The function f is defined by $$f : x \mapsto \frac{3(x+1)}{2x^2 + 7x - 4} \cdot \frac{1}{x + 4} \quad x \in \mathbb{R}, \; x > \frac{1}{2}$$ (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 8 - 2012 - Paper 6

To find the inverse of the function ( f(x) ):

1. We set ( y = f(x) ) which gives us: $y = \frac{1}{2x - 1}$
2. Next, we interchange the variables: $x = \frac{1}{2y - 1}$
3. Solving for ( y ): $2y - 1 = \frac{1}{x}$
   $2y = \frac{1}{x} + 1$
   $y = \frac{1 + x}{2x}$

Thus, ( f^{-1}(x) = \frac{1 + x}{2x} ).

The domain of ( f^{-1}(x) ) requires that the input to the original function is non-zero:

From ( f^{-1}(x) = \frac{1 + x}{2x} ), the function is undefined when the denominator equals zero:

Set ( 2x = 0 ) which gives ( x = 0 ). Hence, the domain is: $x \in \mathbb{R} \setminus \{0\}.$

We need to find the function ( g(x) = \ln(x + 1) ) and set up the equation:

1. Start with: $f(g(x)) = \frac{1}{7}$ Plugging in ( g(x) ): $f(\ln(x + 1)) = \frac{1}{7}$
2. Substitute ( f^{-1} ) for each side: $x + 1 = e^7 - 1$
3. Thus, the solution for ( g(x) = \frac{1}{7} ) in terms of ( e ): $x = e^7 - 1.$