题目 设 $x_{1},x_{2}$ 是 $f(x)=\dfrac{a}{3}x^{3}+\dfrac{b-1}{2}x^{2}+x(a,b\in\mathbb R,a>0)$ 的两个极值点，$f'(x)$ 是 $f(x)$ 的导函数． 问题1 如果 $x_{1}<2<x_{2}<4$，求 $f'(-2)$ 的取值范围； 答案1 $(3,+\infty)$ 解析1 对 $f(x)$ 求导可得$$f'(x)=ax^{2}+(b-1)x+1,$$由题意 $x_{1},x_{2}$ 是方程 $f'(x)=0$ 的两根．<br/>由 $x_{1}<2<x_{2}<4$，且 $a>0$ 得$$\begin{cases}f'(2)<0,\\ f'(4)>0,\end{cases}$$即$$\begin{cases}4a+2b-1<0,\\ 16a+4b-3>0.\end{cases}$$由该不等式组所表示的平面区域可求得$$4a-2b>0,$$因为$$\begin{split}f'(2)&=4a-2(b-1)+1\\ &=4a-2b+3,\end{split}$$故\[f'(-2)=4a-2b+3>3,\]所以 $f'(-2)$ 的取值范围是 $(3,+\infty)$． 备注1 问题2 如果 $0<x_{1}<2$，$x_{2}-x_{1}=2$，求证：$b<\dfrac{1}{4}$． 答案2 略 解析2 方程 $ax^{2}+(b-1)x+1=0$ 的两根为 $x_{1},x_{2}$，由根与系数的关系得$$\begin{cases}x_{1}+x_{2}=-\dfrac{b-a}{a},\\ x_{1}x_{2}=\dfrac{1}{a},\end{cases}$$由于 $x_{1}x_{2}\ne 0$，两式相除得\[-(b-1)=\dfrac{x_{1}+x_{2}}{x_{1}x_{2}}=\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}},\]即$$b=-\dfrac{1}{x_{1}}-\dfrac{1}{x_{2}}+1.$$由条件 $x_{2}=x_{1}+2$ 可得$$b=\varphi(x_{1})=-\dfrac{1}{x_{1}}-\dfrac{1}{x_{1}+2}+1,$$易知当 $x_{1}\in(0,2)$ 时，$$\varphi(x_{1})<\varphi(2)=\dfrac{1}{4},$$故 $b$ 的取值范围是 $\left(-\infty,\dfrac{1}{4}\right)$，得证． 备注2 问题3 如果 $a\geqslant 2$，且 $x_{2}-x_{1}=2$，$x\in (x_{1},x_{2})$ 时，函数 $g(x)=-f'(x)+2(x_{2}-x)$ 的最大值为 $h(a)$，求 $h(a)$ 的最小值． 答案3 略 解析3 因为 $f'(x)=0$ 的两根是 $x_{1},x_{2}$，故可设 $f'(x)=a(x-x_{1})(x-x_{2})$，所以\[\begin{split}g(x)&=-f'(x)+2(x_{2}-x_{1})\\ &=-a(x-x_{1})(x-x_{2})+2(x_{2}-x)\\ &=a(x_{2}-x)\left(x-x_{1}+\dfrac{2}{a}\right),\end{split}\]由于 $x\in(x_{1},x_{2})$，因此$$x_{2}-x>0 , x-x_{1}>0,$$又 $a\geqslant 2$，可知$$x-x_{1}+\dfrac{2}{a}>0,$$故\[\begin{split}g(x)&=a(x_{2}-x_{1})\left(x-x_{1}+\dfrac{2}{a}\right)\\&\leqslant a\left[\dfrac{(x_{2}-x)+\left(x-x_{1}+\dfrac{2}{a}\right)}{2}\right]^{2}\\&=a\left(1+\dfrac{1}{a}\right)^{2}\\&=a+\dfrac{1}{a}+2,\end{split}\]当且仅当 $x_{2}-x=x-x_{1}+\dfrac{2}{a}$，即 $x=x_{1}+1-\dfrac{1}{a}$ 时取等号．<br/>所以 $h(a)=a+\dfrac{1}{a}+2$，$a\in[2,+\infty)$．<br/>当 $a\in(2,+\infty)$ 时，$$h'(a)=1-\dfrac{1}{a^{2}}>0,$$所以 $h(a)$ 在 $(2,+\infty)$ 内是增函数，又因为 $h(a)$ 在 $[2,+\infty)$ 上连续，故 $h(a)$ 在 $[2,+\infty)$ 上是增函数，所以$$h(a)_{\min}=h(2)=\dfrac{9}{2}.$$ 备注3