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Chapter 4: Problem 33

Graph two periods of the given cosecant or secant function. $$y=2 \sec x$$

Short Answer

Expert verified
The graph of \(y=2\sec x\) from \(-2\pi\) to \(2\pi\) consists of a series of U-shaped curves, spaced by vertical asymptotes. The function touches y-axis at \(y=2\) and \(y=-2\) and has vertical asymptotes on either sides of y-axis at \(x=\pi/2\) and \(x=-\pi/2\). The pattern repeats every \(2\pi\).

Step by step solution

01

Identify Properties of Function

Identify the key properties of the function \(y=2 \sec x\). This is a secant function with an amplitude of 2, period of \(2\pi\) (same as cosine function) and no phase shift.
02

Plot Critical Points

The critical points of cosine function come at its peak, mid-point and trough i.e. these are the points where \(\cos x = 1, 0, -1\) respectively. For the secant function, these will correspond to points \((x=2n\pi, y=2)\), \((x=(2n+1)\pi/2, y=\text{undefined})\) and \((x=(2n+1)\pi, y=-2)\) where \(n\) is an integer.
03

Draw Vertical Asymptotes

Draw vertical asymptotes at the points where secant function is undefined i.e. \(x = (2n+1)\pi/2\). These lines help to clearly denote where the secant function is undefined.
04

Sketch the Function

Using the critical points and asymptotes, sketch the secant function. The graph will dip downwards at \((2n+1)\pi, -2\) and rise upwards at \((2n+1)\pi, 2\). It will approach the vertical asymptotes and never cross them, since secant function is undefined there.
05

Finish the Graph

Repeat these steps from \(x=-2\pi\) to \(x=2\pi\). This completes the graphing of two periods of the given secant function.

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