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 function is decreasing on B. There is no solution. Which statement regarding the local extrema of the function is true? A. The function has a local minimum at \( x=0 \) and a local maximum at \( x=12 \). B. The function has a local minimum at \( x=12 \) and a local maximum at \( x=0 \). C. The function has no local extrema.

To find the critical numbers for the function \( f(x) = \frac{x^{2}}{x-6} \), we start by determining its derivative and where it is equal to zero or undefined. 1. **Find the derivative:** Using the quotient rule, we have: \[ f'(x) = \frac{(x-6)(2x) - x^2(1)}{(x-6)^2} = \frac{2x^2 - 12x - x^2}{(x-6)^2} = \frac{x^2 - 12x}{(x-6)^2} = \frac{x(x-12)}{(x-6)^2} \] 2. **Set the derivative to zero:** The critical numbers are found by setting the numerator equal to zero: \[ x(x-12) = 0 \Rightarrow x = 0 \text{ or } x = 12 \] Additionally, the derivative is undefined at \( x = 6 \), which is also important. 3. **Determine the sign of the derivative:** Choose test points in the intervals created by the critical numbers \( 0, 6, \) and \( 12 \) for \( f'(x) \): - For \( x < 0 \), choose \( x = -1 \): \( f'(-1) = \frac{-1(-1-12)}{(-1-6)^2} > 0 \) (increasing) - For \( 0 < x < 6 \), choose \( x = 1 \): \( f'(1) = \frac{1(1-12)}{(1-6)^2} < 0 \) (decreasing) - For \( 6 < x < 12 \), choose \( x = 7 \): \( f'(7) = \frac{7(7-12)}{(7-6)^2} < 0 \) (decreasing) - For \( x > 12 \), choose \( x = 13 \): \( f'(13) = \frac{13(13-12)}{(13-6)^2} > 0 \) (increasing) 4. **Intervals of increase and decrease:** - \( f(x) \) is increasing on \( (-\infty, 0) \) and \( (12, \infty) \). - \( f(x) \) is decreasing on \( (0, 6) \) and \( (6, 12) \). 5. **Local extrema:** - At \( x = 0 \): Local maximum since the function changes from increasing to decreasing. - At \( x = 12 \): Local minimum since the function changes from decreasing to increasing. Summary: - The function is decreasing on \( (0, 6) \) and \( (6, 12) \). - The function has a local minimum at \( x=12 \) and a local maximum at \( x=0 \). Thus, the correct choice is: A. The function is decreasing on \( (0, 6) \) and \( (6, 12) \). And regarding the local extrema: A. The function has a local minimum at \( x=0 \) and a local maximum at \( x=12 \).