《(全國(guó)通用版)2018-2019版高中數(shù)學(xué) 第一章 導(dǎo)數(shù)及其應(yīng)用 1.3 導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用 1.3.1 函數(shù)的單調(diào)性與導(dǎo)數(shù)(一)學(xué)案 新人教A版選修2-2》由會(huì)員分享,可在線閱讀,更多相關(guān)《(全國(guó)通用版)2018-2019版高中數(shù)學(xué) 第一章 導(dǎo)數(shù)及其應(yīng)用 1.3 導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用 1.3.1 函數(shù)的單調(diào)性與導(dǎo)數(shù)(一)學(xué)案 新人教A版選修2-2(14頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、13.1函數(shù)的單調(diào)性與導(dǎo)數(shù)(一)學(xué)習(xí)目標(biāo)1.理解導(dǎo)數(shù)與函數(shù)的單調(diào)性的關(guān)系.2.掌握利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的方法.3.能利用導(dǎo)數(shù)求不超過(guò)三次多項(xiàng)式函數(shù)的單調(diào)區(qū)間知識(shí)點(diǎn)一函數(shù)的單調(diào)性與導(dǎo)函數(shù)的關(guān)系思考觀察圖中函數(shù)f(x),填寫(xiě)下表導(dǎo)數(shù)值切線的斜率傾斜角曲線的變化趨勢(shì)函數(shù)的單調(diào)性f(x)0k0銳角上升遞增f(x)0k0,則f(x)在這個(gè)區(qū)間內(nèi)單調(diào)遞增;(2)如果f(x)0,解集在定義域內(nèi)的部分為增區(qū)間;(4)解不等式f(x)0,解集在定義域內(nèi)的部分為減區(qū)間1函數(shù)f(x)在定義域上都有f(x)0.()類(lèi)型一函數(shù)圖象與導(dǎo)數(shù)圖象的應(yīng)用例1已知函數(shù)yf(x)的定義域?yàn)?,5,部分對(duì)應(yīng)值如下表f(x)的導(dǎo)函數(shù)
2、yf(x)的圖象如圖所示.x1045f(x)1221給出下列關(guān)于函數(shù)f(x)的說(shuō)法:函數(shù)yf(x)是周期函數(shù);函數(shù)f(x)在0,2上是減函數(shù);如果當(dāng)x1,t時(shí),f(x)的最大值是2,那么t的最大值為4;當(dāng)1a2時(shí),函數(shù)yf(x)a有4個(gè)零點(diǎn)其中正確說(shuō)法的個(gè)數(shù)是()A4 B3C2 D1考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)根據(jù)導(dǎo)函數(shù)的圖象確定原函數(shù)圖象答案D解析依題意得,函數(shù)f(x)不可能是周期函數(shù),因此不正確;當(dāng)x(0,2)時(shí),f(x)0,因此函數(shù)f(x)在0,2上是減函數(shù),正確;當(dāng)x1,t時(shí),若f(x)的最大值是2,則結(jié)合函數(shù)f(x)的可能圖象分析可知,此時(shí)t的最大值是5,因此不正確;注意到f(2
3、)的值不明確,結(jié)合函數(shù)f(x)的可能圖象分析可知,將函數(shù)f(x)的圖象向下平移a(1a0,則yf(x)在(a,b)上單調(diào)遞增;如果f(x)0,則yf(x)在這個(gè)區(qū)間上單調(diào)遞減;若恒有f(x)0,則yf(x)是常數(shù)函數(shù),不具有單調(diào)性(2)函數(shù)圖象變化得越快,f(x)的絕對(duì)值越大,不是f(x)的值越大跟蹤訓(xùn)練1已知yxf(x)的圖象如圖所示(其中f(x)是函數(shù)f(x)的導(dǎo)函數(shù)),則所給四個(gè)圖象中,yf(x)的圖象大致是()考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)根據(jù)導(dǎo)函數(shù)圖象確定原函數(shù)圖象答案C解析當(dāng)0x1時(shí),xf(x)0,f(x)1時(shí),xf(x)0,f(x)0,故yf(x)在(1,)上為增函數(shù)故選C.類(lèi)
4、型二利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間例2求下列函數(shù)的單調(diào)區(qū)間(1)yx2ln x;(2)yx(b0)考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求不含參數(shù)函數(shù)的單調(diào)區(qū)間解(1)函數(shù)yx2ln x的定義域?yàn)?0,),又y.若y0,即解得x1;若y0,即解得0x0,則(x)(x)0,所以x或x.所以函數(shù)的單調(diào)遞增區(qū)間為(,),(,)令f(x)0,則(x)(x)0,所以x0,函數(shù)在解集所表示的定義域內(nèi)為增函數(shù)(4)解不等式f(x)0,函數(shù)在解集所表示的定義域內(nèi)為減函數(shù)跟蹤訓(xùn)練2函數(shù)f(x)(x22x)ex(xR)的單調(diào)遞減區(qū)間為_(kāi)考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求不含參數(shù)函數(shù)的單調(diào)區(qū)間答案(2,2)解析
5、由f(x)(x24x2)ex0,即x24x20,解得2x0,得x1,由f(x)0,得0x0時(shí),f(x),a0,0.由f(x)0,得x1,由f(x)0,得0x0,所以f(x)在(,)上單調(diào)遞增若a0,則當(dāng)x(,ln a)時(shí),f(x)0.所以f(x)在(,ln a)上單調(diào)遞減,在(ln a,)上單調(diào)遞增綜上所述,當(dāng)a0時(shí),函數(shù)f(x)在(,)上單調(diào)遞增;當(dāng)a0時(shí),f(x)在(,ln a)上單調(diào)遞減,在(ln a,)上單調(diào)遞增.1函數(shù)f(x)xln x()A在(0,6)上是增函數(shù)B在(0,6)上是減函數(shù)C在上是減函數(shù),在上是增函數(shù)D在上是增函數(shù),在上是減函數(shù)考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)利用導(dǎo)數(shù)值
6、的正負(fù)號(hào)判定函數(shù)的單調(diào)性答案A2若函數(shù)f(x)的圖象如圖所示,則導(dǎo)函數(shù)f(x)的圖象可能為()考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)根據(jù)原函數(shù)圖象確定導(dǎo)函數(shù)圖象答案C解析由f(x)的圖象可知,函數(shù)f(x)的單調(diào)遞增區(qū)間為(1,4),單調(diào)遞減區(qū)間為(,1)和(4,),因此,當(dāng)x(1,4)時(shí),f(x)0,當(dāng)x(,1)或x(4,)時(shí),f(x)0,即ln x10,得x.故函數(shù)f(x)的單調(diào)遞增區(qū)間為.4若函數(shù)f(x)x3bx2cxd的單調(diào)遞減區(qū)間為1,2,則b_,c_.考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)已知單調(diào)區(qū)間求參數(shù)值答案6解析f(x)3x22bxc,由題意知,f(x)0即3x22bxc0的兩根為1和2.
7、由得5試求函數(shù)f(x)kxln x的單調(diào)區(qū)間考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求含參數(shù)函數(shù)的單調(diào)區(qū)間解函數(shù)f(x)kxln x的定義域?yàn)?0,),f(x)k.當(dāng)k0時(shí),kx10,f(x)0時(shí),由f(x)0,即0,解得0x0,即0,解得x.當(dāng)k0時(shí),f(x)的單調(diào)遞減區(qū)間為,單調(diào)遞增區(qū)間為.綜上所述,當(dāng)k0時(shí),f(x)的單調(diào)遞減區(qū)間為(0,);當(dāng)k0時(shí),f(x)的單調(diào)遞減區(qū)間為,單調(diào)遞增區(qū)間為.1導(dǎo)數(shù)的符號(hào)反映了函數(shù)在某個(gè)區(qū)間上的單調(diào)性,導(dǎo)數(shù)絕對(duì)值的大小反映了函數(shù)在某個(gè)區(qū)間或某點(diǎn)附近變化的快慢程度2利用導(dǎo)數(shù)求函數(shù)f(x)的單調(diào)區(qū)間的一般步驟:(1)確定函數(shù)f(x)的定義域;(2)求導(dǎo)數(shù)f(
8、x);(3)在函數(shù)f(x)的定義域內(nèi)解不等式f(x)0和f(x)0,所以在(4,5)上,f(x)是增函數(shù)2函數(shù)yf(x)的圖象如圖所示,則導(dǎo)函數(shù)yf(x)的圖象可能是()考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)根據(jù)原函數(shù)圖象確定導(dǎo)函數(shù)圖象答案D解析函數(shù)f(x)在(0,),(,0)上都是減函數(shù),當(dāng)x0時(shí),f(x)0,當(dāng)x0時(shí),f(x)0),函數(shù)在(,0)上單調(diào)遞減,在(0,a)上單調(diào)遞增,在(a,)上單調(diào)遞減,故選C.4函數(shù)f(x)xex的一個(gè)單調(diào)遞增區(qū)間是()A1,0 B2,8C1,2 D0,2考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求不含參數(shù)的函數(shù)的單調(diào)區(qū)間答案A解析因?yàn)閒(x)(1x)ex0,又因
9、為ex0,所以x0,yxex在(0,)內(nèi)為增函數(shù)6.函數(shù)f(x)的導(dǎo)函數(shù)f(x)的圖象如圖所示,若ABC為銳角三角形,則下列不等式一定成立的是()Af(cos A)f(cos B)Bf(sin A)f(sin B)Df(sin A)f(cos B)考點(diǎn)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性題點(diǎn)比較函數(shù)值的大小答案D解析根據(jù)圖象知,當(dāng)0x0,f(x)在區(qū)間(0,1)上是增函數(shù)ABC為銳角三角形,A,B都是銳角且AB,則0BA,則sinsin A,0cos Bsin Af(cos B)7定義在R上的函數(shù)f(x),若(x1)f(x)2f(1)Bf(0)f(2)2f(1)Cf(0)f(2)2f(1)Df(0)f(2
10、)與2f(1)大小不定考點(diǎn)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性題點(diǎn)比較函數(shù)值的大小答案C解析(x1)f(x)1時(shí),f(x)0,x0,則f(x)在(1,)上單調(diào)遞減,在(,1)上單調(diào)遞增,f(0)f(1),f(2)f(1),則f(0)f(2)2f(1)二、填空題8若函數(shù)f(x)的導(dǎo)函數(shù)為f(x)x24x3,則函數(shù)f(x1)的單調(diào)遞減區(qū)間是_考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求不含參數(shù)的函數(shù)的單調(diào)區(qū)間答案(0,2)解析由f(x)x24x3,f(x1)(x1)24(x1)3x22x,令f(x1)0,解得0x2,所以f(x1)的單調(diào)遞減區(qū)間是(0,2)9.在R上可導(dǎo)的函數(shù)f(x)的圖象如圖所示,則關(guān)于x的不
11、等式xf(x)0的解集為_(kāi)考點(diǎn)函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系題點(diǎn)利用單調(diào)性確定導(dǎo)數(shù)值的正負(fù)號(hào)答案(,1)(0,1)解析由xf(x)0可得,或由題圖可知當(dāng)1x1時(shí),f(x)0,當(dāng)x1時(shí),f(x)0,則或解得0x1或x1,xf(x)0,解得x0,故f(x)的單調(diào)遞增區(qū)間為(0,)11已知函數(shù)f(x)2x3ax21(a為常數(shù))在區(qū)間(,0),(2,)上單調(diào)遞增,且在區(qū)間(0,2)上單調(diào)遞減,則a的值為_(kāi)考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)已知單調(diào)區(qū)間求參數(shù)值答案6解析由題意得f(x)6x22ax0的兩根為0和2,可得a6.12定義在R上的函數(shù)f(x)滿足f(1)1,f(x)2x1的x的取值范圍是_考點(diǎn)利用導(dǎo)數(shù)
12、研究函數(shù)的單調(diào)性題點(diǎn)構(gòu)造法的應(yīng)用答案(,1)解析令g(x)f(x)2x1,則g(x)f(x)2g(1)0時(shí),x0,即f(x)2x1的解集為(,1)三、解答題13已知函數(shù)f(x)x3bx2cxd的圖象經(jīng)過(guò)點(diǎn)P(0,2),且在點(diǎn)M(1,f(1)處的切線方程為6xy70.(1)求函數(shù)yf(x)的解析式;(2)求函數(shù)yf(x)的單調(diào)區(qū)間考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求不含參數(shù)的函數(shù)的單調(diào)區(qū)間解(1)由yf(x)的圖象經(jīng)過(guò)點(diǎn)P(0,2),知d2,f(x)x3bx2cx2,f(x)3x22bxc.由在點(diǎn)M(1,f(1)處的切線方程為6xy70,知6f(1)70,即f(1)1.又f(1)6,即解得
13、bc3,故所求函數(shù)解析式是f(x)x33x23x2.(2)f(x)3x26x3.令f(x)0,得x1;令f(x)0,得1x0,試討論f(x)的單調(diào)性考點(diǎn)利用導(dǎo)函數(shù)求函數(shù)的單調(diào)區(qū)間題點(diǎn)利用導(dǎo)數(shù)求含參數(shù)的函數(shù)的單調(diào)區(qū)間解f(x)的定義域?yàn)?0,),f(x)1.令g(x)x2ax2,其判別式a28.(1)當(dāng)0,即0a0,都有f(x)0,此時(shí)f(x)是(0,)上的單調(diào)遞增函數(shù);(2)當(dāng)0,即a2時(shí),當(dāng)且僅當(dāng)x時(shí),有f(x)0,對(duì)定義域內(nèi)其余的x都有f(x)0,此時(shí)f(x)也是(0,)上的單調(diào)遞增函數(shù);(3)當(dāng)0,即a2時(shí),方程g(x)0有兩個(gè)不同的實(shí)根:x1,x2,0x1x2.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x(0,x1)x1(x1,x2)x2(x2,)f(x)00f(x)即f(x)在和上單調(diào)遞增;在上單調(diào)遞減.14