Let's solve the problem step by step: Given: \[ f(x) = -3\cos^2(x) \quad \text{on} \quad [-\pi, \pi] \] We are to find the intervals where \( f(x) \) is **increasing** and **decreasing**. --- ### Step 1: Find the derivative \[ f(x) = -3\cos^2(x) \] Let’s use the chain rule: \[ f'(x) = -3 \cdot 2\cos(x)\cdot(-\sin(x)) = 6\cos(x)\sin(x) \] \[ f'(x) = 6\sin(x)\cos(x) \] Recall the double angle identity: \( \sin(2x) = 2\sin(x)\cos(x) \): \[ f'(x) = 3\sin(2x) \] --- ### Step 2: Find critical points Set \( f'(x) = 0 \): \[ 3\sin(2x) = 0 \implies \sin(2x) = 0 \] \[ 2x = n\pi \implies x = \frac{n\pi}{2} \] where \( n \) is an integer. Within \([-\pi, \pi]\), the points are: \[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \] --- ### Step 3: Test intervals The intervals are: - \( (-\pi, -\frac{\pi}{2}) \) - \( (-\frac{\pi}{2}, 0) \) - \( (0, \frac{\pi}{2}) \) - \( (\frac{\pi}{2}, \pi) \) Pick a test point in each interval and plug into \( f'(x) \): #### a) \( x = -\frac{3\pi}{4} \) in \( (-\pi, -\frac{\pi}{2}) \): \[ f'(-\frac{3\pi}{4}) = 3 \sin(-\frac{3\pi}{2}) = 3 \cdot 1 = 3 \quad (\text{positive, so increasing}) \] #### b) \( x = -\frac{\pi}{4} \) in \( (-\frac{\pi}{2}, 0) \): \[ f'(-\frac{\pi}{4}) = 3 \sin(-\frac{\pi}{2}) = 3 \cdot (-1) = -3 \quad (\text{negative, so decreasing}) \] #### c) \( x = \frac{\pi}{4} \) in \( (0, \frac{\pi}{2}) \): \[ f'(\frac{\pi}{4}) = 3 \sin(\frac{\pi}{2}) = 3 \cdot 1 = 3 \quad (\text{positive, so increasing}) \] #### d) \( x = \frac{3\pi}{4} \) in \( (\frac{\pi}{2}, \pi) \): \[ f'(\frac{3\pi}{4}) = 3 \sin(\frac{3\pi}{2}) = 3 \cdot (-1) = -3 \quad (\text{negative, so decreasing}) \] --- ### Step 4: Write intervals - **Increasing:** \( (-\pi, -\frac{\pi}{2}) \) and \( (0, \frac{\pi}{2}) \) - **Decreasing:** \( (-\frac{\pi}{2}, 0) \) and \( (\frac{\pi}{2}, \pi) \) --- ### Step 5: Select the correct answer This matches **choice A**. **Final Answer:** > **A.** The function is increasing on the open interval(s) \( (-\pi, -\frac{\pi}{2}), (0, \frac{\pi}{2}) \) and decreasing on the open interval(s) \( (-\frac{\pi}{2}, 0), (\frac{\pi}{2}, \pi) \).

Let's analyze the function \( f(x) = -3 \cos^2(x) \) step by step to determine where it is increasing or decreasing on the interval \([- \pi, \pi]\). --- ### **Step 1: Find the derivative \(f'(x)\)** Given: \[ f(x) = -3 \cos^2(x) \] Using the chain rule: \[ f'(x) = -3 \cdot 2 \cos(x) \cdot (-\sin(x)) = 6 \cos(x) \sin(x) \] Recall the double angle identity: \[ \sin(2x) = 2 \sin(x) \cos(x) \] So, \[ f'(x) = 6 \cos(x) \sin(x) = 3 \times 2 \cos(x) \sin(x) = 3 \sin(2x) \] **Thus:** \[ \boxed{ f'(x) = 3 \sin(2x) } \] --- ### **Step 2: Find critical points where \(f'(x) = 0\)** Set: \[ 3 \sin(2x) = 0 \implies \sin(2x) = 0 \] Solutions: \[ 2x = n\pi \implies x = \frac{n\pi}{2} \] Within \([- \pi, \pi]\): - For \(n = -2, -1, 0, 1, 2\): \[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \] --- ### **Step 3: Test the sign of \(f'(x)\) in each interval** Intervals between critical points: - \((- \pi, -\frac{\pi}{2})\) - \((- \frac{\pi}{2}, 0)\) - \((0, \frac{\pi}{2})\) - \((\frac{\pi}{2}, \pi)\) Choose test points: | Interval | Test Point | Compute \(f'(x) = 3 \sin(2x)\) | Sign of \(f'(x)\) | Behavior | |------------|--------------|------------------------------|------------------|----------| | \((- \pi, -\frac{\pi}{2})\) | \(x = -\frac{3\pi}{4}\) | \(2x = -\frac{3\pi}{2}\) | \(\sin(-\frac{3\pi}{2}) = 1\) | \(3 \times 1 = 3 > 0\) | Increasing | | \((- \frac{\pi}{2}, 0)\) | \(x = -\frac{\pi}{4}\) | \(2x = -\frac{\pi}{2}\) | \(\sin(-\frac{\pi}{2}) = -1\) | \(3 \times -1 = -3 < 0\) | Decreasing | | \((0, \frac{\pi}{2})\) | \(x = \frac{\pi}{4}\) | \(2x = \frac{\pi}{2}\) | \(\sin(\frac{\pi}{2})=1\) | \(3 \times 1=3 > 0\) | Increasing | | \((\frac{\pi}{2}, \pi)\) | \(x= \frac{3\pi}{4}\) | \(2x= \frac{3\pi}{2}\) | \(\sin(\frac{3\pi}{2})=-1\) | \(3 \times -1 = -3<0\) | Decreasing | --- ### **Step 4: Write the intervals** - **Increasing** on: \[ (-\pi, -\frac{\pi}{2}) \quad \text{and} \quad (0, \frac{\pi}{2}) \] - **Decreasing** on: \[ (-\frac{\pi}{2}, 0) \quad \text{and} \quad (\frac{\pi}{2}, \pi) \] --- ### **Final conclusion:** The function \(f(x)\) is increasing on \(\boxed{(-\pi, -\frac{\pi}{2}) \text{ and } (0, \frac{\pi}{2})}\), and decreasing on \(\boxed{(-\frac{\pi}{2}, 0) \text{ and } (\frac{\pi}{2}, \pi)}\). This matches **choice A**. --- ### **Answer:** **A.** The function is increasing on the open intervals \(\boxed{(-\pi, -\frac{\pi}{2})}\) and \(\boxed{(0, \frac{\pi}{2})}\), and decreasing on the open intervals \(\boxed{(-\frac{\pi}{2}, 0)}\) and \(\boxed{(\frac{\pi}{2}, \pi)}\).

Let's analyze the function \(f(x) = -3 \cos^2(x)\) carefully to determine where it is increasing or decreasing on the interval \([- \pi, \pi]\). --- ### **Step 1: Find the derivative \(f'(x)\)** Given: \[ f(x) = -3 \cos^2(x) \] Apply the chain rule: \[ f'(x) = -3 \times 2 \cos(x) \times (-\sin(x)) = 6 \cos(x) \sin(x) \] Recall the double angle identity: \[ \sin(2x) = 2 \sin(x) \cos(x) \] Express \(f'(x)\) using this: \[ f'(x) = 6 \cos(x) \sin(x) = 3 \times 2 \cos(x) \sin(x) = 3 \sin(2x) \] **Therefore:** \[ \boxed{ f'(x) = 3 \sin(2x) } \] --- ### **Step 2: Find critical points where \(f'(x) = 0\)** Set the derivative equal to zero: \[ 3 \sin(2x) = 0 \implies \sin(2x) = 0 \] Solutions: \[ 2x = n \pi \implies x = \frac{n \pi}{2} \] Within \([- \pi, \pi]\), the values of \(x\) are: \[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \] --- ### **Step 3: Determine the sign of \(f'(x)\) in each interval** Intervals between critical points: - \((- \pi, -\frac{\pi}{2})\) - \((- \frac{\pi}{2}, 0)\) - \((0, \frac{\pi}{2})\) - \((\frac{\pi}{2}, \pi)\) Choose test points in each interval: | Interval | Test Point | Compute \(f'(x) = 3 \sin(2x)\) | Sign of \(f'(x)\) | Behavior | |------------|--------------|------------------------------|------------------|----------| | \((- \pi, -\frac{\pi}{2})\) | \(x = -\frac{3\pi}{4}\) | \(2x = -\frac{3\pi}{2}\) | \(\sin(-\frac{3\pi}{2})=1\) | \(3 \times 1=3 > 0\) | Increasing | | \((- \frac{\pi}{2}, 0)\) | \(x = -\frac{\pi}{4}\) | \(2x=-\frac{\pi}{2}\) | \(\sin(-\frac{\pi}{2})=-1\) | \(3 \times -1=-3<0\) | Decreasing | | \((0, \frac{\pi}{2})\) | \(x= \frac{\pi}{4}\) | \(2x= \frac{\pi}{2}\) | \(\sin(\frac{\pi}{2})=1\) | \(3 \times 1=3>0\) | Increasing | | \((\frac{\pi}{2}, \pi)\) | \(x= \frac{3\pi}{4}\) | \(2x= \frac{3\pi}{2}\) | \(\sin(\frac{3\pi}{2})=-1\) | \(3 \times -1=-3<0\) | Decreasing | --- ### **Step 4: Summarize the behavior** - **Increasing** on: \[ \boxed{(-\pi, -\frac{\pi}{2}) \quad \text{and} \quad (0, \frac{\pi}{2})} \] - **Decreasing** on: \[ \boxed{(-\frac{\pi}{2}, 0) \quad \text{and} \quad (\frac{\pi}{2}, \pi)} \] --- ### **Final answer:** The function \(f(x)\) **increases** on \(\boxed{(-\pi, -\frac{\pi}{2}) \text{ and } (0, \frac{\pi}{2})}\), and **decreases** on \(\boxed{(-\frac{\pi}{2}, 0) \text{ and } (\frac{\pi}{2}, \pi)}\). This corresponds to **Choice A**. --- ### **Summary of options:** - **A.** The function is increasing on \((- \pi, -\frac{\pi}{2})\) and \((0, \frac{\pi}{2})\), decreasing on \((- \frac{\pi}{2}, 0)\) and \((\frac{\pi}{2}, \pi)\). **(Correct)** - **B.** The function is increasing on certain intervals, and never decreasing. **(Incorrect)** - **C.** The function is decreasing on certain intervals, and never increasing. **(Incorrect)** - **D.** The function is never increasing or decreasing. **(Incorrect)** **Thus, the best choice is A.**