16導(dǎo)數(shù)的單調(diào)性練習(xí)

1、若函數(shù)f(x)=sinx-萩則函數(shù)f(x)在區(qū)間(0,兀上的單調(diào)增區(qū)間為()-兀-兀0,gB.予兀7171C.0,方D.0,31. 函數(shù)f(x)=x3+ax2+bx+c,其中a,b,c為實數(shù)，當(dāng)a2-3b<0時，f(x)在R上()A.是增函數(shù)A. 是減函數(shù)B. 是常函數(shù)C. 既不是增函數(shù)也不是減函數(shù)已知函數(shù)y=f(x)在-4,6內(nèi)可導(dǎo)，其圖象如圖，記y=f(x)的導(dǎo)函數(shù)為y=f'x),則不等式f'x)<0的解集為()A-4,1U業(yè)，6A.337RB.3,0U3,5C.-4，-4U1,733D.-4,-3U0,1U5,62. (2019石家莊二中月考)已知函數(shù)f(x)

2、=寸x+lnx,則下列選項正確的是()A.f(e)<f(兀)<2)B.f(兀fe)<f(2)C. f(e)<f(2)<f(兀)D. f(2)<f(e)<f(兀)3. 若函數(shù)f(x)=kxInx在區(qū)間(1,+°°)上單調(diào)遞增，貝Uk的取值范圍是()A.(8,2B.(8,1C.2,+8)D.1,+8)4. 函數(shù)f(x)=(x2+x+1)ex(xR)的單調(diào)減區(qū)間為7.函數(shù)y=f(x)的導(dǎo)函數(shù)的圖象如圖所示，給出下列判斷:1, 函數(shù)y=f(x)在區(qū)間一3,2上單倜遞增; 函數(shù)y=f(x)在區(qū)間一；,3上單調(diào)遞減; 函數(shù)y=f(x)在區(qū)間(4

3、,5)上單調(diào)遞增.上述判斷正確的是(填序號).3(2019泰州期末檢測)右函數(shù)f(x)=(x2+mx)e的單倜減區(qū)間是一分,1,則頭數(shù)m的值為Inx8. 證明：函數(shù)f(x)=在區(qū)間(0,2)上單倜遞增.x9. 已知函數(shù)f(x)=x2+ax-Inx,若f(x)為單調(diào)遞減函數(shù)，求實數(shù)a的取值范圍.B能力提升10. 已知函數(shù)f(x)=x2-ax+3在(0,1)上為減函數(shù)，函數(shù)g(x)=x2alnx在(1,2)上為增函數(shù)，貝Ua=()A.1B.2C.0D.V212.已知函數(shù)f(x)的定義域為R,f(x)為f(x)的導(dǎo)函數(shù)，且f(x)(x1)f(x)0,則A.f(1)0B.f(x)0C.f(x)0D.(x1)f(x)013.已知函數(shù)f(x)是R上的偶函數(shù)，且在(0,+8)上有f'x)>0,若f(-1)=0,則關(guān)于x的不等式xf(x)<0的解集是.14.已知函數(shù)f(x)=x3+ax2+7(aR).試討論f(x)的單調(diào)性.,-(x1)215.已知函數(shù)f(x)=lnx2(1)求函數(shù)f(x)的單調(diào)遞增區(qū)間;證明：當(dāng)x