To solve the problem, we need to find the absolute maximum and minimum values of the given function:

$g(x) = x \sqrt{25 - x^2} \quad \text{on the interval} \quad [0, 5].$

Steps to find the absolute extrema:

1. Domain:
   The domain of $g(x)$ is restricted by the square root. The term under the square root, $25 - x^2$, must be non-negative: $25 - x^2 \geq 0 \quad \Rightarrow \quad x^2 \leq 25 \quad \Rightarrow \quad -5 \leq x \leq 5.$ Since the problem is asking for the interval $[0, 5]$, the function is valid within this interval.

2. Critical Points:
   To find the critical points, we need the derivative of $g(x)$ and set it equal to zero.
   Using the product and chain rules, the derivative of $g(x) = x \sqrt{25 - x^2}$ is: $g'(x) = \sqrt{25 - x^2} + x \cdot \frac{-x}{\sqrt{25 - x^2}}.$ Simplifying this expression: $g'(x) = \frac{(25 - x^2) - x^2}{\sqrt{25 - x^2}} = \frac{25 - 2x^2}{\sqrt{25 - x^2}}.$ Set $g'(x) = 0$ to find the critical points: $\frac{25 - 2x^2}{\sqrt{25 - x^2}} = 0 \quad \Rightarrow \quad 25 - 2x^2 = 0 \quad \Rightarrow \quad x^2 = \frac{25}{2} \quad \Rightarrow \quad x = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \approx 3.54.$

3. Evaluate at Critical Points and Endpoints:

   • At $x = 0$: $g(0) = 0 \times \sqrt{25 - 0^2} = 0.$
   • At $x = 5$: $g(5) = 5 \times \sqrt{25 - 5^2} = 5 \times \sqrt{0} = 0.$
   • At $x = \frac{5}{\sqrt{2}} \approx 3.54$: $g\left(\frac{5}{\sqrt{2}}\right) = \frac{5}{\sqrt{2}} \times \sqrt{25 - \left(\frac{5}{\sqrt{2}}\right)^2} = \frac{5}{\sqrt{2}} \times \sqrt{25 - \frac{25}{2}} = \frac{5}{\sqrt{2}} \times \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}} = \frac{25}{2}.$

4. Conclusion:

   • The absolute minimum is $g(0) = 0$.
   • The absolute maximum is $g\left(\frac{5}{\sqrt{2}}\right) = \frac{25}{2} = 12.5$.

Thus, the absolute maximum is 12.5, and the absolute minimum is 0.

Would you like more details on any of these steps or concepts?

Here are 5 related questions:

1. How do you find the domain of functions involving square roots?
2. What are the necessary conditions to apply the derivative test for critical points?
3. Why are both endpoints and critical points checked for absolute extrema?
4. How do you simplify complex derivatives using the product rule?
5. Can there be a maximum value outside the critical points and endpoints?

Tip: Always check the function's behavior at the boundary of the interval as well as at critical points to ensure all possible extrema are considered.