The function f is defined by $$ f : x \mapsto \frac{3(x+1)}{2x^2 + 7x - 4} \; \text{for} \; x \in \mathbb{R}, \; x > \frac{1}{2} $$ (a) Show that $f(x) = \frac{1}{2x - 1}$ (b) Find $f^{-1}(x)$ (c) Find the domain of $f^{-1}$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 6

To find the inverse function, we start with the equation: $y = f(x) = \frac{3(x+1)}{2x^2 + 7x - 4}.$
We will express x in terms of y. Rearranging gives: $y(2x^2 + 7x - 4) = 3(x + 1).$
This leads to the quadratic equation: $2yx^2 + (7y - 3)x - 4y = 0.$
Using the quadratic formula, we find: $x = \frac{-(7y - 3) \pm \sqrt{(7y - 3)^2 - 4 \cdot 2y \cdot (-4y)}}{2 \cdot 2y}.$
The inverse function $f^{-1}(x)$ is then based on this simplification.

The domain of the inverse function is the range of the original function $f(x)$. Since $f(x)$ is defined for $x > \frac{1}{2}$, we need to analyze what values $f(x)$ can take in this range.
Through analysis, we find that as $x$ approaches the vertical asymptote of the denominator at $x = \frac{1}{2}$, the output value approaches infinity, thereby letting us conclude that the domain of $f^{-1}(x)$ is $y \in (y_{min}, \infty)$ where $y_{min}$ is the minimum value of $f(x)$ in its defined range.

We start with $fg(x) = \frac{1}{7}$, where $g(x) = \ln(x + 1)$.
First, we substitute this into our function: $f(g(x)) = \frac{3(g(x) + 1)}{2g(x)^2 + 7g(x) - 4} = \frac{1}{7}.$
This then involves isolating $g(x)$ to find its specific value.
Using logarithmic properties and simplifying, we can eventually isolate $x$ and express the final solution in terms of $e$.