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 $f^{-1}$ (d) Find the solution of $fg(x) = \frac{1}{7}$, giving your answer in terms of $e$. - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 6

To find the inverse function $f^{-1}(x)$, start with: $y = \frac{1}{2x - 1}$

Swapping $x$ and $y$ gives: $x = \frac{1}{2y - 1}$

Rearranging this equation leads to: $2y - 1 = \frac{1}{x}$

Thus, the inverse is: $2y = 1 + \frac{1}{x}$

Consequently,Dividing by 2: $y = \frac{1}{2} \left( 1 + \frac{1}{x} \right)$

This implies: $f^{-1}(x) = \frac{1 + \frac{1}{x}}{2}$.

The domain of the inverse function $f^{-1}(x)$ is determined by the range of $f(x)$. Since $f(x)$ can take on any value in the interval $(0, \infty)$ for $x > \frac{1}{2}$, the domain of $f^{-1}(x)$ is also: $(0, \infty)$.

To solve $fg(x) = \frac{1}{7}$ with $g(x) = \ln(x+1)$:

We begin by rewriting the equation as: $f(g(x)) = \frac{1}{7}$

Substituting $g(x)$: $f(\ln(x+1)) = \frac{1}{7}$

Since $f(x) = \frac{1}{2x - 1}$, we substitute again: $\frac{1}{2 \ln(x+1) - 1} = \frac{1}{7}$

Cross-multiplying leads to: $7 = 2 \ln(x+1) - 1$

Solving for $\ln(x+1)$ gives: $2 \ln(x+1) = 8 \Rightarrow \ln(x+1) = 4$

Exponentiating both sides yields: $x + 1 = e^4 \Rightarrow x = e^4 - 1$

Thus, the solution is: $x = e^4 - 1$.