Problem 63 - Precalculus: Functions and Graphs

Question

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: \([-2\pi, -\frac{3\pi}{2}) \cup (-\frac{\pi}{2}, \frac{\pi}{2})\); Decreasing: \([-\frac{3\pi}{2}, -\frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2})\).

Step 1: Understand the Secant Function

The secant function, denoted as \( \sec(x) \), is the reciprocal of the cosine function, i.e., \( \sec(x) = \frac{1}{\cos(x)} \). The function is undefined where \( \cos(x) = 0 \), which happens at \( x = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.

Step 2: Determine Critical Points

To find where the secant function is increasing or decreasing, consider the derivative of \( \sec(x) \). The critical points occur where the derivative is zero or undefined. The derivative of \( \sec(x) \) is \( \sec(x)\tan(x) \), which is undefined at the same points as \( \sec(x) \) itself. The critical points in the interval \([-2\pi, 2\pi]\) are \(-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\).

Step 3: Test Intervals Using the Derivative

Use the intervals defined by the critical points and evaluate the sign of the derivative \( \sec(x)\tan(x) \) to determine where \( \sec(x) \) is increasing or decreasing. Choose test points in:1. \([-2\pi, -\frac{3\pi}{2})\), 2. \(-\frac{3\pi}{2}, -\frac{\pi}{2})\), 3. \(-\frac{\pi}{2}, \frac{\pi}{2})\), 4. \(\frac{\pi}{2}, \frac{3\pi}{2})\), 5. \(\frac{3\pi}{2}, 2\pi]\).Evaluate \( \sec(x)\tan(x) \) in each interval to determine the sign.

Step 4: Identify Increasing Intervals

The secant function \( \sec(x) \) is increasing where \( \sec(x)\tan(x) > 0 \). By testing points from each interval:- In \([-2\pi, -\frac{3\pi}{2})\) and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \( \sec(x)\tan(x) \) is positive.Hence, \( \sec(x) \) is increasing in these intervals.

Step 5: Identify Decreasing Intervals

The secant function \( \sec(x) \) is decreasing where \( \sec(x)\tan(x) < 0 \). By testing points from each interval:- In \([-\frac{3\pi}{2}, -\frac{\pi}{2})\) and \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), \( \sec(x)\tan(x) \) is negative.Hence, \( \sec(x) \) is decreasing in these intervals.

Key Concepts

Trigonometric Functions

Trigonometric functions are foundational in both mathematics and real-world applications. They describe relationships involving angles and lengths in triangles. Among popular trigonometric functions are sine, cosine, and tangent, as well as their reciprocals: cosecant, secant, and cotangent. Understanding how these functions work is crucial for solving a variety of problems in math and physics.

• Sine \(\sin(x)\) gives the y-coordinate on the unit circle, and it varies between -1 and 1.
• Cosine \(\cos(x)\) provides the x-coordinate on the unit circle, and also ranges between -1 and 1.
• Tangent \(\tan(x)\) is the ratio of sine to cosine \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), and it can take any real value.

The secant function, denoted as \(\sec(x)\), is particularly interesting because it is the reciprocal of the cosine function. This means it is defined as \(\sec(x) = \frac{1}{\cos(x)}\). However, it becomes undefined where cosine equals zero, specifically at angles of \(x = \frac{\pi}{2} + n\pi\), where \(n\) is an integer. Crafting a mental picture of these undefined points helps significantly, particularly because it shows disruptions in the graph, such as vertical asymptotes, contributing to a better understanding of the behavior and properties of the secant function. Understanding these trigonometric functions will help you analyze and determine the behavior of functions like secant over various intervals.

Calculus

Calculus provides us with powerful tools to analyze functions, especially concerning their rates of change. The core concepts include derivatives and integrals, which allow us to delve deeper into understanding the behavior and characteristics of mathematical functions.The derivative of a function describes how it changes at any point; this rate of change is critical in identifying increasing or decreasing behavior. For \(\sec(x)\), we look for where its derivative is zero or undefined to find critical points. The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\), illustrating how the secant function's growth is influenced by both secant and tangent properties.

• Differentiate \(\sec(x)\) to derive \(\sec(x)\tan(x)\), highlighting points of change.
• Understand that critical points occur where the derivative \(\sec(x)\tan(x)\) is zero or undefined.
• Analyze these critical points to establish intervals of increase and decrease.

In essence, calculus allows us to find these critical points and test intervals between them, enhancing our understanding of the original function's dynamics. By evaluating the sign of \(\sec(x)\tan(x)\) within these distinct intervals, we can deduce where the function increases or decreases, revealing the behavior of \(\sec(x)\) over its domain.

Critical Points

Critical points in mathematics are points on a graph at which a function's derivative is zero or undefined. Understanding these points is vital as they often indicate where a function changes direction from increasing to decreasing or vice versa. This concept is key when analyzing any function, especially to understand local maximums, minimums, or points of inflection.For the secant function \(\sec(x)\), critical points are found by solving for where its derivative \(\sec(x)\tan(x)\) equals zero or is undefined. These occur at the points \(x = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\) within the interval of \([-2\pi, 2\pi]\).

• Identify these critical points to understand potential shifts in the function's behavior.
• Use critical points to segment the domain into intervals where the function might be consistently increasing or decreasing.
• Test these intervals to confirm where \(\sec(x)\) is actually increasing or decreasing.

Critically examining these points not only helps map out the secant function's graph but also deepens your understanding of the function's properties. By confidently tackling these critical areas, you gain insight into the fluctuations and patterns of the function, enabling you to predict its behavior across its domain effectively.