问题及解答

设 $f(x)$ 在 $[0,1]$ 上三阶连续可导, 且满足 $f(0)=1$, $f(1)=2$, $f'(\frac{1}{2})=0$. 

证明: 存在 $\xi\in(0,1)$, 使得 $f'''(\xi)=24$.

$f(x)$ 在 $[0,1]$ 上三阶连续可导, 将其在 $x=\frac{1}{2}$ 处展开. 存在 $\eta\in(x,\frac{1}{2})$ 或 $(\frac{1}{2},x)$, 有

\[
f(x)=f(\frac{1}{2})+\frac{f'(\frac{1}{2})}{1!}(x-\frac{1}{2})+\frac{f''(\frac{1}{2})}{2!}(x-\frac{1}{2})^2+\frac{f'''(\eta)}{3!}(x-\frac{1}{2})^3
\]

于是对于 $x=0$ 和 $x=1$, 分别存在 $\xi_1\in(0,\frac{1}{2})$ 和 $\xi_2\in(\frac{1}{2},1)$ 使得

\[
\begin{aligned}
f(0)&=f(\frac{1}{2})+\frac{f'(\frac{1}{2})}{1!}(0-\frac{1}{2})+\frac{f''(\frac{1}{2})}{2!}(0-\frac{1}{2})^2+\frac{f'''(\xi_1)}{3!}(0-\frac{1}{2})^3,\\
f(1)&=f(\frac{1}{2})+\frac{f'(\frac{1}{2})}{1!}(1-\frac{1}{2})+\frac{f''(\frac{1}{2})}{2!}(1-\frac{1}{2})^2+\frac{f'''(\xi_2)}{3!}(1-\frac{1}{2})^3.\\
\end{aligned}
\]

注意到 $f(0)=1$, $f(1)=2$, 及 $f'(\frac{1}{2})=0$, 将上面两式化简, 得

\[
\begin{aligned}
1&=f(\frac{1}{2})+\frac{1}{8}\cdot f''(\frac{1}{2})-\frac{1}{48}\cdot f'''(\xi_1),\qquad(1)\\
2&=f(\frac{1}{2})+\frac{1}{8}\cdot f''(\frac{1}{2})+\frac{1}{48}\cdot f'''(\xi_2).\qquad(2)\\
\end{aligned}
\]

两式相减（(2)-(1)）, 得

\[
1=\frac{1}{48}\bigl(f'''(\xi_2)+f'''(\xi_1)\bigr),
\]

推出

\[
f'''(\xi_1)+f'''(\xi_2)=48.
\]

由题设, $f'''(x)\in C[0,1]$, 故由介值定理, 存在 $\xi\in(\xi_1,\xi_2)$, 使得

\[
f'''(\xi)=\frac{f'''(\xi_1)+f'''(\xi_2)}{2}=24.
\] 

Note. $\min\{f'''(\xi_1),f'''(\xi_2)\}\leqslant\dfrac{f'''(\xi_1)+f'''(\xi_2)}{2}\leqslant\max\{f'''(\xi_1),f'''(\xi_2)\}$.