(1)设g(x)=ax2+bx+c,g(x)的图象经过坐标原点,所以c=0. ∵g(x+1)=g(x)+2x+1∴a(x+1)2+b(x+1)=ax2+bx+2x+1 即:ax2+(2a+b)x+a+b=ax2+(b+2)x+1 ∴a=1,b=0,g(x)=x2; (2)函数f(x)=mx2-ln(x+1)的定义域为(-1,+∞).f′(x)=2mx-=, 令k(x)=2mx2+2mx-1,k(x)=2m(x+)2--1,k(x)max=k(-)=--1, ∵-2<m<0,∴k(x)max=--1<0,k(x)=2mx2+2mx-1<0在(-1,+∞)上恒成立, 即f′(x)<0,当-2<m<0时,函数f(x)在定义域(-1,+∞)上单调递减. (3)当m=1时,f(x)=x2-ln(x+1).,令h(x)=x3-f(x)=x3-x2+ln(x+1), 则h′(x)=在[0,+∞)上恒正, ∴h(x)在[0,+∞)上单调递增,当x∈(0,+∞)时,恒有h(x)>h(0)=0., 即当x∈(0,+∞)时,有x3-x2+ln(x+1)>0,ln(x+1)>x2-x3, 对任意正整数n,取x=得ln(+1)>-. |