## Thomas' Calculus 13th Edition

$\frac{1}{2}$
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= $\frac{1}{2}$