A function g is defined by $g(x)=\begin{cases} \dfrac{2}{1+x^2} &\text{if } x \le 0, \\ 0 &\text{if } x > 0. \end{cases}$

1. State the requirement for an inverse function and show how $g$ violates it.

For a function to have an inverse, it must be a one-to-one function.
From the definition, $g(x) = 0$ for all $x > 0$.
Because there are multiple values of $x$ (e.g., $x=1$ and $x=2$) that produce the same $y$-value of $0$, $g$ is not a one-to-one function.

1. Determine the requirement for an inverse function to exist

For $g^{-1}$ to exist, $g(x)$ must be a one-to-one function.
From the given definition and graph, $g(x)$ is strictly increasing for $x \le 0$, making it one-to-one on this interval. However, for $x > 0$, $g(x) = 0$, which is a constant and therefore many-to-one.

2. State the greatest value of $k$ based on the domain restriction

To ensure the function remains one-to-one, the domain $x \le k$ cannot include any part of $x > 0$. Therefore, the maximum possible value for $k$ is $0$.

1. Use the value of $k$ from part (b) and set up the equation to make $x^2$ the subject.

From part (b), the greatest possible value of $k$ is $0$, so the domain of $g$ is $x \le 0$.
Let $y = \frac{2}{1+x^2}$.
$1 + x^2 = \frac{2}{y}$
$x^2 = \frac{2}{y} - 1$

2. Take the square root, selecting the negative root due to the restricted domain, to find $g^{-1}(x)$.

Since the domain of $g$ is restricted to $x \le 0$, $x$ must be non-positive, so we take the negative square root:
$x = -\sqrt{\frac{2}{y} - 1}$
Substituting $x$ for $y$ to express the inverse function:
$g^{-1}(x) = -\sqrt{\frac{2}{x} - 1}$

3. Determine the domain of $g^{-1}$ by finding the range of $g$ for $x \le 0$.

The domain of $g^{-1}$ is equal to the range of $g$ for $x \le 0$.
As $x \to -\infty$, $g(x) \to 0$.
At $x = 0$, $g(0) = \frac{2}{1+0^2} = 2$.
The function is continuous and increasing on this restricted domain, so the range of $g$ is $0 < y \le 2$.
Therefore, the domain of $g^{-1}$ is $0 < x \le 2$.

1. Sketch the graph of $y=g(x)$ for the restricted domain.

Based on the restriction $x \le 0$, we draw the portion of the curve $y = \dfrac{2}{1+x^2}$ that lies in the second quadrant and the $y$-axis. It passes through the point $(0, 2)$ and smoothly approaches the negative $x$-axis ($y = 0$) as an asymptote when $x \to -\infty$.

2. Sketch the graph of the inverse function $y=g^{-1}(x)$.

The graph of $y = g^{-1}(x)$ is the reflection of the graph of $y = g(x)$ across the line $y = x$.
Reflecting the coordinates, the point $(0, 2)$ maps to $(2, 0)$, and the horizontal asymptote $y = 0$ (for $x \to -\infty$) maps to the vertical asymptote $x = 0$ (for $y \to -\infty$). We draw this reflected curve in the fourth quadrant.

3. State the equation of the line of reflection.

By the fundamental property of inverse functions, a function and its inverse are symmetrical to each other with respect to the line $y = x$. Therefore, this is the required line of reflection.