The functions f and g are defined by $f: x \mapsto \ln(2x - 1), \quad x \in \mathbb{R}, \; x > \frac{1}{2},$ g: x \mapsto \frac{2}{x - 3}, \quad x \in \mathbb{R}, \; x \neq 3.$ (a) Find the exact value of fg(4) - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 5

To find the inverse of $f(x) = \ln(2x - 1)$, we first set $y = \ln(2x - 1)$ and solve for $x$:

$y = \ln(2x - 1) \Rightarrow e^y = 2x - 1 \Rightarrow 2x = e^y + 1 \Rightarrow x = \frac{e^y + 1}{2}.$
The inverse function is therefore:

$f^{-1}(x) = \frac{e^x + 1}{2}.$
The domain of $f^{-1}(x)$ is all real numbers, as the range of $f(x)$ is $(-\infty, \infty)$.

The function $g(x) = \frac{2}{x - 3}$ has a vertical asymptote at $x = 3$, where the function is undefined. The graph of $y = |g(x)|$ is above the x-axis for all $x \neq 3$.

To sketch:

• The vertical asymptote is $x = 3$.
• As $x \to 3^-$, $g(x) \to -\infty$, hence $|g(x)| \to +\infty$.
• As $x \to 3^+$, $g(x) \to +\infty$, hence $|g(x)| \to +\infty$.
• The y-intercept occurs when $x = 0$:

$g(0) = \frac{2}{0 - 3} = -\frac{2}{3} \Rightarrow |g(0)| = \frac{2}{3}.$
Thus, the coordinates of the point where the graph crosses the y-axis are $(0, \frac{2}{3}).$

To solve the equation:

$\frac{2}{|x - 3|} = 3,$
multiply both sides by $|x - 3|$ (which is positive because absolute values are non-negative):

$2 = 3|x - 3| \Rightarrow |x - 3| = \frac{2}{3}.$
This gives us two cases:

1. $x - 3 = \frac{2}{3} \Rightarrow x = \frac{2}{3} + 3 = \frac{11}{3}.$
2. $x - 3 = -\frac{2}{3} \Rightarrow x = -\frac{2}{3} + 3 = \frac{7}{3}.$
   Thus, the exact values of x are $\frac{11}{3}$ and $\frac{7}{3}$.