Identify the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) \[ h(x)=108 x-x^{3} \] increasing decreasing Need Help? Readit

To find where the function \( h(x) = 108x - x^3 \) is increasing or decreasing, we first need to compute its derivative: \[ h'(x) = 108 - 3x^2. \] Setting the derivative equal to zero to find critical points gives us: \[ 108 - 3x^2 = 0 \] \[ 3x^2 = 108 \] \[ x^2 = 36 \] \[ x = 6 \quad \text{or} \quad x = -6. \] Next, we can test intervals around these critical points, such as \( (-\infty, -6) \), \( (-6, 6) \), and \( (6, \infty) \): 1. For \( x < -6 \) (e.g., \( x = -7 \)): \[ h'(-7) = 108 - 3(-7)^2 = 108 - 147 = -39 \quad (\text{decreasing}) \] 2. For \( -6 < x < 6 \) (e.g., \( x = 0 \)): \[ h'(0) = 108 - 3(0)^2 = 108 \quad (\text{increasing}) \] 3. For \( x > 6 \) (e.g., \( x = 7 \)): \[ h'(7) = 108 - 3(7)^2 = 108 - 147 = -39 \quad (\text{decreasing}) \] Based on these tests, we find that the function \( h(x) \) is: - **Increasing** on the interval \( (-6, 6) \) - **Decreasing** on the intervals \( (-\infty, -6) \) and \( (6, \infty) \) So, the final answer in interval notation is: Increasing: \( (-6, 6) \) Decreasing: \( (-\infty, -6) \) and \( (6, \infty) \)