5. Which of the following is true for f(x) = x² -4 ? X-2 There is a non-removable discontinuity at x = 2. %D There is a removable discontinuity at x = 2. %3D The function is continuous for all real numbers.
5. Which of the following is true for f(x) = x² -4 ? X-2 There is a non-removable discontinuity at x = 2. %D There is a removable discontinuity at x = 2. %3D The function is continuous for all real numbers.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Question 5:**
Which of the following is true for \( f(x) = \frac{x^2 - 4}{x - 2} \)?
- [ ] There is a non-removable discontinuity at \( x = 2 \).
- [ ] There is a removable discontinuity at \( x = 2 \).
- [ ] The function is continuous for all real numbers.
**Discussion:**
To determine which statement is correct, we need to analyze the function \( f(x) = \frac{x^2 - 4}{x - 2} \).
First, simplify the function:
- \( x^2 - 4 \) can be factored as \( (x - 2)(x + 2) \).
- Thus, the function becomes \( f(x) = \frac{(x - 2)(x + 2)}{x - 2} \).
For \( x \neq 2 \), this simplifies to \( f(x) = x + 2 \). However, at \( x = 2 \), the expression \( \frac{0}{0} \) indicates a discontinuity.
Since this discontinuity is caused by a factor that can be canceled (the \( x - 2 \) in both the numerator and the denominator), it is a removable discontinuity.
Therefore, the correct choice is:
- [ ] There is a removable discontinuity at \( x = 2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bf8d9f8-9f3c-4f11-9f6a-4feeeac00eb3%2Ff8ff6d5a-b554-4e1c-8825-8224f9fb1f99%2Fsiz5g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 5:**
Which of the following is true for \( f(x) = \frac{x^2 - 4}{x - 2} \)?
- [ ] There is a non-removable discontinuity at \( x = 2 \).
- [ ] There is a removable discontinuity at \( x = 2 \).
- [ ] The function is continuous for all real numbers.
**Discussion:**
To determine which statement is correct, we need to analyze the function \( f(x) = \frac{x^2 - 4}{x - 2} \).
First, simplify the function:
- \( x^2 - 4 \) can be factored as \( (x - 2)(x + 2) \).
- Thus, the function becomes \( f(x) = \frac{(x - 2)(x + 2)}{x - 2} \).
For \( x \neq 2 \), this simplifies to \( f(x) = x + 2 \). However, at \( x = 2 \), the expression \( \frac{0}{0} \) indicates a discontinuity.
Since this discontinuity is caused by a factor that can be canceled (the \( x - 2 \) in both the numerator and the denominator), it is a removable discontinuity.
Therefore, the correct choice is:
- [ ] There is a removable discontinuity at \( x = 2 \).
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