Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 197: 1



Work Step by Step

f(a)= f(0)= 0+0-1= -1. f(b)= f(1)= 1+2-1 = 2. Hence $\frac{f(b)-f(a)}{b-a}=\frac{2-(-1)}{1-0}=3$. MVT states that there is a point c ε (0,1) such that f'(c) =3. f'(c)= 2c+2= 3 ⇒ c= $\frac{1}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.