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

First, find the derivative of the function: \[ h(x) = 108x - x^3 \] Differentiate with respect to \(x\): \[ h'(x) = 108 - 3x^2 \] Next, set the derivative equal to zero to find the critical points: \[ 108 - 3x^2 = 0 \quad \implies \quad 3x^2 = 108 \quad \implies \quad x^2 = 36 \quad \implies \quad x = \pm 6 \] Now, analyze the sign of \(h'(x)\) on the intervals determined by the critical points: 1. For \(x < -6\), choose a test point such as \(x = -7\): \[ h'(-7) = 108 - 3(-7)^2 = 108 - 3(49) = 108 - 147 = -39 \quad (\text{negative}) \] Thus, \(h(x)\) is decreasing on \((-\infty, -6)\). 2. For \(-6 < x < 6\), choose a test point such as \(x = 0\): \[ h'(0) = 108 - 3(0)^2 = 108 \quad (\text{positive}) \] Thus, \(h(x)\) is increasing on \((-6, 6)\). 3. For \(x > 6\), choose a test point such as \(x = 7\): \[ h'(7) = 108 - 3(7)^2 = 108 - 3(49) = 108 - 147 = -39 \quad (\text{negative}) \] Thus, \(h(x)\) is decreasing on \((6, \infty)\). The intervals in interval notation are: - Increasing on: \((-6, 6)\) - Decreasing on: \((-\infty, -6) \cup (6, \infty)\)