Let f be a real-valued function,defined on R | vedclass

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(A) Given $f(x) = 3 \log_{e} \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$.We can rewrite this as $f(x) = 3 \log_{e} |x-1| - 3 \log_{e} |x+1| - \fra

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$(-\infty, -1) \cup \left[ \frac{1}{2}, 1 \right) \cup (1, \infty)$

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(A) Given $f(x) = 3 \log_{e} \left| \frac{x-1}{x+1} ight| - \frac{2}{x-1}$.We can rewrite this as $f(x) = 3 \log_{e} |x-1| - 3 \log_{e} |x+1| - \frac{2}{x-1}$.Now,differentiate $f(x)$ with respect to $x$:$f'(x) = 3 \left( \frac{1}{x-1} ight) - 3 \left( \frac{1}{x+1} ight) + \frac{2}{(x-1)^2}$.Simplify the expression:$f'(x) = 3 \left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} ight) + \frac{2}{(x-1)^2} = 3 \left( \frac{2}{x^2-1} ight) + \frac{2}{(x-1)^2}$.$f'(x) = \frac{6}{(x-1)(x+1)} + \frac{2}{(x-1)^2} = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{6x - 6 + 2x + 2}{(x-1)^2(x+1)} = \frac{8x - 4}{(x-1)^2(x+1)} = \frac{4(2x-1)}{(x-1)^2(x+1)}$.For $f(x)$ to be increasing,$f'(x) \geq 0$.Since $(x-1)^2 > 0$ for all $x 
eq 1$,the sign of $f'(x)$ depends on $\frac{2x-1}{x+1}$.Using the sign scheme for $\frac{2x-1}{x+1} \geq 0$:The critical points are $x = \frac{1}{2}$ and $x = -1$.Testing intervals: $(-\infty, -1)$,$(-1, \frac{1}{2}]$,and $[\frac{1}{2}, \infty)$.$f'(x) > 0$ for $x \in (-\infty, -1) \cup [\frac{1}{2}, 1) \cup (1, \infty)$.

Let $f$ be a real-valued function,defined on $R - \{-1, 1\}$ and given by $f(x) = 3 \log_{e} \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$. In which of the following intervals is the function $f(x)$ increasing?

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