The function \(f(x)=x^3-3x\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Increasing in \((-\infty, -1] \cup [1, \infty)\) and decreasing in \((-1, 1)\)
The sign of the derivative decides where f increases or decreases.
\( f'(x) = 3x^2-3 = 3(x-1)(x+1) \)
\( f'(x) = 0 \) at \( x = -1,\ 1 \)
For \(x \lt -1\) and for \(x \gt 1\) both factors have the same sign, so \(f'(x) \gt 0\); f is increasing on \((-\infty, -1]\) and on \([1, \infty)\).
For \(-1 \lt x \lt 1\) the factors have opposite signs, so \(f'(x) \lt 0\); f is decreasing on \((-1, 1)\).
Option (4) joins the two intervals of increase with \(\cup\); it means f is increasing on each of these intervals, not across the gap between them (\(f(-1) = 2\) is larger than \(f(1) = -2\)).
Why the other options are incorrect
Option (1) Increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\) — Incorrect. \(f'(0.5) = -2.25 \lt 0\), so f is decreasing just to the right of 0.
Option (2) Decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\) — Incorrect. \(f'(2) = 9 \gt 0\), so f is increasing on \((1, \infty)\).
Option (3) Decreasing in \((-\infty, -1] \cup [1, \infty)\) and increasing in \((-1, 1)\) — Incorrect. The roles are reversed: \(f'(0) = -3 \lt 0\), so f decreases on \((-1, 1)\).