The function \(f:R \to R\) (where \(R\) is set of real numbers) defined as \(f(x) = x^2 + 2x\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decreasing in \((-\infty, -1]\)
The sign of the derivative decides where f increases or decreases:
\(f'(x) = 2x + 2 = 2(x + 1)\)
\(f'(x) \lt 0 \text{ for } x \lt -1, \qquad f'(x) \gt 0 \text{ for } x \gt -1\)
So f decreases on \((-\infty, -1)\) and increases on \((-1, \infty)\). At x = −1 the derivative is 0, but a zero derivative at a single endpoint does not stop f from decreasing up to that point (for \(x_1 \lt x_2 \le -1\), \(f(x_1) \gt f(x_2)\)). So f is decreasing in \((-\infty, -1]\).
Why the other options are incorrect
Option (2) increasing in \((-\infty, -1]\) — Incorrect. \(f'(x) \lt 0\) there; for example f(−3) = 3 is greater than f(−1) = −1.
Option (3) decreasing in \((-\infty, 2]\) — Incorrect. The interval goes past the turning point x = −1, and f increases on [−1, 2]: f(−1) = −1, f(2) = 8.
Option (4) increasing on \(R\) — Incorrect. \(f'(x) \lt 0\) for every x < −1.