Question

The function \( f(x) = \log x - \frac{2x}{x+2} \) is increasing for all:

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

Hint:

To determine where a function is increasing, find its derivative and solve the inequality \( f'(x)>0 \).

Answer 1

The Correct Option is D

Step 1: Find the derivative of the function.
We are given: \[ f(x) = \log x - \frac{2x}{x + 2} \] To find where the function is increasing, we first compute the derivative of \( f(x) \): \[ f'(x) = \frac{1}{x} - \frac{d}{dx} \left( \frac{2x}{x + 2} \right) \] Using the quotient rule to differentiate \( \frac{2x}{x+2} \): \[ \frac{d}{dx} \left( \frac{2x}{x+2} \right) = \frac{(x+2)(2) - 2x(1)}{(x+2)^2} = \frac{2x + 4 - 2x}{(x+2)^2} = \frac{4}{(x+2)^2} \] Thus, the derivative is: \[ f'(x) = \frac{1}{x} - \frac{4}{(x+2)^2} \] Step 2: Analyze the sign of the derivative.
For the function to be increasing, we need \( f'(x)>0 \): \[ \frac{1}{x} - \frac{4}{(x+2)^2}>0 \] This inequality holds for \( x>0 \), because the first term \( \frac{1}{x} \) is positive for \( x>0 \) and the second term is always positive. Step 3: Conclusion.
Thus, the function is increasing for \( \boxed{x \in (0, \infty)} \).