Question

If $f(x) = \log(1 + x) - \frac{2x}{2 + x}$, then $f(x)$ is increasing in ______.

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

Hint:

Always establish the natural domain of the function before analyzing intervals of increase/decrease. Even if a derivative test suggests a larger interval, the function cannot increase where it does not mathematically exist!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A function is increasing where its derivative $f'(x) > 0$. First, note the domain of $\log(1+x)$ is $x > -1$.

Step 2: Formula Application:
$f'(x) = \frac{1}{1+x} - \left[ \frac{(2+x)(2) - 2x(1)}{(2+x)^2} \right]$.

Step 3: Explanation:
$f'(x) = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$. $f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+x^2+4x-4-4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$. For $f'(x) > 0$, since $x^2$ and $(2+x)^2$ are always positive, we only need $1+x > 0$, which means $x > -1$.

Step 4: Final Answer:
The function is increasing in $(-1, \infty)$.