The function f is defined by $f : x \mapsto \frac{2(x - 1)}{x^2 - 2x - 3} + \frac{1}{x - 3}, \ x > 3.$ (a) Show that $f(x) = \frac{1}{x + 1}, \ x > 3.$ (b) Find the range of $f.$ (c) Find $f^{-1}(x).$ State the domain of this inverse function - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 5

To find the range of $f$, we note from part (a) that:

$f(x) = \frac{1}{x + 1}, \ x > 3.$

As $x$ approaches $3$, $f(x)$ approaches $\frac{1}{4}$. As $x$ approaches infinity, $f(x)$ approaches $0.$ Hence, the range of $f$ is:

$\text{Range of } f = \left(0, \frac{1}{4}\right).$

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

1. Rearranging gives:

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

Thus, the inverse function is:

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

2. To find the domain of the inverse, we note that $y$ can take values in the range of $f$, hence the domain of $f^{-1}(x)$ is:

   $\text{Domain of } f^{-1}(x) = \left(0, \frac{1}{4}\right).$

Given $g(x) = 2x - 3$, we start with:

$f(g(x)) = f(2x - 3) = \frac{1}{(2x - 3) + 1} = \frac{1}{2x - 2} = \frac{1}{8}.$

Solving this gives:

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

Thus, the solution is:

$x = 5.$