求函数f(x)=x^3+ax^2+bx+c,过曲线y=f(x)上的点P(1,f(1))的切线方程为为y=3x+1若y=f(x)在x=-2时有极值

⑴ f′＝3x²+2ax+b f′?1?＝3+2a+b＝3① 点P(1,f(1))的切线为y=3x+1 f?1?＝4＝1+a+b+c② y=f(x)在x=-2时有极值 12-2a+b＝0③ ①+③ b＝-6 ①a＝3 ②c＝6 ∴f?x?＝x³+3x²-6x+6 ⑵ x²+2x-2＝0 x＝-1±√3 单调区间: [-3,-1-√3],?, [-1-√3,-1+√3]? [