Question

If f : R → (0, 2) defined by $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1$ is invertible, find f⁻¹.

Answer 1

Injectivity of f :
Let x and y be two elements of domain (R), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{e^x-e^{-x}}{e^x-e^{-x}}+1=\frac{e^y-e^{-y}}{e-e^{-y}}+1 \\ \Rightarrow \frac{e^x-e^{-x}}{e^x-e^{-x}}=\frac{e^y-e^{-y}}{e-e^{-y}} \\ \Rightarrow \frac{e^{-x}\left(e^{2x}-1\right)}{e^{-x}\left(e^{2x}+1\right)}=\frac{e^{-y}\left(e^{2y}-1\right)}{e^{-y}\left(e^{2y}+1\right)} \\ \Rightarrow \frac{\left(e^{2x}-1\right)}{\left(e^{2x}+1\right)}=\frac{\left(e^{2y}-1\right)}{\left(e^{2y}+1\right)} \\ \Rightarrow \left(e^{2x}-1\right)\left(e^{2y}+1\right)=\left(e^{2x}+1\right)\left(e^{2y}-1\right) \\ \Rightarrow e^{2x+2y}+e^{2x}-e^{2y}-1=e^{2x+2y}-e^{2x}+e^{2y}-1 \\ \Rightarrow 2\times e^{2x}=2\times e^{2y} \\ \Rightarrow e^{2x}=e^{2y} \\ \Rightarrow 2x=2y \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain $\left(0,2\right)$, such that f(x) = y.

$$\begin{gathered} \frac{e^x-e^{-x}}{e^x+e^{-x}}+1=y \\ \Rightarrow \frac{e^{-x}\left(e^{2x}-1\right)}{e^{-x}\left(e^{2x}+1\right)}+1=y \\ \Rightarrow \frac{e^{-x}\left(e^{2x}-1\right)}{e^{-x}\left(e^{2x}+1\right)}=y-1 \\ \Rightarrow e^{2x}-1=\left(y-1\right)\left(e^{2x}+1\right) \\ \Rightarrow e^{2x}-1=y\times e^{2x}+y-e^{2x}-1 \\ \Rightarrow e^{2x}=y\times e^{2x}+y-e^{2x} \\ \Rightarrow e^{2x}\left(2-y\right)=y \\ \Rightarrow e^{2x}=\frac{y}{2-y} \\ \Rightarrow 2x={\log }_e\left(\frac{y}{2-y}\right) \\ \Rightarrow x=\frac{1}{2}{\log }_e\left(\frac{y}{2-y}\right)\in R\left(\text{domain}\right) \end{gathered}$$

So, f is onto.
$\therefore$ f is a bijection and, hence, it is invertible.

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow \frac{e^y-e^{-y}}{e^y+e^{-y}}+1=x \\ \Rightarrow \frac{e^{-y}\left(e^{2y}-1\right)}{e^{-y}\left(e^{2y}+1\right)}+1=x \\ \Rightarrow \frac{e^{-y}\left(e^{2y}-1\right)}{e^{-y}\left(e^{2y}+1\right)}=x-1 \\ \Rightarrow e^{2y}-1=\left(x-1\right)\left(e^{2y}+1\right) \\ \Rightarrow e^{2y}-1=x\times e^{2y}+x-e^{2y}-1 \\ \Rightarrow e^{2y}=x\times e^{2y}+x-e^{2y} \\ \Rightarrow e^{2y}\left(2-x\right)=x \\ \Rightarrow e^{2y}=\frac{x}{2-x} \\ \Rightarrow 2y={\log }_e\left(\frac{x}{2-x}\right) \\ \Rightarrow y=\frac{1}{2}{\log }_e\left(\frac{x}{2-x}\right)\in R\left(\text{domain}\right) \\ \Rightarrow y=\frac{1}{2}{\log }_e\left(\frac{x}{2-x}\right)=f^{-1}\left(x\right)[\text{from}\left(1\right)] \\ \text{So,}{f}^{-1}\left(x\right)=\frac{1}{2}{\log }_e\left(\frac{x}{2-x}\right) \end{gathered}$$