Problem 24 - Thomas Calculus in SI Units

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 extrema are at \((0,0)\) and \((2\pi,2\pi)\), inflection point at \(x=\pi\).

Step 1: Find the First Derivative

To find local extrema, we need to first find the derivative of the function. Starting with \( y = x - \sin x \), we differentiate to get \( y' = 1 - \cos x \).

Step 2: Solve for Critical Points

Set the derivative equal to zero to find critical points: \( 1 - \cos x = 0 \). This simplifies to \( \cos x = 1 \). The solutions in the interval \( 0 \leq x \leq 2\pi \) are \( x = 0 \) and \( x = 2\pi \).

Step 3: Evaluate Function at Critical Points

Evaluate the function at the critical points: \( y(0) = 0 - \sin 0 = 0 \) and \( y(2\pi) = 2\pi - \sin 2\pi = 2\pi \). These correspond to the points \((0, 0)\) and \((2\pi, 2\pi)\).

Step 4: Determine the Nature of Critical Points

Since \( y' = 1 - \cos x \) and \( \cos x \) ranges only between \(-1\) and \(1\), \( y' > 0 \) for all \( x eq 0, 2\pi \). Therefore, the function is increasing on the interval \( 0 < x < 2\pi \), so \((0, 0)\) is a local minimum and \((2\pi, 2\pi)\) is a local maximum.

Step 5: Find the Second Derivative

To find inflection points, take the second derivative: \( y'' = \sin x \).

Step 6: Solve for Potential Inflection Points

Set the second derivative equal to zero: \( \sin x = 0 \). Solutions within the interval \( 0 \leq x \leq 2\pi \) are \( x = 0, \pi, 2\pi \).

Step 7: Test for Inflection Points

A change in the sign of \( y'' \) around \( x = \pi \) confirms an inflection point because \( y'' \) transitions from positive to negative or vice versa.

Key Concepts

Extrema

In calculus, extrema refer to the points on a graph where a function reaches its maximum or minimum value. These values can be local extrema or absolute (global) extrema.

• **Local extrema:** A function exhibits local extreme values at points where the function changes direction from increasing to decreasing or vice versa. In simpler words, it peaks up or down in a neighbourhood. Finding these involves identifying points where the derivative equals zero or where the derivative does not exist.
• **Absolute extrema:** These occur at the highest or lowest points of the entire graph within a specified domain. These are the true maximum and minimum values of the function within the given interval.

For the function given, **y = x - \sin x**, we found that there is a local minimum at \( (0, 0) \) and a local maximum at \( (2\pi, 2\pi) \). This is determined by considering where the first derivative \( y' = 1 - \cos x \) changes sign.Understanding extrema helps to visualize the general shape of a function’s graph and predict its peaks and valleys.

Critical Points

Critical points of a function occur where the first derivative equals zero or does not exist. These points are important because they provide valuable information that helps to determine the presence of potential turning points for extrema.

• To find these for **y = x - \sin x**, we found the derivative \( y' = 1 - \cos x \). Setting this to zero (since \( y' \) never fails to exist for all x in our interval), we solved \( \cos x = 1 \) which gives critical points at \( x = 0 \) and \( x = 2\pi \) within the domain \( 0 \leq x \leq 2\pi \).
• After finding critical points, we evaluate the original function at these points to locate extrema: \( y(0) = 0 \) and \( y(2\pi) = 2\pi \).

Critical points serve as potential indicators for local maxima or minima, making them essential in analyzing and graphing functions.

Inflection Points

Inflection points are where a function changes its concavity, meaning it shifts from being concave up to concave down, or vice versa. This typically indicates a change in the curve's shape, providing insight into more nuanced behavior beyond just increasing or decreasing.

• We determine inflection points by examining the second derivative. For our function, the second derivative is \( y'' = \sin x \).
• Set \( y'' = \sin x = 0 \) to find potential inflection points, resulting in \( x = 0, \pi, \) and \( 2\pi \) within the specified interval.
• To confirm these candidates as true inflection points, observe the change in sign of \( y'' \) around these points. Specifically, at \( x = \pi \), the sign of \( y'' \) changes, verifying it as a point of inflection.

Inflection points are vital for deeper insights into a function's graph since they signify dynamic changes in the curve's form, allowing for comprehensive analysis of function behaviour and graphical interpretations.