某国统计局开发了两个数学函数来分析外籍劳工政策。第一个函数 $f$ 根据允许进入该国的外籍劳工数量，来对该国基础设施（住房、交通、公用事业和医疗服务等）所承受的压力进行建模，其定义如下： $f(x) = (x - 5)^3 + 200, \quad 0 \le x \le 15$ 其中 $x$ 表示允许进入该国的外籍劳工人数（单位：万），而 $f(x)$ 表示该国基础设施承受的压力。 第二个函数 $g$ 根据该国基础设施承受的压力，对该国本地居民的幸福指数进行建模（从 0（最不幸福）到 1（最幸福）），定义如下： $g(w) = \ln\left(e - \frac{w}{1000}\right), \quad 0 \le w \le 1000(e - 1)$ 其中 $w$ 表示该国基础设施承受的压力，而 $g(w)$ 表示该国本地居民的幸福指数。 第三个函数 h 根据允许进入该国的外籍劳工人数（单位：万） x，对该国的国内生产总值（GDP）进行建模，定义如下： $h(x) = 400 - (x - 10)^2, \quad 0 \le x \le 15$ 其中 h(x) 表示以十亿美元为单位的 GDP。

1. Find the range of $f(x)$ for the given domain.

The function $f(x) = (x - 5)^3 + 200$ is strictly increasing on $0 \le x \le 15$ because its derivative $f'(x) = 3(x - 5)^2 \ge 0$.

Minimum value of $f(x)$ is $f(0) = (0 - 5)^3 + 200 = 75$.
Maximum value of $f(x)$ is $f(15) = (15 - 5)^3 + 200 = 1000 + 200 = 1200$.

Thus, the range of $f$ is $\text{R}_f = [75, 1200]$.

2. Compare the range of $f$ with the domain of $g$ to show existence.

The domain of $g$ is given as $\text{D}_g = [0, 1000(e - 1)]$.

We observe that $1000(e - 1) \approx 1718.3$.
Since $0 \le 75$ and $1200 \le 1000(e - 1)$, we have $[75, 1200] \subseteq [0, 1000(e - 1)]$.

Therefore, $\text{R}_f \subseteq \text{D}_g$, which satisfies the condition for the composite function $gf$ to exist.

1. Identify the values of $x$ representing the given number of foreign workers.

Since $x$ is defined as the number of foreign workers in ten thousands:
For $70,000$ workers, $x = \frac{70000}{10000} = 7$.
For $110,000$ workers, $x = \frac{110000}{10000} = 11$.
The interval for $x$ is $7 \le x \le 11$.

2. Calculate the corresponding range of values for the strain on infrastructure, $w = f(x)$.

Since $f'(x) = 3(x-5)^2 \ge 0$, $f(x) = (x - 5)^3 + 200$ is strictly increasing for $x \ge 5$. The minimum and maximum values of $w$ in the interval $7 \le x \le 11$ therefore occur at the endpoints:
When $x = 7$, $w = f(7) = (7 - 5)^3 + 200 = 8 + 200 = 208$.
When $x = 11$, $w = f(11) = (11 - 5)^3 + 200 = 216 + 200 = 416$.
So, the amount of strain $w$ falls in the range $208 \le w \le 416$.

3. Determine the range of the happiness index by evaluating $g(w)$ at these boundaries.

The happiness index is given by $g(w) = \ln\left(e - \frac{w}{1000}\right)$. Because the term $-\frac{w}{1000}$ decreases as $w$ increases, $g(w)$ is a strictly decreasing function.
Evaluating at the endpoints of $w$:
$g(208) = \ln\left(e - 0.208\right) \approx 0.92040$
$g(416) = \ln\left(e - 0.416\right) \approx 0.83390$
The maximum value is $0.920$ and the minimum is $0.834$.
The range of the happiness index is therefore $0.834 \le gf(x) \le 0.920$ (to 3 s.f.).

1. Find the derivative of the composite happiness function with respect to $x$.

Let the happiness index be $H(x) = gf(x) = \ln\left(e - \frac{(x-5)^3 + 200}{1000}\right)$.

Using the chain rule to differentiate with respect to $x$:
$\frac{dH}{dx} = \frac{1}{e - \frac{(x-5)^3 + 200}{1000}} \times \left( -\frac{3(x-5)^2}{1000} \right)$
$\frac{dH}{dx} = -\frac{3(x-5)^2}{1000e - ((x-5)^3 + 200)}$

2. Analyse the sign of the derivative to conclude the behaviour of the function.

For the domain $0 \le x \le 15$:
The numerator term $-3(x-5)^2 \le 0$ because $(x-5)^2 \ge 0$.
The denominator $1000e - ((x-5)^3 + 200) > 0$ because it is proportional to the positive argument of the natural logarithm, $e - \frac{f(x)}{1000} > 0$.

Since the numerator is non-positive and the denominator is positive, we have:
$\frac{dH}{dx} \le 0$
As the first derivative is negative (or zero at $x=5$), the happiness index decreases as $x$ increases.

1. Set up the compound inequality for the target happiness index.

$0.7 \le g(f(x)) \le 0.9$
$0.7 \le \ln\left(e - \frac{f(x)}{1000}\right) \le 0.9$

2. Solve the inequality to find the required range for $f(x)$.

$e^{0.7} \le e - \frac{f(x)}{1000} \le e^{0.9}$
$e^{0.7} - e \le -\frac{f(x)}{1000} \le e^{0.9} - e$
$-1000(e^{0.7} - e) \ge f(x) \ge -1000(e^{0.9} - e)$
$1000(e - e^{0.9}) \le f(x) \le 1000(e - e^{0.7})$
$258.68 \le f(x) \le 704.53$

3. Solve for the corresponding range of $x$ (number of foreign workers).

$258.68 \le (x - 5)^3 + 200 \le 704.53$
$58.68 \le (x - 5)^3 \le 504.53$
$\sqrt[3]{58.68} \le x - 5 \le \sqrt[3]{504.53}$
$3.886 \le x - 5 \le 7.961$
$8.886 \le x \le 12.961$

4. Determine the maximum and minimum values of the GDP function $h(x)$ over this restricted domain.

$h(x) = 400 - (x - 10)^2$
The curve is a downward-opening parabola with a maximum vertex at $x = 10$.
Since $x = 10$ lies within the interval $[8.886, 12.961]$, the maximum GDP is exactly:
$h(10) = 400$
The minimum GDP over this interval occurs at the endpoint furthest from the vertex ($x = 12.961$):
$h(12.961) = 400 - (12.961 - 10)^2$
$h(12.961) = 400 - (2.961)^2 \approx 391.23$

5. State the final range for the GDP.

$391 \le \text{GDP} \le 400$