The interval in which the function \(f(x) = 2x^3 + 3x^2 - 12x + 1\) is strictly increasing, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((-\infty,-2)\cup (1,\infty)\)
f is strictly increasing where \(f'(x) \gt 0\):
\(f'(x) = 6x^2 + 6x - 12\)
\(= 6(x^2 + x - 2)\)
\(= 6(x + 2)(x - 1)\)
The product \((x+2)(x-1)\) is positive when both factors have the same sign:
\(f'(x) \gt 0\) for \(x \lt -2\) or \(x \gt 1\)
\(f'(x) \lt 0\) for \(-2 \lt x \lt 1\)
So f is strictly increasing in \((-\infty,-2)\cup(1,\infty)\).
Why the other options are incorrect
Option (2) \((-\infty,1)\) — Incorrect. It contains x = 0, where \(f'(0) = -12 \lt 0\).
Option (3) \((-\infty,-1)\cup (2,\infty)\) — Incorrect. It contains x = −1.5, where \(f'(-1.5) = 6(0.5)(-2.5) \lt 0\).
Option (4) \((-2,1)\) — Incorrect. Here \(f'(x) \lt 0\), so this is the interval where f is strictly decreasing.