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Chapter 1: Problem 71

Describe the interval(s) on which the function is continuous. $$ f(x)=\sec \frac{\pi x}{4} $$

Short Answer

Expert verified
The function \( f(x) = \sec(\frac{\pi x}{4}) \) is continuous on the intervals where \(x \neq (2n+1)\), where \(n\) is an integer.

Step by step solution

01

Identify the form of the given function

The given function \( f(x) = \sec(\frac{\pi x}{4}) \) is in the form of a secant function.
02

Understand the relationship with the cosine function

Recognize where the cosine function is equals zero because secant x is equal to 1/cos(x), and it is undefined when cos(x)=0.
03

Determine where Cosine function equals zero

The cosine function equals zero at \(x = \frac{(2n+1)\pi}{2}\), where \(n\) is an integer.
04

Apply transformations

We have a horizontal stretching by a factor of 4 in our secant function, so we multiply the \(x\) from our cosine function by 4, giving us \(x = 2n+1\), where \(n\) is an integer.
05

Identify the intervals of continuity

Putting it all together, the function \( f(x) = \sec(\frac{\pi x}{4}) \) is continuous over the interval \(x \neq (2n+1)\), where \(n\) is an integer.

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Key Concepts

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

Secant Function
The secant function, often written as \(\sec(x)\), is a trigonometric function related to the cosine function. It is defined as the reciprocal of the cosine function, which means \( \sec(x) = \frac{1}{\cos(x)} \). This means that wherever the cosine function is zero, the secant function is undefined. The secant function has some unique properties:
  • It has vertical asymptotes where the cosine function equals zero.
  • It's periodic, with a period of \(2\pi\).
Understanding the behavior of the secant function is crucial when determining where it is continuous and when it becomes undefined.
Cosine Relationship
The cosine function, \(\cos(x)\), plays a pivotal role in the behavior of the secant function. Since \(\sec(x) = \frac{1}{\cos(x)}\), the relationship between these two functions is that secant will be undefined whenever cosine is zero. The cosine function equals zero at specific points, described by the formula:
  • \(x = \frac{(2n+1)\pi}{2}\)
where \(n\) is an integer. This formula signifies the points on the x-axis where cosine equals zero, creating discontinuities for the secant function. The periodic nature of the cosine function creates a repeating pattern for these points.
Continuous Intervals
To determine where the secant function is continuous, one must locate the points where its reciprocal, the cosine function, does not equal zero. This gives us the intervals between the discontinuities:
  • The secant function is continuous everywhere except at the points where \(\cos(x) = 0\).
  • For the given function \(f(x) = \sec(\frac{\pi x}{4})\), the transformation affects where these discontinuities occur.
Using the transformation described in this problem, our continuous intervals for \(f(x)\) exclude where \(x = 2n+1\), simplified from the transformation of the cosine zeros.
Mathematical Transformations
Mathematical transformations allow us to manipulate functions, helping us understand their behavior in different scenarios. In the exercise, the secant function argument undergoes a transformation from a basic linear transformation:
  • \(\sec(\frac{\pi x}{4})\) implies a horizontal stretch by a factor of 4.
  • The sequence \(x = \frac{(2n+1)\pi}{2}\) transforms into \(\frac{\pi x}{4} = \frac{(2n+1)\pi}{2}\).
Simplifying this equation, we find \(x = 2n+1\), where \(n\) is an integer. This describes the locations of the discontinuities in the transformed secant function, \(f(x)\). So, the function is continuous for all real numbers except these specific x-values.

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Most popular questions from this chapter

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x-3}{x^{2}-9} $$

Find the constant \(a\), or the constants \(a\) and \(b\), such that the function is continuous on the entire real line. $$ g(x)=\left\\{\begin{array}{ll} \frac{x^{2}-a^{2}}{x-a}, & x \neq a \\ 8, & x=a \end{array}\right. $$

Describe the interval(s) on which the function is continuous. $$ f(x)=x \sqrt{x+3} $$

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x-1}{x^{2}+x-2} $$

Prove that if \(\lim _{x \rightarrow c} f(x)=0\), then \(\lim _{x \rightarrow}|f(x)|=0\).
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