Given: `f(x)=2x^2-3x+1,[0,2]`

The function is a continuous polynomial on the closed interval [0,2].

The function is differentiable on the open interval (0,2).

`f'(x)=4x-3`

`f(c)=4c-3`

`[f(2)-f(0)]/(2-0)=[3-1]/2=1`

`4c-3=1`

`4c=4`

`c=1`

The answer is **c=1** which is in the interval [0,2].