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Essential Calculus: Early Transcendentals
Found in: Page 215
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

To determine the values of \(c\) that satisfies the conclusion of the Mean Value Theorem for the interval \((1,7)\) using the given graph of the function.

The values of \(c\)that satisfies the conclusion of the Mean Value Theorem for the interval \((1,7)\) by the given graph of the function is \(\underline {c = 0.3,3,6.3} \).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given fucntion

\(f(x)\)is a continuous function on \((1,7)\) and \(f(x)\) is differentiable in \((1,7)\).

Step 2: Concept of Mean Value Theorem

Assume \(f\) to be a function satisfying the following properties.

(1) \(f\)Continuous on the interval \((a,b)\).

(2) \(f\) Is differentiable on the interval \((a,b)\).

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

Step 3: Check the conditions of Mean Value Theorem

Here, in the given graph is \(a = 1,\;b = 7\).

Also, from the given graph \(f(x)\) is a continuous function on \(\left( {1,7} \right)\) and \(f(x)\) is differentiable in \((1,7)\).

Clearly, from the graph it can be seen that \(f(1) = 0.5;\;f(7) = 1.5\).

Thus, \({f^\prime }(c)\) is evaluated as shown below.

\(\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(7) - f(1)}}{{7 - 1}}\\ &= \frac{{1.5 - 0.5}}{{7 - 1}}\\ &= \frac{1}{6}\\ &= 0.167\end{aligned}\)

It can be observed from the graph that the values of \(c = 0.3,3,6.3\).

Therefore, the values of \(c\) that satisfies the conclusion of the Mean Value Theorem for the interval \((1,7)\) by the given graph of the function is .

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