Question

The function \( f(x) = \dfrac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1)\cos x \) :

• A. increases in [1/2, ∞)
• B. decreases in [1/2, ∞)
• C. increases in (-∞, 1/2]
• D. decreases in (-∞, 1/2]

Hint:

To find intervals of monotonicity, find $f'(x)$ and determine where $f'(x) \geq 0$ (increasing) or $f'(x) \leq 0$ (decreasing).

Answer 1

The Correct Option is A

Step 1: $f'(x) = \frac{12x^2 - 6x}{6} - 2\cos x + 2\cos x - (2x-1)\sin x$.
Step 2: $f'(x) = (2x^2 - x) - (2x-1)\sin x = (2x-1)(x - \sin x)$.
Step 3: Since $x>\sin x$ for $x>0$, the sign of $f'(x)$ depends on $(2x-1)$.
Step 4: $f'(x) \geq 0$ for $x \geq 1/2$.