Question

The inverse function of $f\left(x\right) = \frac{8^{2x}-8^{-2x}}{8^{2x} + 8^{-2x}}, x\epsilon \left(-1, 1\right),$ is __________.

• A. $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1-x}{1+x}\right)$
• B. $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{4} log_{e} \left(\frac{1+x}{1-x}\right)$
• D. $\frac{1}{4} log_{e} \left(\frac{1- x}{1+ x}\right)$

Answer 1

Answer: (B)

$f \left(x\right)=y=\frac{8^{4x}-1}{8^{4x}+1}=1-\frac{2}{8^{4x}+1}$
so, $8^{4x}+1=\frac{2}{1-y} \Rightarrow 8^{4x}=\frac{1+y}{1-y}$
$\Rightarrow x=\ell n\left(\frac{1+y}{1-y}\right)\times \frac{1}{4\ell n8}=f ^{-1}\left(y\right)$
Hence, $f ^{-1}\left(x\right)=\frac{1}{4}log_{g}\,e\ell n\left(\frac{1+x}{1-x}\right)$