Problem 66 - Calculus Early Transcendentals

Question

Find the relative extrema in the interval \(0

Step-by-Step Solution

Answer

Relative maxima: \( x = \frac{\pi}{3} \), relative minima: \( x = \frac{5\pi}{3} \).

Step 1: Find the derivative of the function

To find the relative extrema, we need to determine where the derivative of the function is zero or undefined. We start by finding the derivative of \( f(x) = \frac{\sin x}{2 - \cos x} \) using the quotient rule. The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). For our function, \( u = \sin x \) and \( v = 2 - \cos x \). So, we find: \( u' = \cos x \) and \( v' = \sin x \). Now substitute:\[ f'(x) = \frac{ \cos x (2 - \cos x) - \sin x (\sin x) }{ (2 - \cos x)^2 } = \frac{ 2 \cos x - \cos^2 x - \sin^2 x }{ (2 - \cos x)^2 } \]

Step 2: Simplify the derivative

Recognize that \( \cos^2 x + \sin^2 x = 1 \). Use this identity to simplify the derivative expression:\[ f'(x) = \frac{ 2 \cos x - 1 }{ (2 - \cos x)^2 } \]

Step 3: Find critical points

Critical points occur where the derivative is zero or undefined. Set the numerator of \( f'(x) \) to zero:\[ 2 \cos x - 1 = 0 \] Solving for \( \cos x \), we get \( \cos x = \frac{1}{2} \). In the interval \( 0 < x < 2\pi \), this happens at \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). Check where the denominator is zero: \( 2 - \cos x eq 0 \), ensuring it's not undefined for these \( x \).

Step 4: Test intervals around critical points

Use test points in the intervals determined by the critical points to identify the sign of \( f'(x) \). Choose test points \( x = \frac{\pi}{4}, \pi, \frac{3\pi}{2} \) and evaluate the derivative:- For \( x \) in \( (0, \frac{\pi}{3}) \), choose \( x = \frac{\pi}{4} \): \( f'(x) > 0 \)- For \( x \) in \( (\frac{\pi}{3}, \frac{5\pi}{3}) \), choose \( x = \pi \): \( f'(x) < 0 \)- For \( x \) in \( (\frac{5\pi}{3}, 2\pi) \), choose \( x = \frac{3\pi}{2} \): \( f'(x) > 0 \)This indicates a relative maximum at \( x = \frac{\pi}{3} \) and a relative minimum at \( x = \frac{5\pi}{3} \).

Step 5: Verify with graph

Plot \( f(x) = \frac{\sin x}{2 - \cos x} \) using a graphing utility, and observe the behavior of the function at critical points. The graph should confirm a peak at \( x = \frac{\pi}{3} \) and a trough at \( x = \frac{5\pi}{3} \), which matches our analytical findings.

Key Concepts

Calculus

Calculus is a field of mathematics that explores changes between values related to functions. It's essential to learn calculus because it provides the foundation for advanced topics in both mathematics and science. Calculus primarily consists of two branches: **differential calculus** and **integral calculus**. Both branches work together to give you tools for solving real-world problems involving rates of change and accumulation.

By understanding calculus, you can explore how functions behave by examining their graphs, finding limits, derivatives, and integrals. For instance, when looking for relative extrema of a function, you'll need calculus to find where a function reaches its highest or lowest values, known as relative maxima and minima. These concepts are vital for fields like engineering, physics, economics, and biology, where you model physical systems and predict their behavior.

Derivative

A derivative is a measure of how a function changes as its input changes. In simple terms, it represents the slope of the function at any given point. The derivative is a fundamental tool in calculus that allows you to determine instantaneous rates of change.

To find a derivative, you typically need to understand the rules of differentiation, such as the power rule, product rule, chain rule, and quotient rule. Once you find the derivative of a function, you can use it to find critical points, where the derivative equals zero or is undefined. These critical points are essential in determining where the function might have relative extrema, such as a peak or a valley on its graph.

Quotient Rule

The quotient rule is a method for finding the derivative of a quotient of two functions. When you have a function that's written as a fraction, like \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule provides a formula to differentiate it easily.

The quotient rule states that if you have two differentiable functions, \( u(x) \) and \( v(x) \), then the derivative of their quotient is given by \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) with respect to \( x \), respectively.

Using this rule is essential when differentiating functions like \( \frac{\sin x}{2 - \cos x} \). Applying the quotient rule helps you find critical points by allowing you to work with both the numerator and the denominator during differentiation.

Critical Points

Critical points of a function occur where its derivative is either zero or undefined. These points are significant because they can indicate potential relative maxima, minima, or saddle points. To determine the nature of these critical points, you'll often perform additional analysis, such as using the first derivative test or the second derivative test.

In solving problems, identifying critical points often involves setting the derivative of the function equal to zero and solving for \( x \). Moreover, it's crucial to check if the function might be undefined at those points, ensuring they fit the criteria. Once you find these critical points within a given interval, test the intervals around them to determine the behavior of the function - whether it's increasing or decreasing.

For example, with the function \( f(x) = \frac{\sin x}{2 - \cos x} \), critical points in the interval \( 0 < x < 2\pi \) help identify the function's relative extrema, guiding you to understand how the function behaves across its domain.