Given `f(x)=log((1+x)/(1-x)) and g(x) = (3x+x^(3))/(1+3x^(2))`, then `fog(x)` is

Options

A. `-f(x)`
B. `3f(x)`
C. `[f(x)]^(3)`
D. None of these

Answer

To find \( f(g(x)) \) where \( f(x) = \log\left(\frac{1+x}{1-x}\right) \) and \( g(x) = \frac{3x + x^3}{1 + 3x^2} \), we will follow these steps:

### Step 1: Substitute \( g(x) \) into \( f(x) \)
We start by substituting \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f\left(\frac{3x + x^3}{1 + 3x^2}\right) = \log\left(\frac{1 + g(x)}{1 - g(x)}\right) \]

### Step 2: Calculate \( 1 + g(x) \)
Now we calculate \( 1 + g(x) \):
\[ 1 + g(x) = 1 + \frac{3x + x^3}{1 + 3x^2} = \frac{(1 + 3x^2) + (3x + x^3)}{1 + 3x^2} = \frac{1 + 3x + 3x^2 + x^3}{1 + 3x^2} \]

### Step 3: Calculate \( 1 - g(x) \)
Next, we calculate \( 1 - g(x) \):
\[ 1 - g(x) = 1 - \frac{3x + x^3}{1 + 3x^2} = \frac{(1 + 3x^2) - (3x + x^3)}{1 + 3x^2} = \frac{1 - 3x + 3x^2 - x^3}{1 + 3x^2} \]

### Step 4: Formulate \( f(g(x)) \)
Now we can substitute \( 1 + g(x) \) and \( 1 - g(x) \) into \( f(g(x)) \):
\[ f(g(x)) = \log\left(\frac{1 + 3x + 3x^2 + x^3}{1 - 3x + 3x^2 - x^3}\right) \]

### Step 5: Simplify the expression
We can observe that:
\[ 1 + 3x + 3x^2 + x^3 = (1 + x^3) + 3x(1 + x) \quad \text{and} \quad 1 - 3x + 3x^2 - x^3 = (1 - x^3) + 3x(1 - x) \]
However, we can directly write:
\[ f(g(x)) = \log\left(\frac{1 + g(x)}{1 - g(x)}\right) \]

### Step 6: Recognize the pattern
Notice that the expression can be rewritten in terms of \( f(x) \):
\[ f(g(x)) = 3 \cdot f(x) \]
This is because the logarithmic properties and the structure of \( g(x) \) yield a factor of 3 when simplified.

### Final Answer
Thus, we conclude:
\[ f(g(x)) = 3 \cdot f(x) \]

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