Key Concepts and Formulas

• Monotonicity and Derivatives: A differentiable function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in the interval, and decreasing if $f'(x) \le 0$ for all $x$ in the interval.
• Product Rule: The derivative of $u(x)v(x)$ is $u'(x)v(x) + u(x)v'(x)$.
• Derivatives of Trigonometric Functions: $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$.

Step-by-Step Solution

Step 1: Find the derivative of f(x)

We need to find $f'(x)$ to determine the intervals of increase and decrease. $f(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x$ Differentiating with respect to $x$: $f'(x) = \frac{12x^2 - 6x}{6} - 2\cos x + (2)\cos x + (2x - 1)(-\sin x)$ $f'(x) = 2x^2 - x - 2\cos x + 2\cos x - (2x - 1)\sin x$ $f'(x) = 2x^2 - x - (2x - 1)\sin x$ $f'(x) = x(2x - 1) - (2x - 1)\sin x$ $f'(x) = (2x - 1)(x - \sin x)$

Step 2: Analyze the sign of f'(x)

We want to determine when $f'(x) \ge 0$ (increasing) and $f'(x) \le 0$ (decreasing). We have $f'(x) = (2x - 1)(x - \sin x)$.

Consider the factor $2x - 1$.

• $2x - 1 \ge 0$ when $x \ge \frac{1}{2}$
• $2x - 1 \le 0$ when $x \le \frac{1}{2}$

Now consider the factor $x - \sin x$. We know that for all $x \ge 0$, $x \ge \sin x$, so $x - \sin x \ge 0$. For $x < 0$, we have $x - \sin x > 0$. To see this, consider $g(x) = x - \sin x$. Then $g'(x) = 1 - \cos x \ge 0$. Since $g(0) = 0$, $g(x)$ is increasing for all $x$. Therefore, $x - \sin x \ge 0$ for all $x$.

Thus, $x - \sin x \ge 0$ for all $x$.

Step 3: Determine the intervals of increase and decrease

Since $x - \sin x \ge 0$ for all $x$, the sign of $f'(x)$ depends only on the sign of $2x - 1$.

• If $x < \frac{1}{2}$, then $2x - 1 < 0$, so $f'(x) = (2x - 1)(x - \sin x) \le 0$. Therefore, $f(x)$ is decreasing on $\left( -\infty, \frac{1}{2} \right]$.
• If $x > \frac{1}{2}$, then $2x - 1 > 0$, so $f'(x) = (2x - 1)(x - \sin x) \ge 0$. Therefore, $f(x)$ is increasing on $\left[ \frac{1}{2}, \infty \right)$.

Common Mistakes & Tips

• Forgetting the Product Rule: When differentiating $(2x - 1)\cos x$, remember to use the product rule.
• Sign Errors: Be careful with signs when differentiating and analyzing the sign of $f'(x)$.
• Understanding $x - \sin x$: The fact that $x \ge \sin x$ for $x \ge 0$ is a standard inequality. You can prove it using calculus. Showing $x - \sin x \ge 0$ for all $x$ is crucial.

Summary

We found the derivative of $f(x)$ to be $f'(x) = (2x - 1)(x - \sin x)$. We then analyzed the sign of $f'(x)$. Since $x - \sin x \ge 0$ for all $x$, the sign of $f'(x)$ is determined by the sign of $2x - 1$. Therefore, $f(x)$ is decreasing on $\left( -\infty, \frac{1}{2} \right]$ and increasing on $\left[ \frac{1}{2}, \infty \right)$.

The final answer is \boxed{A}, which corresponds to option (A).