Find the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and the local extrema. \( f(x)=x^{3}-27 x+4 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 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.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is decreasing on (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.

To determine the intervals on which the function \( f(x) = x^3 - 27x + 4 \) is increasing or decreasing, we first need to find its derivative \( f'(x) \) and analyze its critical points. ### Step 1: Find the derivative The derivative of the function is given by: \[ f'(x) = \frac{d}{dx}(x^3 - 27x + 4) = 3x^2 - 27 \] ### Step 2: Set the derivative to zero to find critical points To find the critical points, we set the derivative equal to zero: \[ 3x^2 - 27 = 0 \] Solving for \( x \): \[ 3x^2 = 27 \\ x^2 = 9 \\ x = \pm 3 \] Thus, the critical points are \( x = -3 \) and \( x = 3 \). ### Step 3: Test intervals around the critical points We will test the sign of \( f'(x) \) in the intervals \( (-\infty, -3) \), \( (-3, 3) \), and \( (3, \infty) \). 1. **Interval \( (-\infty, -3) \)**: - Choose \( x = -4 \): \[ f'(-4) = 3(-4)^2 - 27 = 48 - 27 = 21 > 0 \] Thus, \( f(x) \) is increasing on \( (-\infty, -3) \). 2. **Interval \( (-3, 3) \)**: - Choose \( x = 0 \): \[ f'(0) = 3(0)^2 - 27 = -27 < 0 \] Thus, \( f(x) \) is decreasing on \( (-3, 3) \). 3. **Interval \( (3, \infty) \)**: - Choose \( x = 4 \): \[ f'(4) = 3(4)^2 - 27 = 48 - 27 = 21 > 0 \] Thus, \( f(x) \) is increasing on \( (3, \infty) \). ### Step 4: Identify local extrema - At \( x = -3 \), since \( f(x) \) changes from increasing to decreasing, there is a local maximum. - At \( x = 3 \), since \( f(x) \) changes from decreasing to increasing, there is a local minimum. ### Summary of Results - The function is increasing on the intervals \( (-\infty, -3) \) and \( (3, \infty) \). - The function is decreasing on the interval \( (-3, 3) \). - There is a local maximum at \( x = -3 \) and a local minimum at \( x = 3 \). ### Final Answers 1. The function is increasing on: \[ (-\infty, -3) \cup (3, \infty) \] 2. The function is decreasing on: \[ (-3, 3) \] 3. Local extrema: - Local maximum at \( x = -3 \) - Local minimum at \( x = 3 \)