Calculus: Early Transcendentals 8th Edition

$x = 1$
Since $f$ is a polynomial, we know that it is continuous on $[0,2]$ and differentiable on $(0,2)$. Thus the Mean Value Theorem can be applied. $f'(c) = \frac{f(b) - f(a)}{b-a}$ $f'(c) = \frac{f(2) - f(0)}{2-0}$ $f'(c) = \frac{2(2^{2})-3(2)+1 - (2(0^{2})-3(0)+1)}{2}$ $f'(c) = \frac{8 - 6 + 1 - 1}{2}$ $f'(c) = \frac{2}{2}$ $f'(c) = 1$ $f(c) = 2x^{2}-3x+1$ $f'(c) = 4x -3$ $1 = 4x - 3$ $1 + 3 = 4x$ $4 = 4x$ Solve for x: $1 = x$