Problem 25 - Thomas Calculus in SI Units

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 maxima at \(x = \frac{4\pi}{3}, \frac{5\pi}{3}\); absolute minimum at \(x = 0\). Inflection points at \(x = \frac{\pi}{2}, \frac{3\pi}{2}\).

Step 1: Find the derivative of the function

To identify extreme points and inflection points, we first need the derivative of the function. The given function is \(y = \sqrt{3}x - 2\cos x\). The derivative of \(y\) with respect to \(x\) is \(y' = \sqrt{3} + 2\sin x\). This derivative will help us locate critical points.

Step 2: Determine critical points

Critical points occur where the derivative is zero or undefined. Set the derivative to zero and solve for \(x\): \[ \sqrt{3} + 2\sin x = 0 \] \[ \sin x = -\frac{\sqrt{3}}{2} \] The sine function equals \(-\frac{\sqrt{3}}{2}\) at \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\) within the interval \([0, 2\pi]\). These are the critical points.

Step 3: Determine the nature of critical points

To classify each critical point, use the second derivative test. First, find the second derivative of \(y\):\(y'' = 2\cos x\).Evaluate \(y''\) at the critical points \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\):- For \(x = \frac{4\pi}{3}\), \(y'' = 2\cos\left(\frac{4\pi}{3}\right) = 2\left(-\frac{1}{2}\right) = -1\), indicating a local maximum.- For \(x = \frac{5\pi}{3}\), \(y'' = 2\cos\left(\frac{5\pi}{3}\right) = 2\left(-\frac{1}{2}\right) = -1\), indicating a local maximum.

Step 4: Evaluate boundary points for absolute extremes

Next, evaluate the function at the endpoints \(x = 0\) and \(x = 2\pi\):- At \(x = 0\), \(y = \sqrt{3}(0) - 2\cos(0) = -2\).- At \(x = 2\pi\), \(y = \sqrt{3}(2\pi) - 2\cos(2\pi) = 2\pi\sqrt{3} - 2\).Comparing these with the. values at the critical points, since both critical points are local maxima, \(x = 0\) gives the absolute minimum.

Step 5: Find inflection points by inspecting second derivative

Inflection points occur where the second derivative changes sign. The second derivative is \(y'' = 2\cos x\). Set \(y'' = 0\) and solve:\[ 2\cos x = 0 \]\[ \cos x = 0 \]The cosine function equals 0 at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). These are potential inflection points.

Key Concepts

Understanding Critical Points

Critical points are important elements in calculus used to determine where a function might have maximum or minimum values. They occur where the first derivative of a function is zero or undefined. These points help identify possible peaks and valleys in the curve of a function.

In the exercise, the function given is \(y = \sqrt{3}x - 2\cos x\). To find critical points, we need to calculate the derivative of \(y\), specifically looking for where it equals zero. For this function, the first derivative, \(y'\), is \(\sqrt{3} + 2\sin x\).

Setting this derivative equal to zero, we solve \(\sqrt{3} + 2\sin x = 0\) to find that \(\sin x = -\frac{\sqrt{3}}{2}\). Within the interval \([0, 2\pi]\), this leads us to the critical points at \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\). Understanding how to find these points is crucial for analyzing the behavior of a function.

Exploring Inflection Points

Inflection points indicate where a function's curve changes concavity – that is, where it switches from curving upward to curving downward, or vice versa. This change is found by examining the second derivative and identifying where it changes sign.

For the given function, the second derivative \(y''\) is calculated as \(y'' = 2\cos x\). To find potential inflection points, we set \(y'' = 0\), yielding \(2\cos x = 0\). Solving this equation shows us that \(\cos x = 0\) at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).

These are the points where the function may change its concavity. Understanding these points provides deeper insights into the overall shape and behavior of the function's graph.

Applying the Second Derivative Test

The second derivative test is a simple and powerful tool for classifying the type of critical point we have found. It helps determine whether a critical point is a local maximum, a local minimum, or neither.

In our exercise, we identified critical points at \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\). To apply the second derivative test, we calculate \(y''(x) = 2\cos x\) at these points. Evaluating the second derivative at these points, we found:

• At \(x = \frac{4\pi}{3}\), \(y'' = -1\), indicating a local maximum since \(y'' < 0\).
• At \(x = \frac{5\pi}{3}\), \(y'' = -1\), indicating another local maximum since \(y'' < 0\).

This test solidifies our understanding of the behavior near these critical points, confirming their classification and providing clarity when sketching or analyzing the graph.