已知f(x)=x^2-2x+m且|x-m|
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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
那么|f(x)-f(m)|<3|(x+m-2)|<3×(2|m|+5)= 6|m|+15
得证
太难了
|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
所以|(x-m)||(x+m-2)|<3|(x+m-2...
全部展开
|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
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
收起