将方程为 $y=f(x)$ 的曲线先关于 y 轴进行反射，接着沿 x 轴负方向平移 1 个单位，最后在平行于 y 轴的方向上进行拉伸（缩放因子为 2）。所得曲线的方程由 $y=g(x)$ 给出，其中 $g(x)=\dfrac{1}{\sqrt{4-x^{2}}}\;,\quad -2<x<0$。

1. Relate $g(x)$ to $f(x)$ by applying the given sequence of transformations.

Let the original curve be $y = f(x)$.

1. Reflection in the $y$-axis: Replace $x$ with $-x$, giving $y = f(-x)$.
2. Translation of $1$ unit in the negative $x$-axis (left shift by 1): Replace $x$ with $x+1$, giving $y = f(-(x+1)) = f(-x-1)$.
3. Scaling with factor 2 parallel to the $y$-axis: Replace $y$ with $\frac{y}{2}$ (or simply multiply the function by $2$), giving $y = 2f(-x-1)$.

Thus, the resulting function is $g(x) = 2f(-x-1)$.

2. Use the given equation for $g(x)$ to find the algebraic rule for $f(x)$.

We are given $g(x) = \dfrac{1}{\sqrt{4-x^2}}$.
Substitute this into our transformation relationship:
$2f(-x-1) = \dfrac{1}{\sqrt{4-x^2}}$
$f(-x-1) = \dfrac{1}{2\sqrt{4-x^2}}$
To find $f(x)$, we perform a change of variable. Let $u = -x-1$, which rearranges to $x = -u-1$.
Substituting $x$ gives:
$f(u) = \dfrac{1}{2\sqrt{4-(-u-1)^2}}$
$f(u) = \dfrac{1}{2\sqrt{4-(u+1)^2}}$

3. Determine the original domain of $f(x)$ corresponding to the transformed domain of $g(x)$.

The domain of $g(x)$ is given as $-2 < x < 0$.
Apply the same variable substitution $x = -u-1$ to the domain inequalities:
$-2 < -u-1 < 0$
Add $1$ to all parts:
$-1 < -u < 1$
Multiply by $-1$ (and flip the inequality signs):
$1 > u > -1 \implies -1 < u < 1$

Replacing $u$ with $x$, we obtain the fully defined original function:
$f(x) = \dfrac{1}{2\sqrt{4-(x+1)^2}} \quad \text{for } -1 < x < 1$

1. Determine the endpoints, asymptotes, and shape of $y = g(x)$.

Given $g(x) = \dfrac{1}{\sqrt{4-x^2}}$ for $-2 < x < 0$:
As $x \to -2^+$, the denominator approaches $0$ and $g(x) \to \infty$, giving a vertical asymptote at $x = -2$.
As $x \to 0^-$, $g(x) \to \dfrac{1}{2}$, giving an open endpoint at $(0, \dfrac{1}{2})$.
For $-2 < x < 0$, $g(x) > 0$ and the graph is strictly decreasing, lying entirely in the second quadrant.

2. Determine the corresponding features of the inverse function $y = g^{-1}(x)$.

The graph of $y = g^{-1}(x)$ is the reflection of $y = g(x)$ across the line $y = x$.

• The vertical asymptote $x = -2$ reflects to a horizontal asymptote $y = -2$.
• The open endpoint $(0, \dfrac{1}{2})$ reflects to an open endpoint at $(\dfrac{1}{2}, 0)$.
• The resulting graph lies entirely in the fourth quadrant and is strictly decreasing.

3. Sketch both graphs on the same diagram.

1. Draw and label the axes and the reference line $y = x$.
2. Sketch $y = g(x)$ in the second quadrant, labeling the vertical asymptote $x = -2$ and the open circle at $(0, \dfrac{1}{2})$.
3. Sketch $y = g^{-1}(x)$ in the fourth quadrant, labeling the horizontal asymptote $y = -2$ and the open circle at $(\dfrac{1}{2}, 0)$.
   Both curves should exhibit clear symmetry about the line $y = x$.

1. Find the $x$-coordinate on $g(x)$ that corresponds to the boundary $x=1$ on $g^{-1}(x)$.

Let $g(x) = 1$.
$\frac{1}{\sqrt{4-x^2}} = 1$
$4 - x^2 = 1 \implies x^2 = 3$
Since the domain of $g$ is $-2 < x < 0$, we have $x = -\sqrt{3}$.
Additionally, as $x \to 0^-$, $g(x) \to \frac{1}{2}$, which indicates that the $x$-axis boundary ($y=0$ for $g^{-1}(x)$) corresponds to the $y$-axis asymptote boundary ($x=0$ for $g(x)$).

2. Relate the required area under $g^{-1}(x)$ to an area involving $g(x)$ using symmetry.

By the symmetry of inverse functions about the line $y = x$, the region bounded by $y = g^{-1}(x)$, the $x$-axis ($y=0$), and $x = 1$ is geometrically identical to the region bounded by $x = g^{-1}(y)$ (which is $y = g(x)$), the $y$-axis ($x=0$), and $y = 1$.

From the graph of $y=g(x)$, this is the area enclosed by the rectangle (from $x = -\sqrt{3}$ to $x = 0$ and $y=0$ to $y = 1$) minus the area under the curve $y = g(x)$ over the same $x$-interval:
$\text{Required Area} = \left( 1 \times \sqrt{3} \right) - \int_{-\sqrt{3}}^{0} g(x) \,dx$
$= \sqrt{3} - \int_{-\sqrt{3}}^{0} \frac{1}{\sqrt{4-x^2}} \,dx$

3. Evaluate the definite integral of $g(x)$.

Using the standard formula $\int \frac{1}{\sqrt{a^2-x^2}} \,dx = \arcsin\left(\frac{x}{a}\right)$:
$\int_{-\sqrt{3}}^{0} \frac{1}{\sqrt{4-x^2}} \,dx = \left[ \arcsin\left(\frac{x}{2}\right) \right]_{-\sqrt{3}}^{0}$
$= \arcsin(0) - \arcsin\left(-\frac{\sqrt{3}}{2}\right)$
$= 0 - \left(-\frac{\pi}{3}\right) = \frac{\pi}{3}$

4. Calculate the exact final area.

Substitute the evaluated integral back into the area equation:
$\text{Required Area} = \sqrt{3} - \frac{\pi}{3}$