(x0,y0), (x1,y1)是数据集上相异的两点,且x0<x1。y=fx在xi上的函数值和一阶导数值分别为yi=fxi和mi=f'xi (i=0,1),求三次多项式H3x,使其满足H3xi=yiH3'xi=mi  （i=0,1）。
构造三次埃尔米特插值多项式
H3x=y0a0x+y1a1x+m0β0x+m1β1x

条件 函数	函数值	导数值
	x0	x1	x0	x1
a0x	1	0	0	0
a1x	0	1	0	0
β0x	0	0	1	0
β1x	0－	0	0	1

由  a0x1=a0'x1=0
可写成 a0x=[a+b(x-x0)](x-x1)2
由a0x0=1 得 a=1(x0-x1)2
再由a0'(x0) = 0得b = - 2(x0-x1)3，所以

a0x = [1+2x-x0x1-x0]( x-x1x0-x1)2
同理（将x0<->x1）

a1x = [1+2x-x0x1-x0]( x-x1x0-x1)2
同样由β0(x0) = β0(x1)= β0'(x1) = 0，可令

β0(x) = c(x-x0)( x-x1)2
再由β0'(x0) = 1，得c = 1(x0-x1)2

β0(x) = (x-x0) ( x-x1x0-x1)2 ，β1(x) = (x-x1) ( x-x0x1-x0)2
a0(x) = [1+2x-x0x1-x0] ( x-x1x0-x1)2 ，β0(x) = (x-x0) ( x-x1x0-x1)2
a1(x) = [1+2x-x1x0-x1] ( x-x0x1-x0)2 ，β1(x) = (x-x1) ( x-x0x1-x0)2
即

a0(x) = [1+2l1(x)]l02(x)
β0(x) = [x-x0]l02(x)

a1(x) = [1+2l0(x)]l12(x)
β1(x) = [x-x1]l12(x)
l02(x) ，l12(x) 为以(x0，y0)，(x1，y1) 插值点得Lagrange一次基函数。

可得满足条件得三次埃尔米特插值多项式为
H3x = y0a0(x) + y1a1(x) + m0β0x+m1β1x

 = y0[1+2x-x0x1-x0] ( x-x1x0-x1)2 + y1[1+2x-x1x0-x1] ( x-x0x1-x0)2 + m0(x-x0) ( x-x1x0-x1)2 + m1(x-x1) ( x-x0x1-x0)2