Question

規(guī)定:兩個(gè)連續(xù)函數(shù)(圖象不間斷)f(x),G(x)在閉區(qū)間［a,b］上都有意義,我們稱函數(shù)|f(x)-G(x)|在［a,b］上的最大值叫做函數(shù)f(x)與G(x)在［a,b］上的“絕對(duì)差”.

(1)試求函數(shù)f(x)=x²與G(x)=x(x-2)(x-4)在閉區(qū)間［-3,3］上的“絕對(duì)差”；

(2)設(shè)函數(shù)f(x)=x²及函數(shù)h_m(x)=(a+b)x+m都定義在已知區(qū)間［a,b］上,記f(x)與h_m(x)的“絕對(duì)差”為D(m).若D(m)的最小值是D(m₀),則稱f(x)可用h_m0(x)“替代”,試求m₀的值,使f(x)可用h_m0(x)“替代”.

Answer 1

解:(1)記F(x)=f(x)-g(x),?

則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“絕對(duì)差”為12.                                                                                      ?

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

從而令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)=                                                      ?

∴當(dāng)M=M₀=-時(shí),D(M₀)最小.?

故當(dāng)M₀=-(a²+6AB+b²)時(shí),f(x)可用h(x)“替代”.