The department of statistics of a country has developed two mathematical functions to analyse the foreign worker policy. The first function $f$ models the amount of strain to the country's infrastructure (housing, transportation, utilities and access to medical care etc.) based on the number of foreign workers allowed into the country and is defined as follows: $f(x) = (x - 5)^3 + 200, \quad 0 \le x \le 15$ where $x$ denotes the number of foreign workers, in ten thousands, allowed into the country and $f(x)$ denotes the amount strain to the country's infrastructure. The second function $g$ models the happiness index, from 0 (least happy) to 1 (most happy), of the country's local population based on the amount of strain to the country's infrastructure and is defined as follows: $g(w) = \ln\left(e - \frac{w}{1000}\right), \quad 0 \le w \le 1000(e - 1)$ where $w$ denotes the amount of strain to the country's infrastructure and $g(w)$ denotes the happiness index of the country's local population. A third function h models the gross domestic product (GDP) of the country based on the number of foreign workers (in ten thousands), x, allowed into the country and is defined as follows: $h(x) = 400 - (x - 10)^2, \quad 0 \le x \le 15$ where h(x) denotes the GDP in billions of dollars.

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.

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