函数 f 定义如下： $f : x \mapsto x^2 - 3x$，对于 $|x| < a$。 (i) 写出使反函数 $f^{-1}$ 存在的最大 $a$ 值。由此求出 $f^{-1}(x)$，并写出在此 $a$ 值下 $f^{-1}$ 的定义域。 在问题的其余部分中，使用 $a = 1$。 函数 g 在区间 A 上定义如下： g : $x \mapsto \ln(3x+2)$，对于 $x \in A$。 (ii) 一位学生建议了以下 2 个可能的区间 A。

1. Determine the largest value of $a$ for $f$ to be a one-to-one function.

By completing the square, $f(x) = x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$.
The graph of $y = f(x)$ is a parabola with its line of symmetry at $x = \frac{3}{2}$.
For the function to have an inverse, it must be a one-to-one function, meaning its domain $|x| < a$ (or $-a < x < a$) cannot contain points on both sides of the line of symmetry.
The rightmost limit of the interval is $x = a$, which cannot exceed $\frac{3}{2}$.
Therefore, the largest value of $a$ is $\frac{3}{2}$.

2. Find the inverse function $f^{-1}(x)$ by making $x$ the subject.

Let $y = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$.
$\left(x - \frac{3}{2}\right)^2 = y + \frac{9}{4}$
$x - \frac{3}{2} = \pm \sqrt{y + \frac{9}{4}}$
$x = \frac{3}{2} \pm \sqrt{y + \frac{9}{4}}$
Since the domain is $-\frac{3}{2} < x < \frac{3}{2}$, we have $x < \frac{3}{2}$, so we must reject the positive root.
$x = \frac{3}{2} - \sqrt{y + \frac{9}{4}}$
Therefore, $f^{-1}(x) = \frac{3}{2} - \sqrt{x + \frac{9}{4}}$.

3. Determine the domain of $f^{-1}$ by evaluating the range of $f$.

The domain of $f^{-1}$ is exactly the range of $f$ for $x \in \left(-\frac{3}{2}, \frac{3}{2}\right)$.
Since $f$ is strictly decreasing on this interval:
The lower bound is as $x \to \frac{3}{2}$: $f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) = -\frac{9}{4}$.
The upper bound is as $x \to -\frac{3}{2}$: $f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) = \frac{9}{4} + \frac{9}{2} = \frac{27}{4}$.
Thus, the range of $f$ is $-\frac{9}{4} < y < \frac{27}{4}$.
The domain of $f^{-1}$ is $-\frac{9}{4} < x < \frac{27}{4}$.

1. State the domain of $f$ and the condition for $fg$ to exist.

The domain of $f$ is given by $|x| < 1$, so $D_f = (-1, 1)$.
For the composite function $fg$ to exist, the range of $g$ must be a subset of the domain of $f$, i.e., $R_g \subseteq D_f$.

2. Determine if interval (a) satisfies the condition.

For interval (a), $A = \left(-\dfrac{2}{3}, -\dfrac{1}{3}\right)$.
As $x \to -\dfrac{2}{3}^+$, $3x+2 \to 0^+$, so $g(x) \to -\infty$.
At $x = -\dfrac{1}{3}$, $g\left(-\dfrac{1}{3}\right) = \ln(1) = 0$.
Since $g(x) = \ln(3x+2)$ is strictly increasing, the range of $g$ is $R_g = (-\infty, 0)$.
Because $(-\infty, 0) \not\subseteq (-1, 1)$, $fg$ does not exist for interval (a).

3. Determine if interval (b) satisfies the condition.

For interval (b), $A = \left(-\dfrac{1}{2}, 0\right)$.
At $x = -\dfrac{1}{2}$, $g\left(-\dfrac{1}{2}\right) = \ln\left(\dfrac{1}{2}\right) = -\ln 2$.
At $x = 0$, $g(0) = \ln 2$.
Thus, $R_g = (-\ln 2, \ln 2)$.
Since $e > 2$, we know $\ln 2 < 1$ (approx. $0.693 < 1$).
This means $(-\ln 2, \ln 2) \subseteq (-1, 1)$, satisfying $R_g \subseteq D_f$.
Therefore, $fg$ exists for interval (b).

4. Find the range of the composite function $fg$.

To find the range of $fg$, we evaluate $f(u) = u^2 - 3u$ over the domain $u \in R_g = (-\ln 2, \ln 2)$.
The turning point of the quadratic function $f(u)$ is at $u = \dfrac{3}{2}$.
Since the interval $(-\ln 2, \ln 2)$ lies entirely to the left of the turning point ($u < 1.5$), $f(u)$ is strictly decreasing on this interval.
Lower bound: $f(\ln 2) = (\ln 2)^2 - 3\ln 2$.
Upper bound: $f(-\ln 2) = (-\ln 2)^2 - 3(-\ln 2) = (\ln 2)^2 + 3\ln 2$.
Therefore, the range of $fg$ is $\left( (\ln 2)^2 - 3\ln 2, (\ln 2)^2 + 3\ln 2 \right)$.