Identify the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) \[ y=x \sqrt{100-x^{2}} \] increasing decreasing

To determine where the function \( y = x \sqrt{100 - x^2} \) is increasing or decreasing, we can find its derivative. The function is defined for \( -10 < x < 10 \). 1. Take the derivative using the product and chain rules: \[ y' = \sqrt{100 - x^2} + x \cdot \frac{d}{dx}(100 - x^2)^{1/2} \] After simplifying, we find that: \[ y' = \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \] Setting \( y' = 0 \) to find critical points leads to \( 100 - 2x^2 = 0 \), meaning \( x = -5 \) or \( x = 5 \). 2. Analyzing intervals: - For \( x \in (-10, -5) \): Choose \( x = -6 \), \( y' > 0 \) (increasing). - For \( x \in (-5, 5) \): Choose \( x = 0 \), \( y' < 0 \) (decreasing). - For \( x \in (5, 10) \): Choose \( x = 6 \), \( y' > 0 \) (increasing). Thus, the function is: - Increasing on the intervals \( (-10, -5) \) and \( (5, 10) \). - Decreasing on the interval \( (-5, 5) \). To sum it up: Increasing: \( (-10, -5) \cup (5, 10) \) Decreasing: \( (-5, 5) \)