Problem 80 - Precalculus

Question

For what \(x\) -values does the graph of \(y=-A \sec \left(\frac{\pi}{2} x\right)\) lie above the \(x\) -axis? (Assume \(A>0 .\) )

Step-by-Step Solution

Answer

The graph lies above the x-axis for \( x \in (1 + 4k, 3 + 4k) \) for any integer \( k \).

Step 1: Understanding the Secant Function

The secant function, denoted as \( \sec(x) \), is the reciprocal of the cosine function. Therefore, for values of \( x \) where \( \sec(x) > 0 \), \( \cos(x) \) must be positive, i.e., \( \cos(x) > 0 \). This is important for determining when the graph of \( y = -A \sec\left(\frac{\pi}{2} x\right) \) is negative and thus, below the x-axis since it is defined as \(-A \sec\left(\frac{\pi}{2} x\right)\).

Step 2: Analyzing the Sign of the Secant Expression

For \( y = -A \sec\left(\frac{\pi}{2} x\right) \), to find when the graph is above the x-axis, it should be \(-A \sec\left(\frac{\pi}{2} x\right) > 0\). Since \( A > 0 \), this inequality depends on \( \sec\left(\frac{\pi}{2} x\right) < 0 \). Therefore, we look at when \( \cos\left(\frac{\pi}{2} x \right) < 0 \).

Step 3: Interval Calculation for Cosine Function

The cosine function is negative in the intervals: \((\pi/2 + 2k\pi, 3\pi/2 + 2k\pi)\) for integer \(k\). We need \(\frac{\pi}{2} x = \theta\) to be in these intervals, implying \(\theta = \frac{\pi}{2} x\).

Step 4: Solving the Inequalities

Convert the intervals \(\frac{\pi}{2} x \in \left(\frac{\pi}{2} + 2k\pi, \frac{3\pi}{2} + 2k\pi\right)\) to \( x \) values by solving the inequalities: \( \frac{\pi}{2} x > \frac{\pi}{2} + 2k\pi \Rightarrow x > 1 + 4k \) and \( \frac{\pi}{2} x < \frac{3\pi}{2} + 2k\pi \Rightarrow x < 3 + 4k \).

Step 5: Finding Final Solution Intervals

Combine the inequalities obtained; \( x \) is in the interval \((1 + 4k, 3 + 4k)\) for integer \( k \). Thus, for the given function, the x-values for which the graph lies above the x-axis are \( x \in (1 + 4k, 3 + 4k) \) for any integer \( k \).

Key Concepts

Secant Function

In trigonometry, the secant function is an important concept. The secant of an angle, denoted as \( \sec(x) \), is the reciprocal of the cosine function. In mathematical terms, \( \sec(x) = \frac{1}{\cos(x)} \). Because it is the reciprocal of cosine, its behavior is closely linked to the cosine graph. Since \( \cos(x) \) can vary from -1 to 1, the secant function can become very large as \( \cos(x) \) approaches zero. This results in vertical asymptotes on the graph of the secant function wherever the cosine function is zero. Understanding this relationship is crucial when determining the behavior and properties of functions that include or are influenced by the secant function.
For example, in the exercise given, we have to look at where the secant function is negative, as it will help us figure out when the entire expression \(-A \sec\left(\frac{\pi}{2} x \right)\) is positive.

Inequalities

Inequalities are mathematical expressions that show the relative size or order of two values. They are essential for solving problems involving conditions that define a range of possible values. The inequality \( -A \sec\left(\frac{\pi}{2} x\right) > 0 \) indicates that we are interested in conditions when this expression is positive. Simplifying the inequality, considering \( A > 0 \), we have \( \sec\left(\frac{\pi}{2} x\right) < 0 \).
Understanding these inequalities involves analyzing when a transformed trigonometric function like secant becomes negative, which directly corresponds to when the cosine function is negative. Hence, solving this type of inequality is vital in interval calculations for trigonometric functions.

Cosine Function

The cosine function is one of the primary trigonometric functions and is fundamental in understanding the behavior of trigonometric equations. Specifically, the cosine function \( \cos(\theta) \) provides a measure of the horizontal position on the unit circle as \( \theta \) varies. Its standard graph repeats every \( 2\pi \) due to its periodic nature.
For our problem, the cosine function needs to be negative. This occurs in two specific intervals: \((\frac{\pi}{2} + 2k\pi, \frac{3\pi}{2} + 2k\pi)\). These intervals tell us precisely where the cosine function drops below zero during its periodic cycles. Comprehending where the cosine function is negative is essential for determining solution intervals for related trigonometric problems.

Interval Calculation

Interval calculation is a core part of solving trigonometric inequalities. Using the known negative intervals of the cosine function, the goal is to find the corresponding \( x \) values. The process includes solving inequalities such as \( \frac{\pi}{2} x > \frac{\pi}{2} + 2k\pi \) and \( \frac{\pi}{2} x < \frac{3\pi}{2} + 2k\pi \), where \( k \) is any integer.
By manipulating these inequalities, we deduce that \( x \) must lie within \( (1 + 4k, 3 + 4k) \). This approach allows us to convert the behavior of a function on the unit circle to corresponding real-number intervals, providing clear solutions for trigonometric expressions over different periods.