Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Suppose \(f\) is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in \(( - b\;,\;b)\) such that \({f^\prime }(c) = \frac{{f(b)}}{b}\).

It is proved that for every positive number \(b\), there exists a number \(c\) in \(( - b,b)\) such that \({f^\prime }(c) = \frac{{f(b)}}{b}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given function \(f\) is an odd function and differentiable everywhere.

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 that for every positive number \(b\), there exists a number \(c\) in \(( - b\;,\;b)\)

It is given that \(f\) is differentiable everywhere.

Thus, \(f\) is differentiable on the interval, \(( - b,b)\).

Since \(f\) is differentiable everywhere, \(f\) must be continuous everywhere.

Therefore, \(f\) is continuous on \(( - b,b)\) and differentiable on, \(( - b,b)\).

Then by Mean value theorem there exist a number \(c\), in \(( - b,b)\) such that:

\({f^\prime }(c) = \frac{{f(b) - f( - b)}}{{b - ( - b)}}\)

\({f^\prime }(c) = \frac{{f(b) - f( - b)}}{{2b}}\) …… (1)

Since \(f\) is an odd function, \(f( - b) = - f(b)\).

Substitute \(f( - b) = - f(b)\) in equation (1).

\(\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(b) - ( - f(b))}}{{2b}}\\{f^\prime }(c) &= \frac{{2f(b)}}{{2b}}\\{f^\prime }(c) &= \frac{{f(b)}}{b}\end{aligned}\)

Hence, it is proved that for every positive number \(b\), there exists a number \(c\) in \(( - b,b)\) such that \({f^\prime }(c) = \frac{{f(b)}}{b}\).

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Essential Calculus: Early Transcendentals

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Short Answer

Suppose \(f\) is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in \(( - b\;,\;b)\) such that \({f^\prime }(c) = \frac{{f(b)}}{b}\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to prove that for every positive number \(b\), there exists a number \(c\) in \(( - b\;,\;b)\)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Suppose \(f\) is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in \(( - b\;,\;b)\) such that \({f^\prime }(c) = \frac{{f(b)}}{b}\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to prove that for every positive number \(b\), there exists a number \(c\) in \(( - b\;,\;b)\)

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