Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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### Question 4 Which of the following is the jump discontinuity of \( f(x) = \frac{|x - 1|}{x^2 + x - 2} \)? - A. \( x = 1 \) - B. \( x = 2 \) - C. \( x = -2 \) - D. \( x = -1 \) - E. \( x = 0 \) In this question, we are asked to identify the jump discontinuity of the given function \( f(x) = \frac{|x - 1|}{x^2 + x - 2} \). To understand the nature of discontinuities better, let's recall that a jump discontinuity occurs at a point where the left-hand limit and the right-hand limit of the function exist but are not equal to each other. That is, the function "jumps" from one value to another at this point. Consider the denominator of the function, \( x^2 + x - 2 \). We can factorize it to find the points where the denominator becomes zero: \[ x^2 + x - 2 = (x + 2)(x - 1). \] The function \( f(x) \) will be undefined wherever the denominator is zero, that is at \( x = -2 \) and \( x = 1 \). To determine the type of discontinuity at these points, we must carefully analyze the behavior of the function around them. By exploring the behavior near \( x = 1 \), we observe that the absolute value in the numerator \( |x - 1| \) results in different expressions depending on whether \( x \) is slightly less than or slightly greater than 1, thus creating a jump discontinuity. Therefore, the correct answer is: - **A. \( x = 1 \)** For further understanding, students are encouraged to explore and graph the function to visualize the discontinuities.
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Transcribed Image Text:### Question 4 Which of the following is the jump discontinuity of \( f(x) = \frac{|x - 1|}{x^2 + x - 2} \)? - A. \( x = 1 \) - B. \( x = 2 \) - C. \( x = -2 \) - D. \( x = -1 \) - E. \( x = 0 \) In this question, we are asked to identify the jump discontinuity of the given function \( f(x) = \frac{|x - 1|}{x^2 + x - 2} \). To understand the nature of discontinuities better, let's recall that a jump discontinuity occurs at a point where the left-hand limit and the right-hand limit of the function exist but are not equal to each other. That is, the function "jumps" from one value to another at this point. Consider the denominator of the function, \( x^2 + x - 2 \). We can factorize it to find the points where the denominator becomes zero: \[ x^2 + x - 2 = (x + 2)(x - 1). \] The function \( f(x) \) will be undefined wherever the denominator is zero, that is at \( x = -2 \) and \( x = 1 \). To determine the type of discontinuity at these points, we must carefully analyze the behavior of the function around them. By exploring the behavior near \( x = 1 \), we observe that the absolute value in the numerator \( |x - 1| \) results in different expressions depending on whether \( x \) is slightly less than or slightly greater than 1, thus creating a jump discontinuity. Therefore, the correct answer is: - **A. \( x = 1 \)** For further understanding, students are encouraged to explore and graph the function to visualize the discontinuities.
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