Find the critical numbers, the intervals on which \( f(x) \) is increasi \( f(x)=\frac{x^{2}}{x-6} \) Select the correct choice below and, if necessary, fill in the ans A. The cntical number(s) of the function are \( x=\square \) (Use a cormiria to separate answers as neerled.) B. There is no solution. Select the correct choico below and, if necessary, fill in the arist A. The function is increasing on \( \square \). B. There is no solution.

**Critical Numbers** The function is defined as \[ f(x)=\frac{x^2}{x-6}. \] Its derivative is computed using the quotient rule: \[ f'(x)=\frac{(x-6)(2x)-x^2}{(x-6)^2}=\frac{2x^2-12x-x^2}{(x-6)^2}=\frac{x^2-12x}{(x-6)^2}=\frac{x(x-12)}{(x-6)^2}. \] Setting \( f'(x)=0 \) leads to \[ x(x-12)=0, \] which gives \( x=0 \) and \( x=12 \). Also, \( f'(x) \) is undefined at \( x=6 \), but since \( f(x) \) is undefined at \( x=6 \), it is not considered as a critical number. Thus, the critical numbers are: \( 0, 12 \). **Intervals of Increase** Since the denominator \((x-6)^2\) is always positive (except at \( x=6 \)), the sign of \( f'(x) \) depends solely on the numerator \( x(x-12) \). We analyze the sign on the following intervals (while noting the domain excludes \( x=6 \)): 1. For \( x \in (-\infty, 0) \), choose \( x=-1 \): \[ (-1)(-1-12)=(-1)(-13)=13>0. \] Thus, \( f'(x)>0 \). 2. For \( x \in (0,6) \), choose \( x=1 \): \[ (1)(1-12)=1\cdot(-11)=-11<0. \] Thus, \( f'(x)<0 \). 3. For \( x \in (6,12) \), choose \( x=10 \): \[ (10)(10-12)=10\cdot(-2)=-20<0. \] Thus, \( f'(x)<0 \). 4. For \( x \in (12,\infty) \), choose \( x=13 \): \[ (13)(13-12)=13\cdot1=13>0. \] Thus, \( f'(x)>0 \). Therefore, \( f(x) \) is increasing on \[ (-\infty, 0) \quad \text{and} \quad (12, \infty). \] **Summary Answers** A. The critical number(s) of the function are \( 0,12 \). B. The function is increasing on \( (-\infty, 0) \cup (12, \infty) \).