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Chapter 4: Problem 25

Sketch one full period of the graph of each function. $$y=2 \sec x$$

Short Answer

Expert verified
To sketch \(y = 2 \sec x\), first plot \(y = 2 \cos x\), identify the points of discontinuity at \(x = \pi/2\) and \(x = 3\pi/2\), and then plot the secant function following the shape of the cosine function, approaching but not reaching the points of discontinuity.

Step by step solution

01

Plot the related cosine function

Begin by sketching one period of the function \(y = 2 \cos x\). The cosine function has a period of \(2\pi\), an amplitude of 2, and undergoes a full cycle in the range \(x = 0\) to \(x = 2\pi\). Plot the maximum at (0,2), the minimum at (\(\pi\),-2), and the zeroes at \(\pi/2\) and \(3\pi/2\).
02

Identify the points of discontinuity

The secant function, \(y = 2 \sec x\), is undefined wherever the cosine function is zero. These are the points of discontinuity, located at \(x = \pi/2\) and \(x = 3\pi/2\) within the period of \(2\pi\). Draw vertical dashed lines at these points.
03

Sketch the secant graph

The graph of the secant function will approach but never reach the points of discontinuity, following the shape of the cosine graph. Plot the positive part of the secant function from 0 to \(\pi/2\) and from \(3\pi/2\) to \(2\pi\), and the negative part of the secant function from \(\pi/2\) to \(3\pi/2\).

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