For the function f(x) = -2x^3+3x^2+36x-10 , which of the following is…

The correct answer is Option (2) → (B), (C) and (D) only

f is increasing where f′(x) > 0 and decreasing where f′(x) < 0.

    \(f'(x) = -6x^2 + 6x + 36\)

    \( = -6(x^2 - x - 6)\)

    \( = -6(x - 3)(x + 2)\)

For \(x \lt -2\), both brackets are negative, their product is positive, and \(f'(x) \lt 0\).

For \(-2 \lt x \lt 3\), the product is negative, and \(f'(x) \gt 0\).

For \(x \gt 3\), both brackets are positive, and \(f'(x) \lt 0\).

So f is decreasing in \((-\infty, -2)\), increasing in \((-2, 3)\) and decreasing in \((3, \infty)\).

Checking the statements

(A) f is increasing in \((-\infty, -2)\) — Incorrect. There \(f'(x) \lt 0\); for example \(f'(-3) = -36\).

(B) f is increasing in \((-2, 3)\) — Correct. There \(f'(x) \gt 0\); for example \(f'(0) = 36\).

(C) f is decreasing in \((-\infty, -2)\) — Correct. There \(f'(x) \lt 0\); for example \(f'(-3) = -36\).

(D) f is decreasing in \((3, \infty)\) — Correct. There \(f'(x) \lt 0\); for example \(f'(4) = -36\).