Problem 96 - University Calculus: Early Transcendentals

Question

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=x^{3 / 4}-\sin x+\frac{1}{2}, \quad[0,2 \pi]$$

Step-by-Step Solution

Answer

Find and compare function values at critical and endpoint points to determine absolute extrema.

Step 1: Plot the Function

Create a graph of the function \( f(x) = x^{3/4} - \sin x + \frac{1}{2} \) over the interval \([0, 2\pi]\). Use a graphing calculator or a computer algebra system (CAS) to visualize how the function behaves within this domain.

Step 2: Find Interior Points Where Derivative is Zero

Compute the derivative of the function, \( f'(x) = \frac{3}{4}x^{-1/4} - \cos x \), and solve for \( f'(x) = 0 \) to find critical points within the interval. Use a computer algebra system to approximate the solutions of the equation, as it may not yield simple analytical solutions.

Step 3: Find Interior Points Where Derivative Does Not Exist

Determine where \( f'(x) \) might not exist. The critical point for the term \( \frac{3}{4}x^{-1/4} \) is at \( x = 0 \) since the term becomes undefined at this point. However, since \( x = 0 \) is an endpoint, it is important only to note if it causes issues within the interval.

Step 4: Evaluate Function at Critical and Endpoint Values

Calculate \( f(x) \) at each point found in steps 2 and 3, as well as at the endpoints of the interval, \( f(0) \) and \( f(2\pi) \). Compute these values to find potential extrema.

Step 5: Identify Absolute Extrema

Compare all calculated function values from step 4. The largest value is the absolute maximum, and the smallest value is the absolute minimum over the interval \([0, 2\pi]\).

Key Concepts

Absolute Extrema

Absolute extrema of a function refer to the highest and lowest values the function attains over a particular interval. In other words, if you picture the graph of a function over an interval, the absolute maximum is the tallest peak, while the absolute minimum is the lowest point. Finding these extreme values involves evaluating the function not only at critical points but also at the endpoints of the interval. This ensures that any potential maximum or minimum is considered, since extrema could occur at these boundary points.

• Check both endpoints of the interval.
• Evaluate the function at all these points.
• Compare values to find the absolute max and min.

By combining values from endpoints and critical points, you are covering all possible scenarios for where the extreme values could occur. This method covers the interval comprehensively, ensuring you identify true absolute extrema.

Critical Points

Critical points are the values of x where the derivative of a function is zero or does not exist. These points are interesting because they might indicate locations where the function reaches either a maximum or a minimum within its interval.To find critical points:

• Calculate the derivative of the function, often denoted as \( f'(x) \).
• Solve the equation \( f'(x) = 0 \) to locate where the slope of the function is zero.
• Identify where \( f'(x) \) does not exist to spot potential sharp corners or vertical tangents in the function graph.

Once the critical points on the interval are identified, they give clues about where the function might switch from increasing to decreasing, or vice versa, hinting at peaks and valleys.

Derivative

The derivative of a function gives insight into the rate of change and the slope of the function at any given point. It is a powerful tool in calculus because analyzing it can reveal where the function is increasing or decreasing.Derivatives are calculated with respect to a variable (often \( x \)), and they can allow us to:

• Predict where changes occur.
• Find critical points by setting the derivative to zero.
• Identify points of non-differentiability that can indicate sharp changes.

For the given function \( f(x) = x^{3/4} - \sin x + \frac{1}{2} \), the derivative \( f'(x) = \frac{3}{4}x^{-1/4} - \cos x \) is instrumental in analyzing how the function behaves over \([0, 2\pi]\). Watching these signs and zero points helps to map out the function's behavior clearly.

Interval Analysis

Interval analysis involves examining how a function behaves over a specified domain. It's crucial for finding absolute extrema since it ensures we consider all relevant segments of the function.When performing interval analysis, focus on:

• Defining the precise boundaries of the interval (e.g., \([0, 2\pi]\)).
• Evaluating both the endpoints and critical points within these boundaries.
• Using algebraic or numerical methods to solve for any points of interest within the interval.

This comprehensive examination allows us to identify where certain features, such as extreme maximums or minimums, occur within the function's graph. Understanding the interval ensures nothing is overlooked, offering a complete picture of the function’s behavior.