Question

If f : R → (−1, 1) defined by $f\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$ is invertible, find f⁻¹.

Answer 1

Injectivity of f:
Let x and y be two elements of domain (R), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{{10}^x-{10}^{-x}}{{10}^x-{10}^{-x}}=\frac{{10}^y-{10}^{-y}}{{10}^y-{10}^{-y}} \\ \Rightarrow \frac{{10}^{-x}\left({10}^{2x}-1\right)}{{10}^{-x}\left({10}^{2x}+1\right)}=\frac{{10}^{-y}\left({10}^{2y}-1\right)}{{10}^{-y}\left({10}^{2y}+1\right)} \\ \Rightarrow \frac{\left({10}^{2x}-1\right)}{\left({10}^{2x}+1\right)}=\frac{\left({10}^{2y}-1\right)}{\left({10}^{2y}+1\right)} \\ \Rightarrow \left({10}^{2x}-1\right)\left({10}^{2y}+1\right)=\left({10}^{2x}+1\right)\left({10}^{2y}-1\right) \\ \Rightarrow {10}^{2x+2y}+{10}^{2x}-{10}^{2y}-1={10}^{2x+2y}-{10}^{2x}+{10}^{2y}-1 \\ \Rightarrow 2\times {10}^{2x}=2\times {10}^{2y} \\ \Rightarrow {10}^{2x}={10}^{2y} \\ \Rightarrow 2x=2y \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity of f:
Let y is in the co domain (R), such that f(x) = y

$$\begin{gathered} \Rightarrow \frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}=y \\ \Rightarrow \frac{{10}^{-x}\left({10}^{2x}-1\right)}{{10}^{-x}\left({10}^{2x}+1\right)}=y \\ \Rightarrow {10}^{2x}-1=y\times {10}^{2x}+y \\ \Rightarrow {10}^{2x}\left(1-y\right)=1+y \\ \Rightarrow {10}^{2x}=\frac{1+y}{1-y} \\ \Rightarrow 2x=\log \left(\frac{1+y}{1-y}\right) \\ \Rightarrow x=\frac{1}{2}\log \left(\frac{1+y}{1-y}\right)\in R\left(\text{domain}\right) \end{gathered}$$

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

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow \frac{{10}^y-{10}^{-y}}{{10}^y+{10}^{-y}}=x \\ \Rightarrow \frac{{10}^{-y}\left({10}^{2y}-1\right)}{{10}^{-y}\left({10}^{2y}+1\right)}=x \\ \Rightarrow {10}^{2y}-1=x\times {10}^{2y}+x \\ \Rightarrow {10}^{2y}\left(1-x\right)=1+x \\ \Rightarrow {10}^{2y}=\frac{1+x}{1-x} \\ \Rightarrow 2y=\log \left(\frac{1+x}{1-x}\right) \\ \Rightarrow y=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right) \\ \text{So,}{f}^{-1}\left(x\right)=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)[\text{from}\left(1\right)] \end{gathered}$$