函数 f 定义为 $f: x \mapsto \dfrac{1}{1 - x^2}$，$-1 < x \le 0$。 (i) 定义 $f^{-1}$。 (ii) 在同一坐标系中画出 $y = ff^{-1}(x)$ 和 $y = f^{-1}f(x)$ 的图像简图。由此解方程 $ff^{-1}(x) = f^{-1}f(x)。$ 图 1 显示了图像 $y = \tfrac{1}{2} + e^{-x}$，$x \ge 0$，该图像经历了如下所示的一系列两次几何变换。$P_1$ 和 $P_2$ 是点 $P$ 在每次变换后对应的点。

1. Find the equation of the curve in Figure 2

The curve in Figure 2 has horizontal asymptote $y = -\tfrac{1}{2}$ and approaches it from below, so it has the form $y = -\tfrac{1}{2} - A e^{-x}$ (the $x\ge 0$ part is preserved by a vertical transformation).

Using $P_1\left(1,\ -\tfrac{1}{2}-\tfrac{1}{2e}\right)$:
$-\tfrac{1}{2} - A e^{-1} = -\tfrac{1}{2} - \tfrac{1}{2e} \ \Rightarrow\ A = \tfrac{1}{2}.$
So the Figure 2 curve is $y = -\tfrac{1}{2} - \tfrac{1}{2}e^{-x}$, $x \ge 0$.

2. Apply the horizontal translation (Figure 2 to Figure 3) to get h(x)

From $P_1(1,\cdot)$ to $P_2(0,\cdot)$ the $x$-coordinate decreases by $1$ with the $y$-value unchanged, so Figure 3 is Figure 2 translated $1$ unit in the negative $x$-direction. Replace $x$ by $x+1$:
$h(x) = -\tfrac{1}{2} - \tfrac{1}{2}e^{-(x+1)}.$
The domain $x \ge 0$ shifts to $x \ge -1$.

Check: $h(0) = -\tfrac{1}{2} - \tfrac{1}{2}e^{-1} = -\tfrac{1}{2} - \tfrac{1}{2e}$, matching $P_2$.

1. State the range of $h$ and the domain of $f$.

From part (iii), the function is $h(x) = -\frac{1}{2} - \frac{1}{2} e^{-(x+1)}$ for $x \ge -1$.
The minimum value occurs at $x = -1$ where $h(-1) = -1$, and the horizontal asymptote is $y = -\frac{1}{2}$.
Thus, the range of $h$ is $\mathrm{R}_h = \left[-1, -\frac{1}{2}\right)$.
From the question, the domain of $f$ is $\mathrm{D}_f = (-1, 0]$.

2. Check the condition for the existence of $fh$ and conclude.

For the composite function $fh$ to exist, we must have $\mathrm{R}_h \subseteq \mathrm{D}_f$.
Notice that $-1 \in \mathrm{R}_h$ (since $h(-1) = -1$), but $-1 \notin \mathrm{D}_f$ because the interval is strictly greater than $-1$.
Since $\mathrm{R}_h \not\subseteq \mathrm{D}_f$, the composite function $fh$ does not exist.

1. State the condition for the existence of the composite function $\mathrm{fh}$.

For the composite function $\mathrm{fh}$ to exist, the range of the inner function $h$ must be a subset of the domain of the outer function $f$.
We are given $D_f = (-1, 0]$.
Thus, we require $h(x) \in (-1, 0]$, which means $-1 < h(x) \le 0$.

2. Solve the inequality to find the maximal domain of $h$.

From part (iii), the function is $h(x) = -\frac{1}{2} - \frac{1}{2}e^{-(x+1)}$.
Substitute $h(x)$ into the inequality:
$-1 < -\frac{1}{2} - \frac{1}{2}e^{-(x+1)} \le 0$
Multiply throughout by $-2$ (which flips the inequalities):
$2 > 1 + e^{-(x+1)} \ge 0$
Subtract 1:
$1 > e^{-(x+1)} \ge -1$
Since $e^{-(x+1)} > 0$ for all real $x$, the right side of the inequality ($e^{-(x+1)} \ge -1$) is always satisfied.
Solving the left inequality:
$e^{-(x+1)} < 1$
$-(x+1) < 0$
$x > -1$
Thus, the maximal domain of $h$ for $\mathrm{fh}$ to exist is $x > -1$.

3. Find the restricted range of $h$ using the maximal domain.

We evaluate the range of $h(x) = -\frac{1}{2}\left(1 + e^{-(x+1)}\right)$ for $x > -1$.
As $x$ increases from $-1$ to $\infty$, the term $e^{-(x+1)}$ strictly decreases from $1$ to $0$.
Therefore, $h(x)$ strictly increases over this interval.
Lower limit (as $x \to -1^+$): $h(x) \to -\frac{1}{2}(1 + e^0) = -1$.
Upper limit (as $x \to \infty$): $h(x) \to -\frac{1}{2}(1 + 0) = -\frac{1}{2}$.
Hence, the new restricted range of $h$ is $R_h = \left(-1, -\frac{1}{2}\right)$.

4. Determine the range of the composite function $\mathrm{fh}$.

The range of $\mathrm{fh}$ is the set of values $f(y)$ where $y \in R_h = \left(-1, -\frac{1}{2}\right)$.
Given $f(y) = \frac{1}{1 - y^2}$:
As $y \to -1^+$, $y^2 \to 1^-$, so $1 - y^2 \to 0^+$, which gives $f(y) \to \infty$.
At the upper bound $y \to -\frac{1}{2}^-$, $y^2 \to \frac{1}{4}^+$, so $1 - y^2 \to \frac{3}{4}^-$.
Thus, $f(y) \to \frac{1}{3/4} = \frac{4}{3}$.
Since $f$ is strictly decreasing for negative values of $x$ (as $y^2$ is strictly decreasing), $f(y)$ takes all values between $\frac{4}{3}$ and $\infty$.
The range of $\mathrm{fh}$ is therefore $\left(\frac{4}{3}, \infty \right)$.