已知函数f（x）=ax^3+bx^2+cx

(1) y=g(x)=ax^2+bx+(k+1)lnx+c则,
g'(x)=2ax+b+(k+1)/x=[2ax^2+bx+(k+1)]/x
令g'(x)>=0,（递增区间）
1,当a=0,b>=0k+1=0时,解得x属于R,当a=0,b=0,k+10时 解得 x>=0或x