The following are questions from a mathematics exam. Please solve them.

For $f(x) = 1 - 2x$:
This is a linear function, which is defined for all real numbers.
The domain of $f$ is $\mathbb{R}$ or $(-\infty, \infty)$.

For $g(x) = \frac{1}{x+3}$:
This is a rational function, which is defined when its denominator is not zero.
So, $x+3 \neq 0 \implies x \neq -3$.
The domain of $g$ is $\{x \in \mathbb{R} \mid x \neq -3\}$ or $(-\infty, -3) \cup (-3, \infty)$.

Assume $g(a) = g(b)$.

$$\frac{1}{a+3} = \frac{1}{b+3}$$

Cross-multiply:

$$b+3 = a+3$$

Subtract 3 from both sides:

$$b = a$$

Since $g(a) = g(b)$ implies $a = b$, the function $g(x)$ is injective (or one-to-one).

Let $y = g(x) = \frac{1}{x+3}$.
To find the inverse, swap $x$ and $y$:

$$x = \frac{1}{y+3}$$

Multiply both sides by $(y+3)$:

$$x(y+3) = 1$$ $$xy + 3x = 1$$

Isolate $y$:

$$xy = 1 - 3x$$ $$y = \frac{1 - 3x}{x}$$

So, the inverse function is $g^{-1}(x) = \frac{1 - 3x}{x}$.

Given $f(x) = 1 - 2x$ and $g(x) = \frac{1}{x+3}$.
Substitute $g(x)$ into $f(x)$:

$$f(g(x)) = 1 - 2\left(\frac{1}{x+3}\right)$$ $$f(g(x)) = 1 - \frac{2}{x+3}$$

Combine terms with a common denominator:

$$f(g(x)) = \frac{x+3}{x+3} - \frac{2}{x+3}$$ $$f(g(x)) = \frac{x+3-2}{x+3}$$ $$f(g(x)) = \frac{x+1}{x+3}$$

Set the expressions for $f(g(x))$ and $g^{-1}(x)$ equal:

$$\frac{x+1}{x+3} = \frac{1-3x}{x}$$

Cross-multiply:

$$x(x+1) = (1-3x)(x+3)$$

Expand both sides:

$$x^2 + x = x + 3 - 3x^2 - 9x$$ $$x^2 + x = -3x^2 - 8x + 3$$

Rearrange into a standard quadratic equation $ax^2 + bx + c = 0$:

$$x^2 + 3x^2 + x + 8x - 3 = 0$$ $$4x^2 + 9x - 3 = 0$$

For $4x^2 + 9x - 3 = 0$, we have $a=4$, $b=9$, $c=-3$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

$$x = \frac{-9 \pm \sqrt{9^2 - 4(4)(-3)}}{2(4)}$$ $$x = \frac{-9 \pm \sqrt{81 + 48}}{8}$$ $$x = \frac{-9 \pm \sqrt{129}}{8}$$

The answer is left in surd form.

i. State the domain of each function
The domain of $f$ is $\boxed{\mathbb{R}}$.
The domain of $g$ is $\boxed{\{x \in \mathbb{R} \mid x \neq -3\}}$.

ii. Show that $g(a) = g(b) \implies a = b$, what conclusion can you draw from the result
As shown in Step 2, $g(a) = g(b) \implies a = b$.
Conclusion: The function $g(x)$ is injective (or one-to-one).

iii. Find the inverse function $g^{-1}$ of $g$ and hence the value of $x$ for which $f(g(x)) = g^{-1}x$ leaving your answer in surd form.
The inverse function $g^{-1}(x)$ is $\boxed{\frac{1 - 3x}{x}}$.
The values of $x$ for which $f(g(x)) = g^{-1}(x)$ are $\boxed{\frac{-9 \pm \sqrt{129}}{8}}$.