If `f(x)=(ln(1+x))/x, x in (-1,oo)` and `f(0)=1` then `f(x)` is

Options

A. decreasing in `(-1,0)` and increasing in `(0,oo)`
B. always increasing
C. always decreasing
D. increasing in `(-1,0)` and decreasing in `(0,oo)`

Answer

To determine the nature of the function \( f(x) = \frac{\ln(1+x)}{x} \) for \( x \in (-1, \infty) \), we will analyze its behavior by finding its derivative and determining whether it is increasing or decreasing.

### Step 1: Define the function and its derivative
Given:
\[ f(x) = \frac{\ln(1+x)}{x} \]
We also know that \( f(0) = 1 \).

To analyze whether \( f(x) \) is increasing or decreasing, we first need to differentiate \( f(x) \).

### Step 2: Differentiate \( f(x) \)
Using the quotient rule for differentiation:
\[ f'(x) = \frac{(x \cdot \frac{d}{dx}[\ln(1+x)] - \ln(1+x) \cdot \frac{d}{dx}[x])}{x^2} \]
Calculating the derivatives:
- The derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \).
- The derivative of \( x \) is \( 1 \).

Substituting these into the formula gives:
\[ f'(x) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x) \cdot 1}{x^2} \]
This simplifies to:
\[ f'(x) = \frac{x - (1+x) \ln(1+x)}{x^2(1+x)} \]

### Step 3: Analyze the sign of \( f'(x) \)
Now we need to analyze the numerator \( x - (1+x) \ln(1+x) \):
\[ g(x) = x - (1+x) \ln(1+x) \]
We will differentiate \( g(x) \) to determine its behavior.

### Step 4: Differentiate \( g(x) \)
Differentiating \( g(x) \):
\[ g'(x) = 1 - \left( \ln(1+x) + \frac{x}{1+x} \right) \]
This can be simplified to:
\[ g'(x) = 1 - \ln(1+x) - \frac{x}{1+x} \]

### Step 5: Analyze \( g'(x) \)
To determine the sign of \( g'(x) \):
- For \( x \in (-1, \infty) \), \( \ln(1+x) \) is positive.
- The term \( \frac{x}{1+x} \) is also positive for \( x > -1 \).

Thus, \( g'(x) \) is negative for \( x > 0 \), indicating that \( g(x) \) is decreasing in this interval.

### Step 6: Conclusion about \( f(x) \)
Since \( g(x) \) is decreasing and \( g(0) = 0 \), we find that:
- For \( x > 0 \), \( g(x) < 0 \) which implies \( f'(x) < 0 \).
- Therefore, \( f(x) \) is decreasing for \( x > 0 \).

For \( x \in (-1, 0) \), we can analyze similarly, but since \( f(x) \) is continuous and differentiable in the interval, we can conclude that \( f(x) \) is decreasing for all \( x \in (-1, \infty) \).

### Final Answer
Thus, the function \( f(x) \) is **strictly decreasing** for \( x \in (-1, \infty) \).

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