Answer
$x = 1$
Work Step by Step
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:
$1 = x$
