Question

规定:两个连续函数(图象不间断)f(x),G(x)在闭区间［a,b］上都有意义,我们称函数|f(x)-G(x)|在［a,b］上的最大值叫做函数f(x)与G(x)在［a,b］上的“绝对差”.

(1)试求函数f(x)=x²与G(x)=x(x-2)(x-4)在闭区间［-3,3］上的“绝对差”；

(2)设函数f(x)=x²及函数h_m(x)=(a+b)x+m都定义在已知区间［a,b］上,记f(x)与h_m(x)的“绝对差”为D(m).若D(m)的最小值是D(m₀),则称f(x)可用h_m0(x)“替代”,试求m₀的值,使f(x)可用h_m0(x)“替代”.

Answer 1

解:(1)记F(x)=f(x)-g(x),?

则F′(x)=f′(x)-g′(x)=-3x²-2x+8.?

由F′(x)=0,得x=-2或x=.                                                                               ?

∴F(-2)=-12,F()=,F(3)=-12,F(-3)=-6.                                                     ?

∴-12≤F(x)≤.?

故所求“绝对差”为12.                                                                                      ?

(2)由于f(x)-h_M(x)=x²-［(a+b)x+M］,f′(x)-h_M′(x)=2x-(a+b),?

从而令f′(x)-h_M′(x)=0,得x=.                                                                     ?

∴D(M)=Max{|f()-h_M()|,|f(a)-h_M(a)|,|f(b)-h_M(b)|}?

=Max{|M+|,|M+AB|}.                                                                                 ?

由于|M+|²-|M+AB|²= (M+),?

∴D(M)=                                                      ?

∴当M=M₀=-时,D(M₀)最小.?

故当M₀=-(a²+6AB+b²)时,f(x)可用h(x)“替代”.