The function `f(x)=log (1+x)-(2+x)` is increasing in

Options

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

Answer

To determine the intervals where the function \( f(x) = \log(1+x) - (2+x) \) is increasing, we need to follow these steps:

### Step 1: Differentiate the Function
First, we find the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}[\log(1+x)] - \frac{d}{dx}[2+x] \]

Using the derivative rules, we have:

- The derivative of \( \log(1+x) \) is \( \frac{1}{1+x} \).
- The derivative of \( 2 \) is \( 0 \).
- The derivative of \( x \) is \( 1 \).

Thus, we can write:

\[ f'(x) = \frac{1}{1+x} - 1 \]

### Step 2: Set the Derivative Greater than Zero
To find where the function is increasing, we set the derivative greater than zero:

\[ f'(x) > 0 \]

This gives us:

\[ \frac{1}{1+x} - 1 > 0 \]

### Step 3: Solve the Inequality
Rearranging the inequality, we have:

\[ \frac{1 - (1+x)}{1+x} > 0 \]

This simplifies to:

\[ \frac{-x}{1+x} > 0 \]

### Step 4: Analyze the Sign of the Expression
To analyze the sign of \( \frac{-x}{1+x} \), we need to consider the numerator and denominator separately:

1. **Numerator**: \( -x > 0 \) implies \( x < 0 \).
2. **Denominator**: \( 1+x > 0 \) implies \( x > -1 \).

### Step 5: Combine the Inequalities
From the two conditions, we have:

- \( x < 0 \)
- \( x > -1 \)

Combining these gives us the interval:

\[ -1 < x < 0 \]

### Conclusion
Thus, the function \( f(x) = \log(1+x) - (2+x) \) is increasing in the interval \( (-1, 0) \).