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\((\dfrac{1}{e}, + \infty )\)

原创文章,作者:leopold,如若转载,请注明出处:https://www.math211.com/2021/02/10/76/

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