Problem 25 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Local minima at \( x = \frac{4\pi}{3} \); absolute minimum at \( x = 0 \); inflection points at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Step 1: Find the Derivative

To identify local extreme points, first find the derivative of the function. The given function is \( y = \sqrt{3} x - 2 \cos x \). Its derivative is \( y' = \sqrt{3} + 2 \sin x \).

Step 2: Set the Derivative to Zero

Set the derivative equal to zero to find critical points: \( \sqrt{3} + 2 \sin x = 0 \). Solving for \( x \), we get \( \sin x = -\frac{\sqrt{3}}{2} \). These occur at \( x = \frac{4\pi}{3} \) and \( x = \frac{5\pi}{3} \) in the interval \( 0 \leq x \leq 2\pi \).

Step 3: Second Derivative Test

Find the second derivative to use the second derivative test. The second derivative is \( y'' = 2 \cos x \). Evaluate \( y'' \) at the critical points. At \( x = \frac{4\pi}{3} \), \( y'' = 2 \cos \left(\frac{4\pi}{3}\right) = 1 \), which is positive, indicating a local minimum. At \( x = \frac{5\pi}{3} \), \( y'' = 2 \cos \left(\frac{5\pi}{3}\right) = 1 \), which is also positive.

Step 4: Find Inflection Points

An inflection point occurs where the second derivative changes sign. Solve \( 2 \cos x = 0 \) for \( x \); this occurs at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \) within the interval. Both are inflection points.

Step 5: Evaluate at Endpoints

Evaluate the function at the interval endpoints \( x = 0 \) and \( x = 2\pi \). At \( x = 0 \), \( y = -2 \), and at \( x = 2\pi \), \( y = 2\sqrt{3} - 2 \). The absolute minimum is at \( x = 0 \) and \( x = \frac{4\pi}{3} \), and the absolute maximum is at \( x = 2\pi \).

Step 6: Graph the Function

Using all the calculated points and behavior of the function, sketch the graph of \( y = \sqrt{3} x - 2 \cos x \) in the interval \( 0 \leq x \leq 2\pi \). The local and absolute extrema and inflection points should be clearly marked.

Key Concepts

Extreme Points

When analyzing a function, finding extreme points is essential. These are locations on the graph where the function either reaches a local or absolute maximum or minimum.
To locate these points, we start by calculating the first derivative of the function and setting it equal to zero. This step identifies the critical points, which are potential candidates for extreme points.

• Local extreme points are the peaks or valleys in a restricted region of the graph.
• Absolute extreme points represent the highest and lowest values over the entire interval being considered.

In our function, by solving the derivative equation, we find critical points at \( x = \frac{4\pi}{3} \) and \( x = \frac{5\pi}{3} \).
These points need to be tested further to determine whether they are actual local minima or maxima using additional steps like the second derivative test, which reveals their nature.

Inflection Points

Inflection points are places on the graph where the curvature changes direction. These points are crucial as they indicate where the graph shifts from concave up to concave down or vice versa. To find inflection points, we analyze the second derivative of the function.
For our function, the second derivative is derived as \( y'' = 2 \cos x \). Inflection points occur when this second derivative changes sign.
By setting the second derivative to zero and solving for \( x \), we find the inflection points at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \) for the given interval. This shift in sign at these points hints at a change in the concavity of the function, and they are marked accordingly on the graph when visualizing the function's behavior.

Derivative

Derivatives are a central concept in calculus, representing the rate of change of a function with respect to a variable. In simpler terms, a derivative tells us how steep a curve is at a given point, acting as a mathematical tool that helps us find slopes and rates.
For our specific function \( y = \sqrt{3}x - 2\cos x \), the derivative \( y' = \sqrt{3} + 2\sin x \) reveals how \( y \) changes as \( x \) changes.

• It helps find critical points where the function could have peaks or valleys (extreme points).
• It aids in determining whether these points are wordy indeed maxima or minima using further analysis.

In our exercise, we set the derivative to zero to identify points of interest that need further exploration to distinguish their role as extreme points on the function's graph.

Trigonometric Functions

Trigonometric functions are fundamental in understanding periodic behaviors of functions across many fields of mathematics and real life applications. In our function, the cosine function, \( \cos x \), plays a significant role.
The presence of \( \cos x \) in the function \( y = \sqrt{3}x - 2\cos x \) introduces periodic variations, influencing the shape and characteristics of the graph across the specified interval.
Trigonometric identities help simplify calculations and provide insights into the behavior of the function.

• They assist during differentiation, which is evident when we find out how \( \cos x \) contributes to the derivatives.
• They explain how periodic changes can result in multiple peaks and valleys, due to the inherent properties of the cosine function.

In conclusion, understanding trigonometric functions aids in predicting and verifying the behavior of the function, providing a deeper insight into its graphical representation.