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 $f(x) = \frac{1}{x+1}, \quad x > 3.$ (b) Find the range of f - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 5

To find the range of $f$, we analyze the function:

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

As $x$ approaches $3$, $f(x)$ approaches rac{1}{4}. As $x$ increases to $\infty$, $f(x)$ approaches $0$. Thus, the range of $f$ is:

$\left(0, \frac{1}{4}\right).$

Let $y = f(x) = \frac{1}{x+1}.$ To find the inverse, solve for $x$:

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

Thus, the inverse function is:

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

The domain of $f^{-1}(x)$ is derived from the range of $f$, which is:

$\left(0, \frac{1}{4}\right).$

To solve for $fg(x) = \frac{1}{8}$, we first find $g(x)$:

$g(x) = 2x - 3.$

Then we have:

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

Setting this equal to rac{1}{8} gives:

$\frac{1}{2x - 2} = \frac{1}{8} \Rightarrow 8 = 2x - 2 \Rightarrow 2x = 10 \Rightarrow x = 5.$

Therefore the solution is:

$x = 5.$