(x-1)^3=x^3-3x^2+3x-1.
x^3=(x-1)^3+3x^2-3x+1
F(x)=x^3 -3bx+2b
=(x-1)^3+3x^2-3x+1+3bx+2b
=(x-1)^3+3x^2+3(b-1)x+2b+1
=(x-1)^3+3[x^2+(b-1)x+(b-1)^2/4]-3(b-1)^2/4+2b+1
=(x-1)^3+3[x+(b-1)/2]^2-(3b^2-6b+3)/4+2b+1
=(x-1)^3+3[x+(b-1)/2]^2-(3b^2-6b+3-8b-4)/4
要使F(x)在[1,2]内恒有正值,(x-1)^3在[1,2]内恒有正值,3[x+(b-1)/2]^2恒有正值.-(3b^2-6b+3-8b-4)/4>=0
即:(3b^2-6b+3-8b-4)/4=<0
3b^2-6b+3-8b-4=<0
3b^2-14b-1=<0
b^2-14b/3-1/3=<0
(b^2-14b/3+196/36)-196/36-1/3=<0
(b-14/6)^2=<52/9
14/6-2根号13/3=
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