Homework Help is Here – Start Your Trial Now!
Math
Calculus
Use the Mean Value Theorem from §4.2 to prove that | sin(b) – sin(a)| < 16 – a| for any interval [a, b].

Use the Mean Value Theorem from §4.2 to prove that | sin(b) – sin(a)| < 16 – a| for any interval [a, b].

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
**Applying the Mean Value Theorem to Prove Inequality** To prove the inequality: \[ | \sin(b) - \sin(a) | \leq |b - a| \] for any interval \([a, b]\), we will utilize the Mean Value Theorem (MVT) as discussed in Section 4.2. ### Mean Value Theorem Statement The MVT states that if a function \(f\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c \in (a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] ### Proof Applying the MVT Consider the function \(f(x) = \sin(x)\), which is continuous and differentiable everywhere. By the MVT, there exists a point \(c \in (a, b)\) such that: \[ \sin'(c) = \frac{\sin(b) - \sin(a)}{b - a} \] We know that \(\sin'(x) = \cos(x)\). Therefore: \[ \cos(c) = \frac{\sin(b) - \sin(a)}{b - a} \] Taking the absolute value of both sides: \[ |\cos(c)| = \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \] We also know that the maximum value of \(|\cos(x)|\) is 1 for all \(x\). Thus, \[ |\cos(c)| \leq 1 \] This implies: \[ \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \leq 1 \] Multiplying through by \(|b - a|\), we get: \[ |\sin(b) - \sin(a)| \leq |b - a| \] which completes the proof.
Transcribed Image Text:**Applying the Mean Value Theorem to Prove Inequality** To prove the inequality: \[ | \sin(b) - \sin(a) | \leq |b - a| \] for any interval \([a, b]\), we will utilize the Mean Value Theorem (MVT) as discussed in Section 4.2. ### Mean Value Theorem Statement The MVT states that if a function \(f\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c \in (a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] ### Proof Applying the MVT Consider the function \(f(x) = \sin(x)\), which is continuous and differentiable everywhere. By the MVT, there exists a point \(c \in (a, b)\) such that: \[ \sin'(c) = \frac{\sin(b) - \sin(a)}{b - a} \] We know that \(\sin'(x) = \cos(x)\). Therefore: \[ \cos(c) = \frac{\sin(b) - \sin(a)}{b - a} \] Taking the absolute value of both sides: \[ |\cos(c)| = \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \] We also know that the maximum value of \(|\cos(x)|\) is 1 for all \(x\). Thus, \[ |\cos(c)| \leq 1 \] This implies: \[ \left| \frac{\sin(b) - \sin(a)}{b - a} \right| \leq 1 \] Multiplying through by \(|b - a|\), we get: \[ |\sin(b) - \sin(a)| \leq |b - a| \] which completes the proof.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Laplace Transformation
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