The function f is defined by $$f : x \mapsto \frac{2(x-1)}{x^2 - 2x - 3} + \frac{1}{x-3}, \quad x > 3.$$ (a) Show that $f(x) = \frac{1}{x+1}, \quad x > 3$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

The range of $f$ can be determined as follows. Given that $f(x) = \frac{1}{x+1}$ for $x > 3$, we find that as $x$ approaches $3$, $f(x)$ approaches $0$. And as $x$ increases without bound, $f(x)$ approaches $1/4$.
Thus, the range of $f$ is $\left(0, \frac{1}{4}\right)$.

Let $y = f(x)$, then we have:

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

Rearranging gives us:

$x + 1 = \frac{1}{y} \Rightarrow x = \frac{1}{y} - 1.$

Thus, the inverse function is:

$f^{-1}(x) = \frac{1}{x} - 1.$

The domain of $f^{-1}(x)$ stems from the range of $f$, which we found to be $\left(0, \frac{1}{4}\right)$. Hence the domain of $f^{-1}(x)$ is also $\left(0, \frac{1}{4}\right)$.

Starting with our defined functions:

We have $fg(x) = f(2x - 3) = \frac{1}{(2x - 3) + 1}.$

Setting this equal to $\frac{1}{8}$, we equate:

$\frac{1}{2x - 2} = \frac{1}{8}.$

Cross-multiplying gives us:

$8 = 2x - 2 \Rightarrow 2x = 10 \Rightarrow x = 5.$

The solution is $x = 5$.