For what values of x is the following identit | vedclass

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(D) The identity is given by $	anh^{-1} x = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} ight)$.For the logarithmic function $\log_e(u)$ to be defined

Vedclass · Accepted Answer

$(-1, 1)$

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(D) The identity is given by $	anh^{-1} x = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} ight)$.For the logarithmic function $\log_e(u)$ to be defined,we must have $u > 0$.Therefore,we require $\frac{1+x}{1-x} > 0$.Let $f(x) = \frac{1+x}{1-x}$. The critical points are $x = -1$ and $x = 1$.Testing the intervals:For $x For $-1  0$.For $x > 1$,$f(x) At $x = 1$ and $x = -1$,the expression is undefined.Thus,the identity is valid only for $x \in (-1, 1)$.

For what values of $x$ is the following identity valid and holds? $\tanh^{-1}(x) = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} \right)$.

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