第三章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用第2節(jié)導(dǎo)數(shù)與函數(shù)的單調(diào)性學(xué)習(xí)導(dǎo)航站核心知識庫：重難考點(diǎn)總結(jié)，梳理必背知識、歸納重點(diǎn)考點(diǎn)1函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系★★★☆☆考點(diǎn)2利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的步驟★★★☆☆（星級越高，重要程度越高）限時【變式訓(xùn)練】挑戰(zhàn)場：感知真題，檢驗(yàn)成果，考點(diǎn)追溯【知識梳理】1.函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系★★★☆☆條件恒有結(jié)論函數(shù)y=f(x)在區(qū)間(a,b)上可導(dǎo)f'(x)>0f(x)在(a,b)上單調(diào)遞增f'(x)<0f(x)在(a,b)上單調(diào)遞減f'(x)=0f(x)在(a,b)上是常數(shù)函數(shù)2.利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的步驟★★★☆☆第1步,確定函數(shù)的定義域;第2步,求出導(dǎo)函數(shù)f'(x)的零點(diǎn);第3步,用f'(x)的零點(diǎn)將f(x)的定義域劃分為若干個區(qū)間,列表給出f'(x)在各區(qū)間上的正負(fù),由此得出函數(shù)y=f(x)在定義域內(nèi)的單調(diào)性.【名師點(diǎn)撥】1.若函數(shù)f(x)在(a,b)上單調(diào)遞增,則x∈(a,b)時,f'(x)≥0恒成立;若函數(shù)f(x)在(a,b)上單調(diào)遞減,則x∈(a,b)時,f'(x)≤0恒成立.2.若函數(shù)f(x)在(a,b)上存在單調(diào)遞增區(qū)間,則x∈(a,b)時,f'(x)>0有解;若函數(shù)f(x)在(a,b)上存在單調(diào)遞減區(qū)間,則x∈(a,b)時,f'(x)<0有解.【教材回歸】概念思考辨析+教材經(jīng)典改編1.思考辨析(在括號內(nèi)打“√”或“×”)(1)函數(shù)f(x)在(a,b)內(nèi)單調(diào)遞增,那么一定有f'(x)>0.()(2)在(a,b)內(nèi)f'(x)≤0且f'(x)=0的根為有限個,則f(x)在(a,b)內(nèi)單調(diào)遞減.()(3)若函數(shù)f(x)在定義域上都有f'(x)>0,則f(x)在定義域上一定單調(diào)遞增.()(4)函數(shù)f(x)=x-sinx在R上是增函數(shù).()2.(人教B選修三P95A組T1改編)已知函數(shù)f(x)的定義域?yàn)閇0,2],且y=f'(x)的圖象如圖所示,則f(x)的單調(diào)遞增區(qū)間是,單調(diào)遞減區(qū)間是.

3.(人教A選修二P101習(xí)題T3改編)函數(shù)f(x)=x3+2x2-4x的單調(diào)遞增區(qū)間是.

4.(蘇教選修一P213【典例】2改編)若函數(shù)f(x)=x3+ax2-ax在R上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是.

【考向核心題型】考點(diǎn)一不含參函數(shù)的單調(diào)性【典例】1(1)設(shè)函數(shù)f(x)在定義域內(nèi)可導(dǎo),f(x)的圖象如圖所示,則其導(dǎo)函數(shù)f'(x)的圖象可能是()(2)(2025·浙江名校聯(lián)考)函數(shù)f(x)=ln(2x-1)-x2+x的單調(diào)遞增區(qū)間是()A.(0,1) B.1C.1?22,【變式訓(xùn)練】1(1)下列函數(shù)在(0,+∞)上單調(diào)遞增的是()A.f(x)=sin2x B.f(x)=xexC.f(x)=x3-x D.f(x)=-x+lnx(2)設(shè)f(x)=2x2-x3,則f(x)的單調(diào)遞減區(qū)間是()A.0,43 C.(-∞,0) D.(-∞,0)和4考點(diǎn)二含參函數(shù)的單調(diào)性【典例】2已知函數(shù)f(x)=2ax3-3(a+1)x2+6x+1(a∈R),試討論函數(shù)f(x)的單調(diào)性.【變式訓(xùn)練】2(2021·全國乙卷節(jié)選)討論函數(shù)f(x)=x3-x2+ax+1的單調(diào)性.考點(diǎn)三函數(shù)單調(diào)性的應(yīng)用角度1由單調(diào)性求參數(shù)【典例】3(多選)(2025·茂名模擬)若f(x)=-13x3+12x2+2x+1是區(qū)間(m-1,m+4)上的單調(diào)函數(shù),則實(shí)數(shù)A.-4 B.-3 C.3 D.4角度2比較大小【典例】4已知函數(shù)f(x)=xsinx,x∈R,則fπ5,f(1),f?π3的大小關(guān)系為角度3解不等式【典例】5(2025·西安模擬)已知函數(shù)f(x)=(ex+e-x)x2,若滿足f(log3m)-e-1e<0,則實(shí)數(shù)mA.0,13 C.(0,3) D.(3,+∞)【變式訓(xùn)練】3(1)(2025·長沙模擬)若函數(shù)f(x)=ex-a+1-x在區(qū)間(0,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為()A.(-∞,-1] B.(-∞,1)C.[0,+∞) D.(-∞,1](2)已知函數(shù)f(x)=lnx-xex,設(shè)a=f32,b=f(2),c=fA.a>b>c B.b>a>cC.c>b>a D.c>a>b(3)不等式x<sinπx4+16的解集為【限時訓(xùn)練】（限時：60分鐘）一、單選題1.函數(shù)f(x)=(x-3)ex的單調(diào)遞減區(qū)間是()A.(-∞,2) B.(0,3)C.(1,4) D.(2,+∞)2.函數(shù)y=f(x)的導(dǎo)函數(shù)y=f'(x)的圖象如圖所示,則函數(shù)y=f(x)的圖象可能是()3.已知函數(shù)f(x)=2x-sinx,則下列結(jié)論正確的是()A.f(2.7)<f(π)<f(e) B.f(π)<f(e)<f(2.7)C.f(e)<f(2.7)<f(π) D.f(2.7)<f(e)<f(π)4.已知函數(shù)f(x)=x3+2x-sinx,若f(2a2)+f(a-1)≤0,則實(shí)數(shù)a的取值范圍為()A.?∞,?12B.?C.(-∞,-1]∪1D.?1,5.(2025·安徽名校聯(lián)考)已知函數(shù)f(x)=sinx+acosx在π4,π2上單調(diào)遞減A.(2-1,+∞) B.[1,+∞)C.(1-2,+∞) D.[-1,+∞)6.函數(shù)f(x)的圖象如圖所示,設(shè)f(x)的導(dǎo)函數(shù)為f'(x),則f(x)·f'(x)>0的解集為()A.(1,6) B.(1,4)C.(-∞,1)∪(6,+∞) D.(1,4)∪(6,+∞)7.已知函數(shù)f(x)=(1-x)lnx+ax在(1,+∞)上不單調(diào),則a的取值范圍是()A.(0,+∞) B.(1,+∞)C.[0,+∞) D.[1,+∞)8.已知f(x)=(a2-1)ex-1-12x2,若不等式f(lnx)<f(x-1)在(1,+∞)上恒成立,則aA.[-1,1]B.[-2,2]C.[0,1]D.(-∞,-2]∪[2,+∞)二、多選題9.(2025·晉城模擬)若一個函數(shù)在區(qū)間D上的導(dǎo)數(shù)值恒大于0,則該函數(shù)在D上純粹遞增,若一個函數(shù)在區(qū)間D上的導(dǎo)數(shù)值恒小于0,則該函數(shù)在D上純粹遞減,則()A.函數(shù)f(x)=x2-2x在[1,+∞)上純粹遞增B.函數(shù)f(x)=x3-2x在[1,2]上純粹遞增C.函數(shù)f(x)=sinx-2x在[0,1]上純粹遞減D.函數(shù)f(x)=ex-3x在[0,2]上純粹遞減10.若函數(shù)f(x)=12x2-9lnx在區(qū)間[m-1,m+1]上單調(diào),則實(shí)數(shù)mA.1 B.2 C.3 D.411.已知函數(shù)f(x)=x?1xlnx,且a=f23,b=f45,c=f(A.a>b B.b>aC.c>b D.c>a三、填空題12.已知函數(shù)f(x)=(x?2)ex,x≥0,13.若函數(shù)f(x)=x3+bx2+x恰有三個單調(diào)區(qū)間,則實(shí)數(shù)b的取值范圍為.

14.(2025·保定質(zhì)檢)已知x∈?π2,π2,則不等式esi