Problem 29 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ g(x)=\sqrt{4-x^{2}}, \quad-2 \leq x \leq 1 $$

Step-by-Step Solution

Answer

The absolute maximum is 2 at \( x = 0 \) and the absolute minimum is 0 at \( x = -2 \).

Step 1: Determine the critical points

First, we need to find the derivative of the function to determine critical points. The function is \( g(x) = \sqrt{4 - x^2} \). Differentiate \( g(x) \) using the chain rule:\[ g'(x) = \frac{1}{2\sqrt{4-x^2}} \times (-2x) = -\frac{x}{\sqrt{4-x^2}}.\]Set the derivative equal to zero and solve for \( x \):\[ -\frac{x}{\sqrt{4-x^2}} = 0 \Rightarrow x = 0.\]Thus, \( x = 0 \) is a critical point.

Step 2: Evaluate the function at the critical point and endpoints

Next, we evaluate \( g(x) \) at the critical point \( x = 0 \), and at the endpoints of the interval \( x = -2 \) and \( x = 1 \):- \( g(-2) = \sqrt{4 - (-2)^2} = \sqrt{0} = 0 \),- \( g(0) = \sqrt{4 - 0^2} = \sqrt{4} = 2 \),- \( g(1) = \sqrt{4 - 1^2} = \sqrt{3} \approx 1.732. \)

Key Concepts

Critical Points

In calculus, finding the critical points of a function is an essential step in identifying where the function reaches maximum or minimum values, also known as extrema. Critical points occur where the derivative of the function is either zero or undefined. In this exercise, we work with the function \( g(x) = \sqrt{4 - x^2} \). To find the critical points, we need to differentiate the function. This requires using the chain rule, which allows us to differentiate composite functions effectively.

After differentiating, we obtain \( g'(x) = -\frac{x}{\sqrt{4-x^2}} \). Setting \( g'(x) \) equal to zero helps us find that one of the critical points is \( x = 0 \). This critical point is crucial because it may be where the function reaches its highest or lowest values within the given interval \( -2 \leq x \leq 1 \).

Chain Rule

The chain rule is a powerful technique in calculus used to differentiate composite functions. A function is considered composite if it is built by applying one function to the results of another. In our function \( g(x) = \sqrt{4 - x^2} \), \( \sqrt{4 - x^2} \) is a composite function since it involves an inner function \( (4 - x^2) \) and an outer function (the square root function).

When using the chain rule, we differentiate the outer function first, keeping the inner function unchanged. Then, we multiply the result by the derivative of the inner function. In terms of \( g(x) \), this involves:

• Differentiating the outer function \( \sqrt{u} \) to get \( \frac{1}{2\sqrt{u}} \).
• Multiplying by the derivative of the inner function \( -2x \).

Ultimately, applying the chain rule correctly allows us to find that \( g'(x) = -\frac{x}{\sqrt{4-x^2}} \).

Endpoints Evaluation

Endpoints evaluation involves examining the values of a function at the boundaries of the given interval. These are crucial because, in many cases, the maximum or minimum value of a function could occur at these endpoints, rather than at critical points within the interval.

For this function, we examine \( g(x) \) at the endpoints \( x = -2 \) and \( x = 1 \):

• At \( x = -2 \), \( g(-2) = \sqrt{4 - (-2)^2} = \sqrt{0} = 0 \).
• At \( x = 1 \), \( g(1) = \sqrt{4 - 1^2} = \sqrt{3} \approx 1.732 \).

By evaluating these endpoints, along with the critical point identified earlier, we can fully determine where the function attains its absolute maximum or minimum values within the defined interval. This method ensures no potential extrema are overlooked.

Graphing Functions

Graphing functions provides a visual representation that helps understand where and how extrema occur. By plotting \( g(x) = \sqrt{4 - x^2} \), you can see the curve and observe the points of interest.

• The critical point \( x=0 \) is where the derivative changes sign, suggesting a possible extremum.
• By evaluating this and knowing the function behavior, we can confirm \( (0, 2) \) as the absolute maximum.
• Additionally, graphing shows \( g(x) \) smoothly approaching the endpoints at \( (-2, 0) \) and \( (1, \sqrt{3}) \).

Through graph visualization, understanding of a function’s behavior across its domain becomes evident, and such a holistic view guides us to accurately identify points of absolute extrema. Further, the plotted graph indicates how the cube is symmetrically shaped over the interval, displaying a satisfying geometric curve.