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Chapter 1: Problem 53

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

Short Answer

Expert verified
The function is continuous on the intervals \( (-\infty,1), (1,3), (3,5), \ldots, (-2n-1,-2n+1), \ldots, (2n+3, \infty) \), for all integers \( n \).

Step by step solution

01

Identify where Cosine function is zero

The cosine function is zero at \( (2n+1) \cdot \frac{\pi}{2} \) for all integers \( n \}. These are the points where the secant function will be discontinuous.
02

Adjust for the Scale Factor

The given function is \( \sec \left( \frac{\pi x}{4} \right) \), not \( \sec(x) \). The argument \( x \) in the original cosine function is scaled by a factor of \( \frac{\pi}{4} \) in the given function. So, the points of discontinuity are given by \( x = 2n+1 \), for all integers \( n \)
03

Identify the intervals of continuity

Between each pair of consecutive points of discontinuity, the function is continuous. Hence the solution is given by the intervals \( (-\infty,1), (1,3), (3,5), \ldots, (-2n-1,-2n+1), \ldots, (2n+3, \infty) \), for all integers \( n \)

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