Problem 49 - Thomas Calculus in SI Units

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$f(x)=\sqrt{25-x^{2}}, \quad-5 \leq x \leq 5$$

Step-by-Step Solution

Answer

The local maximum is at \( x=0 \), and the minima are at \( x=-5 \) and \( x=5 \). The absolute maximum is 5, and the minima 0.

Step 1: Identify the Function Type

The function given is the square root function \( f(x) = \sqrt{25-x^2} \). This is defined only for values of \( x \) where \( 25 - x^2 \geq 0 \), which simplifies to \( -5 \leq x \leq 5 \). This function describes the upper half of a circle with radius 5 centered at the origin.

Step 2: Find the Critical Points

To find local extreme values, compute the derivative of \( f(x) \). \( f'(x) = \frac{d}{dx}(25-x^2)^{1/2} = \frac{-x}{\sqrt{25-x^2}} \). Set \( f'(x) \) equal to zero to find critical points: \( \frac{-x}{\sqrt{25-x^2}}=0 \), giving \( x=0 \). So, \( x=0 \) is a critical point within the interval \([-5, 5]\).

Step 3: Determine Local Extremes

Evaluate \( f(x) \) at the critical point and at the endpoints of the domain: \( x = -5, 0, 5 \). Calculate: \( f(-5) = \sqrt{25-(-5)^2} = 0 \), \( f(0) = \sqrt{25-0^2} = 5 \), and \( f(5) = \sqrt{25-5^2} = 0 \). Thus, the local maximum is \( f(0) = 5 \), and local minima are \( f(-5)=0 \) and \( f(5)=0 \).

Step 4: Determine Absolute Extremes

Comparing the local extreme values, it turns out that the highest value is 5 (at \( x=0 \)) and the lowest value is 0 (at both \( x=-5 \) and \( x=5 \)). Therefore, \( f(0) = 5 \) is an absolute maximum, and \( f(-5) = f(5) = 0 \) are absolute minima.

Step 5: Utilize a Graphing Tool

Use a graphing calculator or software to plot \( f(x) = \sqrt{25-x^2} \) over \([-5, 5]\). Confirm that the graph is a semicircle (upper half of a circle), with a peak at \( x=0 \) and touching the x-axis at \( x=-5 \) and \( x=5 \). This visually supports the computed extreme values.

Key Concepts

Local Extreme Values

Local extreme values in calculus refer to points in the domain of a function where the function reaches a local maximum or minimum. These are places where the function changes direction—either from increasing to decreasing or vice versa.

• **Local Maximum:** A local maximum occurs at a point where the function's value is greater than the neighboring values. For the function \(f(x) = \sqrt{25-x^2}\), the local maximum is at \(x = 0\). At this point, the function value is 5, the highest value in the immediate area.
• **Local Minimum:** A local minimum occurs where the function's value is less than the values of nearby points. In our example, the function has local minima at \(x = -5\) and \(x = 5\), where the function value is 0.

Local extremes are important because they help us understand the behavior of a function within a certain area of its domain. By calculating the derivative and finding where it equals zero or does not exist, we locate these critical points.

Critical Points

Critical points are specific places on the graph of a function where its derivative is zero or undefined. These points are significant because they help identify potential local extremes.
To find critical points:

• Compute the derivative of the function.


• Set the derivative equal to zero and solve for \(x\) to find where the slope of the tangent is horizontal.

For the function \(f(x) = \sqrt{25-x^2}\), the derivative is \(f'(x) = \frac{-x}{\sqrt{25-x^2}}\). Setting this equal to zero gives a critical point at \(x = 0\).

Critical points are not always points of local extremes, but they are necessary to consider when trying to find them. Checking the derivative at intervals around these points can determine whether they are maxima, minima, or neither.

Absolute Extremes

Absolute extremes refer to the highest and lowest values a function can attain within a given domain. Unlike local extremes, which are concerned with immediate neighboring values, absolute extremes look at the entire domain.

To find absolute extremes, evaluate the function at:

• Endpoints of the domain.
• All critical points found within the domain.

In the case of the function \(f(x) = \sqrt{25-x^2}\) for \(-5 \leq x \leq 5\):

• The absolute maximum is \(f(0) = 5\).
• The absolute minima are \(f(-5) = 0\) and \(f(5) = 0\).

These absolute extremes signify the definitive bounds of the function values across the specified range. It is crucial to consider both endpoints and critical points as absolute extremes can reside at any of these locations.