§3.2導(dǎo)數(shù)與函數(shù)的單調(diào)性考試要求1.結(jié)合實例，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，會求函數(shù)的單調(diào)區(qū)間(其中多項式函數(shù)一般不超過三次).3.會利用函數(shù)的單調(diào)性判斷大小，求參數(shù)的取值范圍等簡單應(yīng)用．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)在區(qū)間(a，b)上____________f′(x)<0f(x)在區(qū)間(a，b)上____________f′(x)＝0f(x)在區(qū)間(a，b)上是____________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____________；第2步，求出導(dǎo)數(shù)f′(x)的________；第3步，用f′(x)的零點將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．常用結(jié)論上單調(diào)遞減，則當(dāng)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，則當(dāng)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，則當(dāng)x∈(a，b)時，f′(x)<0有解．思考辨析判斷下列結(jié)論是否正確(請在括號中打“√”或“×”)(1)如果函數(shù)f(x)在某個區(qū)間內(nèi)恒有f′(x)＝0，則f(x)在此區(qū)間內(nèi)沒有單調(diào)性．()(2)在(a，b)內(nèi)f′(x)≤0且f′(x)＝0的根有有限個，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)．()教材改編題1．f′(x)是f(x)的導(dǎo)函數(shù)，若f′(x)的圖象如圖所示，則f(x)的圖象可能是()2．函數(shù)f(x)＝x2－2lnx的單調(diào)遞減區(qū)間是()A．(0,1) B．(1，＋∞)C．(－∞，1) D．(－1,1)3．已知函數(shù)f(x)＝xsinx，x∈R，則f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,5)))，f(1)，f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)))的大小關(guān)系為________．(用“<”連接)題型一不含參函數(shù)的單調(diào)性例1(1)函數(shù)f(x)＝eq\f(3lnx,2)＋eq\f(9,2x)＋x的單調(diào)遞減區(qū)間為________________．(2)若函數(shù)f(x)＝eq\f(lnx＋1,ex)，則函數(shù)f(x)的單調(diào)遞增區(qū)間為________________．聽課記錄：______________________________________________________________________________________________________________________________________思維升華確定不含參數(shù)的函數(shù)的單調(diào)性，按照判斷函數(shù)單調(diào)性的步驟即可，但應(yīng)注意兩點，一是不能漏掉求函數(shù)的定義域，二是函數(shù)的單調(diào)區(qū)間不能用并集，要用“逗號”或“和”隔開．跟蹤訓(xùn)練1已知函數(shù)f(x)＝x－lnx－eq\f(ex,x).判斷函數(shù)f(x)的單調(diào)性．________________________________________________________________________________________________________________________________________________________________________________________________________________________題型二含參數(shù)的函數(shù)的單調(diào)性例2已知函數(shù)f(x)＝aex－x，a∈R.(1)當(dāng)a＝1時，求曲線y＝f(x)在點(1，f(1))處的切線方程；(2)試討論函數(shù)f(x)的單調(diào)性．________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華(1)研究含參數(shù)的函數(shù)的單調(diào)性，要依據(jù)參數(shù)對不等式解集的影響進(jìn)行分類討論．(2)劃分函數(shù)的單調(diào)區(qū)間時，要在函數(shù)定義域內(nèi)討論，還要確定導(dǎo)數(shù)為零的點和函數(shù)的間斷點．跟蹤訓(xùn)練2(2023·濰坊模擬)已知函數(shù)f(x)＝ex－ax－a，a∈R，討論f(x)的單調(diào)區(qū)間．________________________________________________________________________________________________________________________________________________________________________________________________________________________題型三函數(shù)單調(diào)性的應(yīng)用命題點1比較大小或解不等式例3(1)(多選)下列不等式成立的是()A．2ln

eq\f(3,2)<eq\f(3,2)ln2 B.eq\r(2)lneq\r(3)<eq\r(3)lneq\r(2)C．5ln4<4ln5 D．π>elnπ(2)函數(shù)f(x)＝ex－e－x＋sinx，則不等式f(x)>0的解集是()A．(0，＋∞) B．(－∞，0)C．(0，π) D．(－π，0)聽課記錄：______________________________________________________________________________________________________________________________________命題點2根據(jù)函數(shù)的單調(diào)性求參數(shù)例4已知函數(shù)f(x)＝lnx－eq\f(1,2)ax2－2x(a≠0)．(1)若f(x)在[1,4]上單調(diào)遞減，求實數(shù)a的取值范圍；________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)若f(x)在[1,4]上存在單調(diào)遞減區(qū)間，求實數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________思維升華由函數(shù)的單調(diào)性求參數(shù)的取值范圍的方法(1)函數(shù)在區(qū)間(a，b)上單調(diào)，實際上就是在該區(qū)間上f′(x)≥0(或f′(x)≤0)恒成立．(2)函數(shù)在區(qū)間(a，b)上存在單調(diào)區(qū)間，實際上就是f′(x)>0(或f′(x)<0)在該區(qū)間上存在解集．跟蹤訓(xùn)練3(1)設(shè)函數(shù)f(x)是定義域為R的奇函數(shù)，且當(dāng)x≥0時，f(x)＝ex－cosx，則不等式f(2x－1)＋f(x－2)>0的解集為()A．(－∞，1