Key Concepts and Formulas

• A function $f(x)$ is increasing where its derivative $f'(x) \ge 0$.
• Derivative of $\log_e |u(x)|$: $\frac{d}{dx} \log_e |u(x)| = \frac{u'(x)}{u(x)}$.
• Quotient rule: $\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Step-by-Step Solution

Step 1: Find the derivative, $f'(x)$.

We are given $f(x) = 3 \log_e \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$. We need to find its derivative.

$f'(x) = \frac{d}{dx} \left[ 3 \log_e \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1} \right]$

$f'(x) = 3 \frac{d}{dx} \left[ \log_e \left| \frac{x-1}{x+1} \right| \right] - 2 \frac{d}{dx} \left[ (x-1)^{-1} \right]$

Let $u(x) = \frac{x-1}{x+1}$. Then $u'(x) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{2}{(x+1)^2}$. So, $\frac{d}{dx} \left[ \log_e \left| \frac{x-1}{x+1} \right| \right] = \frac{u'(x)}{u(x)} = \frac{\frac{2}{(x+1)^2}}{\frac{x-1}{x+1}} = \frac{2}{(x+1)^2} \cdot \frac{x+1}{x-1} = \frac{2}{(x+1)(x-1)}$.

Also, $\frac{d}{dx} \left[ (x-1)^{-1} \right] = -1(x-1)^{-2} = -\frac{1}{(x-1)^2}$.

Therefore, $f'(x) = 3 \left[ \frac{2}{(x+1)(x-1)} \right] - 2 \left[ -\frac{1}{(x-1)^2} \right]$ $f'(x) = \frac{6}{(x-1)(x+1)} + \frac{2}{(x-1)^2}$

Step 2: Simplify $f'(x)$ and find when $f'(x) \ge 0$.

We want to find the intervals where $f'(x) \ge 0$. $f'(x) = \frac{6}{(x-1)(x+1)} + \frac{2}{(x-1)^2} \ge 0$ Find a common denominator: $f'(x) = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} \ge 0$ Simplify the numerator: $f'(x) = \frac{6x - 6 + 2x + 2}{(x-1)^2(x+1)} \ge 0$ $f'(x) = \frac{8x - 4}{(x-1)^2(x+1)} \ge 0$ Factor out a 4: $f'(x) = \frac{4(2x - 1)}{(x-1)^2(x+1)} \ge 0$ Since 4 is positive, we can ignore it: $\frac{2x - 1}{(x-1)^2(x+1)} \ge 0$

Step 3: Analyze the inequality.

We need to determine the sign of $\frac{2x - 1}{(x-1)^2(x+1)}$. The critical points are $x = \frac{1}{2}$, $x = 1$, and $x = -1$. Note that $x=1$ and $x=-1$ are excluded from the domain of the original function. The factor $(x-1)^2$ is always positive (except at $x=1$, where it is zero, and thus excluded). So, we only need to consider the sign of $\frac{2x-1}{x+1}$. We analyze the intervals:

• $(-\infty, -1)$: Choose $x = -2$. Then $\frac{2(-2)-1}{-2+1} = \frac{-5}{-1} = 5 > 0$. So, $f'(x) > 0$ on $(-\infty, -1)$.
• $(-1, \frac{1}{2})$: Choose $x = 0$. Then $\frac{2(0)-1}{0+1} = \frac{-1}{1} = -1 < 0$. So, $f'(x) < 0$ on $(-1, \frac{1}{2})$.
• $(\frac{1}{2}, 1)$: Choose $x = \frac{3}{4}$. Then $\frac{2(\frac{3}{4})-1}{\frac{3}{4}+1} = \frac{\frac{1}{2}}{\frac{7}{4}} = \frac{2}{7} > 0$. So, $f'(x) > 0$ on $(\frac{1}{2}, 1)$.
• $(1, \infty)$: Choose $x = 2$. Then $\frac{2(2)-1}{2+1} = \frac{3}{3} = 1 > 0$. So, $f'(x) > 0$ on $(1, \infty)$.

Therefore, $f(x)$ is increasing on $(-\infty, -1)$, $(\frac{1}{2}, 1)$, and $(1, \infty)$. Since the question asks for intervals where $f'(x) \ge 0$, we include $x = \frac{1}{2}$. Therefore, the intervals are $(-\infty, -1) \cup [\frac{1}{2}, 1) \cup (1, \infty)$, which can be written as $(-\infty, -1) \cup ([\frac{1}{2}, \infty) - \{1\})$.

Common Mistakes & Tips

• Remember to exclude values that make the denominator zero, both in the derivative and the original function.
• When analyzing the inequality, consider the sign of each factor in the numerator and denominator.
• Don't forget to include the point where the numerator is zero if the question asks for intervals where $f'(x) \ge 0$.

Summary

We found the derivative of the given function, simplified it, and analyzed the inequality $f'(x) \ge 0$ to determine the intervals where the function is increasing. We considered the critical points and the sign of the derivative in each interval, remembering to exclude points where the original function or its derivative is undefined. The final answer is $(-\infty, -1) \cup ([\frac{1}{2}, \infty) - \{1\})$.

Final Answer

The final answer is \boxed{(-\infty, -1) \cup \left( {[{1 \over 2},\infty ) - { 1} } \right)}, which corresponds to option (A).