The problem asks to graph the function $f(x) = \frac{1}{12} x \sqrt{12 - x}$ and determine the absolute extrema on the closed interval $[0, 12]$.

Step-by-Step Solution:

1. Domain of the function:
   Since $\sqrt{12 - x}$ is defined only when $12 - x \geq 0$, we have $x \leq 12$. The domain is $[0, 12]$.

2. Critical Points:
   To find the critical points, we first need the derivative of the function $f(x)$. Using the product rule and chain rule: $f(x) = \frac{1}{12} x \sqrt{12 - x}$ Let’s differentiate this:

   $f'(x) = \frac{1}{12} \left( \sqrt{12 - x} + x \cdot \frac{-1}{2\sqrt{12 - x}} \right)$ Simplifying this: $f'(x) = \frac{1}{12} \left( \sqrt{12 - x} - \frac{x}{2\sqrt{12 - x}} \right)$ To find critical points, set $f'(x) = 0$: $\sqrt{12 - x} - \frac{x}{2\sqrt{12 - x}} = 0$ Multiply through by $2\sqrt{12 - x}$ to eliminate the denominator: $2(12 - x) - x = 0$ Simplifying: $24 - 2x - x = 0 \quad \Rightarrow \quad 24 = 3x \quad \Rightarrow \quad x = 8$ So, $x = 8$ is a critical point.

3. Endpoints:
   We need to evaluate $f(x)$ at the endpoints $x = 0$ and $x = 12$, as well as at the critical point $x = 8$.

   • $f(0) = \frac{1}{12} \cdot 0 \cdot \sqrt{12 - 0} = 0$
   • $f(12) = \frac{1}{12} \cdot 12 \cdot \sqrt{12 - 12} = 0$
   • $f(8) = \frac{1}{12} \cdot 8 \cdot \sqrt{12 - 8} = \frac{8}{12} \cdot \sqrt{4} = \frac{8}{12} \cdot 2 = \frac{16}{12} = \frac{4}{3}$

4. Conclusion (Extrema):

   • At $x = 0$, $f(0) = 0$
   • At $x = 8$, $f(8) = \frac{4}{3}$ (maximum)
   • At $x = 12$, $f(12) = 0$

   Therefore, the absolute maximum occurs at $x = 8$, and the absolute minimum occurs at both $x = 0$ and $x = 12$.

Would you like me to show a graph of the function or explain any step in more detail?

Related Questions:

1. How do you apply the product rule and chain rule in this example?
2. Can the function have any other critical points in the interval?
3. What would happen if we extended the interval beyond $[0, 12]$?
4. How would the second derivative help in confirming the nature of the critical points?
5. Could there be another approach to finding the absolute extrema without derivatives?

Tip: Always check endpoints when determining absolute extrema on a closed interval!