## Task B-4.3.

Let $f: \mathbb{R} \rightarrow (-1,1)$,

$$
f(x)=\frac{100^{-x}-100^{x}}{100^{-x}+100^{x}}
$$

Determine $f^{-1}(x)-f^{-1}(1+2 x)$.
## Solution.

Let's determine the inverse function of the function $f$. Write the function $f$ as follows:

$$
y=f(x)=\frac{100^{-x}-100^{x}}{100^{-x}+100^{x}} \cdot \frac{100^{x}}{100^{x}}=\frac{1-100^{2 x}}{1+100^{2 x}}
$$

Further, we have

$$
y+y \cdot 100^{2 x}=1-100^{2 x}
$$

which means $100^{2 x}(y+1)=1-y$, so $100^{2 x}=\frac{1-y}{1+y}$.

From this, we get $x=\frac{1}{2} \log _{100} \frac{1-y}{1+y}=\frac{1}{4} \log \frac{1-y}{1+y}$.

Thus,

$$
f^{-1}(x)=\frac{1}{4} \log \frac{1-x}{1+x}
$$

and

$$
f^{-1}(1+2 x)=\frac{1}{4} \log \frac{1-(1+2 x)}{1+(1+2 x)}=\frac{1}{4} \log \frac{-2 x}{2+2 x}=\frac{1}{4} \log \frac{-x}{1+x}
$$

Finally,

$$
\begin{aligned}
f^{-1}(x)-f^{-1}(1+2 x) & =\frac{1}{4} \log \frac{1-x}{1+x}-\frac{1}{4} \log \frac{-x}{1+x}=\frac{1}{4}\left(\log \frac{1-x}{1+x}-\log \frac{-x}{1+x}\right) \\
& =\frac{1}{4} \log \left(\frac{1-x}{1+x}: \frac{-x}{1+x}\right)=\frac{1}{4} \log \frac{1-x}{-x}=\frac{1}{4} \log \frac{x-1}{x}
\end{aligned}
$$