Rolle’s Theorem

Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) and f(a) = f(b), there exists at least one c in (a,b) such that f’(c) = 0.
Proof:-
Since f(x) is continuous in closed interval [a,b], there exist c,d in [a,b] such that
f(c) = M(the maximum value of f(x))
f(d) = m the minimum value of f(x)

CASE I
If M = m, then f(x) is constant in [a,b] and in this case f’(x) = 0 for all x in (a,b). That is theorem is true in this case.
CASE II
If M = m, then at least one of them is different from f(a) and f(b). Let M = f(a). We shall show that f’(c) = 0 with c in (a,b)
At x = c, with h>0
RHD = lim h 0 f(c+h)-f(c)/h
= lim h 0 f(c+h)-M/h = -ve/+ve <= 0 ……….i
LHD = lim h 0 f(c-h)-f(c)/-h
= lim h 0 f(c-h)-M/-h = -ve/-ve >= 0 ……….ii

Since f’(x) exist in (a,b), I and ii must be equal and this will be possible only when both i and ii are equal to zero.
Hence f’(c) = 0 with c in (a,b).
CASE III
If m is different from f(a) and f(b) we can similarly show that f’(d) = 0 with d in (a,b). This completes the proof.