Math Problem Statement

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.

Solution

To solve this problem, we need to analyze the function g(x)=2xx2g(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)g(x)

The function g(x)=2xx2g(x) = \sqrt{2x - x^2} requires that the expression under the square root be non-negative: 2xx20.2x - x^2 \geq 0. This inequality can be rewritten as: x(2x)0.x(2 - x) \geq 0. This inequality holds when x=0x = 0 or x=2x = 2 (the roots), and xx is between them. Therefore, the interval for xx is: 0x2.0 \leq x \leq 2.

Step 2: Find the Derivative of g(x)g(x)

To find critical points, we calculate the derivative of g(x)g(x). We can use the chain rule here: g(x)=(2xx2)12.g(x) = (2x - x^2)^{\frac{1}{2}}. Using the chain rule, the derivative g(x)g'(x) is: g(x)=12(2xx2)12(22x)=22x22xx2.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)=1x2xx2.g'(x) = \frac{1 - x}{\sqrt{2x - x^2}}.

Step 3: Find Critical Points

To find critical points, set g(x)=0g'(x) = 0: 1x2xx2=0.\frac{1 - x}{\sqrt{2x - x^2}} = 0. This equation is zero when 1x=01 - x = 0, or x=1x = 1.

Step 4: Evaluate g(x)g(x) at the Endpoints and Critical Points

Now, we evaluate g(x)g(x) at x=0x = 0, x=2x = 2, and x=1x = 1 to find the absolute maximum and minimum values.

  1. At x=0x = 0: g(0)=2002=0=0.g(0) = \sqrt{2 \cdot 0 - 0^2} = \sqrt{0} = 0.

  2. At x=2x = 2: g(2)=2222=44=0=0.g(2) = \sqrt{2 \cdot 2 - 2^2} = \sqrt{4 - 4} = \sqrt{0} = 0.

  3. At x=1x = 1: g(1)=2112=21=1=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)g(x) on the interval [0,2][0, 2] is 11 at x=1x = 1.
  • The absolute minimum of g(x)g(x) on the interval [0,2][0, 2] is 00, which occurs at both endpoints x=0x = 0 and x=2x = 2.

Step 6: Graph the Function and Identify Extrema

The function g(x)=2xx2g(x) = \sqrt{2x - x^2} resembles a semicircle since it can be rewritten in the form (x1)2+g(x)2=1(x - 1)^2 + g(x)^2 = 1, which represents a circle centered at (1,0)(1, 0) with radius 11. Thus, the graph will show a semicircle on the interval [0,2][0, 2].

  • Maximum point: (1,1)(1, 1)
  • Minimum points: (0,0)(0, 0) and (2,0)(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)=2xx2g(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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Function Analysis
Derivatives
Extrema
Domain of a Function

Formulas

g(x) = √(2x - x²)
g'(x) = (1 - x) / √(2x - x²)

Theorems

Extreme Value Theorem

Suitable Grade Level

Undergraduate Calculus