Question

Testing for Continuity In Exercises \(77-84\) , describe the interval(s) on which the function is continuous. $$ f(x)=\sec \frac{\pi x}{4} $$

Short answer

The function \(f(x)=\sec\frac{\pi x}{4}\) is continuous on the intervals \((-\infty, 2) \cup (2, 6) \cup (6, 10) \cup \ldots \cup (4n, 4n + 2) \cup \ldots\), where n is an integer.

Step-by-step solution

Recall properties of secant and cosine function

Recall that \(\sec\theta = \frac{1}{\cos\theta}\). So \(\sec\theta\) will be undefined where \(\cos\theta = 0\). Also, the cosine function \(\cos\theta = 0\) at \(\theta = \frac{\pi}{2} + n \pi\), where n is an integer.

Set up the equation

We set \(\frac{\pi x}{4} = \frac{\pi}{2} + n \pi\). This is because the secant function will be undefined where the cosine function equals zero.

Solve the equation

Solving the equation \(\frac{\pi x}{4} = \frac{\pi}{2} + n \pi\) gives \(x = 2 + 4n\), where n is an integer. This means that the function is discontinuous at \(x = 2 + 4n\).

Write out the intervals of continuity

Between any two consecutive points of discontinuity, the function is continuous. This gives the intervals \((-\infty, 2) \cup (2, 6) \cup (6, 10) \cup \ldots \cup (4n, 4n + 2) \cup \ldots\), where n is an integer.

Key concepts

Secant Function

The secant function is an important trigonometric function defined as the reciprocal of the cosine function. In mathematical terms, this is expressed as \( \sec \theta = \frac{1}{\cos \theta} \). Because it is the reciprocal, the secant function will only be defined where \( \cos \theta eq 0 \). This means that if the cosine of an angle is zero, the secant function will become undefined at that point.

Some useful characteristics of the secant function include:

• It has no maximum or minimum values as it can range from negative to positive infinity.
• The secant function is periodic, with a period of \( 2\pi \).
• It has vertical asymptotes where the cosine equals zero because the secant function heads to infinity.

Understanding the secant function is crucial in determining where a function involving secant, such as \( f(x) = \sec \frac{\pi x}{4} \), is undefined and therefore discontinuous.

Cosine Function

The cosine function is one of the foundational trigonometric functions, and it is essential for understanding both the secant and the behavior of many mathematical functions. The cosine function is graphed in a wave-like pattern called a 'cosine wave', which repeats every \( 2\pi \).

Some key properties of the cosine function include:

• It ranges from -1 to 1, meaning the maximum value is 1 and the minimum value is -1.
• It is an even function, which means \( \cos(-\theta) = \cos(\theta) \).
• It reaches zero at angles in the form of \( \theta = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.

Understanding where the cosine function equals zero is particularly crucial when considering discontinuities in functions where \( \sec \theta \) is involved.

Discontinuity

A discontinuity in a function occurs at a point where the function is not continuous. When dealing with the secant function, discontinuities occur where the cosine function equals zero as the secant function becomes undefined there.

For the given function \( f(x) = \sec \frac{\pi x}{4} \), discontinuities occur at points where \( \frac{\pi x}{4} = \frac{\pi}{2} + n \pi \), after solving you get \( x = 2 + 4n \). These are the specific \( x \) values where you will find discontinuities, meaning the function will not be defined.

Recognizing and understanding discontinuities is crucial because it allows you to identify intervals where the function maintains continuity and can be evaluated smoothly.

Intervals of Continuity

In mathematics, understanding intervals of continuity is key to analyzing a function's behavior. For the function \( f(x) = \sec \frac{\pi x}{4} \), the points of discontinuity are found at \( x = 2 + 4n \) where the function is not defined. The intervals of continuity are those segments where the function is continuous, which occurs between these points of discontinuity.

The intervals can be written as \(( - \infty, 2) \cup (2, 6) \cup (6, 10) \cup \ldots \) generally expressed as \((4n, 4n+2)\) for integers \( n \).

Identifying these intervals gives a clear understanding of where the function can be consistently used. In these continuous intervals, there are no breaks, holes, or points of jump, and the function is smooth.