課前基礎(chǔ)鞏固課堂考點(diǎn)探究第16講導(dǎo)數(shù)與函數(shù)的單調(diào)性教師備用習(xí)題作業(yè)手冊(cè)1.借助幾何直觀了解函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系.

2.能利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性.

3.對(duì)于多項(xiàng)式函數(shù),能求不超過三次的多項(xiàng)式函數(shù)的單調(diào)區(qū)間.課標(biāo)要求導(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“y=f(x)單調(diào)遞增(減)”的

條件

函數(shù)的單調(diào)性與導(dǎo)數(shù)遞增遞減課前基礎(chǔ)鞏固?知識(shí)聚焦?≥0≤0充分題組一常識(shí)題1.[教材改編]函數(shù)f(x)=ex-x的單調(diào)遞增區(qū)間是

.2.[教材改編]比較大小:x

lnx(x∈(1,+∞)).

課前基礎(chǔ)鞏固?對(duì)點(diǎn)演練?(0,+∞)

[解析]由f'(x)=ex-1>0,解得x>0,故f(x)的單調(diào)遞增區(qū)間是(0,+∞).>

3.[教材改編]已知f(x)是定義在R上的可導(dǎo)函數(shù),函數(shù)y=ef'(x)的圖像如圖3-16-1所示,則f(x)的單調(diào)遞減區(qū)間是

.

課前基礎(chǔ)鞏固

[解析]因?yàn)楫?dāng)x≤2時(shí),ef'(x)≤1,所以當(dāng)x≤2時(shí),f'(x)≤0,所以f(x)的單調(diào)遞減區(qū)間是(-∞,2].圖3-16-1題組二常錯(cuò)題索引:弄錯(cuò)可導(dǎo)函數(shù)在某區(qū)間上單調(diào)時(shí)導(dǎo)數(shù)滿足的條件致誤;求單調(diào)區(qū)間時(shí)忽略定義域;討論函數(shù)單調(diào)性時(shí)分類標(biāo)準(zhǔn)有誤.4.若函數(shù)f(x)=kx-lnx在區(qū)間(1,+∞)上單調(diào)遞增,則k的取值范圍是

.

課前基礎(chǔ)鞏固[1,+∞)

5.函數(shù)f(x)=x+ln(2-x)的單調(diào)遞增區(qū)間為

.

課前基礎(chǔ)鞏固

6.討論函數(shù)y=ax3-x在R上的單調(diào)性時(shí),對(duì)參數(shù)a應(yīng)分

情況討論.

[解析]

y'=3ax2-1,所以對(duì)a應(yīng)分a>0,a=0,a<0三種情況討論.例1(多選題)已知函數(shù)y=f(x)的導(dǎo)函數(shù)y=f'(x)的圖像如圖3-16-2所示,則下列結(jié)論正確的是(

)A.f(a)<f(b)<f(c)B.f(e)<f(d)<f(c)C.當(dāng)x=c時(shí),f(x)取得極小值D.當(dāng)x=d時(shí),f(x)取得最小值課堂考點(diǎn)探究探究點(diǎn)一導(dǎo)函數(shù)圖像的應(yīng)用[思路點(diǎn)撥]利用導(dǎo)函數(shù)圖像,結(jié)合導(dǎo)函數(shù)的符號(hào)確定函數(shù)的單調(diào)性、極值及函數(shù)值的大小關(guān)系.圖3-16-2AB課堂考點(diǎn)探究[解析]結(jié)合導(dǎo)函數(shù)的圖像,可知f(x)在(-∞,c]上單調(diào)遞增,在(c,e)上單調(diào)遞減,在[e,+∞)上單調(diào)遞增.對(duì)于A,因?yàn)閍<b<c,所以由f(x)的單調(diào)性可知f(a)<f(b)<f(c),故A正確;對(duì)于B,因?yàn)閏<d<e,所以由f(x)的單調(diào)性可知f(c)>f(d)>f(e),故B正確;對(duì)于C,當(dāng)x=c時(shí),f(x)取得極大值,故C錯(cuò)誤;對(duì)于D,由B可知,f(d)不是f(x)的最小值,故D錯(cuò)誤.故選AB.圖3-16-2[總結(jié)反思]函數(shù)的導(dǎo)函數(shù)f'(x)的單調(diào)性不能確定函數(shù)f(x)的單調(diào)性,反而是導(dǎo)函數(shù)f'(x)的正負(fù)區(qū)間可以確定函數(shù)f(x)的單調(diào)區(qū)間,即f'(x)>0?原函數(shù)在對(duì)應(yīng)區(qū)間單調(diào)遞增;f'(x)<0?原函數(shù)在對(duì)應(yīng)區(qū)間單調(diào)遞減.課堂考點(diǎn)探究課堂考點(diǎn)探究變式題

[2021·云南師大附中模擬]已知定義域?yàn)镽的函數(shù)f(x)的導(dǎo)函數(shù)圖像如圖3-16-3所示,則下列函數(shù)值的大小關(guān)系中,一定正確的是(

)

A.f(a)>f(b)>f(0) B.f(0)<f(c)<f(d)C.f(b)<f(0)<f(c) D.f(c)<f(d)<f(e)D[解析]由f(x)的導(dǎo)函數(shù)的圖像可知,f(x)在(a,b),(c,+∞)上單調(diào)遞增,在(b,c)上單調(diào)遞減,所以f(a)<f(b),A錯(cuò)誤;f(b)>f(0)>f(c),B,C錯(cuò)誤;f(c)<f(d)<f(e),D正確.故選D.圖3-16-3例2

(1)函數(shù)f(x)=x2-4lnx在定義域上的單調(diào)性是(

)A.先增后減 B.先減后增 C.單調(diào)遞增 D.單調(diào)遞減課堂考點(diǎn)探究探究點(diǎn)二求函數(shù)的單調(diào)性或單調(diào)區(qū)間(不含參)

B

課堂考點(diǎn)探究

A[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.課堂考點(diǎn)探究課堂考點(diǎn)探究

D

課堂考點(diǎn)探究

C

例3(1)已知函數(shù)f(x)=x2-a(x+alnx)(a≠0),討論f(x)的單調(diào)性.課堂考點(diǎn)探究探究點(diǎn)三討論含參函數(shù)的單調(diào)區(qū)間

課堂考點(diǎn)探究

課堂考點(diǎn)探究②當(dāng)0<a<e時(shí),令f'(x)>0,解得0<x<a或x>e,令f'(x)<0,解得a<x<e,∴f(x)的單調(diào)遞增區(qū)間為(0,a),(e,+∞),單調(diào)遞減區(qū)間為(a,e);③當(dāng)a=e時(shí),f'(x)≥0恒成立,∴f(x)的單調(diào)遞增區(qū)間為(0,+∞),無單調(diào)遞減區(qū)間;④當(dāng)a>e時(shí),令f'(x)>0,解得0<x<e或x>a,令f'(x)<0,解得e<x<a,∴f(x)的單調(diào)遞增區(qū)間為(0,e),(a,+∞),單調(diào)遞減區(qū)間為(e,a).

[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).對(duì)于含有參數(shù)的導(dǎo)數(shù)符號(hào)判定問題,應(yīng)就參數(shù)的范圍討論導(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.課堂考點(diǎn)探究課堂考點(diǎn)探究變式題

(1)已知函數(shù)f(x)=(m-2)x2-(m-4)x-lnx-2,則(

)A.當(dāng)0<m<2時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增B.當(dāng)m<0時(shí),函數(shù)f(x)在(1,+∞)上單調(diào)遞增C.當(dāng)m≥2時(shí),函數(shù)f(x)在(1,+∞)上單調(diào)遞增D.當(dāng)m=0時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增C課堂考點(diǎn)探究

課堂考點(diǎn)探究

課堂考點(diǎn)探究(2)[2021·北京朝陽區(qū)一模節(jié)選]已知函數(shù)f(x)=(ax-1)ex(a∈R),求f(x)的單調(diào)區(qū)間.

例4已知函數(shù)f(x)=x2-4x+(2-a)lnx,a∈R.(1)若f(x)在區(qū)間[2,+∞)內(nèi)單調(diào)遞增,求a的取值范圍;課堂考點(diǎn)探究探究點(diǎn)四已知函數(shù)單調(diào)性確定參數(shù)的取值范圍

課堂考點(diǎn)探究

例4已知函數(shù)f(x)=x2-4x+(2-a)lnx,a∈R.(2)若f(x)存在單調(diào)遞減區(qū)間,求a的取值范圍.[總結(jié)反思](1)f(x)在區(qū)間D上單調(diào)遞增(減),只要滿足f'(x)≥0(f'(x)≤0)在區(qū)間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)三類進(jìn)行討論.課堂考點(diǎn)探究課堂考點(diǎn)探究變式題

(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≤1 C.a≥1 D.a≥2A

課堂考點(diǎn)探究(2)[2021·重慶八中模擬]已知函數(shù)f(x)=x3+mlnx在區(qū)間[2,3]上不具有單調(diào)性,則m的取值范圍是(

)A.(-∞,-81) B.(-24,+∞) C.(-81,-24) D.(-81,+∞)C

課堂考點(diǎn)探究探究點(diǎn)五利用函數(shù)的單調(diào)性解決問題

C

課堂考點(diǎn)探究

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.課堂考點(diǎn)探究課堂考點(diǎn)探究

C

課堂考點(diǎn)探究(2)已知函數(shù)f(x)=e|x|+cosx,則使得f(2x)≤f(x-1)成立的x的取值范圍是

.

【備選理由】例1考查兩個(gè)函數(shù)的導(dǎo)函數(shù)的圖像與原函數(shù)圖像的關(guān)系,需要綜合分析;例2主要考查了利用導(dǎo)數(shù)求解函數(shù)的單調(diào)區(qū)間,解答中應(yīng)熟記導(dǎo)數(shù)與函數(shù)的單調(diào)性的關(guān)系,準(zhǔn)確運(yùn)算是解答的關(guān)鍵,難點(diǎn)是導(dǎo)數(shù)符號(hào)的判斷,著重考查推理與運(yùn)算能力;例3主要考查利用導(dǎo)數(shù)確定函數(shù)單調(diào)性再比較大小;例4考查含參函數(shù)的單調(diào)性問題,需要對(duì)參數(shù)分類討論;例5主要考查根據(jù)函數(shù)的單調(diào)區(qū)間確定參數(shù)的取值范圍,需要轉(zhuǎn)化為利用導(dǎo)數(shù)研究不等式恒成立問題,考查轉(zhuǎn)化思想.教師備用習(xí)題例1[配例1使用]如圖為函數(shù)y=f(x),y=g(x)的導(dǎo)函數(shù)的圖像,那么y=f(x),y=g(x)的圖像可能是()教師備用習(xí)題D[解析]從導(dǎo)函數(shù)的圖像知兩個(gè)函數(shù)的圖像在x=x0處的切線斜率相同,排除A,C;導(dǎo)函數(shù)的函數(shù)值反映的是原函數(shù)的圖像的切線的斜率大小,由題圖可知y=f(x)的導(dǎo)函數(shù)單調(diào)遞減,所以原函數(shù)圖像的切線的斜率隨著x的增大越來越小,排除B.故選D.ABCD

教師備用習(xí)題B

教師備用習(xí)題B

教師備用習(xí)題B

教師備用習(xí)題

例5[配例4使用]已知函數(shù)f(x)=ex-1-axlnx+(a-1)x(x>0).(1)若a=0,求函數(shù)f(x)的單調(diào)區(qū)間;教師備用習(xí)題解:當(dāng)a=0時(shí),f(x)=ex-1-x,f'(x)=ex-1-1,當(dāng)x>1時(shí),f'(x)=ex-1-1>0,所以f(x)在(1,+∞)上單調(diào)遞增;當(dāng)0<x<1時(shí),f'(x)=ex-1-1<0,所以f(x)在(0,1)上單調(diào)遞減.故函數(shù)f(x)的單調(diào)遞增區(qū)間為(1,+∞),單調(diào)遞減區(qū)間為(0,1).例5[配例4使用]已知函數(shù)f(x)=ex-1-axlnx+(a-1)x(x>0).(2)若函數(shù)f(x)為定義域內(nèi)的增函數(shù),求實(shí)數(shù)a的取值集合.教師備用習(xí)題

例5[配例4使用]已知函數(shù)f(x)=ex-1-axlnx+(a-1)x(x>0).(2)若函數(shù)f(x)為定義域內(nèi)的增函數(shù),求實(shí)數(shù)a的取值集合.教師備用習(xí)題(i)當(dāng)0<a<1時(shí),g'(a)=ea-1-1<0,g'(1)=1-a>0,所以?x0∈(a,1),使得g'(x0)=0,且當(dāng)x∈(0,x0)時(shí),g'(x)<0,當(dāng)x∈(x0,+∞)時(shí),g'(x)>0,因此當(dāng)x∈(x0,1)時(shí),g'(x)>0,此時(shí)g(x)<g(1)=0,不符合題意;(ii)當(dāng)a=1時(shí),g'(1)=1-a=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,故g(x)≥g(1)=0,符合題意;(iii)當(dāng)a>1時(shí),g'(a)=ea-1-1>0,g'(1)=1-a<0,所以?x0∈(1,a),使得g'(x0)=0,且當(dāng)x∈(0,x0)時(shí),g'(x)<0,當(dāng)x∈(x0,+∞)時(shí),g'(x)>0,因此當(dāng)x∈(1,x0)時(shí),g'(x)<0,此時(shí)g(x)<g(1)=0,不符合題意.綜上所述,a的取值集合為{1}.基礎(chǔ)熱身

圖K16-1C1234567891011121314151617

BC

1234567891011121314151617

B

1234567891011121314151617

C

1234567891011121314151617

1234567891011121314151617

圖K16-2

1234567891011121314151617

綜合提升

圖K16-3A.

B.

C.

D.

C1234567891011121314151617

1234567891011121314151617

B

1234567891011121314151617

D

1234567891011121314151617

B

1234567891011121314151617

ABD1234567891011121314151617

1234567891011121314151617

1234567891011121314151617

1234567891011121314151617

123456