2024年数学高考大一轮复习第三章 培优课 §3.8　隐零点与极值点偏移问题

§3.8　隐零点与极值点偏移问题

隐零点问题是指对函数的零点设而不求，通过一种整体代换和过渡，再结合题目条件最终解决问题；极值点偏移是指函数在极值点左右的增减速度不一样，导致函数图象不具有对称性，隐零点与极值点偏移问题常常出现在高考数学的压轴题中，这类题往往对思维要求较高，过程较为烦琐，计算量较大，难度大．

题型一　隐零点

例1 (2023·郑州模拟)已知函数f(x)＝ex＋1－＋1，g(x)＝＋2.

(1)求函数g(x)的极值；

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(2)当x>0时，证明：f(x)≥g(x)．

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思维升华 零点问题求解三步曲

(1)用函数零点存在定理判定导函数零点的存在性，列出零点方程f′(x0)＝0，并结合f′(x)的单调性得到零点的取值范围．

(2)以零点为分界点，说明导函数f′(x)的正负，进而得到f(x)的最值表达式．

(3)将零点方程适当变形，整体代入最值式子进行化简证明，有时(1)中的零点范围还可以适当缩小．

跟踪训练1 (2023·潍坊模拟)设函数f(x)＝x－aln x－2.

(1)求f(x)的单调区间；

(2)若a＝1，f′(x)为f(x)的导函数，当x>1时，ln x＋1>(1＋k)f′(x)，求整数k的最大值．

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题型二　极值点偏移

例2 已知函数f(x)＝xe－x.

(1)求函数f(x)的单调区间和极值；

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(2)若x1≠x2且f(x1)＝f(x2)，求证：x1＋x2>2.

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思维升华 极值点偏移问题的解法

(1)(对称化构造法)构造辅助函数：对结论x1＋x2>(<)2x0型，构造函数F(x)＝f(x)－f(2x0－x)；对结论x1x2>(<)x型，构造函数F(x)＝f(x)－f ，通过研究F(x)的单调性获得不等式．

(2)(比值代换法)通过代数变形将所证的双变量不等式通过代换t＝化为单变量的函数不等式，利用函数单调性证明．

跟踪训练2 已知函数f(x)＝ln x－ax＋b(a，b∈R)有两个不同的零点x1，x2.

(1)求f(x)的最值；

(2)证明：x1x2<.

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