Let g(x) be the inverse of f(x)=2x+1−21−x2x+2−x then g(x) be :

y=2x+1−21−x2x+2−x=22x+1−222x+1y22x+1=22x−1222x:y+y=22x⋅2−222x(y−2)=−(2+y)22x=(y+2)(2−y)2x=log2⁡(2+y)(2−y)x=12log2⁡(2+y)(2−y)f−1(x)=12log2⁡(2+x)(2−x)

Let g(x) be the inverse of f(x)=2x+1−21−x2x+2−x then g(x) be :

y=2x+1−21−x2x+2−x=22x+1−222x+1y22x+1=22x−1222x:y+y=22x⋅2−222x(y−2)=−(2+y)22x=(y+2)(2−y)2x=log2⁡(2+y)(2−y)x=12log2⁡(2+y)(2−y)f−1(x)=12log2⁡(2+x)(2−x)

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$Let g (x) be the inverse of f (x) = \frac{2^{x + 1} - 2^{1 - x}}{2^{x} + 2^{- x}} then g (x) be :$

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$\frac{1}{2} \log_{2} (\frac{2 + x}{2 - x})$

$\log_{2} (\frac{2 - x}{2 + x})$

$- \frac{1}{2} \log_{2} (\frac{2 + x}{2 - x})$

$\log_{2} (\frac{2 + x}{2 - x})$

answer is C.

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Detailed Solution

$y = \frac{2^{x + 1} - 2^{1 - x}}{2^{x} + 2^{- x}} = \frac{2^{2 x + 1} - 2}{2^{2 x} + 1}$

$\begin{array}{l} y (2^{2 x} + 1) = (2^{2 x} - 1) 2 \\ 2^{2 x} : y + y = 2^{2 x} \cdot 2 - 2 \\ 2^{2 x} (y - 2) = - (2 + y) \\ 2^{2 x} = \frac{(y + 2)}{(2 - y)} \\ 2 x = \log_{2} \frac{(2 + y)}{(2 - y)} \\ x = \frac{1}{2} \log_{2} \frac{(2 + y)}{(2 - y)} \\ f^{- 1} (x) = \frac{1}{2} \log_{2} \frac{(2 + x)}{(2 - x)} \end{array}$