Let `f:(-1,oo) rarr R` be defined by `f(0)=1` and `f(x)=1/x log_(e)(1+x), x !=0`. Then the function f:

Options

A. increases in `(–1, oo) `
B. decreases in (–1,0) and increases in `(0, oo)`
C. increases in (–1,0) and decreases in `(0, oo)`
D. decreases in `(-1, oo)`.

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 1 & \text{if } x = 0 \\ \frac{1}{x} \log(1+x) & \text{if } x \neq 0 \end{cases} \]

We want to determine the behavior of this function on the interval \( (-1, \infty) \).

### Step 1: Find the derivative of \( f(x) \)

For \( x \neq 0 \), we will differentiate \( f(x) = \frac{1}{x} \log(1+x) \). We can use the quotient rule for differentiation.

Let \( u = \log(1+x) \) and \( v = x \). Then,

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Calculating \( u' \) and \( v' \):

- \( u' = \frac{1}{1+x} \)
- \( v' = 1 \)

Now substituting these into the quotient rule:

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

This simplifies to:

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

### Step 2: Analyze the sign of \( f'(x) \)

To determine where \( f'(x) \) is positive or negative, we need to analyze the expression:

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

We will check the sign of \( f'(x) \) in two intervals: \( (-1, 0) \) and \( (0, \infty) \).

### Step 3: Check the interval \( (0, \infty) \)

Let's evaluate \( f'(x) \) at a point in this interval, say \( x = \frac{1}{2} \):

\[ f'\left(\frac{1}{2}\right) = \frac{1}{1+\frac{1}{2}} - \frac{\log(1+\frac{1}{2})}{\left(\frac{1}{2}\right)^2} \]

Calculating this gives:

\[ f'\left(\frac{1}{2}\right) = \frac{1}{\frac{3}{2}} - \frac{\log\left(\frac{3}{2}\right)}{\frac{1}{4}} = \frac{2}{3} - 4 \log\left(\frac{3}{2}\right) \]

Since \( \log\left(\frac{3}{2}\right) \) is positive, we can estimate \( 4 \log\left(\frac{3}{2}\right) \) and find that \( f'\left(\frac{1}{2}\right) < 0 \). Thus, \( f(x) \) is decreasing on \( (0, \infty) \).

### Step 4: Check the interval \( (-1, 0) \)

Now, let's evaluate \( f'(x) \) at a point in this interval, say \( x = -\frac{1}{2} \):

\[ f'\left(-\frac{1}{2}\right) = \frac{1}{1-\frac{1}{2}} - \frac{\log\left(1-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^2} \]

Calculating this gives:

\[ f'\left(-\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} - \frac{\log\left(\frac{1}{2}\right)}{\frac{1}{4}} = 2 + 4 \cdot 0.693 \]

Since \( \log\left(\frac{1}{2}\right) \) is negative, we find that \( f'\left(-\frac{1}{2}\right) > 0 \). Thus, \( f(x) \) is increasing on \( (-1, 0) \).

### Conclusion

From our analysis, we find that:

- \( f(x) \) is increasing on \( (-1, 0) \)
- \( f(x) \) is decreasing on \( (0, \infty) \)

Thus, the correct option is that \( f(x) \) is decreasing on \( (0, \infty) \) and increasing on \( (-1, 0) \).