14.求次数小于等于3的多项式P(x),使其满足条件 P(0)=0 ,P'(0)=1 ,P(1)=1 ,-|||-.P'(1)=2 。

步骤 1：确定插值条件
题目要求求解一个次数小于等于3的多项式P(x)，使其满足以下条件：
- P(0) = 0
- P'(0) = 1
- P(1) = 1
- P'(1) = 2

步骤 2：构造埃尔米特插值多项式
根据埃尔米特插值公式，我们构造一个满足上述条件的多项式。记 ${x}_{0}=0$ ,${x}_{1}=1$ ,由题设知 $f({x}_{0})=0$ ,$f({x}_{1})=1$ ,$f'({x}_{0})=1$ ,$f'({x}_{1})=2$ ,利用两点的埃尔米特插值公式,有
$P(x)={\alpha }_{0}(x)+{\alpha }_{1}(x)+{\beta }_{0}(x)+{\beta }_{1}(x)$
其中α0(x),α1 (x),β0(x),β1(x)是埃尔米特插值基函数,即
${\alpha }_{0}(x)=(x-{x}_{1}){(\dfrac {x-{x}_{0}}{{x}_{1}-{x}_{0}})}^{2}=(x-1){x}^{2}$
${\alpha }_{1}(x)=(x-{x}_{0}){(\dfrac {x-{x}_{1}}{{x}_{0}-{x}_{1}})}^{2}=x{(x-1)}^{2}$
${\beta }_{0}(x)=(x-{x}_{1}){(\dfrac {x-{x}_{0}}{{x}_{1}-{x}_{0}})}^{2}(2\dfrac {x-{x}_{0}}{{x}_{1}-{x}_{0}}-1)=(x-1){x}^{2}(2x-1)$
${\beta }_{1}(x)=(x-{x}_{0}){(\dfrac {x-{x}_{1}}{{x}_{0}-{x}_{1}})}^{2}(2\dfrac {x-{x}_{1}}{{x}_{0}-{x}_{1}}-1)=x{(x-1)}^{2}(2x-1)$

步骤 3：计算多项式P(x)
将上述基函数代入埃尔米特插值公式，得到
$P(x)={\alpha }_{0}(x)+{\alpha }_{1}(x)+{\beta }_{0}(x)+{\beta }_{1}(x)$
$=(x-1){x}^{2}+x{(x-1)}^{2}+(x-1){x}^{2}(2x-1)+x{(x-1)}^{2}(2x-1)$
$=x^{3}-x^{2}+x^{3}-2x^{2}+x+x^{3}-x^{2}+2x^{3}-2x^{2}$
$=4x^{3}-4x^{2}+x$