The interval in which the function \(f(x) = 2x^3 + 3x^2 - 12x + 1\) is strictly increasing is –
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-\infty, -2) \cup (1, \infty)\)
f is strictly increasing where f’(x) > 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:
\(x \lt -2\) or \(x \gt 1\)
Between −2 and 1, f’(x) < 0 and f decreases. So f is strictly increasing on each of the intervals \((-\infty, -2)\) and \((1, \infty)\), which is what the union in Option (3) lists.