The largest interval, in which the function \(f(x) = x^3 + 2x^2-1\) is increasing, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left(-\infty,\displaystyle\frac{-4}{3}\right]\cup[0,\infty)\)
f is increasing where \(f'(x) \ge 0\):
\(f'(x) = 3x^2 + 4x = x(3x + 4)\)
\( \displaystyle f'(x) = 0 \text{ at } x = -\frac{4}{3} \text{ and } x = 0 \)
Sign of \(f'(x)\) on each part:
\(x \lt -\frac{4}{3}\): both factors negative, \(f'(x) \gt 0\)
\(-\frac{4}{3} \lt x \lt 0\): factors of opposite signs, \(f'(x) \lt 0\)
\(x \gt 0\): both factors positive, \(f'(x) \gt 0\)
So f is increasing on \(\left(-\infty, -\frac{4}{3}\right]\) and on \([0, \infty)\), and decreasing on \(\left[-\frac{4}{3}, 0\right]\). Option (4) lists both of these intervals.
Here f is increasing on each interval separately, not across the gap between them: \(f\left(-\frac{4}{3}\right) = \frac{5}{27}\) is greater than \(f(0) = -1\).
Why the other options are incorrect
Option (1) \((0,\infty)\) — Incorrect. f is also increasing on \(\left(-\infty, -\frac{4}{3}\right]\), so this is only part of the answer.
Option (2) \((-4,4)\) — Incorrect. It contains \(\left(-\frac{4}{3}, 0\right)\), where f decreases; for example \(f'(-1) = -1\).
Option (3) \(\left[\displaystyle\frac{-4}{3},0\right]\) — Incorrect. This is the interval where \(f'(x) \le 0\), so f is decreasing there.