The function `f(x)=log(1+x)-2*x/(2+x)` is increasing on

Options

A. `(-1,oo)`
B. `(-oo,0)`
C. `(-oo,oo)`
D. None of these

Explanation

To determine the intervals where the function \( f(x) = \log(1+x) - \frac{2x}{2+x} \) is increasing, we need to find the derivative \( f'(x) \) and analyze where it is greater than zero.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \log(1+x) \right) - \frac{d}{dx} \left( \frac{2x}{2+x} \right) \]

Using the derivative of \( \log(1+x) \):
\[ \frac{d}{dx} \left( \log(1+x) \right) = \frac{1}{1+x} \]

For the second term, we apply the quotient rule:
\[ \frac{d}{dx} \left( \frac{2x}{2+x} \right) = \frac{(2+x)(2) - 2x(1)}{(2+x)^2} = \frac{2(2+x) - 2x}{(2+x)^2} = \frac{4}{(2+x)^2} \]

Thus, we have:
\[ f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2} \]

### Step 2: Set the derivative greater than zero
To find where \( f(x) \) is increasing, we set \( f'(x) > 0 \):
\[ \frac{1}{1+x} - \frac{4}{(2+x)^2} > 0 \]

### Step 3: Solve the inequality
Rearranging gives:
\[ \frac{1}{1+x} > \frac{4}{(2+x)^2} \]

Cross-multiplying (valid since both sides are positive for \( x > -1 \)):
\[ (2+x)^2 > 4(1+x) \]

Expanding both sides:
\[ 4 + 4x + x^2 > 4 + 4x \]

This simplifies to:
\[ x^2 > 0 \]

### Step 4: Determine the intervals
The inequality \( x^2 > 0 \) holds for all \( x \neq 0 \). However, we also need to consider the domain of \( f(x) \). The function \( \log(1+x) \) is defined for \( x > -1 \). Therefore, we combine these results:

- \( f'(x) > 0 \) for \( x \in (-1, 0) \) and \( x \in (0, \infty) \).
- The function is defined for \( x > -1 \).

Thus, the function \( f(x) \) is increasing on the interval:
\[ (-1, 0) \cup (0, \infty) \]

### Final Answer
The function \( f(x) = \log(1+x) - \frac{2x}{2+x} \) is increasing on the intervals \( (-1, 0) \) and \( (0, \infty) \).

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