The functions f and g are defined below. $f(x)=\begin{cases} \dfrac{2}{2-x} & \text{for } 0\le x<2, \\ \ln(x-2) & \text{for } 2<x\le 4. \end{cases}$ $g(x) = -2x^{2} + x + 1 \quad \text{for } 0\le x<1.$

1. Determine the endpoints and intercepts for both parts of the piecewise function.

For $0 \le x < 2$, $f(x) = \dfrac{2}{2-x}$.
When $x = 0$, $y = \dfrac{2}{2} = 1$. The starting endpoint is $(0, 1)$ (inclusive).

For $2 < x \le 4$, $f(x) = \ln(x-2)$.
Let $y = 0 \implies \ln(x-2) = 0 \implies x - 2 = 1 \implies x = 3$. The $x$-intercept is $(3, 0)$.
When $x = 4$, $y = \ln(4-2) = \ln 2$. The terminating endpoint is $(4, \ln 2)$ (inclusive).

2. Identify the behavior near the common boundary $x = 2$ to find asymptotes.

As $x \to 2^-$, $(2-x) \to 0^+$, so $f(x) = \dfrac{2}{2-x} \to \infty$.
As $x \to 2^+$, $(x-2) \to 0^+$, so $f(x) = \ln(x-2) \to -\infty$.
Thus, the line $x = 2$ is a vertical asymptote for both branches.

3. Combine the analysis into a single sketch.

• Draw the axes and label the vertical asymptote $x = 2$.
• For the domain $0 \le x < 2$, draw a curve from the solid dot $(0, 1)$ going upwards towards the asymptote.
• For the domain $2 < x \le 4$, draw a curve coming up from alongside the vertical asymptote, passing through $(3, 0)$, and finishing at the solid dot $(4, \ln 2)$.

1. Find the range of $g$ by finding its turning point and checking the domain endpoints.

To find the turning point of $g(x) = -2x^2 + x + 1$, we can complete the square or differentiate:
$g'(x) = -4x + 1$
Setting $g'(x) = 0$ gives $x = \frac{1}{4}$. Since the coefficient of $x^2$ is $-2$ (which is negative), this is a maximum point.
The maximum value is:
$g\left(\frac{1}{4}\right) = -2\left(\frac{1}{4}\right)^2 + \frac{1}{4} + 1 = \frac{9}{8}$
Next, evaluate $g(x)$ at the boundaries of its domain $0 \le x < 1$:
$g(0) = -2(0)^2 + 0 + 1 = 1$
As $x \to 1$, $g(x) \to -2(1)^2 + 1 + 1 = 0$.
Since the domain is strictly $x < 1$, the minimum is strictly greater than $0$.
Therefore, the range of $g$ is $0 < g(x) \le \frac{9}{8}$, or $\mathrm{R}_g = (0, \frac{9}{8}]$.

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

From the definition of $f$, its domain $\mathrm{D}_f$ is $0 \le x < 2$ and $2 < x \le 4$, meaning $\mathrm{D}_f = [0, 2) \cup (2, 4]$.
For the composite function $fg$ to exist, we require the range of the inner function $g$ to be a subset of the domain of the outer function $f$, i.e. $\mathrm{R}_g \subseteq \mathrm{D}_f$.
We have $\mathrm{R}_g = (0, \frac{9}{8}]$.
Since $(0, \frac{9}{8}] \subseteq [0, 2) \cup (2, 4]$, the condition $\mathrm{R}_g \subseteq \mathrm{D}_f$ is satisfied.
Hence, the composite function $fg$ exists.

1. Identify the range of $g$ from part (ii) and select the appropriate definition of $f(x)$.

From part (ii), the range of $g$ is $R_g = (0, \frac{9}{8}]$.
Since $(0, \frac{9}{8}] \subset [0, 2)$, the relevant piece of $f$ to apply is $f(x) = \frac{2}{2-x}$.

2. Find the range of $f$ over the interval $(0, \frac{9}{8}]$.

The function $y = \frac{2}{2-x}$ is strictly increasing for $x < 2$.
Evaluate $f(x)$ at the boundaries of the interval $(0, \frac{9}{8}]$:
As $x \to 0^+$, $f(x) \to \frac{2}{2-0} = 1$.
At $x = \frac{9}{8}$, $f\left(\frac{9}{8}\right) = \frac{2}{2-\frac{9}{8}} = \frac{2}{\frac{7}{8}} = \frac{16}{7}$.
Therefore, the range of $fg$ is $(1, \frac{16}{7}]$.