If f(x) = x + log left( frac{x-1}{x+1} right) | vedclass

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(B) To determine the nature of the function $f(x) = x + \log \left( \frac{x-1}{x+1} \right)$,we first find its derivative $f'(x)$.First,note that for

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monotonically increasing function

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(B) To determine the nature of the function $f(x) = x + \log \left( \frac{x-1}{x+1} \right)$,we first find its derivative $f'(x)$.First,note that for the function to be well-defined,we require $\frac{x-1}{x+1} > 0$,which implies $x \in (-\infty, -1) \cup (1, \infty)$.Now,differentiate $f(x)$ with respect to $x$:$f'(x) = 1 + \frac{d}{dx} [\log(x-1) - \log(x+1)]$$f'(x) = 1 + \left( \frac{1}{x-1} - \frac{1}{x+1} \right)$$f'(x) = 1 + \frac{(x+1) - (x-1)}{(x-1)(x+1)}$$f'(x) = 1 + \frac{2}{x^2 - 1}$$f'(x) = \frac{x^2 - 1 + 2}{x^2 - 1} = \frac{x^2 + 1}{x^2 - 1}$Since $x^2 + 1 > 0$ for all real $x$,the sign of $f'(x)$ depends on the denominator $x^2 - 1$.For $x \in (1, \infty)$,$x^2 > 1$,so $x^2 - 1 > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing in $(1, \infty)$.For $x \in (-\infty, -1)$,$x^2 > 1$,so $x^2 - 1 > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing in $(-\infty, -1)$.Therefore,$f$ is a monotonically increasing function in its domain.

If $f(x) = x + \log \left( \frac{x-1}{x+1} \right)$ is a well-defined real-valued function,then $f$ is

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