Problem 80 - Thomas Calculus

Question

In Exercises \(75-80,\) you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. $$\begin{array}{l}{\text { a. Plot the function over the interval to see its general behavior there. }} \\ {\text { b. Find the interior points where } f^{\prime}=0 . \text { (In some exercises, }} \\ {\text { you may have to use the numerical equation solver to approximate }} \\ {\text { a solution.) You may want to plot } f^{\prime} \text { as well. }}\\\\{\text { c. Find the interior points where } f^{\prime} \text { does not exist. }} \\ {\text { d. Evaluate the function at all points found in parts (b) and (c) }} \\ {\text { and at the endpoints of the interval. }} \\ {\text { e. Find the function's absolute extreme values on the interval }} \\ {\text { and identify where they occur. }}\end{array}$$ $$ f(x)=x^{3 / 4}-\sin x+\frac{1}{2}, \quad[0,2 \pi] $$

Step-by-Step Solution

Answer

Absolute extremum occurs at endpoints and critical points; calculate these for exact values.

Step 1: Plot the function

Using a Computer Algebra System (CAS), plot the function \( f(x) = x^{3/4} - \sin x + \frac{1}{2} \) over the interval \([0, 2\pi]\). Observe the general shape to anticipate where the extrema might occur.

Step 2: Determine where the derivative equals zero

Find the derivative \( f'(x) \) using the CAS: \( f'(x) = \frac{3}{4}x^{-1/4} - \cos x \). Solve the equation \( f'(x) = 0 \) within the interval \([0, 2\pi]\). Approximate solutions might be needed due to the complexity of the equation.

Step 3: Determine where the derivative does not exist

Examine the derivative \( f'(x) = \frac{3}{4}x^{-1/4} - \cos x \) to find where it does not exist. Since \( x^{-1/4} \) is undefined at \( x=0 \), identify \( x=0 \) as a point where \( f' \) does not exist.

Step 4: Evaluate the function at critical points and endpoints

Evaluate \( f(x) \) at all the interior points where \( f'(x) = 0 \), \( f'(x) \) does not exist, and at the endpoints \( x=0 \) and \( x=2\pi \). This involves substituting these values into the function.

Step 5: Identify absolute extrema

Compare the values of \( f(x) \) obtained in the previous step. The highest value is the absolute maximum, and the lowest is the absolute minimum, both within the interval \([0, 2\pi]\). Determine where these extrema occur.

Key Concepts

Function Plot

Plotting a function is an essential step when dealing with calculus extremum problems. A function plot visually represents the behavior of a function over a specified interval. For the function \( f(x) = x^{3/4} - \sin x + \frac{1}{2} \), plotting over the interval \([0, 2\pi]\) helps to understand its overall shape and layout. This initial overview gives clues on where potential highs and lows, or extrema, might occur.

• Helps find general trends and patterns.
• Identifies intervals of increase and decrease.
• Predicts locations of possible critical points or extrema.

By studying the graph, you begin to anticipate where you might find important features such as peaks, valleys, or points of inflection. This visual insight is invaluable before delving deeper into derivatives and critical points.

Derivative

The derivative is a core concept in calculus that gives us the slope or rate of change of a function at any point. For our function \( f(x) = x^{3/4} - \sin x + \frac{1}{2} \), the derivative is calculated as \( f'(x) = \frac{3}{4}x^{-1/4} - \cos x \). Understanding the derivative involves:

• Determining how steep or flat the function graph is at particular points.
• Finding the rate at which a function value changes as \( x \) changes.
• Identifying where the derivative equals zero or doesn't exist (critical points).

A derivative set to zero indicates potential maxima or minima, while an undefined derivative can highlight points of discontinuity or cusps. Derivatives provide the mathematical framework to analyze and solve problems concerning slopes and curves.

Critical Points

Critical points are where the derivative of a function equals zero or does not exist. These points are crucial as they can indicate local maxima, minima, or points of inflection. For the function \( f(x) \), solve \( f'(x) = 0 \) to find where potential extrema occur.

• Solutions to \( f'(x) = 0 \) within \([0, 2\pi]\) may require numerical approximations due to complexity.
• Examining where \( f'(x) \) does not exist, such as at \( x=0 \) for this function.

Understanding these points on a graph can help pinpoint significant changes in the function's behavior and suggest where to evaluate for potential extremum.

Extrema

Extrema refer to the maximum and minimum values a function can take on a given interval. Finding these points involves evaluating the function at all critical points and the endpoints of the interval. For \( f(x) \), check:

• Values where \( f'(x) = 0 \)
• Endpoints of the interval \([0, 2\pi]\)
• Any points where \( f'(x) \) doesn't exist (such as \( x=0 \))

After calculating \( f(x) \) at these points, compare the results to determine where the absolute maxima and minima occur. Recognizing these values helps to understand the function's greatest and lowest outputs, revealing the overall behavior across the interval.

CAS (Computer Algebra System)

A Computer Algebra System (CAS) is invaluable in solving complex calculus problems. It can perform symbolic calculations like derivatives and solve equations numerically or symbolically when they become difficult by hand. In this problem:

• The CAS helps to plot the function and its derivative accurately.
• It assists in solving the derivative equation \( f'(x) = 0 \) to find critical points.
• Performs a comprehensive check of values at endpoints and undefined points for minimum and maximum values.

Using a CAS simplifies the analysis process, providing tools for both visual and computational approaches, ensuring results are precise and time-efficient.