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.

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**Step 1. Find the domain.** The function is $$ f(x)=\frac{x^2}{x-6}. $$ The denominator is zero when $$ x-6=0 $$, i.e. when $$ x=6 $$. Thus, the domain of $$ f $$ is all real numbers except $$ x=6 $$. --- **Step 2. Compute the derivative using the quotient rule.** For $$ f(x)=\frac{g(x)}{h(x)} \quad \text{with} \quad g(x)=x^2 \quad \text{and} \quad h(x)=x-6, $$ the derivative is given by $$ f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}. $$ We have: $$ g'(x)=2x \quad \text{and} \quad h'(x)=1. $$ Thus, $$ f'(x)=\frac{2x(x-6)-x^2(1)}{(x-6)^2}. $$ Simplify the numerator: $$ 2x(x-6)-x^2 = 2x^2-12x-x^2 = x^2-12x. $$ Factor the numerator: $$ x^2-12x = x(x-12). $$ So, the derivative is $$ f'(x)=\frac{x(x-12)}{(x-6)^2}. $$ --- **Step 3. Find the critical numbers.** Critical numbers occur when $$ f'(x)=0 $$ or when $$ f'(x) $$ is undefined, provided the point is in the domain of $$ f $$. - Set the numerator equal to zero: $$ x(x-12)=0. $$ This gives $$ x=0 $$ or $$ x=12 $$. - The derivative is undefined at $$ x=6 $$ since the denominator $$(x-6)^2=0$$ there. However, $$ x=6 $$ is not in the domain of $$ f $$, so we do not count it as a critical number. Thus, the critical numbers are: $$ \boxed{x=0,\ x=12}. $$ --- **Step 4. Determine intervals where $$ f(x) $$ is increasing or decreasing.** The derivative is: $$ f'(x)=\frac{x(x-12)}{(x-6)^2}. $$ Note that $$(x-6)^2$$ is always positive for all $$ x \neq 6 $$. The sign of $$ f'(x) $$ is determined by the product $$ x(x-12) $$. Divide the real line (excluding $$ x=6 $$) into intervals based on the zeros of $$ x(x-12) $$ and the discontinuity at $$ x=6 $$: - $$ (-\infty, 0) $$ - $$ (0, 6) $$ - $$ (6, 12) $$ - $$ (12, \infty) $$ Now, analyze the sign of $$ x(x-12) $$ in each interval: 1. **Interval $$ (-\infty, 0) $$:** Choose $$ x=-1 $$. Then, $$ x(-1) < 0 $$ and $$ -1-12=-13 < 0 $$. The product $$ (-1)(-13)=13>0 $$. So, $$ f'(x)>0 $$ and $$ f $$ is increasing. 2. **Interval $$ (0, 6) $$:** Choose $$ x=3 $$. Then, $$ 3>0 $$ and $$ 3-12=-9 < 0 $$. The product $$ 3(-9)=-27<0 $$. So, $$ f'(x)<0 $$ and $$ f $$ is decreasing. 3. **Interval $$ (6, 12) $$:** Choose $$ x=10 $$. Then, $$ 10>0 $$ and $$ 10-12=-2<0 $$. The product $$ 10(-2)=-20<0 $$. So, $$ f'(x)<0 $$ and $$ f $$ is decreasing. 4. **Interval $$ (12, \infty) $$:** Choose $$ x=15 $$. Then, $$ 15>0 $$ and $$ 15-12=3>0 $$. The product $$ 15 \times 3=45>0 $$. So, $$ f'(x)>0 $$ and $$ f $$ is increasing. Thus, $$ f $$ is increasing on: $$ (-\infty, 0) \quad \text{and} \quad (12, \infty) $$ and decreasing on: $$ (0, 6) \quad \text{and} \quad (6, 12). $$ --- **Step 5. Determine the local extrema.** - At $$ x=0 $$: The derivative changes from positive (increasing) to negative (decreasing). This means $$ f $$ has a local maximum at $$ x=0 $$. Compute $$ f(0) $$: $$ f(0)=\frac{0^2}{0-6}=0. $$ So, the local maximum is at $$ (0,0) $$. - At $$ x=12 $$: The derivative changes from negative (decreasing) to positive (increasing). This means $$ f $$ has a local minimum at $$ x=12 $$. Compute $$ f(12) $$: $$ f(12)=\frac{12^2}{12-6}=\frac{144}{6}=24. $$ So, the local minimum is at $$ (12,24) $$. --- **Final Answer:** A. The critical number(s) of the function are $$ x=0,\,12 $$. $$ f(x) $$ is increasing on $$ (-\infty, 0) $$ and $$ (12, \infty) $$ and decreasing on $$ (0,6) $$ and $$ (6,12) $$. The function has a local maximum at $$ (0,0) $$ and a local minimum at $$ (12,24) $$. A. The critical numbers are $$ x=0 $$ and $$ x=12 $$. $$ f(x) $$ is increasing on $$ (-\infty, 0) $$ and $$ (12, \infty) $$, and decreasing on $$ (0,6) $$ and $$ (6,12) $$. There is a local maximum at $$ (0,0) $$ and a local minimum at $$ (12,24) $$.

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