题目

( 2024年深圳一模第$14$题 )

已知函数 $f(x)=a(x-x_1)(x-x_2)(x-x_3)(a > 0)$, 设曲线 $y=f(x)$ 在点 $(x_i, f(x_i))$ 处切线的斜率为 $k_i(i=1,2,3)$, 若 $x_1, x_2, x_3$ 均不相等, 且 $k_2=-2$, 则 $k_1+4 k_3$ 的最小值为$\underline{\hspace{2cm}} $.

解析

先证明一个引理

若一元三次函数有三个零点$x_1,x_2,x_3$，则$$\dfrac{1}{f’(x_1)}+\dfrac{1}{f’(x_2)}+\dfrac{1}{f’(x_3)}=0.$$

证明:不妨设$f(x)=(x-x_1)(x-x_2)(x-x_3)$, 则
$$f’(x)=(x-x_2)(x-x_3)+(x-x_1)(x-x_3)+(x-x_1)(x-x_2)$$则
$$\begin{aligned}
&\dfrac{1}{f’(x_1)}+\dfrac{1}{f’(x_2)}+\dfrac{1}{f’(x_3)}\\
=&\dfrac{1}{(x_1-x_2)(x_1-x_3)}+\dfrac{1}{(x_2-x_1)(x_2-x_3)}+\dfrac{1}{(x_3-x_1)(x_3-x_2)}\\
=&\dfrac{(x_3-x_2)+(x_1-x_3)+(x_2-x_1)}{(x_1-x_2)(x_2-x_3)(x_3-x_1)}\\
=&0.
\end{aligned}$$

由引理$$\begin{aligned}
\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}=0.
\end{aligned}$$
由$k_2=-2$得$$\begin{aligned}
\dfrac{1}{k_1}+\dfrac{1}{k_3}=\dfrac12.
\end{aligned}$$
由$a > 0$可得$x_1,x_3$位于$f(x)$单调递增区间，所以$k_1,k_3\in(0,+\infty)$,所以
$$\begin{aligned}
\dfrac12&=\dfrac{1}{k_1}+\dfrac{1}{k_3}\\
&=\dfrac{1^2}{k_1}+\dfrac{2^2}{4k_3}\\
&\geqslant\dfrac{(1+2)^2}{k_1+4k_3}=\dfrac{9}{k_1+4k_3}\\
\end{aligned}$$
所以$$k_1+4k_3\leqslant 18.$$
当且仅当$k_1=6,k_3=3$取等号，此时可取$f(x)=(x-1) (x-3) (x-4)$.
综上，$k_1+4 k_3$ 的最小值为18.