函数 $f$ 由下式给出： $f : x \mapsto a + \dfrac{3a}{(x+3)(x-1)}$，其中 $a>0$。

1. Find the first derivative of $f(x)$ with respect to $x$.

$f(x) = a + 3a(x^2 + 2x - 3)^{-1}$

Using the chain rule:
$f'(x) = -3a(x^2 + 2x - 3)^{-2}(2x + 2)$
$f'(x) = -\frac{6a(x + 1)}{(x^2 + 2x - 3)^2}$

2. Equate the derivative to zero to find the $x$-coordinate of the stationary point.

At a stationary point, $f'(x) = 0$.
$-\frac{6a(x + 1)}{(x^2 + 2x - 3)^2} = 0$

Since $a > 0$, the denominator cannot be infinite and the fraction equals zero only if the numerator is zero:
$-6a(x + 1) = 0$
$x + 1 = 0$
$x = -1$

1. Find the coordinates of the stationary point using the x-coordinate from part (a).

From part (a), the $x$-coordinate of the stationary point is $x = -1$.
Substitute $x = -1$ into $f(x)$ to find the $y$-coordinate:
$y = a + \frac{3a}{(-1+3)(-1-1)} = a + \frac{3a}{(2)(-2)} = a - \frac{3}{4}a = \frac{1}{4}a$.
Thus, the stationary point is $\left(-1, \frac{1}{4}a\right)$.

2. Determine the equations of the vertical and horizontal asymptotes.

Vertical asymptotes occur where the denominator is zero:
$(x+3)(x-1) = 0 \implies x = -3 \text{ and } x = 1$.

The horizontal asymptote is found by examining the behavior as $x \to \pm\infty$:
$\lim_{x \to \pm\infty} \left(a + \frac{3a}{(x+3)(x-1)}\right) = a + 0 = a$.
Hence, the asymptotes are $x = -3$, $x = 1$, and $y = a$.

3. Find the axial intercepts to guide the shape of the sketch.

Find the $y$-intercept by substituting $x = 0$:
$y = a + \frac{3a}{(3)(-1)} = a - a = 0$.

Find the $x$-intercepts by setting $y = 0$:
$a\left(1 + \frac{3}{x^2+2x-3}\right) = 0$
$\frac{x^2+2x-3+3}{x^2+2x-3} = 0 \implies \frac{x^2+2x}{(x+3)(x-1)} = 0$
$x(x+2) = 0 \implies x = 0 \text{ or } x = -2$.
The graph intercepts the axes at $(0,0)$ and $(-2,0)$.

4. Sketch the graph of y = f(x).

Draw and label the horizontal line $y=a$ and vertical lines $x=-3$ and $x=1$ with dashed lines.
Plot the maximum stationary point at $\left(-1, \frac{1}{4}a\right)$ and the intercepts at $(0,0)$ and $(-2,0)$.

The graph has three distinct branches:

• For $x < -3$: A curve strictly above $y=a$, decreasing from $+\infty$ towards the horizontal asymptote $y=a$ as $x \to -\infty$.
• For $-3 < x < 1$: An inverted U-shape starting from $-\infty$ near $x=-3$, crossing at $(-2,0)$, peaking at $\left(-1, \frac{1}{4}a\right)$, crossing at $(0,0)$, and heading to $-\infty$ near $x=1$.
• For $x > 1$: A curve strictly above $y=a$, decreasing from $+\infty$ near $x=1$ towards $y=a$ as $x \to +\infty$.

1. Identify the interval from the properties of the graph sketched in the previous part

Based on the graph from part (b):

• The graph is increasing ($f'(x) > 0$) for $x < -1$ (where $x \neq -3$).
• The graph is concave downwards (upside-down bowl shape, $f''(x) < 0$) for $-3 < x < 1$.

The intersection of these two conditions is the interval between the vertical asymptote $x = -3$ and the maximum stationary point at $x = -1$.

Therefore, the range is $-3 < x < -1$.

1. State the expression for $f'(x)$ and determine its asymptotes.

From part (a), the first derivative is:
$f'(x) = -\frac{6a(x+1)}{(x+3)^2(x-1)^2}$
The denominator is zero when $x = -3$ or $x = 1$. Thus, the equations of the vertical asymptotes are $x = -3$ and $x = 1$.

As $x \to \pm\infty$, the highest power of $x$ in the denominator is $x^4$ while in the numerator it is $x$. Therefore, $f'(x) \to 0$. The horizontal asymptote is $y = 0$.

2. Find the coordinates of the points where the graph crosses the axes.

To find the $x$-intercept, set $f'(x) = 0$:
$-6a(x+1) = 0 \implies x = -1$
The graph crosses the $x$-axis at $(-1, 0)$.

To find the $y$-intercept, set $x = 0$:
$f'(0) = -\frac{6a(0+1)}{(0+3)^2(0-1)^2} = -\frac{6a}{(9)(1)} = -\frac{2a}{3}$
The graph crosses the $y$-axis at $\left(0, -\frac{2a}{3}\right)$.

3. Analyse the behaviour of $f'(x)$ near the asymptotes to deduce the shape of the graph.

Since the denominator $(x+3)^2(x-1)^2$ is strictly positive for $x \neq -3, 1$ and $a>0$, the sign of $f'(x)$ depends exclusively on $-6a(x+1)$.

• For $x < -1$ ($x \neq -3$), $f'(x) > 0$. Thus, as $x \to -3$ from both the left and the right, $f'(x) \to +\infty$.
• For $x > -1$ ($x \neq 1$), $f'(x) < 0$. Thus, as $x \to 1$ from both the left and the right, $f'(x) \to -\infty$.

With these limits, intercepts, and the $y=0$ horizontal asymptote, the 3 branches of the sketch can be accurately drawn.

1. Determine the range of the function $g$.

For $x \ge 2$, the denominator $(x+3)(x-1)$ is strictly positive and increasing.
Since $a>0$, $g(x) = a + \dfrac{3a}{(x+3)(x-1)}$ is a strictly decreasing function.
At the minimum value of the domain $x=2$, the function attains its maximum value:
$g(2) = a + \dfrac{3a}{(2+3)(2-1)} = a + \dfrac{3a}{5} = \dfrac{8a}{5}$
As $x \to \infty$, the term $\dfrac{3a}{(x+3)(x-1)} \to 0$, so $g(x) \to a$.
Thus, the range of $g$ is $R_g = \left( a, \dfrac{8a}{5} \right]$.

2. Apply the condition for the existence of a composite function to find the condition on $a$.

For the composite function $gg$ to exist, the reason/condition is that the range of the inner function $g$ must be a subset of the domain of the outer function $g$ ($D_g = [2, \infty)$).
$R_g \subseteq D_g \implies \left( a, \dfrac{8a}{5} \right] \subseteq [2, \infty)$
For every element in the interval $\left( a, \dfrac{8a}{5} \right]$ to be greater than or equal to $2$, the infimum (greatest lower bound) of the interval must be at least $2$.
Therefore, we must have:
$a \ge 2$

1. Find the range of the inner function $g(x)$ for $a=5$.

For $a=5$, $g(x) = 5 + \frac{15}{(x+3)(x-1)}$ for $x \ge 2$.
Since $x \ge 2$, the denominator $(x+3)(x-1)$ is positive and increasing, making $g(x)$ a strictly decreasing function.
The maximum value occurs at $x=2$:
$g(2) = 5 + \frac{15}{(5)(1)} = 8$
As $x \to \infty$, $g(x) \to 5$.
Therefore, the range of the inner function $g(x)$ is $R_g = (5, 8]$.
This forms the domain for the outer function in $gg$.

2. Determine the range of $gg(x)$ by evaluating the outer function $g$ over the domain $(5, 8]$.

We now evaluate $g(x)$ for $x \in (5, 8]$.
Since $g$ is strictly decreasing for $x \ge 2$, the minimum value of $gg$ will be at $x=8$ and the supremum will be as $x \to 5^+$.
$g(8) = 5 + \frac{15}{(8+3)(8-1)} = 5 + \frac{15}{77} = \frac{400}{77}$
As $x \to 5^+$,
$g(5) = 5 + \frac{15}{(5+3)(5-1)} = 5 + \frac{15}{32} = \frac{175}{32}$
Thus, the exact range of $gg$ is $[\frac{400}{77}, \frac{175}{32})$.