Find the critical numbers, the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and the local extrema. Do not graph. \( f(x)=\frac{x^{2}}{x-6} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The critical number(s) of the function are \( x=\square \). (Use a comma to separate answers as needed.) B. There is no solution.

To find the critical numbers, we need to find the derivative of \( f(x) \) and determine where it is equal to zero or undefined. 1. **Find the derivative of \( f(x) \)**: Using the quotient rule, the derivative \( f'(x) \) is: \[ f'(x) = \frac{(x-6)(2x) - x^2(1)}{(x-6)^2} = \frac{2x^2 - 12x - x^2}{(x-6)^2} = \frac{x^2 - 12x}{(x-6)^2} = \frac{x(x-12)}{(x-6)^2} \] 2. **Find critical numbers**: Set the numerator equal to zero: \[ x(x-12) = 0 \implies x = 0 \text{ or } x = 12 \] Also, note the derivative is undefined at \( x = 6 \) due to the denominator. So the critical numbers are \( x = 0, 6, \) and \( 12 \). 3. **Determine intervals of increase and decrease**: Test the sign of \( f'(x) \) in the intervals \( (-\infty, 0) \), \( (0, 6) \), \( (6, 12) \), and \( (12, \infty) \): - For \( (-\infty, 0) \): Pick \( x = -1 \) → \( f'(-1) \) is positive. - For \( (0, 6) \): Pick \( x = 1 \) → \( f'(1) \) is negative. - For \( (6, 12) \): Pick \( x = 8 \) → \( f'(8) \) is positive. - For \( (12, \infty) \): Pick \( x = 13 \) → \( f'(13) \) is negative. So: - \( f(x) \) is increasing on \( (-\infty, 0) \) and \( (6, 12) \). - \( f(x) \) is decreasing on \( (0, 6) \) and \( (12, \infty) \). 4. **Local extrema**: - There is a local maximum at \( x = 0 \) (changing from increasing to decreasing). - There is a local minimum at \( x = 12 \) (changing from increasing to decreasing). Putting this all together: A. The critical number(s) of the function are \( x = 0, 6, 12 \).