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Chapter 4: Problem 73

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. $$g(x)=e^{-x^{2} / 2} \sin x$$

Short Answer

Expert verified
As \(x\) increases, the function \(g(x) = e^{-x^{2} / 2} \sin x\) tends towards zero, due to the effect of the damping factor \(e^{-x^{2} / 2}\). The graph of the function shows a declining oscillation, reflecting this behavior.

Step by step solution

01

Identify the function and its damping factor

In the function \(g(x) = e^{-x^{2} / 2} \sin x\), the sine factor, \(\sin x\), oscillates between -1 and 1, while the factor \(e^{-x^{2} / 2}\) 'damps' or reduces this oscillation as \(x\) increases.
02

Graphing the function

Using a graphing utility, input the function \(g(x) = e^{-x^{2} / 2} \sin x\) and graph it. The graph shows an oscillation that diminishes as \(x\) increases, due to the damping factor.
03

Graphing the damping factor

In the same viewing window, graph the damping factor \(e^{-x^{2} / 2}\). This graph will not oscillate, but will show a decrease as \(x\) increases, demonstrating the reducing effect it has on the function \(g(x)\).
04

Describing the behavior of the function

As \(x\) increases, it can be seen from both graphs that the value of the function \(g(x) = e^{-x^{2} / 2} \sin x\) decreases and tends towards zero. This is due to the damping factor \(e^{-x^{2} / 2}\), which itself tends towards zero as \(x\) increases, therefore reducing the magnitude of the \(\sin x\) oscillation.

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Most popular questions from this chapter

Converting to \(\mathrm{D}^{\circ} \mathrm{M}^{\prime} \mathrm{S}^{\prime \prime}\) Form \(\quad\) Convert each angle measure to degrees, minutes, and seconds without using a calculator. Then check your answers using a calculator. (a) \(240.6^{\circ}\) (b) \(-145.8^{\circ}\)

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