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

To find the inverse function, we start with the equation:

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

Exponentiating both sides gives:

$e^y = 2x - 1.$

Rearranging this, we find:

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

Thus, the inverse function is:

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

The domain of f^{-1}(x) is all real numbers,( x ∈ ℝ ).

To sketch the graph of ( y = |g(x)| = \left| \frac{2}{x-3} \right| ), we first identify the vertical asymptote at ( x = 3 ). The graph will approach this line but never touch it. The graph is mirrored in the x-axis for values of ( g(x) < 0 ).

To find where the graph crosses the y-axis, we evaluate ( g(0) ):

$g(0) = \frac{2}{0-3} = -\frac{2}{3}, \text{ so } |g(0)| = \frac{2}{3}.$

Thus, the graph crosses the y-axis at the point (0, ( \frac{2}{3} )).

To solve ( \frac{2}{|x-3|} = 3 ), we first clear the fraction:

$|x - 3| = \frac{2}{3}.$

This gives us two cases to solve:

1. ( x - 3 = \frac{2}{3} )
   ( x = 3 + \frac{2}{3} = \frac{11}{3} )

2. ( x - 3 = -\frac{2}{3} )
   ( x = 3 - \frac{2}{3} = \frac{7}{3} )

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