Rolle's theorem: Statements: If y =f(x) is a real valued function of a real variable such that: 1) f(x) is continuous on [a,b] 2) f(x) is differentiable on (a,b) 3) f(a) = f(b) then there exists a real number c[itex]\in[/itex](a,b) such that f'(c)=0 what if the the f(x) is like the following graph: here there is a point 'c' for which f'(c) =0 but f(a) [itex]\neq[/itex] f(b) So to take such cases in consideration can we make a change to the last statement of Rolle's theorem as: 3)f(c) > [f(a),f(b)] Or f(c)<[f(a),f(b)] are there any exceptions to the above statement?
Yes, that statement works. The easiest way to prove this is to notice that if f(c.) > f(b) > f(a) (for example), then there exists some d such that a<d<c and f(d) = f(b) by the intermediate value theorem. Then applying Rolle's theorem to the interval [d,c] completes the proof. This computer seems to insist on writing f(c.) without the period as f©, hence the strange notation.