微積分問題 洛爾定理

Question

初微中
洛爾定理的證明

請寫詳細

拜託各位大大了

Answer 1

ROLLE"S THEOREM:
let F be a function that satisfies the following there hypotheses:
1.F is continuous on the closed interval[a,b].
2.F isdifferentiable on the open interval [a,b].
3 F(a)=F(b)
then there is a number c in (a,b) such that F'(c)=0
proof
there are three cases:
1.F(x)=k,a constant
then F'(x)=0,so the number c can be taken to be any number in (a,b).
2.F(x)>F(a) for some x in (a,b)
by the extreme value theorem(which we can apply by hypothesis 1),F has a maximum value somewhere in [a,b],since F(a)=F(b),it must attain this maximum value at a number c in the open interval(a,b).then F has a local maximum at c and, by hypothesis 2,F is differentiable at c. therefore,F'(c)=0 by fermat's theorem.
3. F(x) by the extreme value theorem ,F has a minimum value in [a,b]and,since F(a)=F(b),it attains this minimum value at a number c in (a,b).again F'(c)=0 by fermat's theorem.

Answer 2

【Rolle's Thm】                                                                                1) f(x)  is conti. on [a,b]  2) f'(x) exists on (a,b) 3) f(a) = f(b)   Then there is c in (a,b) s.t. f'(c)=0                                                                                PF :                                                                               Assume that f'(x) ≠ 0 for all x in (a,b) . Since [a,b] is compact and f is conti. on [a,b]                                                                                => there exist c and d are in [a,b] s.t. Sup(x in [a,b]) f(x) = f(c) and Inf(x in [a,b]) f(x) = f(d)                                                                                But f'(x) ≠ 0 => c and d are both lie on a or b, which implies that  Sup(x in [a,b]) f(x) = Inf(x in [a,b]) f(x), hence f is a constant function => f'(c) = 0 for all c in (a,b), contradiction.So, the Rolle's Thm. is true.