Homework Help is Here – Start Your Trial Now!
Math
Calculus
x² – 16 tate why f(x) is continuous on [1, 2]. x² – 9

x² – 16 tate why f(x) is continuous on [1, 2]. x² – 9

Calculus: Early Transcendentals
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...
See similar textbooks
icon
Related questions
Question
**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 \([1, 2]\).
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 \([1, 2]\).
Expert Solution
Step 1 definition used

f(x) is continuous on [a,b] if and only if 

1. ∀x0∈(a,b)⊆R, limx→x0f(x)=f(x0)

2. limx→a+f(x)=f(a) and limx→b-f(x)=f(b)

steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Vector Space
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS