If a real-valued function f is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then such a c exists in the open interval (a, b) that f'(c) = 0
Hp.
f(x) continuous on [a,b]
f(x) is differentiable on (a,b)
f(a) = f(b)
Th.
f'(c) = 0 Graphic proof: Since the function is continuous in [a,b] the arc it draws on the graphic must raise from a, reach a maximum in a certain point c and then reach the x axis again in b. It's easy to see that the slope of the tangent in c must be 0 since the line is horizontal. Mathematical proof:
Let's study the increment of the function in a point h.
f(c + h) - f(c) =
f(c + h) - f(c) >= 0 h > 0
Apply the Theorem of sign permanence
lim h->0- [f(c + h) - f(c)]/h
lim h->0+ [f(c + h) - f(c)]/h >= 0 h > 0
Use the definition of differentiability
lim h- > 0 [f(c + h) - f(c)]/h = 0
Use the definition of derivative
f'(c) = 0
Q.E.D.