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由函数的连续性,加上x->0,f(x)=cos(x)+x+o(x)。知道f(0)=0,由f(x)=f(0)+f'(t)*x,其中t是在(0,x)。f(1)=f(0)+f'(t1);f(2)=f(0)+2f'(t1);f(3)=f(0)+3f'(t1);代入上式可知f'(t1)+4f'(t2)+9f'(t3)=0;可知(1/14)f'(t1)+(4/14)f'(t2)+(9/14)f'(t3)=0,知道min(f'(t1),f'(t2),f'(t3))=<(1/14)f'(t1)+(4/14)f'(t2)+(9/14)f'(t3)<=max(f'(t1),f'(t2),f'(t3)),结果得到:min(f'(t1),f'(t2),f'(t3)=<0<=max(f'(t1),f'(t2),f'(t3)),由于f'(x)的连续性可知(大家要注意可导的定义,左导数等于右导数,决定了导函数的连续性),必有一点x,使得f'(x)=0;
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