The interval on which the function \(f(x) = x^3 + 2x^2 - 1\) is decreasing, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \([-4/3, 0)\)
f is decreasing where its derivative is not positive:
\(f'(x) = 3x^2 + 4x = x(3x + 4)\)
\(x(3x + 4) \le 0\)
\( \displaystyle -\frac{4}{3} \le x \le 0 \)
So f is decreasing on \([-4/3, 0]\), and hence on \([-4/3, 0)\), which lies inside it. Outside this interval \(f'(x) \gt 0\), so f is increasing on \((-\infty, -4/3)\) and on \((0, \infty)\).
Why the other options are incorrect
Option (1) \((-\infty, -4/3)\) — Incorrect. f is increasing there; for example \(f'(-2) = 4 \gt 0\).
Option (2) \([0, \infty)\) — Incorrect. f is increasing there; for example \(f'(1) = 7 \gt 0\).
Option (4) \([-4/3, \infty)\) — Incorrect. It includes \((0, \infty)\), where f is increasing.