新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第08講 一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（二）（講）（解析版）

第2講一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（二）本講為重要知識(shí)點(diǎn)，也是導(dǎo)數(shù)中的難點(diǎn)。主要以切線的題型進(jìn)行總結(jié)，也包含了一些隱零點(diǎn)的思想和極值點(diǎn)偏移的思想解決相關(guān)的切線的問題。還是要注意函數(shù)的思想和導(dǎo)數(shù)的幾何意義來理解這類題的核心思想?？键c(diǎn)一由導(dǎo)數(shù)的幾何意義求基礎(chǔ)切線問題導(dǎo)數(shù)的幾何意義函數(shù)f(x)在點(diǎn)x0處的導(dǎo)數(shù)f′(x0)的幾何意義是在曲線y＝f(x)上點(diǎn)P(x0，f(x0))處的切線的斜率．相應(yīng)地，切線方程為y－f(x0)＝f′(x0)(x－x0)．給切點(diǎn)求切線以曲線上的點(diǎn)(x0，f(x0))（已知x0為具體值）為切點(diǎn)的切線方程的求解步驟：①求出函數(shù)f(x)的導(dǎo)數(shù)f′(x)；②求切線的斜率f′(x0)；③寫出切線方程y－f(x0)＝f′(x0)(x－x0)，并化簡．有切線無切點(diǎn)求切點(diǎn)以曲線上的點(diǎn)(x0，f(x0))（x0為未知值，可以設(shè)出來）為切點(diǎn)的切線方程的求解步驟：①求出函數(shù)f(x)的導(dǎo)數(shù)f′(x)；②求切線的斜率f′(x0)；③寫出切線方程y－f(x0)＝f′(x0)(x－x0)，并化簡．無切點(diǎn)求參規(guī)律同上，注意待定系數(shù)法的應(yīng)用。無切點(diǎn)多參思維同上，依舊是設(shè)切點(diǎn)，待定系數(shù)求解方程（組）?？键c(diǎn)二復(fù)雜切線問題“過點(diǎn)”型切線SKIPIF1<0以上是“在點(diǎn)”與“過點(diǎn)”的區(qū)別，判斷切線條數(shù)1.設(shè)點(diǎn)列方程過程同前（求切線過程）2.切線條數(shù)判斷，實(shí)質(zhì)是切點(diǎn)橫坐標(biāo)為變量的函數(shù)（方程）零點(diǎn)個(gè)數(shù)判斷多函數(shù)（多曲線）的公切線1.兩個(gè)曲線有公切線，且切點(diǎn)是同一點(diǎn)2.兩個(gè)曲線有公切線，但是切點(diǎn)不是同一點(diǎn)?？键c(diǎn)三切線的應(yīng)用切線的應(yīng)用：距離最值主要思維：利用平移直線，直到與該函數(shù)切線重合。切線的應(yīng)用：距離公式轉(zhuǎn)化型1.距離公式形式：平方和2.以此還可以類比斜率公式形式切線的應(yīng)用：恒成立求參等應(yīng)用利用切線作為“臨界線”放縮。這類思維，有時(shí)也應(yīng)用于大題的不等式證明，稱之為“切線放縮”。切線的應(yīng)用：零點(diǎn)等對于函數(shù)與直線交點(diǎn)個(gè)數(shù)，可以借助于切線（臨界線）來求解，但是一定要注意函數(shù)一般情況下，是比較簡單的凸凹函數(shù)。如下圖（示意圖），可以講清楚這里邊的“非充要”性高頻考點(diǎn)一由導(dǎo)數(shù)的幾何意義求基礎(chǔ)切線問題例1、已知函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的方程為__________.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則所求切線的方程為SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1、曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為______.【答案】SKIPIF1<0【解析】解：由SKIPIF1<0，得SKIPIF1<0，所以在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，所以所求的切線方程為SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0，例2、曲線SKIPIF1<0在SKIPIF1<0處的切線平行于直線SKIPIF1<0，則SKIPIF1<0點(diǎn)的坐標(biāo)為（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0和SKIPIF1<0D．SKIPIF1<0和SKIPIF1<0【答案】C【詳解】令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0，故選C.【變式訓(xùn)練】2、已知函數(shù)SKIPIF1<0為偶函數(shù)，若曲線SKIPIF1<0的一條切線與直線SKIPIF1<0垂直，則切點(diǎn)的橫坐標(biāo)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0為偶函數(shù)，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0設(shè)切點(diǎn)得橫坐標(biāo)為SKIPIF1<0，則SKIPIF1<0解得SKIPIF1<0，（負(fù)值舍去）所以SKIPIF1<0.故選：D例3、已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0的取值是（）A．-1 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0.故選：SKIPIF1<0.【變式訓(xùn)練】3、若曲線SKIPIF1<0的一條切線是直線SKIPIF1<0，則實(shí)數(shù)b的值為___________【答案】SKIPIF1<0【解析】設(shè)切點(diǎn)為SKIPIF1<0，對函數(shù)SKIPIF1<0求導(dǎo)，得到SKIPIF1<0，又曲線SKIPIF1<0的一條切線是直線SKIPIF1<0，所以切線斜率為SKIPIF1<0，∴SKIPIF1<0，因此SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，代入切線SKIPIF1<0，可得SKIPIF1<0.故答案為：SKIPIF1<0.例4、若直線SKIPIF1<0是曲線SKIPIF1<0的切線，且SKIPIF1<0，則實(shí)數(shù)b的最小值是______.【答案】SKIPIF1<0【解析】SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，由于直線SKIPIF1<0是曲線SKIPIF1<0的切線，設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)b遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)b遞減，∴SKIPIF1<0為極小值點(diǎn)，也為最小值點(diǎn)，∴b的最小值為SKIPIF1<0.故答案為：SKIPIF1<0．【變式訓(xùn)練】4、已知函數(shù)f（x）=axlnx﹣bx（a，b∈R）在點(diǎn)（e，f（e））處的切線方程為y=3x﹣e，則a+b=_____.【答案】0【詳解】∵在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，SKIPIF1<0，代入SKIPIF1<0得SKIPIF1<0①.又SKIPIF1<0②.聯(lián)立①②解得：SKIPIF1<0.SKIPIF1<0.故答案為：0.高頻考點(diǎn)二復(fù)雜切線問題例1、過原點(diǎn)作曲線SKIPIF1<0的切線，則切點(diǎn)的坐標(biāo)為___________，切線的斜率為__________.【答案】SKIPIF1<0SKIPIF1<0解：設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0；SKIPIF1<0；故由題意得，SKIPIF1<0；解得，SKIPIF1<0；故切點(diǎn)坐標(biāo)為SKIPIF1<0；切線的斜率為SKIPIF1<0；故切線方程為SKIPIF1<0，整理得SKIPIF1<0．故答案為：SKIPIF1<0；SKIPIF1<0.【變式訓(xùn)練】1、過點(diǎn)SKIPIF1<0與曲線SKIPIF1<0相切的直線方程為______________.【答案】SKIPIF1<0.【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0切線方程為SKIPIF1<0，SKIPIF1<0切線過點(diǎn)SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即所求切線方程為SKIPIF1<0.故答案為：SKIPIF1<0.例2、已知曲線SKIPIF1<0，則過點(diǎn)SKIPIF1<0可向SKIPIF1<0引切線，其切線條數(shù)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0的坐標(biāo)代入切線方程得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.因此，過點(diǎn)SKIPIF1<0可向SKIPIF1<0引切線，有三條.故選：C.【變式訓(xùn)練】2、已知過點(diǎn)A（a，0）作曲線C：y＝x?ex的切線有且僅有兩條，則實(shí)數(shù)a的取值范圍是（）A．（﹣∞，﹣4）∪（0，+∞） B．（0，+∞）C．（﹣∞，﹣1）∪（1，+∞） D．（﹣∞，﹣1）【答案】A【詳解】設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則切線方程為：SKIPIF1<0，切線過點(diǎn)SKIPIF1<0代入得：SKIPIF1<0SKIPIF1<0，即方程SKIPIF1<0有兩個(gè)解，則有SKIPIF1<0或SKIPIF1<0.故答案為:A.例3、直線SKIPIF1<0與曲線SKIPIF1<0相切也與曲線SKIPIF1<0相切，則稱直線SKIPIF1<0為曲線SKIPIF1<0和曲線SKIPIF1<0的公切線，已知函數(shù)SKIPIF1<0，其中SKIPIF1<0，若曲線SKIPIF1<0和曲線SKIPIF1<0的公切線有兩條，則SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】設(shè)切點(diǎn)求出兩個(gè)函數(shù)的切線方程，根據(jù)這個(gè)兩個(gè)方程表示同一直線，可得方程組，化簡方程組，可以得到變量SKIPIF1<0關(guān)于其中一個(gè)切點(diǎn)橫坐標(biāo)的函數(shù)形式，求導(dǎo)，求出函數(shù)的單調(diào)性，結(jié)合該函數(shù)的正負(fù)性，畫出圖象圖形，最后利用數(shù)形結(jié)合求出SKIPIF1<0的取值范圍.【詳解】設(shè)曲線SKIPIF1<0的切點(diǎn)為：SKIPIF1<0，SKIPIF1<0，所以過該切點(diǎn)的切線斜率為SKIPIF1<0，因此過該切點(diǎn)的切線方程為：SKIPIF1<0；設(shè)曲線SKIPIF1<0的切點(diǎn)為：SKIPIF1<0，SKIPIF1<0，所以過該切點(diǎn)的切線斜率為SKIPIF1<0，因此過該切點(diǎn)的切線方程為：SKIPIF1<0，則兩曲線的公切線應(yīng)該滿足：SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，所以函數(shù)有最大值為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，函數(shù)的圖象大致如下圖所示：要想有若曲線SKIPIF1<0和曲線SKIPIF1<0的公切線有兩條，則SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C【變式訓(xùn)練】3、函數(shù)SKIPIF1<0與SKIPIF1<0有公切線SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為（）A．4 B．2 C．1 D．SKIPIF1<0【答案】A【解析】設(shè)公切線SKIPIF1<0與兩個(gè)函數(shù)SKIPIF1<0與SKIPIF1<0圖象的切點(diǎn)分別為ASKIPIF1<0和BSKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0解得SKIPIF1<0，所以有SKIPIF1<0化簡得SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，則SKIPIF1<0恒成立，即得函數(shù)SKIPIF1<0SKIPIF1<0在定義域上為增函數(shù)，又因SKIPIF1<0，則可解得方程SKIPIF1<0，SKIPIF1<0，則由SKIPIF1<0解得SKIPIF1<0.故選：A.高頻考點(diǎn)三切線的應(yīng)用例1、點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖像上，若滿足到直線SKIPIF1<0的距離為1的點(diǎn)SKIPIF1<0有且僅有1個(gè)，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，設(shè)直線SKIPIF1<0與SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得切點(diǎn)為SKIPIF1<0，由題可知SKIPIF1<0到直線SKIPIF1<0的距離為1，所以SKIPIF1<0，解得SKIPIF1<0，結(jié)合圖象可知，SKIPIF1<0.故選：B.【變式訓(xùn)練】1、點(diǎn)A在直線y＝x上，點(diǎn)B在曲線SKIPIF1<0上，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．2【答案】A【分析】設(shè)平行于直線y＝x的直線y＝x＋b與曲線SKIPIF1<0相切，將題意轉(zhuǎn)化為兩平行線間的距離，由導(dǎo)數(shù)的幾何意義可得SKIPIF1<0的值，進(jìn)而可得結(jié)果.【詳解】設(shè)平行于直線y＝x的直線y＝x＋b與曲線SKIPIF1<0相切，則兩平行線間的距離即為SKIPIF1<0的最小值．設(shè)直線y＝x＋b與曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則由切點(diǎn)還在直線y＝x＋b上可得SKIPIF1<0，由切線斜率等于切點(diǎn)的導(dǎo)數(shù)值可得SKIPIF1<0，聯(lián)立解得m＝1，b＝－1，由平行線間的距離公式可得SKIPIF1<0的最小值為SKIPIF1<0，故選：A.例2、若SKIPIF1<0，則SKIPIF1<0的最小值是A．1 B．2 C．3 D．4【答案】B【詳解】由題意可轉(zhuǎn)化為點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0間的距離最小值的平方，點(diǎn)A在函數(shù)SKIPIF1<0上，點(diǎn)B在函數(shù)SKIPIF1<0上，這兩個(gè)函數(shù)關(guān)于SKIPIF1<0對稱，所以轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0的距離的最小值2倍的平方，此時(shí)SKIPIF1<0，∴SKIPIF1<0斜率為1的切線方程為SKIPIF1<0，它與SKIPIF1<0的距離為SKIPIF1<0.故原式的最小值為2.故選：B.【變式訓(xùn)練】2、若SKIPIF1<0，則SKIPIF1<0的最小值是A．1 B．2 C．3 D．4【答案】B【分析】原題等價(jià)于函數(shù)SKIPIF1<0上的點(diǎn)SKIPIF1<0與函數(shù)SKIPIF1<0上的點(diǎn)SKIPIF1<0間的距離最小值的平方，結(jié)合兩個(gè)函數(shù)關(guān)于SKIPIF1<0對稱，將其轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0的距離的最小值2倍的平方，利用導(dǎo)數(shù)求切線方程最后轉(zhuǎn)化求兩平行線間的距離平方即可.【詳解】由題意可轉(zhuǎn)化為點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0間的距離最小值的平方，點(diǎn)A在函數(shù)SKIPIF1<0上，點(diǎn)B在函數(shù)SKIPIF1<0上，這兩個(gè)函數(shù)關(guān)于SKIPIF1<0對稱，所以轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0的距離的最小值2倍的平方，此時(shí)SKIPIF1<0，∴SKIPIF1<0斜率為1的切線方程為SKIPIF1<0，它與SKIPIF1<0的距離為SKIPIF1<0.故原式的最小值為2.故選：B.例3、已知SKIPIF1<0為實(shí)數(shù)，則“SKIPIF1<0對任意的實(shí)數(shù)SKIPIF1<0恒成立”是“SKIPIF1<0”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】B【分析】先根據(jù)導(dǎo)數(shù)的幾何意義求出直線SKIPIF1<0與曲線SKIPIF1<0相切時(shí)SKIPIF1<0的值，再數(shù)形結(jié)合將SKIPIF1<0對任意的實(shí)數(shù)SKIPIF1<0恒成立轉(zhuǎn)化為SKIPIF1<0，最后判斷充要關(guān)系即可得解.【詳解】設(shè)直線SKIPIF1<0與曲線SKIPIF1<0相切，且切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，SKIPIF1<0，所以切線方程為SKIPIF1<0.數(shù)形結(jié)合可知，SKIPIF1<0對任意的實(shí)數(shù)SKIPIF1<0恒成立等價(jià)于SKIPIF1<0.而由SKIPIF1<0不能得到SKIPIF1<0，故充分性不成立；反之，由SKIPIF1<0可得到SKIPIF1<0，故必要性成立.故選：B.【變式訓(xùn)練】3、