[安乐椅#12] 函数三阶导与极值点偏移

性质： 若 \(f(x)\) 为单峰函数，对于 \(\forall x \in I,f^{\prime\prime\prime}(x)>0\)，且 \(f(x)\) 存在两变号零点 \(x_1,x_2\)，则 \(f^\prime (\frac{x_1+x_2}{2})<0\)

证明： 记 \(m=x_1+x_2\)，不妨设 \(x_1<\frac{m}{2}<x_2\)

令 \(g(x)=f(x)-f(m-x)\)

\(\therefore g^{\prime}(x)=f^{\prime}(x)+f^{\prime}(m-x)\)

\(\space\space\space\space g^{\prime\prime}(x)=f^{\prime}(x)-f^{\prime\prime}(m-x)\)

\(\space\space\space\space g^{\prime\prime\prime}(x)=f^{\prime}(x)+f^{\prime\prime\prime}(m-x)>0\)

\(\therefore g^{\prime\prime}(x) \uparrow\)

又 \(\because g^{\prime\prime}(\frac{m}{2})=0\)

\(\therefore\)

| \(\space\space\space x:\space\space\space\) | \(\space(-\infty,\frac{m}{2})\) | \(\space\space\space\space\space\space\frac{m}{2}\space\space\space\space\space\space\) | \((\frac{m}{2},+\infty)\) |
| \(g^{\prime\prime}(x):\) | \(\space\space\space\space\space\space-\space\space\space\space\space\space\) | \(\space\space\space\space\space\space\space0\space\space\space\space\space\space\space\) | \(\space\space\space\space\space\space+\space\space\space\space\space\space\) |
| \(g^{\prime}(x):\,\) | \(\,\space\space\space\space\space\space\downarrow\space\space\space\space\space\space\) | \(\space\space2f^\prime(\frac{m}{2})\space\space\) | \(\space\space\space\space\space\space\uparrow\space\space\space\space\space\space\) |

\(\therefore g^\prime(x) \ge g^\prime(\frac{m}{2})=2f^\prime(\frac{m}{2})\)

假设 \(f^\prime(\frac{m}{2})\ge 0\)

\(\therefore g^\prime(x) \ge 0\)

\(\therefore g(x) \uparrow\)

又 \(\because g(\frac{m}{2})=0,x_1<\frac{m}{2}\)

\(\therefore g(x_1)<0\)

即 \(f(x_1)-f(m-x_1)<0\)

又 \(\because m-x_1=x_2\)

\(\therefore f(x_1)<f(x_2)\)，矛盾

故假设不成立

\(\therefore f^\prime(\frac{m}{2})<0\)

即 \(f^\prime (\frac{x_1+x_2}{2})<0\)

\(\Box\)