Given `f(x)=log[((1+x))/((1-x))]andg(x)=((3x+x^(2)))/((1+3x^(2)))`, then what is `f[g(x)]` equal to ?

Options

A. `-f(x)`
B. `3[f(x)]`
C. `[f(x)]^(3)`
D. `-3[f(x)]`

Explanation

To find \( f[g(x)] \), we need to substitute \( g(x) \) into the function \( f(x) \).

Given:
- \( f(x) = \log\left(\frac{1+x}{1-x}\right) \)
- \( g(x) = \frac{3x + x^2}{1 + 3x^2} \)

### Step 1: Substitute \( g(x) \) into \( f(x) \)

We need to compute \( f[g(x)] \):
\[ f[g(x)] = f\left(\frac{3x + x^2}{1 + 3x^2}\right) \]

### Step 2: Write \( f(g(x)) \)

Using the definition of \( f(x) \):
\[ f[g(x)] = \log\left(\frac{1 + g(x)}{1 - g(x)}\right) \]

### Step 3: Calculate \( 1 + g(x) \) and \( 1 - g(x) \)

First, calculate \( 1 + g(x) \):
\[ 1 + g(x) = 1 + \frac{3x + x^2}{1 + 3x^2} = \frac{(1 + 3x^2) + (3x + x^2)}{1 + 3x^2} = \frac{1 + 3x^2 + 3x + x^2}{1 + 3x^2} = \frac{1 + 4x^2 + 3x}{1 + 3x^2} \]

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

### Step 4: Substitute back into \( f[g(x)] \)

Now substituting back into the logarithm:
\[ f[g(x)] = \log\left(\frac{1 + 4x^2 + 3x}{1 + 2x^2 - 3x}\right) \]

### Final Answer

Thus, we have:
\[ f[g(x)] = \log\left(\frac{1 + 4x^2 + 3x}{1 + 2x^2 - 3x}\right) \]

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