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Chapter 4: Problem 23

Sketch the graph of the function. (Include two full periods.) $$y=\frac{1}{2} \sec \pi x$$

Short Answer

Expert verified
The graph of \( y = \frac{1}{2} sec( \pi x) \) oscillates between -1/2 and 1/2. It has detached parts forming U shapes centered at x-values that are multiples of 2. It has vertical asymptotes at x = (2n + 1), where n is any integer. It repeats every 2 units.

Step by step solution

01

Understanding the Function

In this case, have \(y = \frac{1}{2} sec( \pi x)\). This function can be rewritten as \( y = \frac{1}{2} (1/ cos( \pi x))\), in order to better understand it. The amplitude is 1/2, which means the graph will oscillate between -1/2 and 1/2. The period is given by \(2\pi / \pi = 2\), which means the function will repeat its values every 2 units in the x-axis.
02

Sketching the Basic Cosine Function

Since \(sec x\) is the reciprocal of \( cos x\), it is helpful to first sketch a cosine graph. Note that the cosine function starts at 1, goes to 0 at \( \pi /2\), drops to -1 at \( \pi\), comes back up to 0 at \( 3 \pi /2\) and 1 at \( 2 \pi\). Because \( \pi x\) is in the argument of the cosine function, we compress the x-axis by a factor of \( \pi\). So, now the cosine function goes from 1 to -1 in just 2 units in the x-axis.
03

Sketching the Secant Function

The secant function is undefined where the cosine function is 0 and it agrees with the cosine function when the cosine function is 1 or -1. We can now convert the values of the cosine function into those of the secant function by taking reciprocals. Remember, approximately where cosine has value 0, secant will be undefined and we will draw vertical asymptotes. We also need to take into account the multiplication by 1/2, making sure that the graph oscillates between -1/2 and 1/2.
04

Extending the Graph

To include two full periods of the graph, repeat the pattern established in the first period over the next interval of length 2.

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Most popular questions from this chapter

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=\frac{1}{4} \sin 6 \pi t$$

Sketch a graph of the function. $$y=2 \arccos x$$

Use a graphing utility to graph the function and the damping factor of the function in the same viewing window. Describe the behavior of the function as \(x\) increases without bound. $$g(x)=e^{-x^{2} / 2} \sin x$$

Use a graphing utility to graph the function and the damping factor of the function in the same viewing window. Describe the behavior of the function as \(x\) increases without bound. $$y=\frac{4}{x}+\sin 2 x, \quad x>0$$

Fill in the blank. If not possible, state the reason. $$\text { As } x \rightarrow-1^{+}, \text {the value of } \arccos x \rightarrow\text { _____ } .$$
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