文档内容
第 6 讲 导数的综合应用
[考情分析] 1.利用导数研究函数的单调性与极值(最值)是高考的常见题型,而导数与函数、
不等式、方程、数列等的交汇命题是高考的热点和难点.2.多以解答题的形式压轴出现,难度
较大.
母题突破 1 导数与不等式的证明
母题 已知函数f(x)=ex-x2.
(1)求曲线f(x)在x=1处的切线方程;
(2)求证:当x>0时,≥ln x+1.
思路分析
❶求切线方程
↓
❷fx≥e-2x+1
↓
❸ex-x2-e-2x-1≥0
↓
❹ex+2-ex-1≥x2
↓
❺≥x≥ln x+1
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[子题1] 已知函数f(x)=ex-ax-a,当a=1时,令g(x)=.求证:当x>0时,g(x)<1.
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[子题2] (2022·德州联考改编)已知函数f(x)=ln .若x∈(0,1),求证:f(x)g(x)(或 f(x)0(或 f(x)-
g(x)<0),进而构造辅助函数h(x)=f(x)-g(x).
(2)适当放缩构造法:一是根据已知条件适当放缩;二是利用常见放缩结论.
(3)构造“形似”函数,稍作变形再构造,对原不等式同结构变形,根据相似结构构造辅助函
数.
1.(2021·全国乙卷改编)设函数f(x)=ln(1-x),函数g(x)=.求证:g(x)<1.
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2.已知函数f(x)=ex-a-ln(x+a).当a≤1时,证明:f(x)>0.
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