If ` f.(x)= log((1+x)/(1-x))`, then

Options

A. `f(x_(1))*f(x_(2))=f(x_(1)+x_(2))`
B. `f(x+2)-2f(x+1)+f(x)=0`
C. `f(x)+f(x+1)=f(x^(2)+x)`
D. `f(x_(1))+f(x_(2)) = f ((x_(1)+x_(2))/(1+x_(1)x_(2)))`

Explanation

To solve the problem where \( f(x) = \log\left(\frac{1+x}{1-x}\right) \), we need to show that:

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

### Step 1: Write down the expressions for \( f(x_1) \) and \( f(x_2) \)

We start with the definitions of \( f(x_1) \) and \( f(x_2) \):

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

### Step 2: Add \( f(x_1) \) and \( f(x_2) \)

Using the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \), we can combine these two logarithms:

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

### Step 3: Simplify the expression inside the logarithm

Now, we need to simplify the product:

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

Expanding both the numerator and the denominator:

- **Numerator**:
\[ (1+x_1)(1+x_2) = 1 + x_1 + x_2 + x_1 x_2 \]

- **Denominator**:
\[ (1-x_1)(1-x_2) = 1 - x_1 - x_2 + x_1 x_2 \]

Thus, we have:

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

### Step 4: Relate this to \( f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right) \)

Now we need to calculate \( f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right) \):

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

Simplifying the fractions:

- **Numerator**:
\[ 1 + \frac{x_1 + x_2}{1 + x_1 x_2} = \frac{(1 + x_1 x_2) + (x_1 + x_2)}{1 + x_1 x_2} = \frac{1 + x_1 + x_2 + x_1 x_2}{1 + x_1 x_2} \]

- **Denominator**:
\[ 1 - \frac{x_1 + x_2}{1 + x_1 x_2} = \frac{(1 + x_1 x_2) - (x_1 + x_2)}{1 + x_1 x_2} = \frac{1 - x_1 - x_2 + x_1 x_2}{1 + x_1 x_2} \]

Thus, we have:

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

### Conclusion

Since both expressions for \( f(x_1) + f(x_2) \) and \( f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right) \) are equal, we conclude that:

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