函数 f 和 g 定义为： $f : x \rightsquigarrow \left\lfloor\dfrac{2x+2\lambda}{x-2}\right\rfloor, \quad x \in \mathbb{R}, \ -\lambda < x < 2$ $g : x \rightsquigarrow x^2 - \lambda , \quad x \in \mathbb{R}, \ -\sqrt{\lambda} < x < \sqrt{\lambda}$ 其中 $\lambda$ 是正常数。

1. Differentiate $f(x)$ to determine its monotonicity.

$f(x) = \frac{2x+2\lambda}{x-2}$
Using the quotient rule:
$f'(x) = \frac{2(x-2) - (2x+2\lambda)(1)}{(x-2)^2} = \frac{-4 - 2\lambda}{(x-2)^2}$
Since $\lambda > 0$, the numerator $-4 - 2\lambda$ is strictly negative. For $x \in (-\lambda, 2)$, the denominator $(x-2)^2$ is strictly positive.
Thus, $f'(x) < 0$ for all $x$ in the domain.

2. Conclude that $f$ is one-to-one and has an inverse.

Since $f'(x) < 0$ for all valid $x$, $f(x)$ is a strictly decreasing function.
This implies that $f$ is a one-to-one function.
Therefore, the inverse function $f^{-1}$ exists.

3. Form an equation to find the inverse by making $x$ the subject.

Let $y = f(x)$.
$y = \frac{2x+2\lambda}{x-2}$
$y(x-2) = 2x+2\lambda$
$xy - 2y = 2x+2\lambda$
$xy - 2x = 2y+2\lambda$
$x(y - 2) = 2y+2\lambda$
$x = \frac{2y+2\lambda}{y-2}$

4. State the expression for $f^{-1}(x)$.

Replacing $y$ with $x$, we obtain the expression for the inverse:
$f^{-1}(x) = \frac{2x+2\lambda}{x-2}$

1. Establish the condition for the equality to hold based on domains.

By definition, $f^{-1}f(x) = x$ for all $x$ in the domain of $f$, and $ff^{-1}(x) = x$ for all $x$ in the domain of $f^{-1}$.
For the equality $f^{-1}f(x) = ff^{-1}(x)$ to hold, the expressions must both be defined. Thus, $x$ must lie in the domain of both composite functions.
This requires $x \in D_f$ and $x \in D_{f^{-1}}$, meaning we must find $D_f \cap D_{f^{-1}}$.

2. Find the domain of the inverse function by finding the range of $f$.

The domain of $f^{-1}$ is exactly the range of $f$ ($R_f$).
Consider $f(x) = \frac{2x+2\lambda}{x-2}$ on the given domain $D_f = (-\lambda, 2)$:
At the lower bound $x \to -\lambda$, $f(x) \to \frac{2(-\lambda)+2\lambda}{-\lambda-2} = 0$.
As $x \to 2^-$, the numerator approaches $4+2\lambda > 0$ (since $\lambda > 0$) and the denominator approaches $0^-$, so $f(x) \to -\infty$.
Since $f$ is continuous and strictly decreasing on this interval (its derivative is $\frac{-4-2\lambda}{(x-2)^2} < 0$), its range is $R_f = (-\infty, 0)$.
Hence, $D_{f^{-1}} = (-\infty, 0)$.

3. Find the intersection of the two domains.

Intersecting $D_f$ and $D_{f^{-1}}$:
$D_f \cap D_{f^{-1}} = (-\lambda, 2) \cap (-\infty, 0)$
Since $\lambda$ is positive, $-\lambda < 0$, yielding the overlapping interval $(-\lambda, 0)$.

1. Determine the range of $g$ over its initial domain to test the condition for the existence of $fg$.

For $x \in (-\sqrt{\lambda}, \sqrt{\lambda})$, we have $0 \le x^2 < \lambda$.
Subtracting $\lambda$ gives $-\lambda \le x^2 - \lambda < 0$.
Thus, the range of $g$ is $R_g = [-\lambda, 0)$.
The domain of $f$ is given as $D_f = (-\lambda, 2)$.

2. Explain why $fg$ does not exist by comparing $R_g$ and $D_f$.

For $fg$ to exist, the range of $g$ must be a subset of the domain of $f$ ($R_g \subseteq D_f$).
We observe that $-\lambda \in R_g$ (when $x=0$), but $-\lambda \notin D_f$ (since the interval is strictly greater than $-\lambda$).
Since $R_g \not\subseteq D_f$, the composite function $fg$ does not exist.

3. Find the rule for $fg$ using the restricted domain of $g$.

With $D_{g_{\text{new}}} = (0, \sqrt{\lambda})$, the new range of $g$ is $(-\lambda, 0)$. Since $(-\lambda, 0) \subseteq (-\lambda, 2)$, $fg$ now exists.
$fg(x) = f(g(x))$
$= f(x^2 - \lambda)$
$= \dfrac{2(x^2 - \lambda) + 2\lambda}{(x^2 - \lambda) - 2}$
$= \dfrac{2x^2}{x^2 - \lambda - 2}$

4. Determine the range of $fg$ by finding the image of the restricted domain under $f$.

The range of $fg$ is the set of values of $f(y)$ for $y \in (-\lambda, 0)$.
Consider $f(y) = \dfrac{2y + 2\lambda}{y - 2} = 2 + \dfrac{2\lambda + 4}{y - 2}$.
For $y \in (-\lambda, 0)$, the denominator $(y - 2)$ is negative. As $y$ increases from $-\lambda$ to $0$, $(y - 2)$ increases (becomes less negative). Thus, $\dfrac{1}{y-2}$ decreases.
Since $\lambda > 0$, $2\lambda + 4 > 0$, meaning $f(y)$ strictly decreases on this interval.
As $y \to -\lambda^+$, $f(y) \to \dfrac{0}{-\lambda - 2} = 0$.
As $y \to 0^-$, $f(y) \to \dfrac{2\lambda}{-2} = -\lambda$.
Therefore, the range of $fg$ is $(-\lambda, 0)$.