Solving the Functional Equation for f(x)

We are given the functional equation:

\[f\left( {{x_1}} \right) - f\left( {{x_2}} \right) = f\left( {\frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}}} \right)\]

This equation holds for all \(x_1, x_2 \in (-1, 1)\). We need to find the function \(f(x)\).

Let's test the given options by substituting them into the functional equation.

Testing Option 1: \(f(x) = \ln\left(\frac{1-x}{1+x}\right)\)

If \(f(x) = \ln\left(\frac{1-x}{1+x}\right)\), let's evaluate the left-hand side (LHS) of the functional equation:

\[f\left( {{x_1}} \right) - f\left( {{x_2}} \right) = \ln\left( {\frac{{1 - {x_1}}}{{1 + {x_1}}}} \right) - \ln\left( {\frac{{1 - {x_2}}}{{1 + {x_2}}}} \right)\]

Using the logarithm property \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \):

\[LHS = \ln\left( {\frac{{\frac{{1 - {x_1}}}{{1 + {x_1}}}}}{{\frac{{1 - {x_2}}}{{1 + {x_2}}}}}} \right) = \ln\left( {\frac{{1 - {x_1}}}{{1 + {x_1}}} \cdot \frac{{1 + {x_2}}}{{1 - {x_2}}}} \right)\]

Now let's evaluate the right-hand side (RHS) of the functional equation. First, consider the argument of \(f\) on the RHS:

\[y = \frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}}\]

Then \(f(y) = \ln\left(\frac{1-y}{1+y}\right)\). We need to substitute the expression for \(y\):

\[1 - y = 1 - \frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}} = \frac{{1 - {x_1}{x_2} - ({x_1} - {x_2})}}{{1 - {x_1}{x_2}}} = \frac{{1 - {x_1}{x_2} - {x_1} + {x_2}}}{{1 - {x_1}{x_2}}} = \frac{{(1 - {x_1}) + {x_2}(1 - {x_1})}}{{1 - {x_1}{x_2}}} = \frac{{(1 - {x_1})(1 + {x_2})}}{{1 - {x_1}{x_2}}}\]

\[1 + y = 1 + \frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}} = \frac{{1 - {x_1}{x_2} + ({x_1} - {x_2})}}{{1 - {x_1}{x_2}}} = \frac{{1 - {x_1}{x_2} + {x_1} - {x_2}}}{{1 - {x_1}{x_2}}} = \frac{{(1 - {x_2}) + {x_1}(1 - {x_2})}}{{1 - {x_1}{x_2}}} = \frac{{(1 - {x_2})(1 + {x_1})}}{{1 - {x_1}{x_2}}}\]

So, the argument of the logarithm on the RHS is:

\[\frac{{1 - y}}{{1 + y}} = \frac{{\frac{{(1 - {x_1})(1 + {x_2})}}{{1 - {x_1}{x_2}}}}}{{\frac{{(1 - {x_2})(1 + {x_1})}}{{1 - {x_1}{x_2}}}}} = \frac{{(1 - {x_1})(1 + {x_2})}}{{(1 - {x_2})(1 + {x_1})}} = \frac{{1 - {x_1}}}{{1 + {x_1}}} \cdot \frac{{1 + {x_2}}}{{1 - {x_2}}}\]

Thus, the RHS is:

\[RHS = \ln\left( {\frac{{1 - {x_1}}}{{1 + {x_1}}} \cdot \frac{{1 + {x_2}}}{{1 - {x_2}}}} \right)\]

Comparing LHS and RHS, we see that:

\[\ln\left( {\frac{{1 - {x_1}}}{{1 + {x_1}}} \cdot \frac{{1 + {x_2}}}{{1 - {x_2}}}} \right) = \ln\left( {\frac{{1 - {x_1}}}{{1 + {x_1}}} \cdot \frac{{1 + {x_2}}}{{1 - {x_2}}}} \right)\]

The LHS equals the RHS. Therefore, the functional equation is satisfied by \(f(x) = \ln\left(\frac{1-x}{1+x}\right)\).

Let's briefly consider why the other options might not work. The argument \( \frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}} \) resembles the formula for \(\tanh(A-B)\) if \(x_1 = \tanh A\) and \(x_2 = \tanh B\). If we let \(x = \tanh t\), then \(t = \text{arctanh } x\). The functional equation becomes \(f(\tanh t_1) - f(\tanh t_2) = f(\tanh(t_1 - t_2))\). If we define \(g(t) = f(\tanh t)\), this becomes \(g(t_1) - g(t_2) = g(t_1 - t_2)\). The continuous solution to this Cauchy-type equation is \(g(t) = ct\) for some constant c. So, \(f(\tanh t) = ct\). Substituting back \(t = \text{arctanh } x\), we get \(f(x) = c \cdot \text{arctanh } x\). We know that \( \text{arctanh } x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right) \). So, \(f(x) = \frac{c}{2} \ln\left(\frac{1+x}{1-x}\right)\). Using the logarithm property \( \ln\left(\frac{a}{b}\right) = - \ln\left(\frac{b}{a}\right) \), this can be written as \(f(x) = -\frac{c}{2} \ln\left(\frac{1-x}{1+x}\right)\). Option 1 is \(f(x) = \ln\left(\frac{1-x}{1+x}\right)\), which fits this form with \(-\frac{c}{2} = 1\), or \(c = -2\). The other options, involving \(\ln\left(\frac{2+x}{1-x}\right)\) or inverse tangent functions, do not fit this derived form and thus do not satisfy the functional equation for all \(x_1, x_2 \in (-1, 1)\).

Thus, the function \(f(x)\) is indeed \(\ln\left( {\frac{{1 - x}}{{1 + x}}} \right)\).

The final answer is Option 1.