
#1
Jan3014, 04:31 AM

P: 296

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? 



#2
Jan3014, 09:09 AM

Mentor
P: 4,499

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 fİ, hence the strange notation. 



#3
Jan3014, 09:54 AM

P: 296





#4
Jan3014, 12:51 PM

Mentor
P: 4,499

Can we write Rolle's Theorem this way?
Yeah good catch




#5
Jan3014, 05:51 PM

P: 296

Thanks



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