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To solve this problem, we need to analyze the function \( g(x) = \sqrt{2x - x^2} \), find its absolute maximum and minimum values on its domain, and then identify where these extrema occur. Here is a step-by-step solution:

### Step 1: Determine the Domain of \( g(x) \)
The function \( g(x) = \sqrt{2x - x^2} \) requires that the expression under the square root be non-negative:
\[
2x - x^2 \geq 0.
\]
This inequality can be rewritten as:
\[
x(2 - x) \geq 0.
\]
This inequality holds when \( x = 0 \) or \( x = 2 \) (the roots), and \( x \) is between them. Therefore, the interval for \( x \) is:
\[
0 \leq x \leq 2.
\]

### Step 2: Find the Derivative of \( g(x) \)
To find critical points, we calculate the derivative of \( g(x) \). We can use the chain rule here:
\[
g(x) = (2x - x^2)^{\frac{1}{2}}.
\]
Using the chain rule, the derivative \( g'(x) \) is:
\[
g'(x) = \frac{1}{2}(2x - x^2)^{-\frac{1}{2}} \cdot (2 - 2x) = \frac{2 - 2x}{2\sqrt{2x - x^2}}.
\]
Simplifying, we get:
\[
g'(x) = \frac{1 - x}{\sqrt{2x - x^2}}.
\]

### Step 3: Find Critical Points
To find critical points, set \( g'(x) = 0 \):
\[
\frac{1 - x}{\sqrt{2x - x^2}} = 0.
\]
This equation is zero when \( 1 - x = 0 \), or \( x = 1 \).

### Step 4: Evaluate \( g(x) \) at the Endpoints and Critical Points
Now, we evaluate \( g(x) \) at \( x = 0 \), \( x = 2 \), and \( x = 1 \) to find the absolute maximum and minimum values.

1. **At \( x = 0 \):**
   \[
   g(0) = \sqrt{2 \cdot 0 - 0^2} = \sqrt{0} = 0.
   \]

2. **At \( x = 2 \):**
   \[
   g(2) = \sqrt{2 \cdot 2 - 2^2} = \sqrt{4 - 4} = \sqrt{0} = 0.
   \]

3. **At \( x = 1 \):**
   \[
   g(1) = \sqrt{2 \cdot 1 - 1^2} = \sqrt{2 - 1} = \sqrt{1} = 1.
   \]

### Step 5: Determine the Absolute Maximum and Minimum
From the values calculated:
- The **absolute maximum** of \( g(x) \) on the interval \( [0, 2] \) is \( 1 \) at \( x = 1 \).
- The **absolute minimum** of \( g(x) \) on the interval \( [0, 2] \) is \( 0 \), which occurs at both endpoints \( x = 0 \) and \( x = 2 \).

### Step 6: Graph the Function and Identify Extrema
The function \( g(x) = \sqrt{2x - x^2} \) resembles a semicircle since it can be rewritten in the form \( (x - 1)^2 + g(x)^2 = 1 \), which represents a circle centered at \( (1, 0) \) with radius \( 1 \). Thus, the graph will show a semicircle on the interval \( [0, 2] \).

- **Maximum point**: \( (1, 1) \)
- **Minimum points**: \( (0, 0) \) and \( (2, 0) \)

Would you like additional details on any step or have questions on this solution?

### Related Questions:
1. How would you find the absolute extrema for a function without a closed interval?
2. What is the significance of a critical point that is not within the domain of the function?
3. How does rewriting a function in a different form (like a circle equation) help in understanding its graph?
4. What is the process for finding relative extrema in addition to absolute extrema?
5. How can the chain rule be applied to functions with nested compositions like \( g(x) = \sqrt{2x - x^2} \)?

**Tip:** When finding extrema on a closed interval, always evaluate the function at the endpoints as well as any critical points within the interval.

Find the absolute maximum and minimum values of the function g(x) = √(2x - x²) on its given interval. Then graph the function and identify the points on the graph where the absolute extrema occur, including their coordinates.

Explore how to find the absolute maximum and minimum values of the function g(x) = √(2x - x²) on its given interval. This problem involves calculating derivatives, evaluating endpoints and critical points, and understanding extrema through graphical representation. Suitable for undergraduate calculus students.

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