數(shù)學(xué)一輪復(fù)習(xí)第三章導(dǎo)數(shù)及其應(yīng)用微專題二導(dǎo)數(shù)中的函數(shù)構(gòu)造問(wèn)題課件

微專題二導(dǎo)數(shù)中的函數(shù)構(gòu)造問(wèn)題函數(shù)與方程思想、轉(zhuǎn)化與化歸思想是高中數(shù)學(xué)思想中比較重要的兩大思想，而構(gòu)造函數(shù)的解題思路恰好是這兩種思想的良好體現(xiàn).解題技法一、利用f

(x)進(jìn)行抽象函數(shù)構(gòu)造(一)利用f

(x)與x構(gòu)造思路點(diǎn)撥出現(xiàn)“＋”法形式，優(yōu)先構(gòu)造F(x)＝xf

(x)，然后利用函數(shù)的單調(diào)性、奇偶性和數(shù)形結(jié)合求解即可.例1設(shè)f

(x)是定義在R上的偶函數(shù)，當(dāng)x<0時(shí)，f

(x)＋xf′(x)<0，且f

(－4)＝0，則不等式xf

(x)>0的解集為_(kāi)_________________.(－∞，－4)∪(0,4)解析構(gòu)造F

(x)＝xf

(x)，則F′(x)＝f

(x)＋xf′(x)，當(dāng)x<0時(shí)，f

(x)＋xf′(x)<0，可以推出當(dāng)x<0時(shí)，F(xiàn)′(x)<0，∴F

(x)在(－∞，0)上單調(diào)遞減.∵f

(x)為偶函數(shù)，x為奇函數(shù)，∴F

(x)為奇函數(shù)，∴F

(x)在(0，＋∞)上也單調(diào)遞減.根據(jù)f

(－4)＝0可得F

(－4)＝0，根據(jù)函數(shù)的單調(diào)性、奇偶性可得函數(shù)圖象(圖略)，根據(jù)圖象可知xf

(x)>0的解集為(－∞，－4)∪(0,4).例2設(shè)f

(x)是定義在R上的偶函數(shù)，且f

(1)＝0，當(dāng)x<0時(shí)，有xf′(x)－f

(x)>0恒成立，則不等式f

(x)>0的解集為_(kāi)_____________________.(－∞，－1)∪(1，＋∞)當(dāng)x<0時(shí)，xf′(x)－f

(x)>0，可以推出當(dāng)x<0時(shí)，F(xiàn)′(x)>0，F(xiàn)

(x)在(－∞，0)上單調(diào)遞增.∵f

(x)為偶函數(shù)，x為奇函數(shù)，∴F

(x)為奇函數(shù)，∴F

(x)在(0，＋∞)上也單調(diào)遞增.根據(jù)f

(1)＝0可得F

(1)＝0，根據(jù)函數(shù)的單調(diào)性、奇偶性可得函數(shù)圖象(圖略)，根據(jù)圖象可知f

(x)>0的解集為(－∞，－1)∪(1，＋∞).F

(x)＝xnf

(x)，F(xiàn)′(x)＝nxn－1f

(x)＋xnf′(x)＝xn－1[nf

(x)＋xf′(x)]；結(jié)論：(1)出現(xiàn)nf

(x)＋xf′(x)形式，構(gòu)造函數(shù)F

(x)＝xnf

(x)；我們根據(jù)得出的結(jié)論去解決例3.例3已知偶函數(shù)f

(x)(x≠0)的導(dǎo)函數(shù)為f′(x)，且滿足f

(－1)＝0，當(dāng)x>0時(shí)，2f

(x)>xf′(x)，則使得f

(x)>0成立的x的取值范圍是______________.(－1,0)∪(0,1)當(dāng)x>0時(shí)，xf′(x)－2f

(x)<0，可以推出當(dāng)x>0時(shí)，F(xiàn)′(x)<0，F(xiàn)

(x)在(0，＋∞)上單調(diào)遞減.∵f

(x)為偶函數(shù)，x2為偶函數(shù)，∴F

(x)為偶函數(shù)，∴F

(x)在(－∞，0)上單調(diào)遞增.根據(jù)f

(－1)＝0可得F

(－1)＝0，根據(jù)函數(shù)的單調(diào)性、奇偶性可得函數(shù)圖象(圖略)，根據(jù)圖象可知f

(x)>0的解集為(－1,0)∪(0,1).(二)利用f

(x)與ex構(gòu)造例4已知f

(x)是定義在(－∞，＋∞)上的函數(shù)，導(dǎo)函數(shù)f′(x)滿足f′(x)<f

(x)對(duì)于x∈R恒成立，則A.f

(2)>e2f

(0)，f

(2019)>e2019f

(0) B.f

(2)<e2f

(0)，f

(2019)>e2019f

(0)C.f

(2)>e2f

(0)，f

(2019)<e2019f

(0) D.f

(2)<e2f

(0)，f

(2019)<e2019f

(0)√則F′(x)<0，F(xiàn)

(x)在R上單調(diào)遞減，根據(jù)單調(diào)性可知選D.F

(x)＝enxf

(x)，F(xiàn)′(x)＝n·enxf

(x)＋enxf′(x)＝enx[f′(x)＋nf

(x)]；結(jié)論：(1)出現(xiàn)f′(x)＋nf

(x)形式，構(gòu)造函數(shù)F

(x)＝enxf

(x)；我們根據(jù)得出的結(jié)論去解決例5，例6.例5若定義在R上的函數(shù)f

(x)滿足f′(x)－2f

(x)>0，f

(0)＝1，則不等式f

(x)>e2x的解集為_(kāi)_______.{x|x>0}函數(shù)f

(x)滿足f′(x)－2f

(x)>0，則F′(x)>0，F(xiàn)

(x)在R上單調(diào)遞增.根據(jù)單調(diào)性得x>0.例6已知函數(shù)f

(x)在R上可導(dǎo)，其導(dǎo)函數(shù)為f′(x)，若f

(x)滿足：(x－1)[f′(x)－f

(x)]>0，f

(2－x)＝f

(x)·e2－2x，則下列判斷一定正確的是A.f

(1)<f

(0) B.f

(2)>e2f

(0)C.f

(3)>e3f

(0) D.f

(4)<e4f

(0)√則x≥1時(shí)F′(x)≥0，F(xiàn)

(x)在[1，＋∞)上單調(diào)遞增.當(dāng)x<1時(shí)F′(x)<0，F(xiàn)

(x)在(－∞，1]上單調(diào)遞減.又由f

(2－x)＝f

(x)e2－2x?F

(2－x)＝F

(x)?F

(x)關(guān)于x＝1對(duì)稱，根據(jù)單調(diào)性和圖象，可知選C.(三)利用f

(x)與sinx，cosx構(gòu)造sinx，cosx因?yàn)閷?dǎo)函數(shù)存在一定的特殊性，所以也是重點(diǎn)考察的范疇，我們一起看看?？嫉膸追N形式.F

(x)＝f

(x)sinx，F(xiàn)′(x)＝f′(x)sinx＋f

(x)cosx；F

(x)＝f

(x)cosx，F(xiàn)′(x)＝f′(x)cosx－f

(x)sinx；根據(jù)得出的關(guān)系式，我們來(lái)看一下例7.√二、具體函數(shù)關(guān)系式構(gòu)造這類題型需要根據(jù)題意構(gòu)造具體的函數(shù)關(guān)系式，通過(guò)具體的關(guān)系式去解決不等式及求值問(wèn)題.A.α>βB.α2>β2C.α<βD.α＋β>0√思路點(diǎn)撥構(gòu)造函數(shù)f

(x)＝xsinx，然后利用函數(shù)的單調(diào)性和數(shù)形結(jié)合求解即可.解析構(gòu)造f

(x)＝xsinx形式，又∵f

(x)為偶函數(shù)，根據(jù)單調(diào)性和圖象可知選