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Chapter 4: Problem 42

Use a graphing utility to graph the function. (Include two full periods.) $$y=\sec \pi x$$

Short Answer

Expert verified
The graph of the function \(y=\sec \pi x\) displays a sequence of U-shaped curves, dipping below the x-axis. The period is \(2/\pi\) and it has vertical asymptotes at \(x =\frac{(2n+1)}{2\pi}\) for integer values of \(n\).

Step by step solution

01

Identify the Properties of the Function

The secant function has period of \(2\pi\), vertical asymptotes wherever cosine is zero, and range of \((-\infty, -1] \cup [1, \infty)\). However, for \(y=\sec \pi x\), the period is \(2/\pi\) because of the coefficient of \(x\). The graph repeats every \(2/\pi\) units and has vertical asymptotes at \(x=\frac{(2n+1)}{2\pi}\), where \(n\) is an integer, corresponding to where \(\cos \pi x = 0\).
02

Identify the Key Points

In one period from \(0\) to \(2/\pi\) there are key points (derived from cosine graph) at \(x=0\) where \(\cos 0 = 1\) so \(y=\sec (0) = 1\), at \(x=1/(2\pi)\) where \(\cos \pi/2 = 0\) so y is undefined and at \(x=2/\pi\) where \(\cos \pi = -1\) so \(y=\sec \pi = -1\). These key points help establish the shape of \(y=\sec \pi x\) graph.
03

Plot the Graph

First, plot the vertical asymptotes as dotted lines at \(x=\frac{(2n+1)}{2\pi}\). Then plot the key points for one period: (0,1), \((1/(2\pi), undefined)\) and \((2/\pi, -1)\). The secant function will dip down from \(1\) at \(x=0\) towards negative infinity at \(x=1/(2\pi)\), then climb from negative infinity back to \(-1\) at \(x=2/\pi\). After establishing the shape for one period, replicate it for at least one period to the left and one to the right of the original period.

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