第16講　導(dǎo)數(shù)與函數(shù)的單調(diào)性（解析版）

第16講導(dǎo)數(shù)與函數(shù)的單調(diào)性基礎(chǔ)知識函數(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

遞增遞減≥0≤0充分分類訓(xùn)練探究點一求函數(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,求函數(shù)f(x)的單調(diào)區(qū)間.例1[思路點撥]當(dāng)a=-1時,f(x)=(x-2)ex+x-lnx,對f(x)求導(dǎo)即可判斷f(x)的單調(diào)性.解:當(dāng)a=-1時,f(x)=(x-2)ex+x-lnx,x∈(0,+∞),則f'(x)=(x-1)ex+1-1x=(x-1)ex+1x,因為x∈(0,+∞),所以ex+1x所以當(dāng)x>1時,f'(x)>0,當(dāng)0<x<1時,f'(x)<0,所以函數(shù)f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,故f(x)的單調(diào)遞減區(qū)間是(0,1),單調(diào)遞增區(qū)間是(1,+∞).[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變式題(1)A(2)(1,+∞)[解析](1)f'(x)=ex+(x-2)ex=(x-1)ex,令f'(x)>0,解得x>1,所以函數(shù)f(x)=(x-2)ex的單調(diào)遞增區(qū)間是(1,+∞),故選A.(2)函數(shù)的定義域為(0,+∞),f'(x)=1x設(shè)k(x)=1x-lnx-1,則k'(x)=-1x2-探究點二討論含參函數(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例2[思路點撥](1)求出f'(x),對a分兩種情況討論,即可確定f(x)的單調(diào)性;(2)對f(x)求導(dǎo)得f'(x)=2x2-ax+1解:(1)由題可知函數(shù)f(x)的定義域為R.f'(x)=x2-a,a∈R,①當(dāng)a≤0時,f'(x)≥0恒成立,所以函數(shù)y=f(x)在R上單調(diào)遞增.②當(dāng)a>0時,令f'(x)=0,得x=-a或x=a.當(dāng)f'(x)>0時,x<-a或x>a,當(dāng)f'(x)<0時,-a<x<a,所以函數(shù)y=f(x)在(-∞,-a)和(a,+∞)上單調(diào)遞增,在(-a,a)上單調(diào)遞減.綜上可知,當(dāng)a≤0時,函數(shù)y=f(x)在R上單調(diào)遞增;當(dāng)a>0時,函數(shù)y=f(x)在(-∞,-a)和(a,+∞)上單調(diào)遞增,在(-a,a)上單調(diào)遞減.(2)因為f(x)=2x2-令h(x)=2x2-ax+1,則h(0)=1>0,若Δ=a2-8≤0,即0<a≤22,則h(x)≥0恒成立,即f'(x)≥0,所以f(x)在(0,+∞)上單調(diào)遞增.若Δ=a2-8>0,即a>22,令h(x)=0,得x1=a-a2-8當(dāng)0<x<x1或x>x2時,h(x)>0,即f'(x)>0;當(dāng)x1<x<x2時,h(x)<0,即f'(x)<0,所以f(x)在(0,x1)和(x2,+∞)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減.綜上所述,當(dāng)0<a≤22時,f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>22時,f(x)在0,a-a2-84和a+a2-84[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.不含參數(shù)的問題直接解導(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.變式題解:(1)函數(shù)f(x)的定義域為(0,+∞),f'(x)=1x-2ax=1當(dāng)a≤0時,1-2ax2>0,f'(x)>0,故f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時,令f'(x)=0,解得x=2a易知當(dāng)x∈0,2a2a時,f'(x)>0,f(x)單調(diào)遞增,當(dāng)x∈2a2a綜上所述,當(dāng)a≤0時,f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時,f(x)在0,2a2a上單調(diào)遞增,在2a2a(2)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).①若a≥0,則當(dāng)x∈(-∞,1)時,f'(x)<0;當(dāng)x∈(1,+∞)時,f'(x)>0.所以f(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.②若a<0,由f'(x)=0得x=1或x=ln(-2a).(i)若a=-e2,則f'(x)=(x-1)(ex所以f(x)在(-∞,+∞)上單調(diào)遞增.(ii)若a>-e2故當(dāng)x∈(-∞,ln(-2a))∪(1,+∞)時,f'(x)>0;當(dāng)x∈(ln(-2a),1)時,f'(x)<0.所以f(x)在(-∞,ln(-2a)),(1,+∞)上單調(diào)遞增,在(ln(-2a),1)上單調(diào)遞減.(iii)若a<-e2故當(dāng)x∈(-∞,1)∪(ln(-2a),+∞)時,f'(x)>0;當(dāng)x∈(1,ln(-2a))時,f'(x)<0.所以f(x)在(-∞,1),(ln(-2a),+∞)上單調(diào)遞增,在(1,ln(-2a))上單調(diào)遞減.綜上所述,當(dāng)a<-e2時,單調(diào)遞增區(qū)間為(-∞,1)和(ln(-2a),+∞),單調(diào)遞減區(qū)間為(1,ln(-2a));當(dāng)a=-e2當(dāng)-e2當(dāng)a≥0時,單調(diào)遞減區(qū)間為(-∞,1),單調(diào)遞增區(qū)間為(1,+∞).探究點三已知函數(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),求實數(shù)a的最大值.例3[思路點撥](1)先對函數(shù)f(x)求導(dǎo),再根據(jù)在x=0處的切線斜率可得到參數(shù)a的值,代入x=0,求出f(0)的值,則b即可得出;(2)根據(jù)函數(shù)f(x)在R上是增函數(shù),可得f'(x)≥0,即3ex-x-a≥0恒成立,再進行參變分離得a≤3ex-x,構(gòu)造函數(shù)g(x)=3ex-x,對g(x)進行求導(dǎo)分析,找出最小值,即實數(shù)a的最大值.解:(1)由題意知,函數(shù)f(x)=3ex-12x2故f'(x)=3ex-x-a,則f'(0)=3-a,由題意知,3-a=2,即a=1,∴f(x)=3ex-12x2∴2×0+b=3,即b=3.∴a=1,b=3.(2)由題意可知f'(x)≥0,即3ex-x-a≥0恒成立,∴a≤3ex-x恒成立.設(shè)g(x)=3ex-x,則g'(x)=3ex-1.令g'(x)=3ex-1=0,解得x=-ln3.令g'(x)<0,解得x<-ln3,令g'(x)>0,解得x>-ln3,∴g(x)在(-∞,-ln3)上單調(diào)遞減,在(-ln3,+∞)上單調(diào)遞增,∴g(x)在x=-ln3處取得極小值,也是最小值,∴g(x)min=g(-ln3)=1+ln3.∴a≤1+ln3,故a的最大值為1+ln3.[總結(jié)反思](1)f(x)在D上單調(diào)遞增(減),只要滿足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)的圖像的對稱軸x=x0與區(qū)間D的相對位置,一般分x0在區(qū)間左側(cè)、內(nèi)部、右側(cè)進行討論.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,+∞)變式題(1)A(2)A[解析](1)由題意可得,f'(x)=a-(e-x+ex)≤0恒成立,即a≤e-x+ex恒成立,∵e-x+ex≥2(當(dāng)且僅當(dāng)x=0時取等號),∴a≤2,故選A.(2)∵f(x)=ex-(a-1)x+1,∴f'(x)=ex-a+1,若f(x)在(0,1)上不單調(diào),則f'(x)在(0,1)上有變號零點,又f'(x)單調(diào)遞增,則f'(0)f'(1)<0,即(1-a+1)(e-a+1)<0,解得2<a<e+1,∴a的取值范圍是(2,e+1).故選A.探究點四函數(shù)單調(diào)性的簡單應(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,+∞)例4[思路點撥](1)求導(dǎo),利用導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系分析可得函數(shù)f(x)在(1,+∞)上單調(diào)遞增,結(jié)合1<20.3<2<log25,即可得結(jié)論;(2)存在x0∈(0,+∞),使得f(x0)≤a,即a≥f(x)min,先證明ex≥x+1恒成立,利用對數(shù)恒等式結(jié)合不等式放縮求出f(x)的最小值,可得a的取值范圍.(1)B(2)A[解析](1)函數(shù)f(x)=exx+12x2-x,f'(x)=ex(x-1)x2+x-1=(x-1)exx2+1,當(dāng)x∈(1,+∞)時,f'(x)>0,所以f(x)在(1,+∞)上單調(diào)遞增,因為1<20.3<2,log(2)構(gòu)造函數(shù)y=ex-x-1,則y'=ex-1,當(dāng)x<0時,y'<0,函數(shù)在(-∞,0)上單調(diào)遞減;當(dāng)x>0時,y'>0,函數(shù)在(0,+∞)上單調(diào)遞增.∴ymin=0,則y≥0恒成立,即ex≥x+1恒成立.當(dāng)x>0時,f(x)=xex-lnx-x=elnxex-lnx-x=elnx+x-lnx-x≥(lnx+x+1)-lnx-x=1,當(dāng)lnx+x=0時取等號,∴f(x)的最小值為1,∴a≥1,故選A.[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題轉(zhuǎn)化為利用導(dǎo)數(shù)研究函數(shù)單調(diào)性的問題,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,+∞)時有解,則實數(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變式題(1)D(2)C[解析](1)由題意可知,f(x)在定義域上單調(diào)遞增,f(0)=1,則由f(ax-ex+1)>1=f(0),得ax-ex+1>0,即ax+1>ex.令g(x)=ax+1,h(x)=ex,則當(dāng)x∈(0,+∞)時,存在g(x)的圖像在h(x)的圖像上方.g(0)=1,h(0)=1,g'(x)=a,h'(x)=ex,則需滿足g'(0)>h'(0),即a>1.故選D.(2)由題可得f'(x)=ax2+bx+c,f'(x)>0的解集為(-3,1),故f'(x)=a(x+3)(x-1),a<0,可得b=2a,c=-3a,∴b<a<c,故選C.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,e1.C[解析]由題意知,函數(shù)f(x)的定義域為(0,+∞),f'(x)=lnx+1,令f'(x)>0,解得x>1e,所以函數(shù)f(x)的單調(diào)遞增區(qū)間為1e,+∞.故選C.2.已知函數(shù)f(x)=x2-4x,當(dāng)x∈(0,+∞)時,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-2lnx2.D[解析]f(x)=x2-4x在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增.A選項,g(x)=x3-2在(0,+∞)上單調(diào)遞增.B選項,g'(x)=(x-1)ex,當(dāng)0<x<1時,g'(x)<0,函數(shù)g(x)單調(diào)遞減;當(dāng)x>1時,g'(x)>0,函數(shù)g(x)單調(diào)遞增.C選項,g'(x)=(2-x)ex,當(dāng)0<x<2時,g'(x)>0,函數(shù)g(x)單調(diào)遞增;當(dāng)x>2時,g'(x)<0,函數(shù)g(x)單調(diào)遞減.D選項,g'(x)=1-2x=x3.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)3.B[解析]∵f(-x)=(-x)2-cos(-x)=x2-cosx=f(x),∴f(x)為偶函數(shù),∴f(0.5)=f(-0.5).f'(x)=2x+sinx,當(dāng)0<x<π2時,f'(x)=2x+sinx>0,∴函數(shù)f(x)在0,π2上單調(diào)遞增,又0<0.5<0.6<π24.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K16-24.A[解析]當(dāng)x<-2時,f'(x)<0,則f(x)單調(diào)遞減;當(dāng)-2<x<0時,f'(x)>0,則f(x)單調(diào)遞增;當(dāng)x>0時,f'(x)<0,則f(x)單調(diào)遞減.所以符合上述條件的只有選項A.故選A.5.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,+∞)5.B[解析]函數(shù)f(x)的定義域為(0,+∞),f'(x)=1x-x=1-x6.使“函數(shù)f(x)=exx在區(qū)間(0,m]上單調(diào)遞減”成立的一個m值是6.12(答案不唯一)[解析]由題意,知f'(x)=(x-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,π27.B[解析]因為f(x)=sin2x+2cosx,所以f'(x)=2cos2x-2sinx=2(1-2sin2x)-2sinx=-2(sinx+1)(2sinx-1),令f'(x)≤0,因為sinx+1>0,所以2sinx-1≥0,即sinx≥12,又x∈0,π2,所以x∈π6,π2,所以函數(shù)f(x)的單調(diào)遞減區(qū)間是π6,π2.故選B.8.設(shè)函數(shù)f(x)滿足f(x)=x[f'(x)-lnx],且在(0,+∞)上單調(diào)遞增,則f1e的取值范圍是(e為自然對數(shù)的底數(shù)) ()A.[-1,+∞) B.1e,+∞C.-∞,1e D.(-∞,-1]8.B[解析]由f(x)=x[f'(x)-lnx],得f'(x)=f(x)x+lnx,因為函數(shù)f(x)在(0,+∞)上單調(diào)遞增,所以f'(x)≥0在(0,+∞)上恒成立,即f(x)x+lnx≥0,所以f(9.已知函數(shù)f(x)=a-2A.a≤0 B.a≤1C.a≤2e D.a≤3e9.C[解析]根據(jù)題意,函數(shù)f(x)=a當(dāng)x≤1時,f(x)=a-2ex-1,為減函數(shù),當(dāng)x>1時,f(x)=xlnx-2x+a,其導(dǎo)數(shù)f'(x)=lnx-1,可知f(x)在(1,e)上單調(diào)遞減,在(e,+∞)上單調(diào)遞增,且f(x)min=f(e)=a-e,f(x)的大致圖像如圖,若函數(shù)y=f(x)與y=f[f(x)]有相同的值域,則需滿足a-e≤e,則a≤2e.故選C.10.(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>210.BCD[解析]因為f(x)=lnx+1x,所以f'(x)=x-1x2.由于函數(shù)的定義域為(0,+∞),故A錯誤;當(dāng)x∈(e,+∞)時,f'(x)>0,所以函數(shù)f(x)在(e,+∞)上單調(diào)遞增,故B正確;令f'(x)>0,得x>1,令f'(x)<0,得0<x<1,所以函數(shù)f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以當(dāng)x=1時,函數(shù)取得最小值f(1)=1,故C正確;不妨令0<m<n,由f(m)=f(n)得n-mmn=lnnm,欲證m+n>2,只需證明(n-m)(m+11.(多選題)對于定義域為R的函數(shù)f(x),若滿足:①f(0)=0;②當(dāng)x∈R且x≠0時,都有xf'(x)>0;③當(dāng)x1<0<x2且|x1|<|x2|時,都有f(x1)<f(x2).則稱f(x)為“偏對稱函數(shù)”,下列函數(shù)是“偏對稱函數(shù)”的是 ()A.f1(x)=-x3+x2B.f2(x)=ex-x-1C.f3(x)=xsinxD.f4(x)=ln11.BD[解析]經(jīng)驗證,f1(x),f2(x),f3(x),f4(x)都滿足條件①.當(dāng)x∈R且x≠0時,都有xf'(x)>0,即當(dāng)x>0時,f'(x)>0,且當(dāng)x<0時,f'(x)<0,即條件②等價于函數(shù)f(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增.x1<0<x2且|x1|<|x2|等價于-x2<x1<0<-x1<x2,故條件③等價于若-x2<x1<0<-x1<x2,則f(x1)<f(x2).A中,f1(x)=-x3+x2,f1'(x)=-3x2+2x,則當(dāng)x≠0時,由xf1'(x)=-3x3+2x2=x2(2-3x)≤0,得x≥23,不符合條件②,故f1(x)不是“偏對稱函數(shù)”.B中,f2(x)=ex-x-1,f2'(x)=ex-1,當(dāng)x>0時,ex>1,f2'(x)>0,當(dāng)x<0時,0<ex<1,f2'(x)<0,∴函數(shù)f2(x)=ex-x-1在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,符合條件②.由f2(x)的單調(diào)性知,當(dāng)-x2<x1<0<-x1<x2時,f2(x1)<f2(-x2),∴f2(x1)-f2(x2)<f2(-x2)-f2(x2)=-ex2+e-x2+2x則F'(x)=-ex-e-x+2<-2ex·e-x+2=0,∴F(x)在(0,+∞)上是減函數(shù),∴F(x)<F(0)=0,∴f2(x1)<f2(x2),符合條件③,故f2(x)是“偏對稱函數(shù)”.C中,f3(x)=xsinx,則f3(-x)=-xsin(-x)=f3(x),則f3(x)是偶函數(shù),而f3'(x)=sinx+xcosx=1+x2sin(x+φ)(tanφ=x),則根據(jù)三角函數(shù)的性質(zhì)可知,當(dāng)x>0時,f3'(x)的符號有正有負,不符合條件②,故f3(x)不是“偏對稱函數(shù)”.D中,f4(x)=ln(-x+1),x≤0,2x,x>0,當(dāng)x<0時,f4'(x)=1x-1<0,當(dāng)x>0時,f4'(x)=2>0,∴函數(shù)f4(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,符合條件②.由單調(diào)性知,當(dāng)-x2<x1<0<-x1<x2時,f4(x1)<f4(-x2),∴f4(x1)-f4(x2)<f12.已知函數(shù)f(x)=ex+ae-x在[0,1]上不單調(diào),則實數(shù)a的取值范圍為.

12.(1,e2)[解析]由題意可得,f'(x)=ex-aex在(0,1)上有變號零點,故方程a=e2x在(0,1)上有解,因為y=e2x在(0,1)上單調(diào)遞增,所以當(dāng)x∈(0,1)時,e2x∈(1,e2),故1<a<e13.已知函數(shù)f(x)=sin