导函参变可分离，特殊一般来讨论

导函参变可分离，特殊一般来讨论。

解超不等造函数，单调零点把题解。

【例题】已知函数\(f(x) = \ln x + a{x^2} – (a + 2)x,a \in R\)．若\(x>1\)时，\(f(x)> – e\)恒成立，求实数\(a\)的取值范围．

【解析】

\(\begin{array}{c}
f'(x) = \dfrac{1}{x} + 2ax – (a + 2) = \dfrac{{2a{x^2} – (a + 2)x + 1}}{x} = \dfrac{{(2x – 1)(ax – 1)}}{x}\\
= (2x – 1)(a – \dfrac{1}{x})(x>1)
\end{array}\)

注意到，当\(x>1\)时，\(2x – 1>0\)，\(0<\dfrac{1}{x}<1\)

（1）（特殊情况）当\(a \le 0\)时，\(a – \dfrac{1}{x}<0\)，从而\(f'(x)<0\)，因此\(f(x)\)在\((1, + \infty )\)上单调递减

粗略考虑当\(x \to + \infty \)时的极限，根据增长速度：指数型快于幂函数型，幂函数型快于对数型，二次型快于一次型。当\(x \to + \infty \)时，\(f(x) \to – \infty \)．

仔细考虑，得找到一个值比\( – e\)小．

具体思路：首先把对数弄掉，可用放缩

\(f(x) = \ln x + a{x^2} – (a + 2)x<x – 1 + a{x^2} – (a + 2)x = a{x^2} – ax – x – 1<a{x^2} – ax = ax(x – 1)\)

只需\(ax(x – 1)< – e\)，即\(x(x – 1)> – \dfrac{e}{a}\)，只需\({(x – 1)^2} \ge – \dfrac{e}{a}\)

即\(x – 1 \ge \sqrt { – \dfrac{e}{a}} \)，即\(x \ge 1 + \sqrt { – \dfrac{e}{a}} \)

①当\(a = 0\)时，\(f(x) = \ln x – x\)，\(f({e^2}) = 2 – {e^2}< – e\)，不满足恒成立；

②当\(a<0\)时，取\({x_0} = 1 + \sqrt { – \dfrac{e}{a}} \)，则\(f({x_0}) = \ln {x_0} + ax_0^2 – (a + 2){x_0}<{x_0} – 1 + ax_0^2 – (a + 2){x_0}\)

\( = ax_0^2 – a{x_0} – {x_0} – 1<ax_0^2 – a{x_0} = a{x_0}({x_0} – 1) = a(1 + \sqrt { – \dfrac{e}{a}} )(1 + \sqrt { – \dfrac{e}{a}} – 1)\)

\(<a\sqrt { – \dfrac{e}{a}} \cdot \sqrt { – \dfrac{e}{a}} = – e\)，不满足恒成立

综上所述，当\(a \le 0\)，不满足恒成立，应舍去．

（2）（特殊情况）当\(a \ge 1\)时，\(a – \dfrac{1}{x}>0\)，从而\(f'(x)>0\)，即\(f(x)\)在\((0, + \infty )\)上单调递增．

从而当\(x>1\)时，利用单调性有\(f(x)>f(1) = -2> – e\)，满足恒成立

（3）（一般情况）当\(0<a<1\)时，\(\dfrac{1}{a}>1\)

当\(1<x<\dfrac{1}{a}\)时，\(f'(x)<0\)，\(f(x)\)单调递减；

当\(x>\dfrac{1}{a}\)时，\(f'(x)>0\)，\(f(x)\)单调递增．

\(\therefore f{(x)_{\min }} = f(\dfrac{1}{a}) = \ln \dfrac{1}{a} + \dfrac{1}{a} – (a + 2)\dfrac{1}{a} = \ln \dfrac{1}{a} – \dfrac{1}{a} – 1> – e\)

构造函数\(\varphi (x) = \ln x – x – 1\)（\(x>1\)），\({\varphi}'(x) = \dfrac{1}{x} – 1<0\)

故\(\varphi (x)\)在\((1, + \infty )\)上单调递减，注意到\(\varphi (e) = – e\)，\(\dfrac{1}{a}>1\)

故\(\varphi (\dfrac{1}{a})>\varphi (e) \Leftrightarrow \dfrac{1}{a}<e\)，从而\(\dfrac{1}{e}<a<1\)

综上所述，\(a\)的取值范围是\((\dfrac{1}{e}, + \infty )\)