Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the $x$-values at which they occur. \[ f(x)=\frac{2 x}{x^{2}+16} ; \quad[-9,9] \] The absolute maximum value is at $\mathrm{x}=$ (Round to two decimal places as needed. Use a comma to separate answers as needed.)

Step 1 ：Given the function \(f(x)=\frac{2x}{x^{2}+16}\) over the interval \([-9,9]\).

Step 2 ：To find the absolute maximum and minimum values of the function, we first need to find the critical points. Critical points are points where the derivative of the function is either zero or undefined.

Step 3 ：Find the derivative of the function \(f'(x) = -\frac{4x^{2}}{(x^{2} + 16)^{2}} + \frac{2}{x^{2} + 16}\).

Step 4 ：Set the derivative equal to zero and solve for x to find the critical points. The critical points are \(x = -4, 4\).

Step 5 ：Evaluate the function at the critical points and the endpoints of the interval. The values are \(-\frac{1}{4}, \frac{1}{4}, -\frac{18}{97}, \frac{18}{97}\).

Step 6 ：The maximum value of the function on the interval \([-9,9]\) is \(\frac{1}{4}\) and it occurs at \(x=4\).

Step 7 ：The minimum value of the function on the interval \([-9,9]\) is \(-\frac{1}{4}\) and it occurs at \(x=-4\).

Step 8 ：Final Answer: The absolute maximum value is at \(x=\boxed{4}\).