The function $f$ is defined by $$f: x \mapsto \frac{3(x+1)}{2x^2+7x-4} \quad \text{for} \quad x \in \mathbb{R}, \; x > \frac{1}{2}$$ (a) Show that $f(x) = \frac{1}{2x-1}$ (b) Find $f^{-1}(x)$ (c) Find the domain of $f^{-1}$ g(x)=\ln(x+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 function $f^{-1}(x)$, we start from the equation:

$y = f(x) = \frac{1}{2x - 1}$

Rearranging this gives:

$2x - 1 = \frac{1}{y}$

Next, we solve for $x$:

$2x = \frac{1}{y} + 1\quad \Rightarrow \quad x = \frac{1 + y}{2}$

Thus, the inverse function is:

$f^{-1}(x) = \frac{1 + x}{2}$

The domain of $f^{-1}(x) = \frac{1 + x}{2}$ is all real numbers since there are no restrictions on the value of $x$, resulting in:

$\text{Domain of } f^{-1} = (-\infty, \infty)$

To solve $fg(x) = \frac{1}{7}$, we substitute:

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

We have $g(x) = \ln(x + 1)$, substituting we find:

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

Setting the equation:

$\frac{1}{2 \ln(x + 1) - 1} = \frac{1}{7}$

Cross-multiplying and solving for $\ln(x + 1)$:

$7 = 2 \ln(x + 1) - 1$ $8 = 2 \ln(x + 1)\quad \Rightarrow \quad \ln(x + 1) = 4$

Exponentiating:

$x + 1 = e^4\quad \Rightarrow \quad x = e^4 - 1$

Thus the solution is:

$x = e^4 - 1$