/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 Determine the period and sketch ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the period and sketch at least one cycle of the graph of each function. State the range of each function. $$y=\frac{1}{2} \sec x$$

Short Answer

Expert verified
The period is \(2\theta\), the range is \((-∞, -\frac{1}{2}] \text{ U } [\frac{1}{2}, ∞)\). Sketch from \(\theta = -\frac{\theta}{2}\) to \(\theta = \frac{\theta}{2}\).

Step by step solution

01

Understand the basic function

The given function is \(y = \frac{1}{2} \text{sec} \theta\). The secant function, \(\text{sec} \theta\), is the reciprocal of the cosine function. Therefore, \(y = \frac{1}{2} \text{sec} \theta\) can be rewritten as \(y = \frac{1}{2\text{cos} \theta}\).
02

Identify the period of the function

The period of the secant function, \(\text{sec} \theta\), is \(2\theta\). Hence, the period of \(y = \frac{1}{2} \text{sec} \theta\) will be \(2\theta\) as well.
03

Determine vertical asymptotes

Since the secant function is undefined where the cosine function is zero, \(y = \frac{1}{2} \text{sec} \theta\) will have vertical asymptotes where \(\theta = \frac{(2n+1)\theta}{2}\), where n is an integer.
04

Range of the function

The range of the secant function is \((-∞, -1] \text{ U } [1, ∞)\). Since we have \(y = \frac{1}{2} \text{sec} \theta\), this scaling factor would shrink the usual secant values, resulting in \((-∞, -\frac{1}{2}] \text{ U } [\frac{1}{2}, ∞)\).
05

Sketch the graph

To sketch one cycle of the graph start from \(\theta = -\frac{\theta}{2}\) to \(\theta = \frac{\theta}{2}\), taking note of where the function is undefined (vertical asymptotes at \(\theta = -\frac{\theta}{2}, 0, \frac{\theta}{2}\)). Label the x-axis with intervals of \(\frac{\theta}{2}\). Plot the points where \(y = \frac{1}{2}\) and \(y = -\frac{1}{2}\), then draw the typical U-shaped secant curves between these points, approaching the asymptotes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertical asymptotes
Vertical asymptotes occur where a function approaches infinity as the input approaches a certain value. For the secant function, these asymptotes happen where the cosine function is zero because the secant function is the reciprocal of the cosine function.

For the function \(y = \frac{1}{2} \sec \theta\), we find vertical asymptotes where \( \cos \theta = 0 \). These occur at \(\theta = \frac{(2n+1)\pi}{2}\) where \( n \) is an integer.

Thus, the asymptotes for \(y = \frac{1}{2} \sec \theta\) exist at:
  • \(\theta = -\frac{\pi}{2}\)
  • \(\theta = \frac{\pi}{2}\)
  • \(\theta = \frac{3\pi}{2}\)
  • and so on.

Understanding these asymptotes is crucial for sketching the graph accurately.
function range
The range of a function is the set of all possible output values. For the secant function \( \sec \theta \), the range is \((-∞, -1] \cup [1, ∞) \).

When we scale the secant function by \( \frac{1}{2} \), the range shrinks proportionally. Thus, the range for \(y = \frac{1}{2} \sec \theta\) changes to \((-∞, -\frac{1}{2}] \cup [\frac{1}{2}, ∞)\). This means the values will be either greater than or equal to \(\frac{1}{2}\), or less than or equal to \(-\frac{1}{2}\).

Specifically, the output values can't lie within \((-\frac{1}{2},\frac{1}{2})\). Understanding the range helps in predicting the behavior of the function and its graph.
graph sketching
When sketching the graph of \(y = \frac{1}{2} \sec \theta\), follow these steps:

- Identify vertical asymptotes: From the earlier section, we know they occur at \(\theta = \frac{(2n+1)\pi}{2}\).
- Determine the range: The function outputs values in the intervals \((-∞, -\frac{1}{2}] \cup [\frac{1}{2}, ∞)\).
- Plot key points: These are at the maximum and minimum values \(y = \frac{1}{2}\) and \(y = -\frac{1}{2}\).

Begin your sketch from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), noting asymptotes at the boundaries and zero. Plot the points at intervals of \(\frac{\pi}{2}\). Draw the typical U-shapes of the secant curve, ensuring they approach but never touch the asymptotes.

This methodical approach ensures a clear and accurate graph of one cycle of the secant function.

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