremum of $\mathrm{g}(x)$ on this interval using a graphing calculator.

Using a graphing calculator to analyse the graph of $y = \frac{\sin x}{x}$ over $x > \frac{2}{3}$ (ensuring the calculator is in radian mode):

• The function initially decreases from the left boundary $x = \frac{2}{3}$. Since $x = \frac{2}{3}$ is excluded from the domain, the upper bound is a supremum evaluated at this boundary:
  $y \to \frac{\sin(2/3)}{2/3} = \frac{3}{2}\sin\left(\frac{2}{3}\right) \approx 0.9275$
• The absolute minimum on this interval occurs at the first stationary point (trough), located at $x \approx 4.4934$. The minimum value is $y \approx -0.2172$.
• As $x \to \infty$, the function oscillates with strictly decaying amplitude towards $y = 0$, meaning no values will fall outside the bounds found above.

3. State the final range of $\mathrm{g}\mathrm{f}$.

Combining the absolute minimum and the supremum, and keeping 3 significant figures for the non-exact bound, the range is:
$\mathrm{R}_{\mathrm{gf}} = \left[ -0.217, \frac{3}{2}\sin\left(\frac{2}{3}\right) \right)$