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Chapter 5: Problem 102

Use a graphing utility to graph two periods of the function. $$y=-2 \cos \left(2 \pi x-\frac{\pi}{2}\right)$$

Short Answer

Expert verified
The transformed cosine function will have maximum points at x = 0 and x = 1, minimum points at x = 0.5 and x = 1.5, and its amplitude is 2, period is 1.

Step by step solution

01

Identify the Transformations

For the given function \(y=-2 \cos \left(2 \pi x-\frac{\pi}{2}\right)\), the multiplicative factor of -2 means the amplitude of the cosine function is increased by a factor of 2 and reflected in the x-axis. The coefficient \(2\pi\) indicates a horizontal compression by a factor of \(1/(2\pi)\), this makes the period of the function \(1/(2\pi)\) instead of \(2\pi\). Finally, the term \(-\pi/2\) inside the cosine function shifts the graph right by \(\pi/2\) units.
02

Determine the Period and Amplitude

The period of the function is given by \(2\pi / |2\pi|\), which equals \(1\). The amplitude is the absolute value of -2, which is 2.
03

Plot the function

Now we can plot the function. Start by plotting a regular cosine function. Now apply our transformations: Stretch it vertically by a factor of 2, compress it horizontally by a factor of \(2\pi\) (or extend the period to 1), shift right by \(\pi/2\) units and reflect in the x-axis. The function will reach maximum at x = 0 and x = 1 (2 peaks for 2 periods) and will have the lowest points at x = 0.5 and x = 1.5.
04

Use a Graphing Utility

Use a graphing utility to draw this function smoothly. Make sure the maximum point reaches 2 and the minimum point reaches -2. And the function should complete two periods within the range x = 0 to x = 2

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Most popular questions from this chapter

For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=0\) and \(b=2\)

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