性质： 若 \(\forall x \in I,f^{\prime\prime\prime}(x)>0\)，且 \(f(x)\) 存在两零点 \(x_1,x_2\)，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 \(\because g^{\prime\prime}(\frac{m}{2})=0\)
\(\therefore\)
| \(x\) | \((-\infty,\frac{m}{2})\) | \(\frac{m}{2}\) | \((\frac{m}{2},+\infty)\) |
| \(g^{\prime\prime}(x)\) | \(-\) | \(0\) | \(+\) |
| \(g^{\prime}(x)\) | \(\downarrow\) | \(2f^\prime(\frac{m}{2})\) | \(\uparrow\) |