Answer
$\frac{1}{2}$
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= $\frac{1}{2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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