证明三次多项式f(x)=ax^3+bx^2+cx+d(a不等于0)有且仅有一个拐点(x0,f(x0)),且若f(x1)=f(x2)=f(x3)=0,则x0=(x1+x2+x3)/3.

证明：f''(x) = 6ax+2b因为, (x0,f(x0))是f(x)的拐点所以, f''(x0) = 0, 即6ax0+2b=0所以x0 = b/(-3a) .(1)由f(x1)=f(x2)=f(x3)=0知x1,x2,x3为f(x)=0的三个根由韦达定理（一元三次）可得x1+x2+x3 = -b/a .(2)(1)(2)...一元三次的伟达定理要怎么证明x1,x2,x3为f(x)=ax^3+bx^2+cx=0的三个根,则f(x)=a(x-x1)(x-x2)(x-x3)=a(x^2-x1x-x2x+x1x2)(x-x3)上式展开得二次项系数为-a(x1+x2+x3)所以b=-a(x1+x2+x3)即,x1+x2+x3=-b/a