Problem 68 - Precalculus Mathematics for Calculus

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: \( \frac{5\pi}{3} + \sqrt{3} \); Minimum: \( \frac{\pi}{3} - \sqrt{3} \).

Step 1: Find the Derivative

To find the maximum and minimum values of the function, we first need to find its critical points by calculating the derivative of the given function. We have:\[ y = x - 2\sin x \]The derivative of the function with respect to \( x \) is:\[ \frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(2\sin x) = 1 - 2\cos x \]

Step 2: Set the Derivative to Zero

Next, we set the derivative equal to zero to find critical points:\[ 1 - 2\cos x = 0 \]Solving for \( \cos x \), we get:\[ \cos x = \frac{1}{2} \]

Step 3: Find Values of x for Critical Points

Find the values of \( x \) within the given interval where \( \cos x = \frac{1}{2} \). The cosine function equals \( \frac{1}{2} \) at:\[ x = \frac{\pi}{3}, \frac{5\pi}{3} \]

Step 4: Evaluate the Function at Critical Points and Endpoints

Now, evaluate the original function \( y = x - 2\sin x \) at the critical points and the endpoints of the interval \( [0, 2\pi] \):1. At \( x = 0 \): \( y = 0 - 2\sin 0 = 0 \)2. At \( x = 2\pi \): \( y = 2\pi - 2\sin(2\pi) = 2\pi \)3. At \( x = \frac{\pi}{3} \): \( y = \frac{\pi}{3} - 2\sin \left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3} \)4. At \( x = \frac{5\pi}{3} \): \( y = \frac{5\pi}{3} - 2\sin \left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sqrt{3} \)

Step 5: Identify Maximum and Minimum Values

Compare the values of \( y \) obtained from Step 4:- \( y = 0 \) at \( x = 0 \)- \( y = 2\pi \) at \( x = 2\pi \)- \( y = \frac{\pi}{3} - \sqrt{3} \) at \( x = \frac{\pi}{3} \)- \( y = \frac{5\pi}{3} + \sqrt{3} \) at \( x = \frac{5\pi}{3} \)The maximum value is \( \frac{5\pi}{3} + \sqrt{3} \) and the minimum value is \( \frac{\pi}{3} - \sqrt{3} \).

Key Concepts

Critical Points

Critical points are where the function's derivative is zero or undefined. These points are significant because they can indicate potential maximum or minimum values of the function.

In the context of finding extreme values, we first find the derivative of the function, then solve for when this derivative equals zero.

• Critical points occur when the rate of change of the function's value with respect to the independent variable is zero.
• For periodic functions like trigonometric ones, this involves solving equations that might appear more than once in an interval.

After finding the derivative, setting it zero leads to the equation \(1 - 2\cos x = 0\). Solving this gives the specific values for \(x\) where critical points exist, in this case, \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) within the interval.

Derivative of a Function

The derivative of a function measures its rate of change at any given point. It tells us how the function's output value changes marginally with a change in input value.

The process of finding a derivative, known as differentiation, involves applying rules for derivatives to each term in the function.

• For instance, the derivative of a simple linear term \(x\) is \(1\) since the slope of a line \(y = x\) is \(1\).
• For trigonometric functions, such as \(\sin x\), derivatives are found using trigonometric identities, leading to the derivative of \(-2 \sin x\) as \(-2 \cos x\).

In this exercise, deriving \(y = x - 2\sin x\) gives us \(\frac{dy}{dx} = 1 - 2\cos x\), which is instrumental in finding critical points.

Evaluation at Endpoints

Evaluating a function at its interval endpoints is a crucial step in identifying absolute maximum or minimum values within the domain. This process confirms the function's behavior at the boundaries, ensuring no extreme values are overlooked.

For intervals \([0, 2\pi]\), we compute the function's value at the beginning and end to compare with values at critical points.

• At \(x = 0\), the function \(y = 0 - 2\sin 0\) gives \(y = 0\).
• At \(x = 2\pi\), the function \(y = 2\pi - 2\sin (2\pi)\) results in \(y = 2\pi\).

These values are then compared to those at critical points, \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\), aiding in determining the true maximum and minimum values of the function within the given interval.