Find the critical numbers of the function. (Enter your answers as a comma-separated list.) \[ \begin{array}{l} f(x)=\frac{81 x}{x^{2}+36} \\ x=\square \end{array} \]
Solution
Step 1 ：Given the function \(f(x)=\frac{81 x}{x^{2}+36}\), we are asked to find the critical numbers of the function. Critical numbers are where the derivative of the function is equal to zero or undefined.
Step 2 ：To find the derivative of the function, we need to use the quotient rule for differentiation. The quotient rule states that the derivative of a quotient u/v is \((vu' - uv')/v^2\), where u' and v' are the derivatives of u and v respectively.
Step 3 ：Applying the quotient rule, we find the derivative of the function to be \(f'(x) = -\frac{162x^2}{(x^2 + 36)^2} + \frac{81}{x^2 + 36}\).
Step 4 ：We then set the derivative equal to zero and solve for x to find the critical numbers. This gives us the critical numbers as -6 and 6.
Step 5 ：Final Answer: The critical numbers of the function are \(\boxed{-6, 6}\).
