Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 222: 1

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

$f(x)=x^{2}+2x-1$ $f'(x)=2x+2$ f is continuous on $[a,b]=[0,1]$, differentiable on $(a,b)=(0,1)$, so, by the Mean Value Theorem, a $c\in(a,b)$ exists such that $f'(c)=\displaystyle \frac{f(1)-f(0)}{1-0}$ $2c+2=(1+2-1)-(0+0-1)$ $2c+2=3$ $2c=1$ $c=\displaystyle \frac{1}{2}\in(0,1)$