设f(x)在[0,1]上可导,f(0)=0,|f"(x)|≤
【正确答案】正确答案:因为f(x)在[0,1]上可导,所以f(x)在[0,1]上连续,从而|f(x)|在[0,1]上连续,故|f(x)|在[0,1]上取到最大值M,即存在x 0 ∈[0,1],使得|f(x 0 )|=M. 当x 0 =0时,则M=0,所以f(x)≡0,x∈[0,1]; 当x 0 ≠0时,M=|f(x 0 )|=|f(x 0 )一f(0)|=|f"(ξ)|x 0 ≤|f"(ξ)|≤
【答案解析】