第第1 1課時(shí)課時(shí) 利用導(dǎo)數(shù)研究函數(shù)的基本利用導(dǎo)數(shù)研究函數(shù)的基本問題問題考情分析考情分析總綱目錄考點(diǎn)一 導(dǎo)數(shù)的運(yùn)算及幾何意義考點(diǎn)二 利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性考點(diǎn)三 利用導(dǎo)數(shù)研究函數(shù)的極值（最值）問題第第1 1課時(shí)利用導(dǎo)數(shù)研究函數(shù)的基本問題課時(shí)利用導(dǎo)數(shù)研究函數(shù)的基本問題考點(diǎn)一 導(dǎo)數(shù)的運(yùn)算及幾何意義1.導(dǎo)數(shù)公式(1)(sin x)=cos x;(2)(cos x)=-sin x;(3)(ax)=axln a(a0);(4)(logax)=(a0,且a1).1lnxa2.導(dǎo)數(shù)的幾何意義函數(shù)f(x)在x0處的導(dǎo)數(shù)是曲線f(x)在點(diǎn)P(x0,f(x0)處的切線的斜率,曲線f(x)在點(diǎn)P處的切線的斜率k=f (x0),相應(yīng)的切線方程為y-f(x0)=f (x0)(x-x0).典型例題典型例題(1)已知f(x)=x3+2xf (3)+ln x,則f (3)=()A. B.- C.9 D.-9(2)(2016課標(biāo)全國,15,5分)已知f(x)為偶函數(shù),當(dāng)x0,則-x0),則f (x)=-3(x0),f (1)=-2,在點(diǎn)(1,-3)處的切線方程為y+3=-2(x-1),即y=-2x-1.1x132832831x方法歸納方法歸納曲線y=f(x)的切線方程的三種類型及求解方法(1)已知切點(diǎn)P(x0,y0),求切線方程求出切線的斜率f (x0),由點(diǎn)斜式寫出方程;(2)已知切線的斜率k,求切線方程設(shè)切點(diǎn)P(x0,y0),通過方程k=f (x0)解得x0,再由點(diǎn)斜式寫出方程;(3)已知切線上一點(diǎn)(非切點(diǎn)),求切線方程設(shè)切點(diǎn)P(x0,y0),利用導(dǎo)數(shù)求得切線斜率f (x0),再由斜率公式求得切線斜率,列方程(組)解得x0,再由點(diǎn)斜式或兩點(diǎn)式寫出方程.跟蹤集訓(xùn)跟蹤集訓(xùn)1.已知函數(shù)f(x)=cos x,則f()+f =()A.- B.- C.- D.- 1x2232131答案答案 Cf(x)=cos x,f (x)=-cos x+(-sin x),f()+f =-+(-1)=-.1x21x1x21232.(2017廣州綜合測(cè)試(一)設(shè)函數(shù)f(x)=x3+ax2,若曲線y=f(x)在點(diǎn)P(x0,f(x0)處的切線方程為x+y=0,則點(diǎn)P的坐標(biāo)為()A.(0,0) B.(1,-1)C.(-1,1) D.(1,-1)或(-1,1)答案答案 D由題意知,f (x)=3x2+2ax,所以曲線y=f(x)在點(diǎn)P(x0,f(x0)處的切線的斜率為f (x0)=3+2ax0,又切線方程為x+y=0,所以x00,且解得或當(dāng)時(shí),點(diǎn)P的坐標(biāo)為(1,-1);當(dāng)時(shí),點(diǎn)P的坐標(biāo)為(-1,1),故選D.20 x20032000321,0,xaxxxax 01,2xa 01,2.xa 01,2xa 01,2xa 考點(diǎn)二 利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性命題點(diǎn)命題點(diǎn)1.判斷函數(shù)的單調(diào)性.2.求函數(shù)的單調(diào)區(qū)間.3.根據(jù)函數(shù)的單調(diào)性求參數(shù)的取值范圍. 導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系(1)f (x)0是f(x)為增函數(shù)的充分不必要條件,如函數(shù)f(x)=x3在(-,+)上單調(diào)遞增,但f (x)0.(2)f (x)0是f(x)為增函數(shù)的必要不充分條件,當(dāng)函數(shù)在某個(gè)區(qū)間內(nèi)恒有f (x)=0時(shí),f(x)為常數(shù)函數(shù),函數(shù)不具有單調(diào)性.典型例題典型例題(2017浙江,20改編)已知函數(shù)f(x)=(x-)e-x.(1)求f(x)的導(dǎo)函數(shù);(2)求f(x)的單調(diào)區(qū)間.解析解析(1)(x-)=1-,(e-x)=-e-x,f (x)=e-x-(x-)e-x=.(2)令f (x)=0,解得x=1或x=.當(dāng)x變化時(shí),f (x),f(x)的情況如下:21x12x21x121x1121x21x(1)( 212)e21xxxx 12x(1)( 212)e21xxxx 52x1f (x) -0+0-f(x)0121,1251,2525,21212e1252ef(x)的增區(qū)間為,減區(qū)間為,.方法歸納方法歸納根據(jù)函數(shù)y=f(x)在(a,b)上的單調(diào)性求參數(shù)范圍的方法(1)若函數(shù)y=f(x)在(a,b)上單調(diào)遞增,轉(zhuǎn)化為f (x)0在(a,b)上恒成立.(2)若函數(shù)y=f(x)在(a,b)上單調(diào)遞減,轉(zhuǎn)化為f (x)0在(a,b)上恒成立.(3)若函數(shù)y=f(x)在(a,b)上單調(diào),轉(zhuǎn)化為f (x)在(a,b)上不變號(hào),即f (x)0恒成立或f(x)0恒成立.(4)若函數(shù)y=f(x)在(a,b)上不單調(diào),轉(zhuǎn)化為f (x)在(a,b)上有變號(hào)零點(diǎn).51,21,125,2跟蹤集訓(xùn)跟蹤集訓(xùn)1.(2017張掖第一次診斷考試)若函數(shù)f(x)=-x2+x+1在區(qū)間上單調(diào)遞減,則實(shí)數(shù)a的取值范圍是 .33x2a1,32答案答案 10,3解析解析由已知得f (x)=x2-ax+1,函數(shù)f(x)在區(qū)間上單調(diào)遞減,f (x)0在區(qū)間上恒成立,即解得a,實(shí)數(shù)a的取值范圍為.1,321,321 0,2 (3)0,ff1110,429310,aa 10310,32.已知函數(shù)f(x)=ax3+x2(aR)在x=-處取得極值.(1)確定a的值;(2)若g(x)=f(x)ex,討論g(x)的單調(diào)性.43解析解析(1)對(duì)f(x)求導(dǎo)得f (x)=3ax2+2x,因?yàn)閒(x)在x=-處取得極值,所以f =0,即3a+2=0,解得a=.(2)由(1)得g(x)=ex,故g (x)=ex+ex=ex434316943123212xx2322xx3212xx3215222xxx令g (x)=0,解得x=0或x=-1或x=-4.當(dāng)x-4時(shí),g (x)0,故g(x)為減函數(shù);當(dāng)-4x0,故g(x)為增函數(shù);當(dāng)-1x0時(shí),g (x)0時(shí),g (x)0,故g(x)為增函數(shù).綜上知, g(x)在(-,-4)和(-1,0)上為減函數(shù),在(-4,-1)和(0,+)上為增函數(shù).考點(diǎn)三 利用導(dǎo)數(shù)研究函數(shù)的極值(最值)問題可導(dǎo)函數(shù)的極值與最值(1)若在x0附近左側(cè)f (x)0,右側(cè)f (x)0,則f(x0)為函數(shù)f(x)的極大值;若在x0附近左側(cè)f (x)0,則f(x0)為函數(shù)f(x)的極小值.(2)設(shè)函數(shù)y=f(x)在a,b上連續(xù),在(a,b)內(nèi)可導(dǎo),則f(x)在a,b上必有最大值和最小值且在極值點(diǎn)或端點(diǎn)處取得.典型例題典型例題(2017北京,19,13分)已知函數(shù)f(x)=excos x-x.(1)求曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程;(2)求函數(shù)f(x)在區(qū)間上的最大值和最小值.解析解析(1)因?yàn)閒(x)=excos x-x,所以f (x)=ex(cos x-sin x)-1, f (0)=0.又因?yàn)閒(0)=1,所以曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程為y=1.(2)設(shè)h(x)=ex(cos x-sin x)-1,則h(x)=ex(cos x-sin x-sin x-cos x)=-2exsin x.當(dāng)x時(shí),h(x)0,所以h(x)在區(qū)間上單調(diào)遞減.0,20,20,2所以對(duì)任意x有h(x)h(0)=0,即f (x)0, f(x)單調(diào)遞增;x(-2,1)時(shí), f (x)0,g(x)0,g(x)在(0,+)上是增函數(shù),函數(shù)g(x)無極值點(diǎn).1x121x2(1)1axa xx當(dāng)a0時(shí),g(x)=-.令g(x)=0,得x=.2(1)1axa xx1(1)a xxax1a當(dāng)x時(shí),g(x)0;當(dāng)x時(shí),g(x)0時(shí),函數(shù)g(x)有極大值-ln a,無極小值.10,a1,a10,a1,a1a1a1a 2a21a1a12a12a1.函數(shù)f(x)=x+的極值情況是()A.當(dāng)x=1時(shí),取極小值2,但無極大值B.當(dāng)x=-1時(shí),取極大值-2,但無極小值C.當(dāng)x=-1時(shí),取極小值-2;當(dāng)x=1時(shí),取極大值2D.當(dāng)x=-1時(shí),取極大值-2;當(dāng)x=1時(shí),取極小值21x隨堂檢測(cè)隨堂檢測(cè)答案答案 D求導(dǎo)得f (x)=1-,令f (x)=0,得x=1,函數(shù)f(x)在區(qū)間(-,-1)和(1,+)上單調(diào)遞增,在(-1,0)和(0,1)上單調(diào)遞減,所以當(dāng)x=-1時(shí),取極大值-2,當(dāng)x=1時(shí),取極小值2.21x2.(2017成都第二次診斷性檢測(cè))若曲線y=f(x)=ln x+ax2(a為常數(shù))不存在斜率為負(fù)數(shù)的切線,則實(shí)數(shù)a的取值范圍是()A. B.C.(0,+) D.0,+)1,21,2答案答案 D f (x)=+2ax=(x0),根據(jù)題意有f (x)0(x0)恒成立,所以2ax2+10(x0)恒成立,即2a-(x0)恒成立,所以a0,故實(shí)數(shù)a的取值范圍為0,+).故選D.1x221axx21x3.已知函數(shù)f(x)=x2+2ax-ln x,若f(x)在區(qū)間上是增函數(shù),則實(shí)數(shù)a的取值范圍為 .121,23答案答案 4,3解析解析由題意知f (x)=x+2a-0在上恒成立,即2a-x+在上恒成立,當(dāng)x時(shí),=,2a,即a.1x1,231x1,231,23max1xx 8383434.(2017貴陽檢測(cè))已知函數(shù)f(x)=-ln x.(1)求f(x)的單調(diào)區(qū)間;(2)求函數(shù)f(x)在上的最大值和最小值(其中e是自然對(duì)數(shù)的底數(shù)).1xx1,ee解析解析(1)f(x)=-ln x=1-ln x,f(x)的定義域?yàn)?0,+).f (x)=-=,由f (x)00 x1,由f (x)1,f(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.(2)由(1)得f(x)在上單調(diào)遞增,在1,e上單調(diào)遞減,f(x)在上的最大值為f(1)=1-ln 1=0.又f=1-e-ln=2-e,f(e)=1-ln e=-,且ff(e),f(x)在上的最小值為f=2-e.綜上,f(x)在上的最大值為0,最小值為2-e.1xx1x21x1x21xx1,1e1,ee111e1e1e1e1e1,ee1e1,ee