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Q. 59

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 $f$ 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 $c \in (a, b)$ that satisfy the conclusion of the Mean Value Theorem.
$f (x) = \sin (x), [a, b] = [0, \frac{π}{2}]$ .

The function $f (x) = \sin (x)$ satisfies the Mean Value Theorem and the value is, $c = \cos^{- 1} (\frac{2}{π})$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

The given function is,

$f (x) = \sin (x), [a, b] = [0, \frac{π}{2}]$ .

Step 2. Proving Mean Value Theorem.

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

The slope of the line from $(0, f (0))$ to $(\frac{π}{2}, f (\frac{π}{2}))$ is:

$\frac{f (\frac{π}{2}) - f (0)}{\frac{π}{2} - 0} = \frac{\sin (\frac{π}{2}) - \sin (0)}{\frac{π}{2} - 0} = \frac{1 - 0}{\frac{π}{2} - 0} = \frac{1}{\frac{π}{2}} = \frac{2}{π}$

By the Mean Value Theorem, there must exist at least one point $c \in (0, \frac{π}{2})$ with $f^{'} (c) = \frac{2}{π}$

We have to find the value of $c$ with $f^{'} (c) = \frac{2}{π}$ we solve it:

$\cos (c) = \frac{2}{π} \Rightarrow c = \cos^{- 1} (\frac{2}{π})$

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