高考二輪復(fù)習(xí)文科數(shù)學(xué)課件培優(yōu)拓展13洛必達(dá)法則速求參數(shù)范圍

培優(yōu)拓展13

洛必達(dá)法則速求參數(shù)范圍規(guī)律方法1.不等式恒成立或能成立題目.能分離參數(shù)成a≥h(x)或a≤h(x),歸結(jié)為求h(x)的某個(gè)最值(或其極限值)問題.常規(guī)方法不易求得最值或其極限值(往往多次求導(dǎo)后仍為超越結(jié)構(gòu)).可考慮在某個(gè)端點(diǎn)或斷點(diǎn)處應(yīng)用洛必達(dá)法則求最值(或極限值).2.使用洛必達(dá)法則時(shí),是對(duì)分子、分母分別求導(dǎo),而不是對(duì)它們的商求導(dǎo),求導(dǎo)之后再求極限得最值.對(duì)點(diǎn)訓(xùn)練1設(shè)函數(shù)f(x)=ex-1-x-ax2.(1)若a=0,求f(x)的單調(diào)區(qū)間;(2)當(dāng)x≥0時(shí),f(x)≥0,求a的取值范圍.解

(1)當(dāng)a=0時(shí),f(x)=ex-1-x,∴f'(x)=ex-1,當(dāng)x>0時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x<0時(shí),f'(x)<0,f(x)單調(diào)遞減,∴函數(shù)f(x)的單調(diào)遞增區(qū)間是(0,+∞),單調(diào)遞減區(qū)間是(-∞,0).(2)當(dāng)x≥0時(shí),f(x)≥0,即ex-1-x≥ax2在[0,+∞)上恒成立.①當(dāng)x=0時(shí),e0-1-0≥a×0恒成立,a∈R;令h(x)=(x-2)ex+x+2(x>0).則h'(x)=(x-1)ex+1(x>0).h'(x)=(x-1)ex+1單調(diào)遞增,∴h'(x)>(0-1)×e0+1=0,∴h(x)=(x-2)ex+x+2單調(diào)遞增,所以h(x)在(0,+∞)上單調(diào)遞增,且h(1)=0,因此當(dāng)x∈(0,1)時(shí),h(x)<0;當(dāng)x∈(1,+∞)時(shí),h(x)>0,所以當(dāng)x∈(0,1)時(shí),g'(x)<0,當(dāng)x∈(1,+∞)時(shí),g'(x)>0,所以g(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.對(duì)點(diǎn)訓(xùn)練2(原創(chuàng)題)已知函數(shù)f(x)=x3+ax2+bx+a2,當(dāng)a=-1時(shí),若?x∈(-∞,0),都有f(x)≤ex恒成立,求b的取值范圍.令h(x)=ex-2x2-x-1(x<0),則h'(x)=ex-4x-1.令m(x)=ex-4x-1(x<0),則m'(x)=ex-4,∴當(dāng)x<0時(shí),m'(x)=ex-4<0,∴h'(x)=ex-4x-1在(-∞,0)單調(diào)遞減,∴?x∈(-∞,0),h'(x)>e0-4×0-1=0,∴h(x)=ex-2x2-x-