Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for the function $f(x)=\frac{6 x}{x^{2}+6}$ on the domain $[0,5]$
Select the correct choice below and, if necessary, fill in the answer boxes to domplete your choice
A. The absolute maximum is $\square$, which occurs at $\mathrm{x}=$
(Round the absolute maximum to two decimal places as needed. Type an exact answer for the value of $\mathrm{x}$ where the maximum occurs. Use a comma to separate answers as needed )
B. There is no absolute maximum

Step 1 ：First, we need to find the derivative of the function \(f(x)=\frac{6x}{x^{2}+6}\).

Step 2 ：Using the quotient rule, we find that \(f'(x) = -\frac{12x^{2}}{(x^{2} + 6)^{2}} + \frac{6}{x^{2} + 6}\).

Step 3 ：Next, we set the derivative equal to zero to find the critical points: \(-\frac{12x^{2}}{(x^{2} + 6)^{2}} + \frac{6}{x^{2} + 6} = 0\).

Step 4 ：Solving this equation gives us the critical points \(x = -\sqrt{6}\) and \(x = \sqrt{6}\).

Step 5 ：We then evaluate the function at the critical points and the endpoints of the interval [0,5].

Step 6 ：The function values at these points are \(-\frac{\sqrt{6}}{2}\), \(\frac{\sqrt{6}}{2}\), 0, and \(\frac{30}{31}\).

Step 7 ：Comparing these values, we find that the absolute maximum is \(\frac{\sqrt{6}}{2}\), which occurs at \(x = \sqrt{6}\).

Step 8 ：Final Answer: The absolute maximum is \(\boxed{\frac{\sqrt{6}}{2}}\), which occurs at \(x=\boxed{\sqrt{6}}\).