If $f(x)=\frac{a^x-a^{-x}}{a^x+a^{-x}}$, where a, $x$ satisfy the necessary conditions, then $f^{-1}(x)=$

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Aicharya · Accepted Answer

$\begin{aligned} & y=\frac{a^x-a^{-x}}{a^x+a^{-x}}=\frac{a^{2 x}-1}{a^{2 x}+1} \ & \Rightarrow y\left(a^{2 x}+1ight)=a^{2 x}-1 \ & \Rightarrow a^{2 x}=\frac{1+y}{1-y} \ & \Rightarrow x=\frac{1}{2} \log _a\left(\frac{1+y}{1-y}ight) \ & \Rightarrow f^{-1}(x)=\frac{1}{2} \log _a\left(\frac{1+x}{1-x}ight)\end{aligned}$

If $f(x)=\frac{a^x-a^{-x}}{a^x+a^{-x}}$, where a, $x$ satisfy the necessary conditions, then $f^{-1}(x)=$

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