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Chapter 17: Problem 437

Sketch the graph of \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{-\mathrm{x} 2} \sin \mathrm{x}\)

Short Answer

Expert verified
To sketch the graph of \(f(x) = e^{-x^2} \sin x\), we use the following information: 1. The function is symmetric with respect to the y-axis due to the exponential component. 2. The function oscillates between -1 and 1 because of the sine component. 3. As x goes to infinity, the exponential component decreases, causing the amplitude of the oscillations to decrease as well. Please use graphing software, such as Desmos, or a graphing calculator to visualize the graph including the mentioned properties.

Step by step solution

01

Analyze the nature of the function and its components

The given function is \(f(x) = e^{-x^2} \sin x\). It is a product of two functions: \(e^{-x^2}\) and \(\sin x\). 1. The \(e^{-x^2}\) function is an exponential function with a negative exponent, which means that it decreases as x increases and approaches 0 as x goes to infinity. It is symmetric with the y-axis. 2. The \(\sin x\) function is a trigonometric function, which oscillates between -1 and 1 with a period of \(2\pi\).
02

Identify the domain and the range of the function

The domain of the function is the set of all possible x values for the function. Since the exponential and the sine functions are defined for all real numbers, the domain of \(f(x)\) is \(\mathbb{R}\). The range of the function is the set of all possible output values. The exponential component, \(e^{-x^2}\), goes from 0 to 1, and the sine component oscillates between -1 and 1. The range of the function will be the product of these intervals: \([-1, 1]\).
03

Determine the critical points of the function

The critical points are the points where the function changes direction or has a local extremum. To find these points, we can analyze the derivatives of the function: 1. Find the first derivative of the function: \(f'(x) = \frac{d}{dx} (e^{-x^2} \sin x)\) 2. Find the zeroes of the first derivative: \(f'(x) = 0\) (This will give us the critical points) 3. Determine if the critical points are local maxima, local minima, or inflection points by analyzing the second derivative. However, finding the derivatives of the function is not trivial and requires a more advanced understanding of calculus. Therefore, we will skip this step in our sketch and instead rely on the general behavior of the given function for our sketch.
04

Plot the function using the information from the previous steps

Using the information from the previous steps, we can sketch a rough plot of the function \(f(x) = e^{-x^2} \sin x\). 1. The function is symmetric with respect to the y-axis due to the exponential component. 2. The function oscillates between -1 and 1 because of the sine component. 3. As x goes to infinity, the exponential component decreases, causing the amplitude of the oscillations to decrease as well. With this information, we can sketch a rough graph of the function. Please note: The sketch cannot be drawn using only text. To visualize the graph, it is recommended to use graphing software, such as Desmos, or a graphing calculator to verify and visualize the mentioned properties and to obtain an accurate graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the function we are analyzing, the exponential component is represented by \(e^{-x^2}\). This is a special kind of exponential function where the exponent itself is a function of \(x\). This means:
  • The base \(e\) is approximately 2.718, known as Euler's number.
  • As \(x\) increases in magnitude (either positively or negatively), \(e^{-x^2}\) decreases towards zero.
This function is symmetric about the y-axis, meaning that the left and right sides of the graph mirror each other. As \(x\) approaches zero, the value of \(e^{-x^2}\) approaches one, contributing significantly to the amplitude of \(f(x)\). This defining quality is important to understand how the function behaves over all real numbers.
Trigonometric Functions
Trigonometric functions are periodic functions often related to the ratios of the sides of a right triangle. The function \(\sin x\) plays a crucial role in our analysis. It is important to note that:
  • The sine function oscillates between -1 and 1.
  • It completes one full oscillation over an interval of \(2\pi\).
In our composite function \(f(x) = e^{-x^2} \sin x\), the sine function controls the oscillatory behavior. The oscillation amplitude might decrease due to the multiplication with the \(e^{-x^2}\) term as \(x\) moves away from the origin. This is because the exponential component acts as a damping factor. Understanding these oscillations helps identify critical points and features of the graph.
Domain and Range
The domain and range are essential to fully understand any function. The domain of a function represents all possible input values, while the range pertains to possible output values.
For the function \(f(x) = e^{-x^2} \sin x\):
  • The domain is all real numbers \((\mathbb{R})\) since both \(e^{-x^2}\) and \(\sin x\) are defined for every real number \(x\).
  • The range is derived from the product of the ranges of the individual functions. \(e^{-x^2}\) varies from 0 to 1, and \(\sin x\) ranges from -1 to 1.
Thus, the output values or the range of \(f(x)\) will always stay within \([-1, 1]\) due to the limitations imposed by the oscillating factor \(\sin x\) and the decreasing nature of \(e^{-x^2}\). Knowing these ranges helps sketch the function accurately.
Critical Points
Critical points in a function are where the function's rate of change (its derivative) becomes zero or undefined, signifying potential local extrema or inflection points. Finding these critical points involves differentiating the function.
For \(f(x) = e^{-x^2} \sin x\):
  • We need to calculate the derivative \(f'(x)\).
  • This derivative can be more complicated but provides crucial information about where the function changes direction.
Though the analytic process requires solving \(f'(x) = 0\), careful graphing can still provide valuable insights into these critical regions. It helps determine which points on the graph correspond to local maxima or minima and how the function behaves around these points. Understanding critical points is useful for illustrating the shape of complex graphs like this one.

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