Problem 28 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=\cos x+\sqrt{3} \sin x, \quad 0 \leq x \leq 2 \pi \end{equation}

Step-by-Step Solution

Answer

Local maxima at \( x = \frac{\pi}{3}, \frac{4\pi}{3} \); inflection points at \( x = \frac{5\pi}{6}, \frac{11\pi}{6} \).

Step 1: Find the First Derivative

To find the critical points where local extrema might occur, we need the first derivative of the function. Differentiate the function:\[ y = \cos x + \sqrt{3} \sin x \]The derivative is:\[ y' = -\sin x + \sqrt{3} \cos x \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points:\[ -\sin x + \sqrt{3} \cos x = 0 \]Rearrange and solve:\[ \sin x = \sqrt{3} \cos x \]\[ \frac{\sin x}{\cos x} = \sqrt{3} \]\[ \tan x = \sqrt{3} \]Therefore, \( x = \frac{\pi}{3} + n\pi \) for any integer n. In the interval \( 0 \leq x \leq 2\pi \), \( x \) can be \( \frac{\pi}{3} \) or \( \frac{4\pi}{3} \).

Step 3: Test for Local Extrema with Second Derivative

Find the second derivative to test concavity:\[ y'' = -\cos x - \sqrt{3} \sin x \]Evaluate at critical points:At \( x = \frac{\pi}{3} \):\[ y''\left(\frac{\pi}{3}\right) = -\frac{1}{2} - \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = -1 < 0 \]Negative value indicates a local maximum.At \( x = \frac{4\pi}{3} \):\[ y''\left(\frac{4\pi}{3}\right) = \frac{1}{2} + \sqrt{3}\left(-\frac{\sqrt{3}}{2}\right) = -1 < 0 \]Negative value indicates another local maximum.

Step 4: Find the Inflection Points

An inflection point occurs where the second derivative changes sign. Set it equal to zero:\[ -\cos x - \sqrt{3} \sin x = 0 \]Solve for \( x \):\[ \cos x = -\sqrt{3} \sin x \]\[ \frac{\cos x}{\sin x} = -\frac{1}{\sqrt{3}} \]\[ \cot x = -\frac{1}{\sqrt{3}} \]\( x = \frac{5\pi}{6} + n\pi \). In the interval, possible inflection points are \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \).

Step 5: Determine Absolute Extrema

Evaluate the original function at the endpoints and critical points to find absolute extrema:At \( x = 0 \):\[ y(0) = 1 \]At \( x = 2\pi \):\[ y(2\pi) = 1 \]Critical points:At \( x = \frac{\pi}{3} \):\[ y\left(\frac{\pi}{3}\right) = \frac{1}{2} + \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 2 \]At \( x = \frac{4\pi}{3} \):\[ y\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - \sqrt{3}\left(\frac{-\sqrt{3}}{2}\right) = 2 \]Thus, absolute maxima occur at \( x = \frac{\pi}{3} \) and \( x = \frac{4\pi}{3} \) with a value of 2.

Key Concepts

Critical Points

Understanding critical points is crucial in calculus. They are the points on the graph of a function where the derivative is either zero or undefined. These points are important because they could represent local maximums, local minimums, or saddle points.

To find critical points, follow these steps:

• First, compute the derivative of the function.
• Set the derivative equal to zero and solve for the variable. This gives us potential critical points.
• Ensure that the solutions fall within the given interval.

In the given exercise, the function is differentiated to get the first derivative. By setting the derivative to zero, we find the critical points at \( x = \frac{\pi}{3} \) and \( x = \frac{4\pi}{3} \). These are the x-values where the original function might change behavior and have local extrema.

Second Derivative Test

The second derivative test helps determine the nature of critical points. It uses the values of the second derivative at critical points to infer whether they are local maxima, local minima, or points of inflection.

Here is how the test works:

• Calculate the second derivative of the function.
• Evaluate the second derivative at each critical point.
• If the second derivative is positive at a critical point, the function has a local minimum there. If it is negative, the function has a local maximum.
• If the second derivative is zero, the test is inconclusive.

For our exercise, the second derivative \( y'' = -\cos x - \sqrt{3} \sin x \) was found and evaluated at \( x = \frac{\pi}{3} \) and \( x = \frac{4\pi}{3} \). In both cases, the second derivative was negative, indicating local maxima at these points.

Inflection Points

Inflection points occur where a function changes concavity, transitioning from concave up to concave down or vice versa. These are the points where the second derivative changes sign.

To find inflection points:

• Set the second derivative equal to zero and solve for the variable.
• Ensure that these points are within the interval considered.
• Verify that the sign of the second derivative changes around these points.

In our exercise, we solved for inflection points by setting the second derivative to zero. This resulted in potential inflection points at \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \). Around these points, the second derivative changes sign, confirming them as inflection points.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially in problems involving periodic behavior.

Key characteristics of these functions include:

• Periodic nature, with sine and cosine functions having a period of \( 2\pi \).
• Amplitudes representing the maximum value reached by the functions, which is 1 for both sine and cosine.
• Phase shifts that move the graph horizontally.

The function in our exercise, \( y = \cos x + \sqrt{3} \sin x \), is essentially a phase-shifted trigonometric function. Analyzing such functions requires understanding their properties, including how they combine to form new functions with unique behaviors in given intervals. This highlights the intricate interplay between simple trigonometric forms and their derivatives, leading to discoveries of critical, inflection, and extreme points.