Problem 63 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 63-66: Find the intervals between \(-2 \pi\) and \(2 \pi\) on which the given function is (a) increasing or (b) decreasing. secant

Step-by-Step Solution

Answer

Increasing at \( (-(\frac{3}{2})\pi, -\pi) \,; (\frac{\pi}{2}, \pi) \,; (\frac{3}{2} \pi, 2\pi) \). Decreasing at \((-2\pi, -(\frac{3}{2})\pi) \,; (-(\frac{\pi}{2}), 0) \,; (\pi, \frac{3}{2} \pi) \).

Step 1: Identify the Function

In this exercise, the function given is the secant function, denoted as \( \sec(x) \). The secant function is defined as \( \sec(x) = \frac{1}{\cos(x)} \). We need to analyze this function in the interval \( [-2\pi, 2\pi] \).

Step 2: Find the Derivative of the Function

To determine where the function is increasing or decreasing, we first find the derivative. The derivative of \( \sec(x) \) is \( \sec(x) \tan(x) \). This derivative will help us determine the behavior of the function over the specified interval.

Step 3: Determine Critical Points

The critical points occur where the derivative is zero or undefined. Solve \( \sec(x) \tan(x) = 0 \). Since \( \sec(x) \) is never zero, solve \( \tan(x) = 0 \). \( \tan(x) = 0 \) at \( x = k\pi \), where \( k \) is an integer. These points are critical within the interval \([-2\pi, 2\pi]\).

Step 4: Analyze The Intervals Between Critical Points

The critical points \( -2\pi, -\pi, 0, \pi, 2\pi \) divide the interval \([-2\pi, 2\pi]\) into sections. Evaluate the sign of \( \sec(x) \tan(x) \) within each section. Note that \( \sec(x) \) is undefined at \( x = \frac{(2k+1)\pi}{2} \) where the cosine is zero, indicating vertical asymptotes.

Step 5: Determine Where the Function is Increasing

For the function to be increasing, the derivative \( \sec(x) \tan(x) > 0 \). This occurs in the intervals \((-(\frac{3}{2})\pi, -\pi)\), \((\frac{\pi}{2}, \pi)\), and \((\frac{3}{2} \pi, 2\pi)\).

Step 6: Determine Where the Function is Decreasing

For the function to be decreasing, the derivative \( \sec(x) \tan(x) < 0 \). This occurs in the intervals \((-2\pi, -(\frac{3}{2})\pi)\), \(-(\frac{\pi}{2}), 0)\), and \((\pi, \frac{3}{2} \pi)\).

Key Concepts

Secant Function

The secant function, represented as \( \sec(x) \), is one of the six fundamental trigonometric functions. It is mathematically defined as the reciprocal of the cosine function: \( \sec(x) = \frac{1}{\cos(x)} \).
This relationship highlights that wherever cosine equals zero, the secant function becomes undefined. This leads to vertical asymptotes in the graph of the secant function, which are key behaviors to understand when analyzing its properties.
Exploring the properties of \( \sec(x) \) is important because it provides insights into cyclical patterns and behaviors in equations, useful in advanced mathematics.

• \( \sec(x) \) has a range of \([1, \infty) \cup (-\infty, -1] \).
• It is periodic with a period of \(2\pi\), repeating its values every \(2\pi\).
• The vertical asymptotes occur where \(\cos(x) = 0\), or precisely at \(x = \frac{(2k+1)\pi}{2} \), where \(k\) is an integer.

Understanding the behavior of \( \sec(x) \) is vital to solving problems involving trigonometric functions, especially in domains such as signal processing and physics.

Interval Analysis

Interval analysis helps to determine where a function is increasing or decreasing on a specific domain. It involves reviewing changes in signs of the derivative within particular ranges, called intervals. Here, we focus on the interval \([-2\pi, 2\pi]\) for the secant function.
When analyzing \( \sec(x) \), we identify that the interval is broken into pieces separated by critical points and vertical asymptotes.

• Critical points occur where the derivative equals zero or is undefined.
• For \( \sec(x) \), vertical asymptotes split intervals because \( \cos(x) \) equals zero.

Each interval has specific behaviors. The presence of vertical asymptotes, for instance, implies a need for care in determining whether sections are increasing or decreasing.Evaluating the derivative on these intervals helps predict function behavior. This is crucial in many mathematical tasks and applications, requiring proficient interval analysis skills.

Derivative of Trigonometric Functions

Understanding the derivative of trigonometric functions is key to analyzing their behavior over various intervals. Here, for \( \sec(x) \), the derivative is found using quotient rules and trigonometric identities and is given by \( \sec(x) \tan(x) \).
Here’s a basic way to understand this derivative:

• Differentiation: \( \frac{d}{dx} \left( \sec(x) = \frac{1}{\cos(x)} \right) \).
• Apply quotient and chain rules to find \( \sec(x) \tan(x) \).

This derivative offers insight into where the original function \( \sec(x) \) increases and decreases:

• If \( \sec(x) \tan(x) > 0 \), the function \( \sec(x) \) is increasing.
• If \( \sec(x) \tan(x) < 0 \), \( \sec(x) \) is decreasing.

Trigonometric derivatives are pivotal in calculus, solving physics problems, and understanding natural phenomena like sound waves and light waves. Analyzing derivatives allows predictions on changes and behaviors, equipping students with necessary analytical tools.