91Ó°ÊÓ

First, we need to find the first derivative of the function \(f(x) = \cos^2 x\) using the chain rule. Let \(u=\cos x\). Then \(f(x) = u^2\). First, we find the derivatives of \(u\) and \(f(u)\) with respect to \(x\): $$ \frac{du}{dx} = -\sin x $$ $$ \frac{df(u)}{du} = 2u $$ Now, we use the chain rule to find the derivative of \(f(x)\) with respect to x: $$ \frac{df(x)}{dx} = \frac{df(u)}{du} \cdot \frac{du}{dx} = 2u(-\sin x)= -2\cos x\sin x $$

To find the critical points, we need to find the values of x where the first derivative is equal to 0 or undefined. In this case, \(f'(x) = -2\cos x\sin x\), which is defined for all x. Therefore, we set \(f'(x)\) to 0 and solve for x: $$ -2\cos x\sin x = 0 $$ To find the solutions, we can split the equation into two cases: 1. \(\cos x = 0\): Satisfying this equation on the interval \([-\pi, \pi]\), we have \(x = -\frac{\pi}{2}, \frac{\pi}{2}\). 2. \(\sin x = 0\): Satisfying this equation on the interval \([-\pi, \pi]\), we have \(x = -\pi, 0, \pi\). Together, these five critical points are \(-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\).

To determine whether the function is increasing or decreasing, we will examine the sign of the first derivative for test points within the intervals formed by the critical points: \((-\pi, -\frac{\pi}{2}), (-\frac{\pi}{2},0), (0, \frac{\pi}{2}), (\frac{\pi}{2}, \pi)\). 1. \(f'(x)\) in \((-\pi, -\frac{\pi}{2})\): Testing \(x = -\frac{3\pi}{4}\), \(f'(-\frac{3\pi}{4})=-2\cos(-\frac{3\pi}{4})\sin(-\frac{3\pi}{4}) > 0\), so \(f(x)\) is increasing in this interval. 2. \(f'(x)\) in \((-\frac{\pi}{2},0)\): Testing \(x=-\frac{\pi}{4}\), \(f'(-\frac{\pi}{4}) = -2\cos(-\frac{\pi}{4})\sin(-\frac{\pi}{4}) < 0\), so \(f(x)\) is decreasing in this interval. 3. \(f'(x)\) in \((0, \frac{\pi}{2})\): Testing \(x=\frac{\pi}{4}\), \(f'(\frac{\pi}{4}) = -2\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}) < 0\), so \(f(x)\) is decreasing in this interval. 4. \(f'(x)\) in \((\frac{\pi}{2}, \pi)\): Testing \(x=\frac{3\pi}{4}\), \(f'(\frac{3\pi}{4}) = -2\cos(\frac{3\pi}{4})\sin(\frac{3\pi}{4}) > 0\), so \(f(x)\) is increasing in this interval.

Based on our analysis, the function \(f(x) = \cos^2 x\) is increasing on the intervals \((-\pi, -\frac{\pi}{2})\) and \((\frac{\pi}{2}, \pi)\), and it is decreasing on the intervals \((-\frac{\pi}{2},0)\) and \((0, \frac{\pi}{2})\).