a. Find the open interval(s) on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. \[ g(x)=x \sqrt{18-x^{2}} \]

Step 1 ：First, we need to find the derivative of the function \(g(x) = x \sqrt{18 - x^2}\).

Step 2 ：Using the product rule and chain rule, we get \(g'(x) = \sqrt{18 - x^2} - \frac{x^2}{\sqrt{18 - x^2}}\).

Step 3 ：To find the critical points, we set \(g'(x) = 0\). Solving the equation \(\sqrt{18 - x^2} - \frac{x^2}{\sqrt{18 - x^2}} = 0\), we get \(x = 0, \pm \sqrt{12}\).

Step 4 ：We also need to consider the endpoints of the domain of \(g(x)\), which are \(-\sqrt{18}\) and \(\sqrt{18}\).

Step 5 ：Now we test the intervals \((- \sqrt{18}, - \sqrt{12})\), \((- \sqrt{12}, 0)\), \((0, \sqrt{12})\), and \((\sqrt{12}, \sqrt{18})\) in the derivative to determine where the function is increasing or decreasing.

Step 6 ：We find that \(g(x)\) is increasing on \((- \sqrt{18}, - \sqrt{12})\) and \((0, \sqrt{12})\), and decreasing on \((- \sqrt{12}, 0)\) and \((\sqrt{12}, \sqrt{18})\).

Step 7 ：To find the local and absolute extreme values, we evaluate \(g(x)\) at the critical points and endpoints. We find that \(g(0) = 0\), \(g(\pm \sqrt{12}) = \pm 12 \sqrt{2}\), and \(g(\pm \sqrt{18}) = 0\).

Step 8 ：Therefore, the local maximum is \(12 \sqrt{2}\) at \(x = \sqrt{12}\), the local minimum is \(-12 \sqrt{2}\) at \(x = - \sqrt{12}\), and the absolute maximum and minimum are \(12 \sqrt{2}\) and \(-12 \sqrt{2}\) respectively.