The function f(x) = frac{4x^3 - 3x^2}{6} - 2 | vedclass

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(A) Given $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{12x^2 - 6x}{6} - 2 \cos x

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increases in $[\frac{1}{2}, \infty)$

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(A) Given $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{12x^2 - 6x}{6} - 2 \cos x + [2 \cos x + (2x - 1)(-\sin x)]$ $f'(x) = (2x^2 - x) - 2 \cos x + 2 \cos x - (2x - 1) \sin x$ $f'(x) = x(2x - 1) - (2x - 1) \sin x$ $f'(x) = (2x - 1)(x - \sin x)$. We know that for $x > 0$,$x > \sin x$,so $(x - \sin x) > 0$. For $x Now,analyze the sign of $f'(x) = (2x - 1)(x - \sin x)$: $1$. If $x \in [\frac{1}{2}, \infty)$,then $(2x - 1) \geq 0$ and $(x - \sin x) > 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is increasing. $2$. If $x \in [0, \frac{1}{2}]$,then $(2x - 1) \leq 0$ and $(x - \sin x) \geq 0$,so $f'(x) \leq 0$. Thus,$f(x)$ is decreasing. $3$. If $x \in (-\infty, 0]$,then $(2x - 1) Comparing with the options,$f(x)$ increases in $[\frac{1}{2}, \infty)$.

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

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