分析

简答题: 设函数f(x)在区间[0,2]上具有连续导数f(0)=f(2)=0,M=max{|f(x)|},x∈[0,2],
证明:(Ⅰ)存在ξ∈(0,2),使得|f'(ξ)|≥M;
(Ⅱ)若对任意的x∈(0,2),|f’(x)|≤M,则M=0.

正确答案
(Ⅰ)由M=max{|f(x)|},x[0,2]知存在c∈(0,2),使|f(c)|=M.若
c∈(0,1],由拉格朗日中值定理得至少存在一点∈(0,c),使

(Ⅱ)若M>0,则c≠0,2.
由f(0)=f(2)=0及罗尔定理知存在∈(0,2),使f’()=0.
/∈(0,c]时,

于是2M 故M=0.
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