函数 $g$ 定义为 $g(x)=\begin{cases} \displaystyle\frac{2}{1+x^2} &\text{if } x\le 0,\\[4pt] 0 &\text{if } x>0. \end{cases}$ 在本题的其余部分中，$g$ 的定义域为 $x \le k$，其中 $k$ 取在第 (b) 问中确定的值。

1. State the reason based on the definition of a one-to-one function

For $x > 0$, $g(x) = 0$. This means multiple values of $x$ map to the same value $y=0$. Hence, $g$ is a many-to-one function. For a function to have an inverse, it must be a one-to-one function.

1. Identify the maximum domain for which the function is one-to-one

For $g^{-1}$ to exist, the function $g$ must be one-to-one.
From the given definition and graph, $g(x) = 0$ for all $x > 0$. Thus, any interval that includes values of $x > 0$ will contain multiple $x$-values mapping to the same $y$-value (many-to-one).
However, for $x \le 0$, $g(x) = \frac{2}{1+x^2}$ is strictly increasing and therefore one-to-one.
Thus, the restricted domain $x \le k$ can at most extend up to $0$, meaning the greatest possible value is:
$k = 0$

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

Let $y = \frac{2}{1+x^2}$.
$1+x^2 = \frac{2}{y}$
$x^2 = \frac{2}{y} - 1$
$x^2 = \frac{2-y}{y}$
$x = \pm\sqrt{\frac{2-y}{y}}$

2. Choose the correct sign for $x$ based on the restricted domain of $g$.

From part (b), the restricted domain of $g$ for an inverse to exist is $x \le 0$ (so $k=0$).
Since $x \le 0$, we must reject the positive root and take the negative root:
$x = -\sqrt{\frac{2-y}{y}}$
Therefore, $g^{-1}(x) = -\sqrt{\frac{2-x}{x}}$.

3. Determine the domain of $g^{-1}$ by finding the range of $g$ over its restricted domain.

The domain of $g^{-1}$ is the range of $g$ for $x \le 0$.
For $x \le 0$, $x^2 \ge 0$, which implies $1+x^2 \ge 1$.
Taking the reciprocal and multiplying by 2 yields:
$0 < \frac{2}{1+x^2} \le 2$
The range of $g$ is $(0, 2]$.
Thus, the domain of $g^{-1}$ is $0 < x \le 2$.

1. Identify and describe the graph of $y = g(x)$ for its restricted domain.

From part (b), the restricted domain for $g(x)$ to have an inverse is $x \le 0$.
Draw the graph of $y = g(x)$ strictly for $x \le 0$. This is a curve entirely in the second quadrant (and positive $y$-axis) that starts at the $y$-intercept $(0, 2)$ with a solid dot, and continuously decreases towards the left, approaching the negative $x$-axis as a horizontal asymptote ($y = 0$ as $x \to -\infty$). Label this curve as $y = g(x)$.

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

The graph of an inverse function is the reflection of the original function's graph across the line $y = x$.
Reflect the curve from Step 1. The new curve will lie entirely in the fourth quadrant (and positive $x$-axis). It starts at the $x$-intercept $(2, 0)$ with a solid dot, and continuously decreases downwards, approaching the negative $y$-axis as a vertical asymptote ($x = 0$ as $y \to -\infty$). Label this curve as $y = g^{-1}(x)$.

3. State the equation of the line of reflection.

By the geometric property of inverse functions, the reflection axis that maps $y = g(x)$ onto $y = g^{-1}(x)$ is the line $y = x$.