The interval in which the function \( f(x)=log_e(2+x)-\displaystyle\frac{2x}{2+x} \) is increasing is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (2,\infty) \)
The function is defined only for \( 2+x\gt 0 \), that is, \( x\gt -2 \).
\( \displaystyle f'(x)=\frac{1}{2+x}-\frac{2(2+x)-2x}{(2+x)^2} \)
\( \displaystyle =\frac{1}{2+x}-\frac{4}{(2+x)^2} \)
\( \displaystyle =\frac{(2+x)-4}{(2+x)^2} \)
\( \displaystyle =\frac{x-2}{(2+x)^2} \)
The denominator is always positive, so the sign of \( f'(x) \) is the sign of \( x-2 \):
\( \displaystyle f'(x)\gt 0\ \text{for}\ x\gt 2 \)
\( \displaystyle f'(x)\lt 0\ \text{for}\ -2\lt x\lt 2 \)
So f is increasing on \( (2,\infty) \).
Why the other options are incorrect
Option (1) \( [-2,\infty) \) — Incorrect. f is not even defined at x = −2, and it decreases on (−2, 2).
Option (3) \( (-\infty,2) \) — Incorrect. On (−2, 2), \( f'(x)\lt 0 \), so f is decreasing there; and f is not defined for x ≤ −2.
Option (4) \( (-\infty,-1] \) — Incorrect. f is not defined for x ≤ −2, and it decreases on (−2, −1].