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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![**Problem Statement:**
State why \( f(x) = \frac{x^2 - 16}{x^2 - 9} \) is continuous on the interval \([1, 2]\).
**Solution Explanation:**
The function \( f(x) \) is defined as a rational function, which is the quotient of two polynomials: \( x^2 - 16 \) (numerator) and \( x^2 - 9 \) (denominator). Rational functions are continuous everywhere in their domain, which includes all real numbers except where the denominator equals zero.
First, we find where the denominator is equal to zero:
\[ x^2 - 9 = 0 \]
Solving for \( x \), we get:
\[ x^2 = 9 \]
\[ x = \pm 3 \]
This means the function is undefined at \( x = 3 \) and \( x = -3 \).
Next, we check the interval \([1, 2]\). The interval does not include \( x = \pm 3 \), therefore the function \( f(x) \) is defined and continuous for every point in \([1, 2]\). Since the function is a rational function and is defined across this entire interval, \( f(x) \) is continuous on \([1, 2]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4f25a94-dbbd-4fea-a743-0f16d24aa2a3%2F4d61339b-e606-472a-963a-74bec3ea3b5a%2Fj94dj1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
State why \( f(x) = \frac{x^2 - 16}{x^2 - 9} \) is continuous on the interval \([1, 2]\).
**Solution Explanation:**
The function \( f(x) \) is defined as a rational function, which is the quotient of two polynomials: \( x^2 - 16 \) (numerator) and \( x^2 - 9 \) (denominator). Rational functions are continuous everywhere in their domain, which includes all real numbers except where the denominator equals zero.
First, we find where the denominator is equal to zero:
\[ x^2 - 9 = 0 \]
Solving for \( x \), we get:
\[ x^2 = 9 \]
\[ x = \pm 3 \]
This means the function is undefined at \( x = 3 \) and \( x = -3 \).
Next, we check the interval \([1, 2]\). The interval does not include \( x = \pm 3 \), therefore the function \( f(x) \) is defined and continuous for every point in \([1, 2]\). Since the function is a rational function and is defined across this entire interval, \( f(x) \) is continuous on \([1, 2]\).
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Step 1 definition used
f(x) is continuous on [a,b] if and only if
1.
2.
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