91Ó°ÊÓ

To find the intervals where the function is increasing or decreasing, we must first find the derivative of the function \(f(x)\). We have \(f(x) = 3\cos(3x)\). Using the chain rule, we have \(f'(x) = -9\sin(3x)\).

To find the critical points, we need to set the derivative equal to zero and solve for \(x\): $$-9\sin(3x) = 0$$ We get: $$\sin(3x) = 0$$ On the interval \([-\pi, \pi]\), the possible solutions for this equation are when \(3x = n\pi\) where \(n\) is an integer. Divide both sides by \(3\) to get: $$x=\frac{n\pi}{3}$$ However, only the values of \(n\) such that \( -\pi \le \frac{n\pi}{3} \le \pi\) should be considered in our analysis. The integers that satisfy this inequality are \(n=-3,-2,-1,0,1,2,3\). So, the critical points on the interval are \(x=-\pi, -\frac{2\pi}{3}, -\frac{\pi}{3}, 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi\).

To determine whether the function is increasing or decreasing, we will analyze the sign of the derivative on the intervals between the critical points. Consider the following intervals: 1. \((-\pi, -\frac{2\pi}{3})\) 2. \((-\frac{2\pi}{3}, -\frac{\pi}{3})\) 3. \((-\frac{\pi}{3}, 0)\) 4. \((0, \frac{\pi}{3})\) 5. \((\frac{\pi}{3}, \frac{2\pi}{3})\) 6. \((\frac{2\pi}{3}, \pi)\) To check the sign, we will substitute a point from each interval into the derivative \(f'(x)\): 1. \(f'(-\frac{5\pi}{6}) = -9\sin(-\frac{5\pi}{2}) = 9 > 0\) 2. \(f'(-\frac{\pi}{2}) = -9\sin(-\pi) = 0\) 3. \(f'(-\frac{\pi}{6}) = -9\sin(-\frac{\pi}{2}) = -9 < 0\) 4. \(f'(\frac{\pi}{6}) = -9\sin(\frac{\pi}{2}) = -9 < 0\) 5. \(f'(\frac{\pi}{2}) = -9\sin(\pi) = 0\) 6. \(f'(\frac{5\pi}{6}) = -9\sin(\frac{5\pi}{2}) = -9 < 0\)

According to the sign of the derivative on the intervals between the critical points, we can conclude the following: - The function is increasing on the interval \((-\pi, -\frac{2\pi}{3})\). - The function is decreasing on the intervals \((-\frac{\pi}{3}, 0)\), \((0, \frac{\pi}{3})\), and \((\frac{2\pi}{3}, \pi)\). Thus, the intervals on which \(f\) is increasing and decreasing are: Increasing: \((-\pi, -\frac{2\pi}{3})\) Decreasing: \((-\frac{\pi}{3}, 0) \cup (0, \frac{\pi}{3}) \cup (\frac{2\pi}{3}, \pi)\)