Question

The interval in which the curve represented by \( f(x) = 2x + \log\left(\frac{x}{2 + x}\right) \) is increasing is

• A. \((- \infty, 0)\)
• B. \((-2, \infty)\)
• C. \((- \infty, -2) \cup (0, \infty)\)
• D. \((-2, 0)\)

Hint:

For logarithmic functions, always check the domain carefully before analyzing monotonicity.

Answer 1

The Correct Option is C

We are given: 
\[ f(x) = 2x + \log\left(\frac{x}{2 + x}\right) \] Step 1: Find the domain 
For \( \log\left(\frac{x}{2+x}\right) \) to be defined, the argument must be positive: 
\[ \frac{x}{2 + x} > 0 \] This inequality holds when: 
- \(x > 0\) and \(2 + x > 0 \Rightarrow x > 0\) - OR \(x < 0\) and \(2 + x < 0 \Rightarrow x < -2\) So domain of \(f(x)\) is: \( (-\infty, -2) \cup (0, \infty) \) 
Step 2: Differentiate \(f(x)\) 
\[ f'(x) = \frac{d}{dx} \left( 2x + \log\left(\frac{x}{2 + x} \right) \right) = 2 + \frac{d}{dx} \left( \log\left(\frac{x}{2 + x} \right) \right) \] Using logarithmic derivative: 
\[ \frac{d}{dx} \left( \log\left(\frac{x}{2 + x} \right) \right) = \frac{(2 + x)\cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(2 + x)}{(2 + x)^2} = \frac{(2 + x)(1) - x(1)}{(2 + x)^2} = \frac{2 + x - x}{(2 + x)^2} = \frac{2}{(2 + x)^2} \] So: 
\[ f'(x) = 2 + \frac{2}{(2 + x)^2} \] Since \((2 + x)^2 > 0\) in the domain, the term \(\frac{2}{(2 + x)^2} > 0\), so: 
\[ f'(x) > 2 > 0 \] Step 3: Conclude increasing intervals 
Since \(f'(x) > 0\) in its domain \((- \infty, -2) \cup (0, \infty)\), function is increasing in this domain. Step 4: Final Answer 
\[ \boxed{(-\infty, -2) \cup (0, \infty)} \]