Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Use the Mean Value Theorem to prove the inequality \(|\sin a - \sin b|\; \le \;|a - b|\) for all \(a\) and \(b\).

The inequality is \(|\sin a - \sin b|\; \le \;|a - b|\) is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given inequality is \(|\sin a - \sin b|\; \le \;|a - b|\).

Step 2: Concept of Mean value theorem

“Let \(f\) be a function that satisfies the following hypothesis:

1. \(f\) is continuous on the closed interval \(\left( {a\;,{\rm{ }}b} \right)\).

2. \(f\) is differentiable on the open interval \(\left( {a,\;b} \right)\).

Then, there is a number \(c\) in \(\left( {a\;,{\rm{ }}b} \right)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Step 3: Calculation to prove the given inequality

Consider the function, \(f(x) = \sin x\).

Obtain the derivative of, \(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}(\sin x)\\{f^\prime }(x) &= \cos x\end{aligned}\)

Replace \(x\) by \(c\).

Then, \({f^\prime }(c) = \cos c\).

Since \(\sin x\) is continuous and differentiable everywhere, by Mean Value Theorem mentioned above, then there exists a number c such that:

\(\cos (c) = \frac{{\sin (b) - \sin (a)}}{{b - a}}\) …… (1)

Take modulus on both sides of the equation.

\(|\cos (c)| = \left| {\frac{{\sin (b) - \sin (a)}}{{b - a}}} \right|\)

Since \( - 1 \le \cos (c) \le 1\), it implies that \(|\cos (c)| \le 1\).

Therefore, \(\left| {\frac{{\sin (b) - \sin (a)}}{{b - a}}} \right| \le 1\).

Since, by the property, \(\left| {\frac{a}{b}} \right| = \frac{{|a|}}{{|b|}}\).

\(\begin{aligned}|\sin (b) &- \sin (a)|\; \le 1 \cdot |b - a|\\|\sin (b) &- \sin (a)|\; \le \;| - (a - b)|\\|\sin (b) &- \sin (a)|\; \le \;|a - b|\end{aligned}\)

Hence, the inequality is proved.

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Essential Calculus: Early Transcendentals

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Short Answer

Use the Mean Value Theorem to prove the inequality \(|\sin a - \sin b|\; \le \;|a - b|\) for all \(a\) and \(b\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to prove the given inequality

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Use the Mean Value Theorem to prove the inequality \(|\sin a - \sin b|\; \le \;|a - b|\) for all \(a\) and \(b\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to prove the given inequality

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