题目内容：

已知函数f(x)=x^3+ax^2+bx+c.曲线y=f(x)在x=1处切线为l：3x-y+1=0.求(1)若x=2/3时.函数f(x)有极值.

优质解答

F’(x) = 3x^2 + 2ax + b
F’(1) = 3+2a+b = 3 (切线斜率=3)
F(1) = 1+a+b+c = 4(切点（1,4）)
f”(x) = 6x+2a
f”(2/3) = 4 + 2a = 0,a = -2
b = 4
c = 1