Answer
$c=1$
Work Step by Step
Here, $f(x)=2x^2-3x+1$ in interval $[0,2]$.
1. Since $f(x)$ is a polynomial, it is continuous over the given interval $[0,2]$.
2. Similarly, since $f(x)$ is a polynomial, it is differentiable on the open interval $(0,2)$.
Thus, there exists a number $c$ in the open interval $(0,2)$ such that $f'(c)=\frac{f(2)-f(0)}{2-0}$.
Here, $f(0)=1$. Similarly,$f(2)=3$. Thus, $$f'(c)=\frac{3-1}{2}=1$$ $$4c-3=1\implies c=1$$