Consider the function f(x)=-2x^3+3x^2+36x-15 , then which of the foll…

The correct answer is Option (1) → (A) and (B) only

The function increases where f′(x) > 0 and decreases where f′(x) < 0.

    \( f'(x)=-6x^{2}+6x+36 \)

    \( =-6\left(x^{2}-x-6\right) \)

    \( =-6(x-3)(x+2) \)

f′(x) = 0 at x = −2 and x = 3.

    \( -2\lt x\lt 3:\ (x-3)(x+2)\lt 0,\ \text{so}\ f'(x)\gt 0 \)
    \( x\lt -2\ \text{or}\ x\gt 3:\ (x-3)(x+2)\gt 0,\ \text{so}\ f'(x)\lt 0 \)

The union in (B) and (C) is read in the usual way: f decreases (or increases) on each of the intervals \( (-\infty,-2) \) and \( (3,\infty) \).

So f is increasing on (−2, 3) and decreasing on each of \( (-\infty,-2) \) and \( (3,\infty) \).

Checking the statements

(A) \( f(x) \) increases on (−2, 3) — Correct. f′(x) > 0 there.

(B) \( f(x) \) decreases on \( (-\infty,-2)\cup(3,\infty) \) — Correct. f′(x) < 0 on both intervals.

(C) \( f(x) \) increases on \( (-\infty,-2)\cup(3,\infty) \) — Incorrect. f′(x) < 0 there, so f decreases.

(D) \( f(x) \) neither increases nor decreases on (−2, 3) — Incorrect. f′(x) > 0 there, so f increases.