The interval in which the function f(x) = 2x^3 + 3x^2 - 12x + 1 is strictly increasing is - Option 1: [-2, 1] Option 2: (-∞, 1] Option 3: (-∞, -2) (1, ∞) Option 4: (-∞, -1] [2, ∞)

Answer: (-∞, -2) (1, ∞). Solution: f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1) For strict increase, f'(x) > 0. (x+2)(x-1) > 0 when x 1. So f is strictly increasing on (-∞, -2) (1, ∞).

Calculus

Application of Derivatives

Easy

The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing is -

$[-2, 1]$

$(-\infty, 1]$

$(-\infty, -2) \cup (1, \infty)$

$(-\infty, -1] \cup [2, \infty)$

✅ Correct Option: 3

$f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1)$

For strict increase, $f'(x) > 0$ .

$(x+2)(x-1) > 0$ when $x < -2$ or $x > 1$ .

So $f$ is strictly increasing on $(-\infty, -2) \cup (1, \infty)$ .

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The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing is -