有二阶导数，则必有原函数连续且可导，一阶导数连续且可导。f'(x)=2ax+b,f''(x)=2a.f(0)=g(0),故(x->0+)limf(x)=c=f(0)=g(0),f'(x->0+)=(x->0+)lim(2ax+b)=b=f'(x->0-)=g'(0),f''(x->0+)=2a=g''(0).

　　a=g''(0)/2,b=g'(0),c=g(0).