已知f(x)=x^2-2x+m且|x-m|<3，求证|f(x)-f(m)|< 6|m|+15

f(m)=m^2-m;
∴|f(x)-f(m)|
=|(x^2-2x+m)-(m^2-m)|
=|x^2-m^2-2x+2m|
=|(x^2-m^2)-2(x-m)|
=|(x+m)(x-m)-2(x-m)|
=|(x-m)(x+m-2)|
<3|(x+m-2)|.
∵|x-m|<3
∴-3<x-m<3;
2m-5<x+m-2<2m+1.
即2m-5<x+m-2<2m+1<2m+5;
于是-2|m|-5<x+m-2<2|m|+5;
∴|(x+m-2)|<2|m|+5
那么|f(x)-f(m)|<3|(x+m-2)|<3×(2|m|+5)= 6|m|+15
得证

|f(x)-f(m)|=|x^2-2x+m-(m^2-2m+m)|
=|(x^2-m^2)-2(x-m)|
=|(x-m)(x+m-2)|
=|(x-m)||(x+m-2)|
|x-m|<3
则m-3<x<m+3
则2m-3<x+m<2m+3
则2m-5<(x+m-2)<2m+1
所以|(x-m)||(x+m-2)|<3|(x+m-2)|<3min（|2m+1|，|2m-5|）
当m>12/7，|2m+1|>|2m-3|.则3min（|2m+1|，|2m-5|）<3(2|m|+1)<6|m|+15
当m<12/7，|2m+1|<|2m-3|.则3min（|2m+1|，|2m-5|）<3(2|m|+5)
综合得：|f(x)-f(m)|< 6|m|+15

太难了