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.

We start with \[ f(x)=\frac{x^2}{x-6} \] and note the domain: \( x\neq6 \). **1. Find the derivative using the quotient rule.** Let \[ u(x)=x^2 \quad \text{and} \quad v(x)=x-6. \] Then \[ u'(x)=2x \quad \text{and} \quad v'(x)=1. \] The quotient rule gives \[ f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{(v(x))^2}=\frac{2x(x-6)-x^2}{(x-6)^2}. \] Simplify the numerator: \[ 2x(x-6)-x^2=2x^2-12x-x^2=x^2-12x. \] Thus, \[ f'(x)=\frac{x^2-12x}{(x-6)^2}=\frac{x(x-12)}{(x-6)^2}. \] **2. Find the critical numbers.** Critical numbers occur when \( f'(x)=0 \) or \( f'(x) \) is undefined (provided they lie in the domain of \( f \)). - \( f'(x)=0 \) when the numerator is zero: \[ x(x-12)=0 \quad\Longrightarrow\quad x=0 \quad \text{or} \quad x=12. \] - \( f'(x) \) is undefined when the denominator is zero, i.e., when \[ (x-6)^2=0 \quad\Longrightarrow\quad x=6, \] but \( x=6 \) is not in the domain of \( f \). Thus, the critical numbers are \( x=0 \) and \( x=12 \). **3. Determine intervals of increase and decrease by testing the sign of \( f'(x) \).** Recall: \[ f'(x)=\frac{x(x-12)}{(x-6)^2}, \] with \((x-6)^2>0\) for \( x\neq6 \). Hence, the sign of \( f'(x) \) depends only on \( x(x-12) \). Consider the following intervals divided by \( x=0 \), \( x=6 \) (vertical asymptote) and \( x=12 \): - **Interval \((-\infty,0)\):** Choose \( x=-1 \). Then \[ (-1)((-1)-12)=(-1)(-13)=13>0. \] So, \( f'(x) > 0 \) and \( f \) is increasing on \((-\infty,0)\). - **Interval \((0,6)\):** Choose \( x=3 \). Then \[ 3(3-12)=3(-9)=-27<0. \] So, \( f'(x) < 0 \) and \( f \) is decreasing on \((0,6)\). - **Interval \((6,12)\):** Choose \( x=10 \). Then \[ 10(10-12)=10(-2)=-20<0. \] So, \( f'(x) < 0 \) and \( f \) is decreasing on \((6,12)\). - **Interval \((12,\infty)\):** Choose \( x=15 \). Then \[ 15(15-12)=15\cdot3=45>0. \] So, \( f'(x) > 0 \) and \( f \) is increasing on \((12,\infty)\). Thus, we have: - Increasing on \((-\infty, 0)\) and \((12, \infty)\). - Decreasing on \((0,6)\) and \((6,12)\). **4. Identify the local extrema.** - At \( x=0 \): \( f'(x) \) changes from positive (increasing on \((-\infty,0)\)) to negative (decreasing on \((0,6)\)). This indicates a local maximum at \( x=0 \). - At \( x=12 \): \( f'(x) \) changes from negative (decreasing on \((6,12)\)) to positive (increasing on \((12,\infty)\)). This indicates a local minimum at \( x=12 \). **5. Answer the multiple–choice questions.** For the behavior of \( f(x) \): - The function is increasing on \((-\infty,0)\cup (12,\infty)\) and decreasing on \((0,6)\cup (6,12)\). Regarding the local extrema: - The function has a local maximum at \( x=0 \) and a local minimum at \( x=12 \). Thus, the correct statement regarding the local extrema is: B. The function has a local minimum at \( x=12 \) and a local maximum at \( x=0 \).