Problem 66 - Algebra and Trigonometry

Question

Find the maximum and minimum values of the function. $$ y=x-2 \sin x, 0 \leq x \leq 2 \pi $$

Step-by-Step Solution

Answer

Maximum is at \(x = \frac{5\pi}{3}\), minimum is at \(x = \frac{\pi}{3}\).

Step 1: Find the derivative of the function

The first step is to find the derivative of the given function with respect to \(x\). The function is \(y = x - 2\sin x\). Using the rules of differentiation, we get:\[dy/dx = 1 - 2\cos x\]

Step 2: Set the derivative to zero to find critical points

To find the critical points where the function could have maximum or minimum values, set the derivative equal to zero and solve for \(x\):\[1 - 2\cos x = 0\]Solving for \(\cos x\), we get:\[\cos x = \frac{1}{2}\]

Step 3: Solve for critical points within the interval

The angles where \(\cos x = \frac{1}{2}\) are \(x = \frac{\pi}{3}, \frac{5\pi}{3}\). These solutions are within the interval \([0, 2\pi]\). Hence, the critical points are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).

Step 4: Evaluate the function at critical points and endpoints

Now, evaluate the function \(y = x - 2\sin x\) at the critical points and the endpoints of the interval \(x = 0, 2\pi\):- At \(x = 0\): \(y = 0 - 2\sin 0 = 0\).- At \(x = \frac{\pi}{3}\): \(y = \frac{\pi}{3} - 2\sin \frac{\pi}{3} = \frac{\pi}{3} - \sqrt{3}\).- At \(x = \frac{5\pi}{3}\): \(y = \frac{5\pi}{3} - 2\sin \frac{5\pi}{3} = \frac{5\pi}{3} + \sqrt{3}\).- At \(x = 2\pi\): \(y = 2\pi - 2\sin 2\pi = 2\pi\).

Step 5: Compare function values to determine extrema

Compare the values of the function at the critical points and endpoints:- \(y(0) = 0\)- \(y\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3}\)- \(y\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sqrt{3}\)- \(y(2\pi) = 2\pi\)The minimum value is \(y\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3}\) and the maximum value is \(y\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sqrt{3}\).

Key Concepts

Derivatives

Derivatives are fundamental in calculus. They measure how a function changes as its input changes. In simpler terms, the derivative tells us the rate at which one quantity changes with respect to another. For the function given:

• \( y = x - 2\sin x \)

We want to know how this function behaves as \( x \) changes. To do this, we calculate the derivative. For our function, the derivative is:

• \( \frac{dy}{dx} = 1 - 2\cos x \)

This expression tells us the slope of the tangent line to the curve at any given point \( x \). Essentially, it highlights how rapidly the function's value is changing at that point.
Finding the derivative involves rules of differentiation, such as the power rule and trigonometric rules. Here, the derivative of a constant is zero, the derivative of \( x \) is 1, and for \( \sin x \), it's \( \cos x \), leading us to focus on each part's rate of change.
Understanding derivatives helps us predict function behavior and form the backbone for finding critical values, which indicate points of interest, such as peaks and troughs, in functions.

Critical Points

Critical points help identify where a function's graph has a peak (maximum), valley (minimum), or changes direction.
To find these points for the function, we set its derivative to zero, because at critical points, the tangent is horizontal, meaning the rate of change is zero:

• \( 1 - 2\cos x = 0 \)

Solving this, we find:

• \( \cos x = \frac{1}{2} \)

Within the interval \([0, 2\pi]\), this occurs at

• \( x = \frac{\pi}{3} \)
• \( x = \frac{5\pi}{3} \)

Critical points are significant because they identify locations where the function might take on maximum or minimum values or change direction. Once identified, these points require further analysis to understand their nature, which often involves the second derivative or investigation of function values at boundaries or other critical points.

Maximum and Minimum Values

Finding the maximum and minimum values of a function within a specific interval is crucial in dealing with real-world problems. These values help in understanding the function's behavior, like finding peak production levels or detecting lowest temperatures.
We evaluate the function at critical points and endpoints. This ensures we cover all possible extremes within the provided interval:

• Endpoints: \( x = 0 \) and \( x = 2\pi \).
• Critical points: \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \).

Calculate \( y \) at these points by substituting back into the function:

• \( y(0) = 0 \)
• \( y\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3} \)
• \( y\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sqrt{3} \)
• \( y(2\pi) = 2\pi \)

Comparing these values:

• Minimum at \( x = \frac{\pi}{3} \): \( y = \frac{\pi}{3} - \sqrt{3} \)
• Maximum at \( x = \frac{5\pi}{3} \): \( y = \frac{5\pi}{3} + \sqrt{3} \)

By identifying and evaluating these critical and endpoint values, we determine where the function reaches its highest and lowest points within the interval, completing the task of analyzing maximum and minimum values.