If for |x|1, tanh ^{-1}left(frac{1}{x}right) | vedclass

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(C) Given that, $	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$. We know that $	anh ^{-1}(u) = \frac{1}{2

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$\frac{2}{3}$

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(C) Given that, $	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$. We know that $	anh ^{-1}(u) = \frac{1}{2} \log _e\left(\frac{1+u}{1-u}ight)$. Substituting $u = \frac{1}{x}$, we get $	anh ^{-1}\left(\frac{1}{x}ight) = \frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}ight) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight)$. Also, $\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight)$ for $|x|>1$. Adding these two expressions: $	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight) + \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight) = \log _e\left(\frac{x+1}{x-1}ight)$. Comparing this with $\log _e(f(x))$, we get $f(x) = \frac{x+1}{x-1}$. Now, substituting $x = -5$: $f(-5) = \frac{-5+1}{-5-1} = \frac{-4}{-6} = \frac{2}{3}$.

If for $|x|>1$, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$, then $f(-5)=$

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