Find the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and the local extrema. \[ f(x)=x^{3}-27 x+4 \] A. The function is increasing on \( (-\infty,-3),(3, \infty) \) (Type your answer in interval notation. Type integers or simplified fractions. Use a comma to separate answers as needed.) B. The function is never increasing Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is decreasing on \( (-3,3) \) (Type your answer in interval notation. Type integers or simplified fractions Use a comma to separate answers as needed.) B. The function is never decreasing Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (Type integers or simplified fractions.) A. The function has a local minimum \( f(\square)=\square \), and no local maximum. B. The function has a local maximum \( f( \) \( \square \) ) \( = \) \( \square \) and a local minimum \( f( \) \( \square \) ) \( = \) \( \square \) C. The function has a local maximum \( f(\square)=\square \), and no local minimum. \( \square \) D. The function has no local extrema.

To analyze the function \( f(x) = x^3 - 27x + 4 \), we will follow these steps: 1. **Find the derivative \( f'(x) \)** to determine the intervals of increase and decrease. 2. **Set \( f'(x) = 0 \)** to find critical points. 3. **Use the first derivative test** to determine the behavior of \( f(x) \) around the critical points. ### Step 1: Find the derivative The derivative of the function is: \[ f'(x) = 3x^2 - 27 \] ### Step 2: Set the derivative to zero Now, we set the derivative equal to zero to find critical points: \[ 3x^2 - 27 = 0 \] \[ x^2 = 9 \] \[ x = 3 \quad \text{or} \quad x = -3 \] ### Step 3: Determine intervals of increase and decrease We will test the intervals determined by the critical points \( x = -3 \) and \( x = 3 \): - Test interval \( (-\infty, -3) \) - Test interval \( (-3, 3) \) - Test interval \( (3, \infty) \) **Choose test points:** - For \( (-\infty, -3) \), choose \( x = -4 \): \[ f'(-4) = 3(-4)^2 - 27 = 48 - 27 = 21 \quad (\text{positive, so } f \text{ is increasing}) \] - For \( (-3, 3) \), choose \( x = 0 \): \[ f'(0) = 3(0)^2 - 27 = -27 \quad (\text{negative, so } f \text{ is decreasing}) \] - For \( (3, \infty) \), choose \( x = 4 \): \[ f'(4) = 3(4)^2 - 27 = 48 - 27 = 21 \quad (\text{positive, so } f \text{ is increasing}) \] ### Summary of intervals - The function is increasing on \( (-\infty, -3) \) and \( (3, \infty) \). - The function is decreasing on \( (-3, 3) \). ### Step 4: Find local extrema To find the local extrema, we evaluate \( f(x) \) at the critical points \( x = -3 \) and \( x = 3 \): 1. Calculate \( f(-3) \): \[ f(-3) = (-3)^3 - 27(-3) + 4 = -27 + 81 + 4 = 58 \] 2. Calculate \( f(3) \): \[ f(3) = (3)^3 - 27(3) + 4 = 27 - 81 + 4 = -50 \] ### Conclusion - The function has a local minimum at \( f(-3) = 58 \) and a local maximum at \( f(3) = -50 \). ### Final Answers A. The function is increasing on \( (-\infty, -3), (3, \infty) \). B. The function is decreasing on \( (-3, 3) \). C. The function has a local maximum \( f(3) = -50 \) and a local minimum \( f(-3) = 58 \).