第16講　導(dǎo)數(shù)與函數(shù)的單調(diào)性（原卷版）

第16講導(dǎo)數(shù)與函數(shù)的單調(diào)性基礎(chǔ)知識(shí)函數(shù)的單調(diào)性與導(dǎo)數(shù)導(dǎo)數(shù)到單調(diào)性單調(diào)遞增在區(qū)間(a,b)上,若f'(x)>0,則f(x)在這個(gè)區(qū)間上單調(diào)

單調(diào)遞減在區(qū)間(a,b)上,若f'(x)<0,則f(x)在這個(gè)區(qū)間上單調(diào)

單調(diào)性到導(dǎo)數(shù)單調(diào)遞增若函數(shù)y=f(x)在區(qū)間(a,b)上單調(diào)遞增,則f'(x)

單調(diào)遞減若函數(shù)y=f(x)在區(qū)間(a,b)上單調(diào)遞減,則f'(x)

“函數(shù)y=f(x)在區(qū)間(a,b)上的導(dǎo)數(shù)大(小)于0”是“其單調(diào)遞增(減)”的條件

分類(lèi)訓(xùn)練探究點(diǎn)一求函數(shù)的單調(diào)區(qū)間(不含參)例1已知函數(shù)f(x)=(x-2)ex-ax+alnx(a∈R).當(dāng)a=-1時(shí),求函數(shù)f(x)的單調(diào)區(qū)間.[總結(jié)反思]確定函數(shù)f(x)單調(diào)區(qū)間的步驟:(1)確定函數(shù)f(x)的定義域.(2)求f'(x).(3)解不等式f'(x)>0,解集在定義域內(nèi)的部分為單調(diào)遞增區(qū)間;解不等式f'(x)<0,解集在定義域內(nèi)的部分為單調(diào)遞減區(qū)間.變式題(1)函數(shù)f(x)=(x-2)ex的單調(diào)遞增區(qū)間為 ()A.(1,+∞) B.(2,+∞)C.(0,2) D.(1,2)(2)若函數(shù)f(x)=lnx+1ex探究點(diǎn)二討論含參函數(shù)的單調(diào)性例2(1)已知函數(shù)f(x)=13x3(2)已知函數(shù)f(x)=2x[總結(jié)反思](1)利用導(dǎo)數(shù)討論函數(shù)單調(diào)性的關(guān)鍵是確定導(dǎo)數(shù)的符號(hào).不含參數(shù)的問(wèn)題直接解導(dǎo)數(shù)大于(或小于)零的不等式,其解集即為函數(shù)的單調(diào)區(qū)間;含參數(shù)的問(wèn)題,應(yīng)就參數(shù)范圍討論導(dǎo)數(shù)大于(或小于)零的不等式的解.(2)所有求解和討論都必須在函數(shù)的定義域內(nèi),不要超出定義域的范圍.變式題(1)已知函數(shù)f(x)=lnx-ax2+a(a∈R),討論f(x)的單調(diào)性.(2)已知函數(shù)f(x)=(x-2)ex+a(x-1)2,a∈R,求f(x)的單調(diào)區(qū)間.探究點(diǎn)三已知函數(shù)單調(diào)性確定參數(shù)的取值范圍例3已知函數(shù)f(x)=3ex-12x2(1)若函數(shù)f(x)的圖像在x=0處的切線(xiàn)方程為y=2x+b,求a,b的值;(2)若函數(shù)f(x)在R上是增函數(shù),求實(shí)數(shù)a的最大值.[總結(jié)反思](1)f(x)在D上單調(diào)遞增(減),只要滿(mǎn)足f'(x)≥0(≤0)在D上恒成立即可.如果能夠分離參數(shù),則分離參數(shù)后可轉(zhuǎn)化為參數(shù)值與函數(shù)最值之間的關(guān)系.(2)二次函數(shù)在區(qū)間D上大于零恒成立,討論的標(biāo)準(zhǔn)是二次函數(shù)的圖像的對(duì)稱(chēng)軸x=x0與區(qū)間D的相對(duì)位置,一般分x0在區(qū)間左側(cè)、內(nèi)部、右側(cè)進(jìn)行討論.變式題(1)若函數(shù)f(x)=ax+e-x-ex在R上單調(diào)遞減,則實(shí)數(shù)a的取值范圍為 ()A.a≤2 B.a≤1C.a≥1 D.a≥2(2)若函數(shù)f(x)=ex-(a-1)x+1在(0,1)上不單調(diào),則實(shí)數(shù)a的取值范圍是 ()A.(2,e+1)B.[2,e+1]C.(-∞,2]∪[e+1,+∞)D.(-∞,2)∪(e+1,+∞)探究點(diǎn)四函數(shù)單調(diào)性的簡(jiǎn)單應(yīng)用例4(1)已知函數(shù)f(x)=exx+12x2-x,若a=f(20.3),b=f(2),c=f(logA.c<b<a B.a<b<cC.c>a>b D.b>c>a(2)已知函數(shù)f(x)=xex-lnx-x,若存在x0∈(0,+∞),使得f(x0)≤a,則a的取值范圍為 ()A.[1,+∞) B.[e-1,+∞)C.[2,+∞) D.[e,+∞)[總結(jié)反思]利用導(dǎo)數(shù)比較大小或解不等式,常常需要把比較大小或求解不等式的問(wèn)題轉(zhuǎn)化為利用導(dǎo)數(shù)研究函數(shù)單調(diào)性的問(wèn)題,再由單調(diào)性比較大小或解不等式.變式題(1)已知函數(shù)f(x)=x3+2x+1,若f(ax-ex+1)>1在x∈(0,+∞)時(shí)有解,則實(shí)數(shù)a的取值范圍為 ()A.[1,e) B.(0,1)C.(-∞,1) D.(1,+∞)(2)已知函數(shù)f(x)=13ax3+12bx2A.a<b<c B.b<c<aC.b<a<c D.a<c<b同步作業(yè)1.函數(shù)f(x)=xlnx的單調(diào)遞增區(qū)間為 ()A.0,1e B.(e,+∞)C.1e,+∞ D.1e,e2.已知函數(shù)f(x)=x2-4x,當(dāng)x∈(0,+∞)時(shí),在下列函數(shù)中,與f(x)的單調(diào)區(qū)間完全相同的是 ()A.g(x)=x3-2B.g(x)=(x-2)exC.g(x)=(3-x)exD.g(x)=x-2lnx3.已知函數(shù)f(x)=x2-cosx,則f(0.6),f(0),f(-0.5)的大小關(guān)系是 ()A.f(0)<f(0.6)<f(-0.5)B.f(0)<f(-0.5)<f(0.6)C.f(0.6)<f(-0.5)<f(0)D.f(-0.5)<f(0)<f(0.6)4.已知函數(shù)f(x)的導(dǎo)函數(shù)f'(x)的圖像如圖K16-1所示,那么函數(shù)f(x)的圖像最有可能是 ()圖K16-1圖K16-25.函數(shù)f(x)=lnx-12x2的導(dǎo)函數(shù)為f'(x),則f'(x)>0的解集為A.(-∞,1) B.(0,1)C.(1,+∞) D.(0,+∞)6.使“函數(shù)f(x)=exx在區(qū)間(0,m]上單調(diào)遞減”成立的一個(gè)m值是7.函數(shù)f(x)=sin2x+2cosx,x∈0,π2的單調(diào)遞減區(qū)間是 ()A.0,π6 B.π6,π2C.0,π3 D.π3,π28.設(shè)函數(shù)f(x)滿(mǎn)足f(x)=x[f'(x)-lnx],且在(0,+∞)上單調(diào)遞增,則f1e的取值范圍是(e為自然對(duì)數(shù)的底數(shù)) ()A.[-1,+∞) B.1e,+∞C.-∞,1e D.(-∞,-1]9.已知函數(shù)f(x)=a-2A.a≤0 B.a≤1C.a≤2e D.a≤3e10.(多選題)已知函數(shù)f(x)=lnx+1x,則A.函數(shù)f(x)的單調(diào)遞減區(qū)間是(-∞,1)B.函數(shù)f(x)在(e,+∞)上單調(diào)遞增C.函數(shù)f(x)的最小值為1D.若f(m)=f(n)(m≠n),則m+n>211.(多選題)對(duì)于定義域?yàn)镽的函數(shù)f(x),若滿(mǎn)足:①f(0)=0;②當(dāng)x∈R且x≠0時(shí),都有xf'(x)>0;③當(dāng)x1<0<x2且|x1|<|x2|時(shí),都有f(x1)<f(x2).則稱(chēng)f(x)為“偏對(duì)稱(chēng)函數(shù)”,下列函數(shù)是“偏對(duì)稱(chēng)函數(shù)”的是 ()A.f1(x)=-x3+x2B.f2(x)=ex-x-1C.f3(x)=xsinxD.f4(x)=ln12.已知函數(shù)f(x)=ex+ae-x在[0,1]上不單調(diào),則實(shí)數(shù)a的取值范圍為.

13.