The mean value theorem requires that a function f(x)f(x) be continuous on a closed interval [a, b][a,b] and differentiable on the open interval (a, b)(a,b). If it is, then we can conclude that there exists c in (a, b)c∈(a,b) such that f'(c) = (f(b)-f(a))/(b-a)
For the given function f(x) = 2x^2-5x+1, we know f is differentiable (and thus continuous) on all of RR, and so must also be continuous on [0, 2] and differentiable on (0, 2). As such, it satisfies the hypothesis of the mean value theorem.
The conclusion of the mean value theorem, then, gives us that for some c in (0, 2), we have
f'(c) = (f(2)-f(0))/(2-0) = (-1-1)/2 = -1
Taking the derivative of f(x), we can use the power rule to find that
f'(x) = 4x-5
So f'(c) = 4c-5
Substituting that in, we get the equation
4c-5 = -1
=> 4c = 4
:. c = 1
So f(x) = 2x^2-5x+1 satisfies the hypothesis of the mean value theorem on [0, 2], and c=1 satisfies the conclusion.
