函数f(x)=x^3+(a+1)*x^2+4/3+1, a为R 讨论函数单调区间

f(x)=x^3+(a+1)*x^2+4/3+1
求导
f(x)'=3x^2+2(a+1)x+4/3
△=4(a+1）^2-16=4a^2+8a-12=4(a+3)(a-1)
⑴ △<=0是，二次函数3x^2+2(a+1)x+4/3>=0,此时f(x)'>=0
所以
当-3<=a<=1时，f(x)=x^3+(a+1)*x^2+4/3+1在R上单调递增

⑵△>0时，即a<-3或a>1,f(x)在（-∞，（-a-1-√(a+3)(a-1))/3】或【(-a-1+√(a+3)(a-1))/3,∞)上单调递增，在（（-a-1-√(a+3)(a-1))/3，（-a-1+√(a+3)(a-1))/3）上单调递减