   Chapter 4.2, Problem 31E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use the Mean Value Theorem to prove the inequality | sin a = sin b |   ≤   | a − b |     for   all   a   and   b

To determine

To prove: The inequality |sinasinb||ab| for all a and b.

Explanation

Mean value Theorem

“If f be a function that satisfies the following hypothesis:

1. f is continuous on the closed interval [a,b] .

2. f is differentiable on the open interval (a,b) .

Then, there is a number c in (a,b) such that f(c)=f(b)f(a)ba .

Or, equivalently, f(b)f(a)=f(c)(ba) ”.

Proof:

Consider the function, f(x)=sinx .

Obtain the derivative of f(x) .

f(x)=ddx(sinx)=cosx

Replace x by c. Then, f(c)=cosc .

Since sinx is continuous and differentiable everywhere, by Mean Value Theorem mentioned above, then there exists a number c such that,

cos(c)=sin(b)sin(a)ba (1)

Take modulus on both sides of the equation (1),

|cos(c)|=|sin(b)sin(a)ba|

Since 1cos(c)1 , it implies that |cos(c)|1

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