2017模拟考试题(4)(韩)
证明:
设
$\displaystyle g(x)=\frac{1}{12}(x-a)^2(x-b)^2,$
则
$\displaystyle g^{(n)}(x)=2!.$
作辅助函数
$\displaystyle F(t)=f(t)g(x)-f(x)g(t),$
先将$x$看作常量
$\displaystyle \because F(a)=F(b)=0,$
由Rolle定理
$\displaystyle \therefore \exists \eta_1\in(a,b),s.t.$
$\displaystyle F'(\eta_1)=f'(\eta_1)g(x)-f(x)g'(\eta_1)=0,$
又
$\displaystyle \because F'(a)=F'(\eta_1)=F'(b)=0,$
再次由Rolle定理
$\displaystyle \therefore \exists \eta_2\in(a,\eta_1),\eta_3\in(\eta_1,b),s.t.$
$\displaystyle F''(\eta_2)=f''(\eta_2)g(x)-f(x)g''(\eta_2)=0,$
$\displaystyle F''(\eta_3)=f''(\eta_3)g(x)-f(x)g''(\eta_3)=0,$
再由Rolle定理
$\displaystyle \therefore \exists \eta_4\in(\eta_2,\eta_3),s.t.$
$\displaystyle F'''(\eta_4)=f'''(\eta_4)g(x)-f(x)g'''(\eta_4)=0,$
由于$\eta_i$的任意性,故有
$\displaystyle \exists \xi=\eta_4=t\in(a,b),$
$\displaystyle \Rightarrow F^{(4)}(\xi)=f^{(4)}(\xi)g(x)-f(x)g^{(4)}(\xi)=0.$
$\displaystyle \therefore f(x)=\frac{f^{(4)}(\xi)}{2!}(x-a)^2(x-b)^2.$