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Q. 57

Expert-verified

Calculus

Found in: Page 249

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Determine whether or not each function f in Exercises 53–60 satisfies the hypotheses of the Mean Value Theorem on the given interval [a, b]. For those that do, use derivatives and algebra to find the exact values of all c ∈ (a, b) that satisfy the conclusion of the Mean Value Theorem.
$f (x) = \ln (x^{2} + 1), [a, b] = [0, 1]$

The function satisfies the Mean value theorem and the value of c is 0.4028.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

$f (x) = \ln (x^{2} + 1), [a, b] = [0, 1]$

Step 2. Proving Mean Value Theorem.

The function $f (x) = \ln (x^{2} + 1)$ is continuous and differentiable on $[0, 1]$ . The Mean Value Theorem applies to this function on the interval $[0, 1]$ .

The slope of the line from (0, f(0)) to (1, f(1)) is:

$\begin{aligned} f^{'} (c) = \frac{f (1) - f (0)}{1 - 0} \\ = \frac{\ln (1^{2} + 1) - \ln (0^{2} + 1)}{1} \\ = \ln 2 - \ln 0 \\ = 0.6931 \end{aligned}$

Now,

$\begin{aligned} f^{'} (x) = \frac{d}{d x} \ln (x^{2} + 1) \\ = \frac{1}{x^{2} + 1} \cdot 2 x \\ = \frac{2 x}{x^{2} + 1} \end{aligned}$

Therefore,

$\begin{aligned} f^{'} (c) = 0.6931 \\ \frac{2 c}{c^{2} + 1} = 0.6931 \\ 2 c = 0.6931 c^{2} + 0.6931 \end{aligned}$

$\begin{array}{r} 0.6931 c^{2} - 2 c + 0.6931 = 0 \\ c^{2} - 2.8855 c + 1 = 0 \end{array}$

role="math" localid="1648380951546" $\begin{aligned} c = \frac{- (- 2.8855) \pm \sqrt{(- 2.8855)^{2} - 4 \times 1 \times 1}}{2 \cdot 1} \\ c = \frac{2.8855 \pm \sqrt{4.3261}}{2} \\ = \frac{2.8855 \pm 2.08}{2} \\ c = 2.48275 o r 0.40275 \end{aligned}$

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