The function f is defined by $f : x \mapsto 4 + \dfrac{1}{(x-1)^2 + 1},\quad x \ge 0$. The function g is defined by g : x \mapsto \sin x,\quad x \in [0,2\pi]$.

1. Find the $y$-intercept of the graph.

Substitute $x = 0$ into $f(x)$:
$f(0) = 4 + \frac{1}{(0-1)^2 + 1} = 4 + \frac{1}{2} = 4.5$
The $y$-intercept is at $(0, 4.5)$.

2. Find the coordinates of the stationary point.

The function $f(x) = 4 + \frac{1}{(x-1)^2 + 1}$ attains a maximum value when the denominator $(x-1)^2 + 1$ is at its minimum.
Since $(x-1)^2 \ge 0$, the minimum of the denominator occurs at $x = 1$.
$f(1) = 4 + \frac{1}{0 + 1} = 5$
Therefore, the stationary point is a maximum at $(1, 5)$.

3. Determine the equation of the horizontal asymptote.

Consider the behaviour of the function as $x \to \infty$.
As $x \to \infty$, $(x-1)^2 \to \infty$, so $\frac{1}{(x-1)^2 + 1} \to 0$.
Thus, $f(x) \to 4$.
The horizontal asymptote is $y = 4$.

4. Sketch the graph of $y = f(x)$ for the given domain.

Plot the $y$-intercept at $(0, 4.5)$ and the maximum turning point at $(1, 5)$. Draw a smooth curve through these points that continually decreases after $x=1$ and approaches the horizontal asymptote $y = 4$. Ensure the graph is only drawn for $x \ge 0$.

1. Identify the requirement to show that an inverse function does not exist.

A function does not have an inverse if it is not one-to-one. To demonstrate this, we need to find two distinct points in the domain $[0, 2\pi]$ that have the same $y$-coordinate.

2. State two points on the graph of $y=g(x)$ that share the same $y$-coordinate.

Using $y = 0$, we have $g(x) = \sin x = 0$.
In the domain $0 \le x \le 2\pi$, $x$ can be $0$, $\pi$, or $2\pi$.
Selecting two distinct values, $g(0) = 0$ and $g(\pi) = 0$.
Thus, $(0, 0)$ and $(\pi, 0)$ are two points on the graph. (Other pairs such as $(\frac{\pi}{6}, \frac{1}{2})$ and $(\frac{5\pi}{6}, \frac{1}{2})$ are also acceptable).

1. Determine the range of $f$.

From the definition of $f(x) = 4 + \dfrac{1}{(x-1)^2 + 1}$ for $x \ge 0$:
The maximum value occurs at $x = 1$, giving $f(1) = 5$.
As $x \to \infty$, the term $\dfrac{1}{(x-1)^2 + 1} \to 0$, so $f(x) \to 4$.
Thus, the range of $f$ is $R_f = (4, 5]$.

2. Show that the range of $f$ is a subset of the domain of $g$.

The domain of $g$ is given as $D_g = [0, 2\pi]$.
Since $\pi > 3$, we have $2\pi > 6$.
Therefore, $0 < 4$ and $5 < 2\pi$, which means $(4, 5] \subseteq [0, 2\pi]$.
Because $R_f \subseteq D_g$, the composite function $gf$ exists.

1. Find the range of the inner function $f(x)$ for $x \ge 0$.

For $f(x) = 4 + \frac{1}{(x-1)^2 + 1}$, the maximum value occurs at the minimum of the denominator, which is at $x = 1$.
$f(1) = 4 + 1 = 5$.
As $x \to \infty$, the fraction approaches $0$, so $f(x) \to 4$.
Thus, the range of $f$ is $R_f = (4, 5]$.

2. Identify the critical points of the outer function $g(x)$ within the interval $(4, 5]$.

The range of the composite function $gf$ is found by applying $g(x) = \sin x$ to the domain $(4, 5]$.
By drawing a sketch of $y = \sin x$ for $x \in [0, 2\pi]$, we locate the interval $(4, 5]$.
The interval contains $x = \frac{3\pi}{2} \approx 4.71$, which is the minimum turning point of the sine graph.
The minimum value in this interval is $g\left(\frac{3\pi}{2}\right) = -1$.

3. Evaluate the function at the endpoints to determine the supremum.

To find the maximum value, we evaluate $g(x)$ at the limits of the interval:
$g(4) = \sin 4 \approx -0.757$
$g(5) = \sin 5 \approx -0.959$
Since $\sin 4 > \sin 5$, the supremum is approached as $x \to 4^+$.
Because $4$ is not included in $R_f$, $\sin 4$ is excluded from the final range of $gf$.