1、第3章 導(dǎo)數(shù)及其應(yīng)用構(gòu)造法解f (x)與f (x)共存問題以抽象函數(shù)為背景，題設(shè)條件或所求結(jié)論中具有f (x)與f (x)共存的不等式，旨在考查導(dǎo)數(shù)運算法則的逆向、變形應(yīng)用能力的客觀題，是近幾年高考中的一個熱點解答這類問題的策略是將f (x)與f (x)共存的不等式與導(dǎo)數(shù)運算法則結(jié)合起來，合理構(gòu)造出相關(guān)的可導(dǎo)函數(shù)，然后利用函數(shù)的性質(zhì)解決問題構(gòu)造yf (x)g(x)型可導(dǎo)函數(shù)定義在(0，)上的函數(shù)f (x)滿足x2f (x)1，f (2)，則關(guān)于x的不等式f (ex)0，即函數(shù)f(x)在(0，)上單調(diào)遞增所求不等式可化為f(ex)f (ex)3，而f(2)f (2)3，所以ex2，解得x，則不等

2、式f (x)，可得f (x)0，即函數(shù)f(x)f (x)x在r上是增函數(shù)又由f (1)1可得f(1)，故f (x)x，整理得f (x)x，即f(x)f(1)由函數(shù)的單調(diào)性可得不等式的解集為(，1)構(gòu)造f (x)g(x)型可導(dǎo)函數(shù)設(shè)函數(shù)f (x)，g(x)分別是定義在r上的奇函數(shù)和偶函數(shù)當x0，且g(3)0，則不等式f (x)g(x)0的解集是()a(3，0)(3，) b(3，0)(0，3)c(，3)(3，) d(，3)(0，3)a解析：構(gòu)造函數(shù)f(x)f (x)g(x)由題意可知，當x0，所以f(x)在(，0)上單調(diào)遞增又因為f (x)，g(x)分別是定義在r上的奇函數(shù)和偶函數(shù)，所以f(x)是

3、定義在r上的奇函數(shù)，從而f(x)在(0，)上單調(diào)遞增而f(3)f (3)g(3)0，所以f(3)f(3)0，結(jié)合圖象(圖略)可知不等式f (x)g(x)0f(x)0的解集為(3,0)(3，)故選a設(shè)定義在r上的函數(shù)f (x)滿足f (x)f (x)3x2ex，且f (0)0，則()af (x)在r上單調(diào)遞減bf (x)在r上單調(diào)遞增cf (x)在r上有最大值df (x)在r上有最小值c解析：構(gòu)造函數(shù)f(x)exf (x)，則有f(x)exf (x)f (x)ex3x2ex3x2，故f(x)x3c(c為常數(shù))，所以f (x).又f (0)0，所以c0，f (x).f (x)，易知f (x)在區(qū)間

4、(，3上單調(diào)遞增，在3，)上單調(diào)遞減，f (x)maxf (3)，無最小值故選c構(gòu)造型可導(dǎo)函數(shù)已知定義域為r的函數(shù)f (x)的圖象是連續(xù)不斷的曲線，且f (2x)f (x)e22x.當x1時，f (x)f (x)，則()af (1)ef (0)bf (3)e4f (1)cf (2)e5f (2)c解析：構(gòu)造函數(shù)g(x)，因為當x1時，f (x)f (x)，所以g(x)0，可得當x1時，g(x)單調(diào)遞增因為f (2x)f (x)e22x，整理得，即g(2x)g(x)，可得函數(shù)圖象關(guān)于x1對稱，則g(1)g(3)，所以，g(2).因為g(2)g(3)g(1)，所以，化簡可得f (2)e3f (1)故選c(2020長沙明德中學(xué)月考)已知f (x)是函數(shù)f (x)的導(dǎo)函數(shù)，且對任意的實數(shù)x都有f (x)ex(2x1)f (x)，f (0)2，則不等式f (x)4ex的解集為()a(2,3) b(3,2)c(，3)(2，) d(，2)(3，)b解析：令g(x)，則g(x)2x1，所以可設(shè)g(x)x2xc.因為