Problem 23 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x+\sin x, \quad 0 \leq x \leq 2 \pi$$

Step-by-Step Solution

Answer

Local max at \(x = \pi\), abs. max at \(x=2\pi\), abs. min at \(x=0\); inflection at \(x=0,\pi,2\pi\).

Step 1: Find the First Derivative

To find the local extreme points, we first need to determine the critical points, which are found by setting the first derivative equal to zero. The function is given as \[ y = x + \sin x \]The first derivative is \[ y' = 1 + \cos x \]To find the critical points, set the first derivative equal to zero:\[ 1 + \cos x = 0 \]This simplifies to \( \cos x = -1 \).

Step 2: Solve for Critical Points

Now we solve \( \cos x = -1 \) within the interval \([0, 2\pi]\):The equation \( \cos x = -1 \) is true when \(x = \pi\). This is the only critical point within the given interval.

Step 3: Find Second Derivative

To determine if the critical point is a local minimum or maximum, or to find inflection points, we use the second derivative. The second derivative of the function is:\[ y'' = -\sin x \]

Step 4: Determine Concavity and Inflection Points

The inflection points are found where the second derivative changes sign. Set the second derivative equal to zero:\[ -\sin x = 0 \]This simplifies to \( \sin x = 0 \). Within the interval \([0, 2\pi]\), \( \sin x = 0 \) at \(x = 0, \pi, 2\pi\). Since the second derivative changes sign at these points, they are inflection points.

Step 5: Evaluate at Endpoints

To identify absolute extreme points, evaluate the function at the critical points and endpoints of the interval. Given points are \(x = 0, \pi, 2\pi\):\( y(0) = 0 + \sin(0) = 0 \)\( y(\pi) = \pi + \sin(\pi) = \pi \)\( y(2\pi) = 2\pi + \sin(2\pi) = 2\pi \)Compare these values: The smallest value is \(0\), the largest is \(2\pi\).

Step 6: Conclusion from Calculations

From the evaluations and calculations:- \(x = \pi\) is a local maximum.- Absolute minimum at \(x = 0\) with value \(0\).- Absolute maximum at \(x = 2\pi\) with value \(2\pi\).- Inflection points at \(x = 0, \pi, 2\pi\).

Key Concepts

Extrema

In calculus, extrema refer to the maximum and minimum values that a function can attain within a specific range. There are two types of extrema: local and absolute. Local extrema are points where the function reaches a peak (local maximum) or a valley (local minimum) within a small interval. Absolute extrema refers to the highest or lowest values that the function takes over its entire domain.
Understanding these concepts is crucial as they provide insights into the behavior of a function.

• Local Maximum: The function value is higher than nearby points, like in our solution where at \( x = \pi \) it achieves a local maximum.
• Absolute Maximum: The largest value in the entire domain, such as \( y(2\pi) = 2\pi \) in our example.
• Absolute Minimum: The smallest value in the entire domain, found here at \( y(0) = 0 \).

Identifying these points helps in graphing functions and analyzing them in real-world contexts, where maxima might represent highest profits and minima lowest costs.

Critical Points

Critical points of a function are where its derivative is zero or undefined. Identifying these points helps locate potential extrema.
To find the critical points, we solve the equation derived from the first derivative set to zero. For instance, with the function \( y = x + \sin x \), the derivative \( y' = 1 + \cos x \) equals zero at \( \cos x = -1 \). This occurs at \( x = \pi \) in our interval of \([0, 2\pi]\).

• Finding critical points helps us identify where the function's slope changes, potentially leading to peaks and valleys.
• They are essential for determining where to find local extrema.

Remember, a critical point does not always guarantee a local maximum or minimum but indicates where they might occur.

Inflection Points

Inflection points are where a curve changes its concavity, meaning the direction of its curvature. An understanding of inflection points is significant as they signify a shift in the curve's behavior.
For our function \( y = x + \sin x \), these are determined by setting the second derivative \( y'' = -\sin x \) to zero, resulting in \( \sin x = 0 \). This occurs at \( x = 0, \pi, 2\pi \) within \([0, 2\pi]\). These points do not imply extrema but instead mark a change in the function's concavity.

• Inflection points can be crucial in understanding how a function's graph bends and twists.
• They can influence how we calculate integrals or understand the motion in physics problems.

Identifying these points aids in sketching accurate graphs and is vital in fields like engineering and economics, where understanding changes in trends is essential.

Concavity

Concavity describes whether a function curves upwards or downwards. Determining concavity is essential for understanding how the function behaves between critical points and at inflection points.
A function is concave up when its second derivative is positive and concave down when the second derivative is negative. In our example, \( y'' = -\sin x \) provides this information:

• Concave Up: Segments where \( \sin x < 0 \).
• Concave Down: Segments where \( \sin x > 0 \).

Between \(0\) and \(2\pi\), the concavity changes at \( x = \pi \), demonstrating shifts in upward and downward curving.
Understanding concavity aids not only in graphing but in optimization problems where we determine the nature of extrema.