Key Concepts and Formulas

• Monotonicity and the First Derivative Test: A function $f(x)$ is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$.
• Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
• Maclaurin Series: $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$ for $|x| < 1$.

Step-by-Step Solution

Step 1: Calculate the first derivative, $f'(x)$ for $x \ne 0$.

We are given $f(x) = \frac{1}{x} \log_e(1+x)$ for $x \ne 0$. We will use the quotient rule to find $f'(x)$. Let $u(x) = \log_e(1+x)$ and $v(x) = x$. Then, $u'(x) = \frac{1}{1+x}$ and $v'(x) = 1$.

Applying the quotient rule: $f'(x) = \frac{\frac{1}{1+x} \cdot x - \log_e(1+x) \cdot 1}{x^2} = \frac{\frac{x}{1+x} - \log_e(1+x)}{x^2} = \frac{x - (1+x)\log_e(1+x)}{x^2(1+x)}$

Step 2: Analyze the sign of $f'(x)$.

To determine the sign of $f'(x)$, we need to analyze the sign of the numerator, $g(x) = x - (1+x)\log_e(1+x)$, since the denominator $x^2(1+x)$ is positive for $x \in (-1, 0) \cup (0, \infty)$.

Let's find the derivative of $g(x)$: $g'(x) = 1 - \left[1 \cdot \log_e(1+x) + (1+x) \cdot \frac{1}{1+x}\right] = 1 - \log_e(1+x) - 1 = -\log_e(1+x)$

Now, let's find the derivative of $g'(x)$: $g''(x) = -\frac{1}{1+x}$ Since $x > -1$, $1+x > 0$, so $g''(x) = -\frac{1}{1+x} < 0$. This means that $g'(x)$ is a decreasing function.

We also have $g'(0) = -\log_e(1+0) = -\log_e(1) = 0$. Since $g'(x)$ is decreasing and $g'(0) = 0$, we have $g'(x) > 0$ for $x \in (-1, 0)$ and $g'(x) < 0$ for $x \in (0, \infty)$.

This means that $g(x)$ is increasing on $(-1, 0)$ and decreasing on $(0, \infty)$. Also, $g(0) = 0 - (1+0)\log_e(1+0) = 0$. Since $g(x)$ is increasing on $(-1, 0)$ and $g(0) = 0$, $g(x) < 0$ for $x \in (-1, 0)$. Since $g(x)$ is decreasing on $(0, \infty)$ and $g(0) = 0$, $g(x) < 0$ for $x \in (0, \infty)$.

Therefore, $g(x) < 0$ for all $x \in (-1, 0) \cup (0, \infty)$.

Since the denominator of $f'(x)$ is positive and the numerator $g(x)$ is negative, $f'(x) < 0$ for all $x \in (-1, 0) \cup (0, \infty)$.

Step 3: Check the behavior at x = 0

We are given $f(0) = 1$. To check continuity at $x=0$, we can use L'Hopital's rule to find $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log_e(1+x)}{x}$. This is of the form $\frac{0}{0}$, so we can apply L'Hopital's rule: $\lim_{x \to 0} \frac{\log_e(1+x)}{x} = \lim_{x \to 0} \frac{\frac{1}{1+x}}{1} = \lim_{x \to 0} \frac{1}{1+x} = 1$ Thus, $f(x)$ is continuous at $x=0$.

Since $f'(x) < 0$ for $x \ne 0$, $f(x)$ is a decreasing function in the interval $(-1, \infty)$.

Common Mistakes & Tips

• Remember to use the quotient rule correctly when finding the derivative.
• Analyzing the sign of the derivative is crucial for determining monotonicity.
• Don't forget to consider the given value of the function at $x=0$ and check for continuity.

Summary

We found the derivative of the given function $f(x)$ and analyzed its sign. The derivative $f'(x)$ is negative for all $x$ in the domain $(-1, \infty)$ except $x=0$. This indicates that the function $f(x)$ is decreasing in the interval $(-1, \infty)$.

Final Answer

The final answer is \boxed{decreases in (–1, $\infty$)}, which corresponds to option (A).