(1)由题意得f′(x)=,g′(x)=x2-x+b,x>-1, 则解得 ∴f(x)=ln(x+1)(x>-1),g(x)=x3-x2+x. 令h(x)=f(x)-g(x) =ln(x+1)-x3+x2-x(x>-1), ∴h′(x)=-x2+x-1=-, ∴h(x)在(-1,0)上单调递增,在(0,+∞)上单调递减, ∴h(x)≤h(0)=0,∴f(x)≤g(x). (2)当x∈(x1,x2)时,由题意得-1<x1<x<x2, ①设u(x)=(x+1)[f(x)-f(x1)]-(x-x1), 则u′(x)=ln(x+1)-ln(x1+1)>0, ∴u(x)>u(x1)=0,即(x+1)[f(x)-f(x1)]-(x-x1)>0, ∴; ②设v(x)=(x+1)[f(x)-f(x2)]-(x-x2), 则v′(x)=ln(x+1)-ln(x2+1)<0, ∴v(x)>v(x2)=0,即(x+1)[f(x)-f(x2)]-(x-x2)>0, ∴, 由①②得. |