Question

If \( f(x)=x+\log \left(\frac{x-1}{x+1}\right) \) is a well-defined real valued function then \( f \) is

• A. monotonically decreasing function
• B. monotonically increasing function
• C. increasing in \( (1, \infty) \) and decreasing in \( (-\infty, -1) \)
• D. decreasing in \( (1, \infty) \) and increasing in \( (-\infty, -1) \)

Hint:

Always determine the domain of a logarithmic function before analyzing its derivative signs, as the sign of the denominator often depends on the domain constraints.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

To determine whether the function is increasing or decreasing, we need to find its derivative \( f'(x) \) and analyze its sign within the function's domain. The domain is determined by the condition \( \frac{x-1}{x+1} \textgreater 0 \), which implies \( x \in (-\infty, -1) \cup (1, \infty) \).
Step 2: Key Formula or Approach:

1. Derivative of \( \log u \) is \( \frac{1}{u} \cdot u' \). 2. Quotient Rule: \( \left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2} \). 3. Monotonicity condition: \( f'(x) \textgreater 0 \) implies increasing, \( f'(x) \textless 0 \) implies decreasing.
Step 3: Detailed Explanation:

Given \( f(x) = x + \log\left(\frac{x-1}{x+1}\right) \). Differentiate with respect to \( x \): \[ f'(x) = 1 + \frac{1}{\left(\frac{x-1}{x+1}\right)} \cdot \frac{d}{dx}\left(\frac{x-1}{x+1}\right) \] \[ f'(x) = 1 + \frac{x+1}{x-1} \cdot \left( \frac{1(x+1) - 1(x-1)}{(x+1)^2} \right) \] \[ f'(x) = 1 + \frac{x+1}{x-1} \cdot \frac{2}{(x+1)^2} \] \[ f'(x) = 1 + \frac{2}{(x-1)(x+1)} \] \[ f'(x) = 1 + \frac{2}{x^2-1} \] Combine the terms: \[ f'(x) = \frac{x^2 - 1 + 2}{x^2 - 1} = \frac{x^2 + 1}{x^2 - 1} \] Now, analyze the sign of \( f'(x) \): - The numerator \( x^2 + 1 \) is always positive. - The denominator \( x^2 - 1 \) is positive in the domain \( (-\infty, -1) \cup (1, \infty) \) because \( |x| \textgreater 1 \implies x^2 \textgreater 1 \). Since both numerator and denominator are positive, \( f'(x) \textgreater 0 \) for all \( x \) in the domain. Therefore, \( f \) is strictly increasing.
Step 4: Final Answer:

The function is a monotonically increasing function.