Let x in R and |x|

Question

(A) Let $y = 	anh^{-1} x$. Then $x = 	anh y$. Using the definition of the hyperbolic tangent function,we have $x = \frac{e^y - e^{-y}}{e^y + e^{-y}}

Vedclass · Accepted Answer

$\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$

Vedclass · Answer

(A) Let $y = 	anh^{-1} x$. Then $x = 	anh y$. Using the definition of the hyperbolic tangent function,we have $x = \frac{e^y - e^{-y}}{e^y + e^{-y}}$. Multiplying both sides by $(e^y + e^{-y})$,we get $x(e^y + e^{-y}) = e^y - e^{-y}$. $xe^y + xe^{-y} = e^y - e^{-y}$. Rearranging the terms,we get $e^y(1 - x) = e^{-y}(1 + x)$. Dividing both sides by $(1 - x)$,we get $e^y = e^{-y} \frac{1 + x}{1 - x}$. Multiplying both sides by $e^y$,we get $e^{2y} = \frac{1 + x}{1 - x}$. Taking the natural logarithm on both sides,we get $2y = \log \left(\frac{1 + x}{1 - x}ight)$. Therefore,$y = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}ight)$.

Let $x \in R$ and $|x| < 1$. Then $\tanh ^{-1} x=$

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