The mean value theorem requires that a function #f(x)# be continuous on a closed interval #[a, b]# and differentiable on the open interval #(a, b)#. If it is, then we can conclude that there exists #c in (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.
