在某一类单中值等式的证明中，借助待定常数法构造函数，并配合\(\text{Rolle}\)中值定理和\(\text{Lagrange}\)中值定理，可以很方便地证明出一些看似复杂的单中值等式证明题，下以几个简单的例子来说明这种方法的操作流程。
问题1：设\(\displaystyle f\left( x \right)\)在\(\displaystyle \left[ a,b \right]\)上连续，在\(\displaystyle \left( a,b \right)\)内有二阶导数，试证存在\(\displaystyle c\in \left( a,b \right)\)使得
\[f\left( b \right) -2f\left( \frac{a+b}{2} \right) +f\left( a \right) =\frac{\left( b-a \right) ^2}{4}f‘‘\left( c \right)
\]
过程如下：设常数\(\displaystyle k\)可使
\[f\left( b \right) -2f\left( \frac{a+b}{2} \right) +f\left( a \right) =\frac{\left( b-a \right) ^2}{4}k
\]
成立，则只需证明\(\displaystyle k=f‘‘\left( c \right)\)即可.
构造函数
\[F\left( x \right) =f\left( x \right) -2f\left( \frac{a+x}{2} \right) +f\left( a \right) -\frac{k\left( x-a \right) ^2}{4}
\]
由\(\displaystyle F\left( a \right) =F\left( b \right)\)及\(\text{Rolle}\)中值定理知存在\(\displaystyle \xi \in \left( a,b \right)\)，可使\(\displaystyle F‘\left( \xi \right) =0\)，即
\[F‘\left( \xi \right) =f‘\left( \xi \right) -f‘\left( \frac{a+\xi}{2} \right) -\frac{k\left( \xi -a \right)}{2}=0
\]
借助\(\text{Lagrange}\)中值定理，可知存在\(\displaystyle c\in \left( \frac{a+\xi}{2},\xi \right) \subset \left( a,b \right)\)，使得
\[\begin{align*}
F‘\left( \xi \right) &=f‘\left( \xi \right) -f‘\left( \frac{a+\xi}{2} \right) -\frac{k\left( \xi -a \right)}{2}
\&=\frac{\xi -a}{2}\left[ f‘‘\left( c \right) -k \right]
\&=0
\end{align*}
\]
即\(\displaystyle k=f‘‘\left( c \right)\)，故原命题成立。
问题2：设\(\displaystyle f\left( x \right)\)在\(\displaystyle \left[ a,b \right]\)上有三阶导数，试证：必存在\(\displaystyle \xi \in \left( a,b \right)\)，使得
\[f\left( b \right) =f\left( a \right) +\frac{1}{2}\left( b-a \right) \left[ f‘\left( a \right) +f‘\left( b \right) \right] -\frac{1}{12}\left( b-a \right) ^3f‘‘‘\left( \xi \right)
\]
过程如下：设常数\(\displaystyle k\)可使
\[f\left( b \right) =f\left( a \right) +\frac{1}{2}\left( b-a \right) \left[ f‘\left( a \right) +f‘\left( b \right) \right] -\frac{k}{12}\left( b-a \right) ^3
\]
成立，则只需证明\(\displaystyle k=f‘‘‘\left( \xi \right)\)即可.
构造函数
\[F\left( x \right) =f\left( x \right) -f\left( a \right) -\frac{1}{2}\left( x-a \right) \left[ f‘\left( a \right) +f‘\left( x \right) \right] +\frac{k}{12}\left( x-a \right) ^3
\]
由\(\displaystyle F\left( a \right) =F\left( b \right) =0\)及\(\text{Rolle}\)中值定理知，存在\(\displaystyle\xi _1\in \left( a,b \right)\)，使得\(\displaystyle F‘\left( \xi _1 \right) =0\)
又
\[F‘\left( x \right) =\frac{1}{2}\left[ f‘\left( x \right) -f‘\left( a \right) \right] -\frac{1}{2}\left( x-a \right) f‘‘\left( x \right) +\frac{k}{4}\left( x-a \right) ^2
\]
由\(\displaystyle F‘\left( a \right) =F‘\left( \xi _1 \right) =0\)及\(\text{Rolle}\)中值定理知，存在\(\displaystyle\xi \in \left( a,\xi _1 \right) \subset \left( a,b \right)\)，使得\(\displaystyle F‘‘\left( \xi \right) =0\)
又
\[\begin{align*}
F‘‘\left( x \right) &=-\frac{1}{2}\left( x-a \right) f‘‘‘\left( x \right) +\frac{k}{2}\left( x-a \right)
\&=\frac{x-a}{2}\left[ k-f‘‘‘\left( x \right) \right]
\end{align*}
\]
则有\(\displaystyle k=f‘‘‘\left( \xi \right)\)，故原命题成立。
问题3：设\(\displaystyle f\left( x \right)\)在包含\(\displaystyle x_0\)的区间\(\displaystyle I\)上二次可微，\(\displaystyle x_0+h\in I\)，\(\displaystyle \lambda \in \left( 0,1 \right)\)，试证：\(\displaystyle \exists \theta \in \left( 0,1 \right)\)，使得
\[f\left( x_0+\lambda h \right) =\lambda f\left( x_0+h \right) +\left( 1-\lambda \right) f\left( x_0 \right) +\frac{\lambda}{2}\left( \lambda -1 \right) h^2f‘‘\left( x_0+\theta h \right)
\]
过程如下：设常数\(\displaystyle k\)可使
\[f\left( x_0+\lambda h \right) =\lambda f\left( x_0+h \right) +\left( 1-\lambda \right) f\left( x_0 \right) +\frac{k\lambda}{2}\left( \lambda -1 \right) h^2
\]
成立，则只要证明\(\displaystyle k=f‘‘\left( x_0+\theta h \right)\)即可.
构造函数
\[F\left( x \right) =f\left( x_0+hx \right) -xf\left( x_0+h \right) -\left( 1-x \right) f\left( x_0 \right) -\frac{kx\left( x-1 \right)}{2}h^2
\]
注意到\(\displaystyle F\left( 0 \right) =F\left( \lambda \right) =F\left( 1 \right) =0\)，函数\(\displaystyle F\left( x \right)\)在区间\(\displaystyle\left[ 0,1 \right]\)上满足\(\text{Rolle}\)中值定理条件
由\(\text{Rolle}\)中值定理知，存在\(\displaystyle \xi _1\in \left( 0,\lambda \right)\)，\(\displaystyle\xi _2\in \left( \lambda ,1 \right)\)，使得
\[F‘\left( \xi _1 \right) =F‘\left( \xi _2 \right) =0
\]
又
\[\begin{equation*}
F‘\left( x \right) =hf‘\left( x_0+hx \right) -f\left( x_0+h \right) +f\left( x_0 \right) -\frac{k\left( 2x-1 \right)}{2}h^2
\F‘‘\left( x \right) =h^2\left[ f‘‘\left( x_0+hx \right) -k \right]
\end{equation*}
\]
函数\(\displaystyle F‘\left( x \right)\)在\(\displaystyle\left[ 0,1 \right]\)上满足\(\text{Rolle}\)中值定理条件
由\(\text{Rolle}\)中值定理知，存在\(\displaystyle\theta \in \left( \xi _1,\xi _2 \right) \subset \left( 0,1 \right)\)，使得\(\displaystyle F‘‘\left( \theta \right) =0\)，即\(\displaystyle k=f‘‘\left( x_0+\theta h \right)\)，故原命题成立。
问题4：设\(\displaystyle f\left( x \right)\)在\(\displaystyle \left[ a,b \right]\)上5次可微，试证：存在\(\displaystyle\xi \in \left( a,b \right)\)，使得
\[f\left( b \right) =f\left( a \right) +\frac{1}{6}\left( b-a \right) \left[ f‘\left( a \right) +f‘\left( b \right) +4f‘\left( \frac{a+b}{2} \right) \right] -\frac{1}{2880}\left( b-a \right) ^5f^{\left( 5 \right)}\left( \xi \right)
\]
过程如下：令\(\displaystyle c>0,h>0\)，做变换\(\displaystyle a=c-h,b=c+h\)，则原命题等价于证明，存在\(\displaystyle\xi \in \left( a,b \right)\)，使得
\[f\left( c+h \right) =f\left( c-h \right) +\frac{h}{3}\left[ f‘\left( c-h \right) +f‘\left( c+h \right) +4f‘\left( c \right) \right] -\frac{h^5}{90}f^{\left( 5 \right)}\left( \xi \right)
\]
成立
设存在常数\(\displaystyle k\)，可使
\[f\left( c+h \right) =f\left( c-h \right) +\frac{h}{3}\left[ f‘\left( c-h \right) +f‘\left( c+h \right) +4f‘\left( c \right) \right] -\frac{kh^5}{90}
\]
成立，则只需证明\(\displaystyle k=f^{\left( 5 \right)}\left( \xi \right)\)即可.
构造函数
\[F\left( x \right) =f\left( c+x \right) -f\left( c-x \right) -\frac{x}{3}\left[ f‘\left( c-x \right) +f‘\left( c+x \right) +4f‘\left( c \right) \right] +\frac{kx^5}{90}
\]
由\(\displaystyle F\left( 0 \right) =F\left( h \right) =0\)及\(\text{Rolle}\)中值定理知，存在\(\displaystyle \xi _1\in \left( 0,h \right)\)，使得\(\displaystyle F‘\left( \xi _1 \right) =0\)
又
\[F‘\left( x \right) =\frac{2}{3}f‘\left( c+x \right) +\frac{2}{3}f‘\left( c-x \right) -\frac{4}{3}f‘\left( c \right) -\frac{x}{3}\left[ f‘‘\left( c+x \right) -f‘‘\left( c-x \right) \right] +\frac{kx^4}{18}
\]
由\(\displaystyle F‘\left( 0 \right) =F‘\left( \xi _1 \right) =0\)及\(\text{Rolle}\)中值定理知，存在\(\displaystyle \xi _2\in \left( 0,\xi _1 \right)\)，使得\(\displaystyle F‘‘\left( \xi _2 \right) =0\)
又
\[F‘‘\left( x \right) =\frac{1}{3}\left[ f‘‘\left( c+x \right) -f‘‘\left( c-x \right) \right] -\frac{x}{3}\left[ f‘‘‘\left( c+x \right) +f‘‘‘\left( c-x \right) \right] +\frac{2kx^3}{9}
\]
由\(\displaystyle F‘‘\left( 0 \right) =F‘‘\left( \xi _2 \right) =0\)及\(\text{Rolle}\)中值定理知，存在\(\displaystyle \xi _3\in \left( 0,\xi _2 \right)\)，使得\(\displaystyle F‘‘‘\left( \xi _3 \right) =0\)
又
\[F‘‘‘\left( x \right) =-\frac{x}{3}\left[ f^{\left( 4 \right)}\left( c+x \right) -f^{\left( 4 \right)}\left( c-x \right) \right] +\frac{2kx^2}{3}
\]
则借助\(\text{Lagrange}\)中值定理知，存在\(\displaystyle \xi \in \left( c-\xi _3,c+\xi _3 \right) \subset \left( c-h,c+h \right)\)，使得
\[\begin{align*}
F‘‘‘\left( \xi _3 \right) &=\frac{\xi _3}{3}\left[ 2k\xi _3-\left( f^{\left( 4 \right)}\left( c+\xi _3 \right) -f^{\left( 4 \right)}\left( c-\xi _3 \right) \right) \right]
\&=\frac{2\xi _{3}^{2}}{3}\left[ k-f^{\left( 5 \right)}\left( \xi \right) \right]
\&=0
\end{align*}
\]
即得到\(\displaystyle k=f^{\left( 5 \right)}\left( \xi \right)\)，故原命题成立。
使用待定常数法证明一类单中值等式问题