Question

For \(f:\mathbb{R}\to(-1,1)\) defined by \(f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\), what is \(f^{-1}(x)\)?

Options

• \(f^{-1}(x)=\log_e\frac{1+x}{1-x}\)

• \(f^{-1}(x)=\frac{1}{2}\log_e\frac{1+x}{1-x}\)

• \(f^{-1}(x)=2\log_e\frac{1+x}{1-x}\)

• \(f^{-1}(x)=\frac{1}{2}\log_e\frac{1-x}{1+x}\)

Answer 1

Solving \(y=\frac{e^x-e^{-x}}{e^x+e^{-x}}\) gives \(e^{2x}=\frac{1+y}{1-y}\). Taking logarithms and solving for \(x\) yields \(f^{-1}(x)=\frac{1}{2}\log_e\frac{1+x}{1-x}\).