Problem 36 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one. \(S(x)=x-\sin x, \quad 0 \leqslant x \leqslant 4 \pi\)

Step-by-Step Solution

Answer

S(x) is increasing on (0, 2π) and (2π, 4π), concavity alternates based on sin x, and has no local extrema or inflection points.

Step 1: Find the First Derivative

To determine intervals of increase or decrease, we start by finding the first derivative of the function. The function is \( S(x) = x - \sin x \). Differentiate to get: \[ S'(x) = 1 - \cos x \]

Step 2: Determine Critical Points

Set the first derivative equal to zero to find critical points: \[ 1 - \cos x = 0 \]\[ \cos x = 1 \]Solving \( \cos x = 1 \), we find \( x = 0, 2\pi, 4\pi \) within the interval \( 0 \leq x \leq 4\pi \).

Step 3: Test Intervals for Increasing or Decreasing Functions

Check intervals between the critical points: - For \( 0 < x < 2\pi \) and \( 2\pi < x < 4\pi \), evaluate \( S'(x) = 1 - \cos x \).- \( \cos x < 1 \) and \( S'(x) > 0 \) implies these intervals are increasing.

Step 4: Find Local Maxima and Minima

At each critical point, determine if it’s a maxima, minima, or neither:- At \( x = 0, 2\pi, 4\pi \), since \( S'(x) \) changes from positive to positive, these are not local maxima or minima.

Step 5: Find the Second Derivative for Concavity

Find the second derivative to determine concavity:\[ S''(x) = \sin x \]

Step 6: Test Intervals for Concavity

Evaluate \( S''(x) \) for the same intervals:- When \( \sin x > 0 \), the graph is concave up.- When \( \sin x < 0 \), the graph is concave down.

Step 7: Find Inflection Points

Inflection points occur where \( S''(x) = 0 \) and changes sign:- Set \( \sin x = 0 \) implies \( x = 0, \pi, 2\pi, 3\pi, 4\pi \) as potential inflection points.

Step 8: Summary and Sketching the Graph

Use the intervals and behaviors found:- Increasing on \((0, 2\pi)\) and \((2\pi, 4\pi)\).- Concave up where \( \sin x > 0 \), concave down where \( \sin x < 0 \).- Sketch the graph accordingly, noting increasing behavior and no local maxima/minima.

Key Concepts

Critical Points

Critical points in calculus are values of \( x \) where the first derivative of a function equals zero or is undefined. These points help in identifying where the function might have a local maximum, local minimum, or a point of inflection. They are key to understanding the graph's behavior over an interval.
For the function \( S(x) = x - \sin x \), the first derivative is \( S'(x) = 1 - \cos x \). Setting the derivative to zero gives us:

• \( 1 - \cos x = 0 \)
• \( \cos x = 1 \)

Solving \( \cos x = 1 \) identifies the critical points as \( x = 0, 2\pi, \text{and } 4\pi \) over the interval \( 0 \leq x \leq 4\pi \). These points are important to check among the intervals to determine where the function is increasing or decreasing.

Derivative Calculation

Derivative calculation is crucial for understanding the rate of change of a function. It forms the backbone of identifying the behavior of the function, especially in determining increase, decrease, and concavity.
For \( S(x) = x - \sin x \), the first derivative is calculated as \( S'(x) = 1 - \cos x \). This derivative is used to determine:

• Intervals of increase and decrease of the function.
• Critical points where the function changes direction or rate of change.

Steps to find the first derivative include finding critical points by setting \( S'(x) = 0 \) and testing the intervals between critical points for sign changes in \( S'(x) \), indicating where the function might increase or decrease.
Thus, the derivative's sign tells us a lot about the nature of the function over different intervals.

Concavity and Inflection Points

Concavity refers to the direction in which a function curves. It is determined using the second derivative of a function. Inflection points are where the concavity changes from up to down or vice versa.
Calculating the second derivative for \( S(x) = x - \sin x \), we have \( S''(x) = \sin x \). We then test the sign of \( S''(x) \) over the range to determine whether \( S(x) \) is concave up or down:

• When \( \sin x > 0 \), the graph is concave up.
• When \( \sin x < 0 \), the graph is concave down.

To find inflection points, find where the second derivative equals zero and changes sign. Solving \( \sin x = 0 \) gives potential inflection points at \( x = 0, \pi, 2\pi, 3\pi, 4\pi \). The nature of these points is verified by observing a change in sign of \( S''(x) \) around these values. Concavity helps us understand the shape and direction, and inflection points allow us to see where these changes occur.