The functions f and g are defined by f : x ↦ ln(2x-1), x ∈ ℝ, x > rac{1}{2}, g : x ↦ rac{2}{x-3}, x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 5

To find the inverse function f^{-1}(x), we start with:

$y = \ln(2x-1).$

Exponential both sides gives:

$e^y = 2x - 1,$ which can be rearranged to:

$x = \frac{e^y + 1}{2}.$

Thus, the inverse function is:

$f^{-1}(x) = \frac{e^x + 1}{2},$ with the domain of (f^{-1}(x)) being all real numbers ( \mathbb{R} ) since the range of f is ( \mathbb{R} ).

To sketch the graph of (y = |g(x)|), we first determine the vertical asymptote and y-intercept:

1. Vertical Asymptote: Set the denominator equal to zero: $x - 3 = 0 \Rightarrow x = 3.$

2. Y-intercept: Evaluate (g(0):) $g(0) = \frac{2}{0-3} = -\frac{2}{3},$ therefore (|g(0)| = \frac{2}{3}.$$

The graph will show a vertical asymptote at (x = 3) and will cross the y-axis at (y = \frac{2}{3}).

To solve (\frac{2}{|x-3|} = 3), we begin by manipulating the equation:

1. Multiply both sides by (|x-3|): $2 = 3|x-3|.$

2. Solve for (|x-3|): $|x-3| = \frac{2}{3}.$

3. Two cases arise:

   • Case 1: (x - 3 = \frac{2}{3}) \Rightarrow (x = 3 + \frac{2}{3} = \frac{11}{3}.
   • Case 2: (x - 3 = -\frac{2}{3}) \Rightarrow (x = 3 - \frac{2}{3} = \frac{7}{3}.

Thus, the exact values of (x) are (\frac{11}{3}) and (\frac{7}{3}).