In the case where #f(a)=f(b)# so that #f(c)=f(a)=f(b)#, then Rolle's Theorem (or the Mean Value Theorem ) can be applied in a simple way to reach the conclusion. Otherwise, you need the Mean Value Theorem in addition to the fact that derivatives satisfy the "Intermediate Value Property" (even when they are not continuous).
Explanation:
First note that the fact that #2f(c)=f(a)+f(b)# implies that #f(c)=(f(a)+f(b))/2#, meaning that #f(c)# is the average of #f(a)# and #f(b)#. If #f(a)=f(b)#, then #f(c)=f(a)=f(b)#. If #f(a) != f(b)#, then #f(c)# is strictly between #f(a)# and #f(b)#.
If #f(a)=f(b)=f(c)#, then Rolle's Theorem implies the conclusion immediately: there exists a number #alpha# with #a < alpha < c# such that #f'(alpha)=0#.
We now consider the case where #f(a) < f(b)# so that #f(a) < f(c) < f(b)#. The case where #f(a) > f(b)# is similar.
By the Mean Value Theorem, there exists numbers #beta# and #gamma# with #a < beta < b# and #b < gamma < c# such that #f'(beta)=(f(b)-f(a))/(b-a) > 0# and #f'(gamma)=(f(c)-f(b))/(c-b) < 0#.
Even though we have not assumed #f'# is continuous, it is a well-known fact that it satisfies the "Intermediate Value Property" anyway (by Darbaux's Theorem ). This is enough to conclude that there exists a number #alpha# with #beta < alpha < gamma# such that #f'(alpha) = 0#. (If #f'# is assumed to be continuous, then the theorem to apply is the Intermediate Value Theorem ).
Here's parts 1 and 2 of a lecture I gave that might be helpful as a starting point to review these concepts:
