If `f(x)=x.log((1-x)/(1+x)),` determine whether the function `f` is even or odd or neither.

Answer

To determine whether the function \( f(x) = x \log\left(\frac{1-x}{1+x}\right) \) is even, odd, or neither, we will follow these steps:

### Step 1: Define the function
The function is given as:
\[ f(x) = x \log\left(\frac{1-x}{1+x}\right) \]

### Step 2: Calculate \( f(-x) \)
To check if the function is even or odd, we first need to find \( f(-x) \):
\[ f(-x) = -x \log\left(\frac{1-(-x)}{1+(-x)}\right) = -x \log\left(\frac{1+x}{1-x}\right) \]

### Step 3: Simplify \( f(-x) \)
Using the logarithmic identity \( \log\left(\frac{a}{b}\right) = -\log\left(\frac{b}{a}\right) \), we can rewrite \( f(-x) \):
\[ f(-x) = -x \cdot \left(-\log\left(\frac{1-x}{1+x}\right)\right) = x \log\left(\frac{1-x}{1+x}\right) \]
This means:
\[ f(-x) = f(x) \]

### Step 4: Determine if the function is even or odd
Since \( f(-x) = f(x) \), we conclude that the function is **even**.

### Final Conclusion
Therefore, the function \( f(x) = x \log\left(\frac{1-x}{1+x}\right) \) is an **even function**.

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