The function \(f(x) = 2\log_e(x-2) - x^2 + 4x + 1,\ (x \gt 2)\) is increasing on the interval:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (2, 3)
f is increasing where f′(x) > 0, within its domain x > 2.
\( \displaystyle f'(x) = \frac{2}{x - 2} - 2x + 4 = \frac{2}{x - 2} - 2(x - 2) \)
\( \displaystyle = \frac{2 - 2(x - 2)^2}{x - 2} \)
\( \displaystyle = \frac{2(3 - x)(x - 1)}{x - 2} \)
For x > 2, both \(x - 1\) and \(x - 2\) are positive, so the sign of f′(x) is the sign of \(3 - x\):
\(2 \lt x \lt 3\): \(f'(x) \gt 0\), f is increasing
\(x \gt 3\): \(f'(x) \lt 0\), f is decreasing
Why the other options are incorrect
Option (1) (1, 2) — Incorrect. f is not defined there, since \(\log_e(x - 2)\) needs x > 2.
Option (2) (2, 6) — Incorrect. f decreases on the part (3, 6); for example \(f'(4) = 1 - 8 + 4 = -3\).
Option (4) \((3, \infty)\) — Incorrect. There \(3 - x \lt 0\), so \(f'(x) \lt 0\) and f is decreasing.