DerivativeApplications

# Rolle's Theorem

Let f continuous in [a,b] and derivable in (a,b),

*Demonstration*: let h = f(a) = f(b)

Case 1: if f(x) = h in [a,b] then f is constant and f’(x) = 0 in (a,b)

Case 2: if f(x) > h in some x in (a,b), there is a maximum, M, in the interval in (a,b) (Bolz-W. Th.) M > h, then

Case 3: analogous to case 2

QED

**Exercise**: Let the function f(x) = 2 - |x|. As you can see f(-2) = f(2) but it doesn't exist any c in the interval (-2,2)/ f'(c) = 0. Justify why it doesn't contradict Rolle's Theorem.

Solution: Because f is not derivable in x = 0

Licensed under the Creative Commons Attribution Non-commercial Share Alike 3.0 License