The functions f and g are defined by f : x ↦ ln(2x − 1), x ∈ ℝ, x > \frac{1}{2}, g : x ↦ \frac{2}{x − 3}, x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 5

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

Let y = f(x) = ln(2x - 1).

Now, swapping x and y gives:

x = ln(2y - 1).

To find y, we exponentiate both sides:

e^x = 2y - 1.

Rearranging gives:

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

Thus, the inverse function is:

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

The domain of f^{-1} is all real numbers, \mathbb{R}.

To sketch y = |g(x)|, we first note that g(x) has a vertical asymptote at x = 3 because g(x) is undefined there.

The graph of g(x) = \frac{2}{x - 3} is a hyperbola with two branches:

• As x approaches 3 from the left, g(x) approaches -∞.
• As x approaches 3 from the right, g(x) approaches +∞.

For |g(x)|:

• The part where g(x) is negative flips above the x-axis.

The vertical asymptote is represented by the equation x = 3.

The graph crosses the y-axis when x = 0:

g(0) = \frac{2}{0 - 3} = \frac{2}{-3} = -\frac{2}{3},

so |g(0)| = \frac{2}{3}.

To solve \frac{2}{|x - 3|} = 3, we first cross-multiply:

2 = 3|x - 3|.

Dividing both sides by 3 gives:

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

This results in two cases:

1. x - 3 = \frac{2}{3} → x = \frac{2}{3} + 3 = \frac{11}{3}.
2. x - 3 = -\frac{2}{3} → x = -\frac{2}{3} + 3 = \frac{7}{3}.

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