2025高考數(shù)學(xué)二輪復(fù)習(xí)-微培優(yōu)1 函數(shù)圖象的公切線問(wèn)題【課件】

微培優(yōu)1函數(shù)圖象的公切線問(wèn)題2025求解公切線問(wèn)題一般思路:曲線的公切線問(wèn)題,主要利用導(dǎo)數(shù)的幾何意義進(jìn)行解決,關(guān)鍵是抓住切線的斜率進(jìn)行轉(zhuǎn)化.主要應(yīng)用在求公切線方程,切線有關(guān)的參數(shù).常用到以下三個(gè)基本關(guān)系:①切點(diǎn)處的導(dǎo)數(shù)是切線的斜率;②切點(diǎn)在切線上;③切點(diǎn)在曲線上.具體做法一:設(shè)公切線在y=f(x)上的切點(diǎn)P1(x1,f(x1)),在y=g(x)上的切點(diǎn)P2(x2,g(x2)),則f'(x1)=g'(x2)=具體做法二:設(shè)曲線C1:y=f(x)在點(diǎn)P1(x1,f(x1))處的切線為l1:y-f(x1)=f'(x1)(x-x1),整理可得y=f'(x1)·x-f'(x1)·x1+f(x1).設(shè)曲線C2:y=g(x)在點(diǎn)P2(x2,g(x2))處的切線為l2:y-g(x2)=g'(x2)(x-x2),整理得y=g'(x2)·x-g'(x2)·x2+g(x2),由于l1,l2是相同的直線,從而求出公切線方程,公切線的條數(shù)等價(jià)于該方程組的解的個(gè)數(shù).需要注意,做法一直接利用斜率公式,因此不包含P1與P2重合的情況,做法二包含了P1與P2重合的情況,求解時(shí)要根據(jù)題意靈活選用.角度一求兩曲線的公切線方程D(2)(2024·四川綿陽(yáng)高三模擬)已知f(x)=ex-1(e為自然對(duì)數(shù)的底數(shù)),g(x)=lnx+1,則f(x)與g(x)的公切線的方程為

.

y=ex-1或y=x角度二求與公切線有關(guān)的參數(shù)值或范圍問(wèn)題例2(2024·新高考Ⅰ,13)若曲線y=ex+x在點(diǎn)(0,1)處的切線也是曲線y=ln(x+1)+a的切線,則a=

.

ln2解析

由y=ex+x,得y'=ex+1.當(dāng)x=0時(shí),y'=2.∴曲線y=ex+x在點(diǎn)(0,1)處的切線方程為y-1=2(x-0),即y=2x+1.∴直線y=2x+1是曲線y=ln(x+1)+a的切線.例3(2022·全國(guó)甲,文20)已知函數(shù)f(x)=x3-x,g(x)=x2+a,曲線y=f(x)在點(diǎn)(x1,f(x1))處的切線也是曲線y=g(x)的切線.(1)若x1=-1,求a;(2)求a的取值范圍.解

(1)∵f'(x)=3x2-1,∴f'(-1)=2.當(dāng)x1=-1時(shí),f(-1)=0,故y=f(x)在點(diǎn)(-1,0)處的切線方程為y=2x+2.又y=2x+2與y=g(x)相切,將直線y=2x+2代入g(x)=x2+a,得x2-2x+a-2=0.由Δ=4-4(a-2)=0,得a=3.角度三求公切線條數(shù)問(wèn)題例4(2024·廣西南寧一模(1)問(wèn))已知函數(shù)f(x)=ex,g(x)=lnx,若直線l與函數(shù)f(x)和g(x)的圖象均相切,試討論直線l的條數(shù).

令t(x)=-xex+1,則t'(x)=-(x+1)ex,當(dāng)x>-1時(shí),t'(x)<0,t(x)=h'(x)單調(diào)遞減,當(dāng)x<-1時(shí),t'(x)>0,t(x)=h'(x)單調(diào)遞增,又當(dāng)x<0時(shí),t(x)>0,t(1)=1-e<0,故存在唯一的x0∈(0,1),t(x0)=0,故當(dāng)x∈(-∞,x0)時(shí),t(x)=h'(x)>0,h(x)單調(diào)遞增,當(dāng)x∈(x0,+∞)時(shí),t(x)=h'(x)<0,h(x)單調(diào)遞減,h(x)max=h(x0)>h(0)=2>0,又h(2)=3-e2<0,h(-2)=-1+3e-2<0,因此h(x)存在兩個(gè)零點(diǎn),故直線l的條數(shù)為2.針對(duì)訓(xùn)練1.(2024·福建模擬預(yù)測(cè))已知直線y=kx+b既是曲線y=lnx的切線,也是曲線y=-ln(-x)的切線,則(

)A解析

設(shè)直線與y=ln

x的切點(diǎn)為(x1,ln

x1),且x1>0,與y=-ln(-x)的切點(diǎn)為(x2,-ln(-x2)),且x2<0,2.已知函數(shù)f(x)=x3-x+a的圖象關(guān)于原點(diǎn)對(duì)稱,則與曲線y=f(x)和y=x2+均相切的直線l有(

)A.1條

B.2條

C.3條

D.4條C解析

由題意f(-x)=-f(x),所以a=0,所以f(x)=x3-x,所以f'(x)=3x2-1,所以y=f(x)在點(diǎn)(x1,f(x1))處的切線方程為3.(2024·遼寧二模)已知函數(shù)

的圖象與函數(shù)y2=ax(a>0,且a≠1)的圖象在公共點(diǎn)處有相同的切線,則a=

,切線方程為

.

4.(2024·陜西西安模擬改編)已知函數(shù)f(x)=a-2(a≠0),g(x)=lnx.若函數(shù)y=f(x),y=g(x)的圖象存在兩條公切線,求實(shí)數(shù)a的取值范圍.令h'(x)=0,得x=1,當(dāng)x∈(0,1)時(shí),h'(x)>0,h(x)單調(diào)遞增,當(dāng)x∈(1,+∞)時(shí),h'(x)<0,h(x)單調(diào)遞減,h(x)的最大值為h(1)=4,令h(x)=0,得x=,則當(dāng)x→0時(shí),h(x)→-∞;x→+∞時(shí),h(x)→0,則函數(shù)h(x)的大致圖象如下,故實(shí)數(shù)a的取值范圍為(0,2).5.(2024·湖北武漢高三模擬)已知拋物線C1:y=x2+2x和C2:y=-x2+a,如果直線l同