Question

The function \( f(x) = \log(1 + x) - \frac{2x}{2+x} \) is increasing on

• A. \((-\infty, \infty)\)
• B. \((-5, \infty)\)
• C. \((-\infty, 0)\)
• D. \((-1, \infty)\)

Hint:

Always check the domain of the function first. For \(\log(1+x)\), we need \(x > -1\). Then compute derivative and simplify to \(x^2 > 0\), confirming increase everywhere except a single point.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need the interval where \(f(x)\) is increasing. The function is defined only when the argument of \(\log\) is positive, i.e., \(1+x > 0 \Rightarrow x > -1\). So the domain is \((-1, \infty)\). Options outside this domain are automatically invalid.

Step 2: Key Formula or Approach:
A function is increasing where its derivative \(f'(x) > 0\). Compute \(f'(x)\) using differentiation rules.

Step 3: Detailed Explanation:
\(f(x) = \ln(1+x) - \frac{2x}{2+x}\) (using natural log, any base gives same monotonicity). \[ f'(x) = \frac{1}{1+x} - \frac{d}{dx}\left(\frac{2x}{2+x}\right) \] Using quotient rule: \(\frac{d}{dx}\left(\frac{2x}{2+x}\right) = \frac{2(2+x) - 2x(1)}{(2+x)^2} = \frac{4+2x-2x}{(2+x)^2} = \frac{4}{(2+x)^2}\). Thus \(f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}\). Set \(f'(x) > 0\): \[ \frac{1}{1+x} > \frac{4}{(2+x)^2} \implies (2+x)^2 > 4(1+x) \] Expanding: \(x^2 + 4x + 4 > 4 + 4x \implies x^2 > 0\). This inequality holds for all \(x \neq 0\). Within the domain \(x > -1\), we have \(f'(x) > 0\) for all \(x > -1\) except \(x = 0\) where \(f'(0)=0\). Since the derivative is non‑negative and zero only at an isolated point, \(f\) is increasing on the entire domain \((-1, \infty)\).

Step 4: Final Answer:
The function is increasing on \((-1, \infty)\), which corresponds to option (D).