Problem 123 - Calculus Volume 1

Question

For the following exercises, find the local and/or absolute maxima for the functions over the specified domain. $$y=x+\sin (x) \text { over }[0,2 \pi]$$

Step-by-Step Solution

Answer

The absolute maxima is at \( x = 2\pi \) with \( y = 2\pi \).

Step 1: Find the derivative

To find the local maxima, we first need the critical points of the function. These are found by taking the derivative and setting it equal to zero. The derivative of the function \( y = x + \sin(x) \) is \( y' = 1 + \cos(x) \).

Step 2: Solve for critical points

Set the derivative equal to zero to find the critical points: \[ 1 + \cos(x) = 0 \]This simplifies to \( \cos(x) = -1 \). The only solution to this equation in the interval \([0, 2\pi]\) is \( x = \pi \).

Step 3: Evaluate function at critical points and endpoints

We must evaluate \( y = x + \sin(x) \) at \( x = 0, \pi, 2\pi \) to determine which are maxima. \[ y(0) = 0 + \sin(0) = 0 \]\[ y(\pi) = \pi + \sin(\pi) = \pi \]\[ y(2\pi) = 2\pi + \sin(2\pi) = 2\pi \]

Step 4: Identify local/absolute maxima

The values from Step 3 are: \( y(0) = 0 \), \( y(\pi) = \pi \), and \( y(2\pi) = 2\pi \). Among these, \( y(2\pi) = 2\pi \) is the highest value, so it is the absolute maxima over \([0, 2\pi]\).

Key Concepts

Derivative

The concept of a derivative is crucial when studying calculus, as it allows us to find rates of change and, in this context, determine critical points for identifying extrema, like maxima or minima.
The derivative of a function essentially describes how the function's output value changes as its input value changes. In simpler terms, it's the function's slope at any given point.
To find the derivative of the function in our exercise, which is given by \( y = x + \sin(x) \), we apply the rule of differentiation. The derivative, \( y' \), comes out to be \( y' = 1 + \cos(x) \).
This derivative tells us how the function climbs or falls at any point \( x \). In our problem, setting the derivative \( y' = 1 + \cos(x) \) to zero helps us find the critical points where the slope of the tangent to the function is flat, indicating potential peaks or troughs.

• **Flat slope**: The incline neither rises nor falls, which usually suggests a potential maximum or minimum.
• **Critical points**: They are the \( x \) values making \( y'=0 \); in our example, this occurs at \( x = \pi \) within the domain \([0, 2\pi]\).

Local Maxima

Local maxima refer to the highest points in a small, localized area of the function's graph, meaning they stand out only in their immediate vicinity.
To determine a local maximum, we examine the function's critical points, which occur where the derivative is zero or undefined.
In the exercise, one identified critical point is \( x = \pi \). At this point, we evaluate the function \( y = x + \sin(x) \), where we find \( y(\pi) = \pi + \sin(\pi) = \pi \).
To verify whether this point is a local maximum, one could also check the second derivative or observe function behavior around \( x = \pi \). Here, since there are no other explicit conditions showing other changes around \( x = \pi \) in regards to peaks or dips within our domain:

• **Isolation**: It needs comparison with nearby points to confirm it's truly a peak.
• **Second derivative test**: We could use it for further verification by examining the concavity.

Although the exercise concludes simply by computing values, such verification can often show clear assurance of a local max status.

Absolute Maxima

Absolute maxima are the overarching highest points across the entire set domain of the function. Such maxima clearly outstand compared to any other function values in the realm we are analyzing.
We determine absolute maxima by evaluating the function at all critical points and endpoints within the domain. In our exercise, through both critical point evaluation and endpoints check:
- At \( x = 0 \): \( y(0) = 0 + \sin(0) = 0 \)- At \( x = \pi \): \( y(\pi) = \pi + \sin(\pi) = \pi \)- At \( x = 2\pi \): \( y(2\pi) = 2\pi + \sin(2\pi) = 2\pi \)
Among these, \( y(2\pi) = 2\pi \) holds the highest value. Therefore, it stands as the absolute maxima over the interval \([0, 2\pi]\). This peak is important for understanding the function's highest achievable value within that domain:

• **Comparison**: Each \( y(x) \) value is critically checked to conclude absolute nature.
• **End behavior**: Often, checking end limits envelops potential peak spots within bounds.

The determination of an absolute maximum speaks of the entire range, again affirming the importance of a thorough examination across endpoints and crucial in-between points.