In the interval left( -frac{pi}{3}, frac{pi}{ | vedclass

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(A) To determine the nature of the function,we find its derivative: $f'(x) = -\frac{1}{2} + \cos x$ For the interval $x \in \left( -\frac{\pi}{3}, \fr

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Increasing

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(A) To determine the nature of the function,we find its derivative: $f'(x) = -\frac{1}{2} + \cos x$ For the interval $x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$,we know that $\cos x$ ranges from $\frac{1}{2}$ to $1$. Specifically,$\frac{1}{2} Subtracting $\frac{1}{2}$ from all parts,we get: $\frac{1}{2} - \frac{1}{2} $0 Since $f'(x) > 0$ for all $x$ in the given interval,the function $f(x)$ is strictly increasing.

In the interval $\left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$,the function $f(x) = -\frac{x}{2} + \sin x$ is:

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