新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第07講 一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（講）（解析版）

第1講一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（一）本講為重要知識(shí)點(diǎn)，也是高中的難點(diǎn)。題型主要圍繞導(dǎo)數(shù)的幾何意義結(jié)合函數(shù)的思想考察?；緯?huì)考察一題關(guān)于函數(shù)本身的基礎(chǔ)題和一道導(dǎo)數(shù)大題，第一問(wèn)對(duì)于幾何意義的考察屬于基礎(chǔ)知識(shí)，必須掌握，第二問(wèn)的題型相對(duì)較多，需要對(duì)于導(dǎo)數(shù)的應(yīng)用和函數(shù)的思想相結(jié)合去理解其中的變形目的?？键c(diǎn)一導(dǎo)數(shù)的概念及運(yùn)算1．導(dǎo)數(shù)的概念一般地，函數(shù)y＝f(x)在x＝x0處的瞬時(shí)變化率SKIPIF1<0為函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)，記作f′(x0)或y′SKIPIF1<0即f′x0＝SKIPIF1<0.稱(chēng)函數(shù)f′(x)＝SKIPIF1<0為f(x)的導(dǎo)函數(shù)．2．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)．3．基本初等函數(shù)的導(dǎo)數(shù)公式基本初等函數(shù)導(dǎo)函數(shù)f(x)＝c(c為常數(shù))f′(x)＝0f(x)＝sinxf′(x)＝cos_xf(x)＝exf′(x)＝SKIPIF1<0f(x)＝lnxf′(x)＝SKIPIF1<0f(x)＝xα(α∈Q*)f′(x)＝αxα－1f(x)＝cosxf′(x)＝－sin_xf(x)＝ax(a>0，a≠1)f′(x)＝axln_af(x)＝logax(a>0，a≠1)f′(x)＝SKIPIF1<04.導(dǎo)數(shù)的運(yùn)算法則(1)[f(x)±g(x)]′＝f′(x)±g′(x)；(2)[f(x)·g(x)]′＝f′(x)g(x)＋f(x)g′(x)；(3)SKIPIF1<0(g(x)≠0)．5.常用結(jié)論1.f′(x0)代表函數(shù)f(x)在x＝x0處的導(dǎo)數(shù)值；(f(x0))′是函數(shù)值f(x0)的導(dǎo)數(shù)，且(f(x0))′＝0.2.SKIPIF1<0′＝－SKIPIF1<0.3.曲線的切線與曲線的公共點(diǎn)的個(gè)數(shù)不一定只有一個(gè)，而直線與二次曲線相切只有一個(gè)公共點(diǎn).4.函數(shù)y＝f(x)的導(dǎo)數(shù)f′(x)反映了函數(shù)f(x)的瞬時(shí)變化趨勢(shì)，其正負(fù)號(hào)反映了變化的方向，其大小|f′(x)|反映了變化的快慢，|f′(x)|越大，曲線在這點(diǎn)處的切線越“陡”.考點(diǎn)二利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性1.函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系函數(shù)y＝f(x)在區(qū)間(a，b)內(nèi)可導(dǎo)，(1)若f′(x)>0，則f(x)在區(qū)間(a，b)內(nèi)是單調(diào)遞增函數(shù)；(2)若f′(x)<0，則f(x)在區(qū)間(a，b)內(nèi)是單調(diào)遞減函數(shù)；(3)若恒有f′(x)＝0，則f(x)在區(qū)間(a，b)內(nèi)是常數(shù)函數(shù)．討論函數(shù)的單調(diào)性或求函數(shù)的單調(diào)區(qū)間的實(shí)質(zhì)是解不等式，求解時(shí)，要堅(jiān)持“定義域優(yōu)先”原則.2.常用結(jié)論匯總——規(guī)律多一點(diǎn)(1)在某區(qū)間內(nèi)f′(x)>0(f′(x)<0)是函數(shù)f(x)在此區(qū)間上為增(減)函數(shù)的充分不必要條件．(2)可導(dǎo)函數(shù)f(x)在(a，b)上是增(減)函數(shù)的充要條件是對(duì)?x∈(a，b)，都有f′(x)≥0(f′(x)≤0)且f′(x)在(a，b)上的任何子區(qū)間內(nèi)都不恒為零．考點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的極值最值1．函數(shù)的極值(1)函數(shù)的極小值：函數(shù)y＝f(x)在點(diǎn)x＝a的函數(shù)值f(a)比它在點(diǎn)x＝a附近其他點(diǎn)的函數(shù)值都小，f′(a)＝0；而且在點(diǎn)x＝a附近的左側(cè)f′(x)＜0，右側(cè)f′(x)＞0，則點(diǎn)a叫做函數(shù)y＝f(x)的極小值點(diǎn)，f(a)叫做函數(shù)y＝f(x)的極小值．(2)函數(shù)的極大值：函數(shù)y＝f(x)在點(diǎn)x＝b的函數(shù)值f(b)比它在點(diǎn)x＝b附近其他點(diǎn)的函數(shù)值都大，f′(b)＝0；而且在點(diǎn)x＝b附近的左側(cè)f′(x)＞0，右側(cè)f′(x)＜0，則點(diǎn)b叫做函數(shù)y＝f(x)的極大值點(diǎn)，f(b)叫做函數(shù)y＝f(x)的極大值．極小值點(diǎn)、極大值點(diǎn)統(tǒng)稱(chēng)為極值點(diǎn)，極大值和極小值統(tǒng)稱(chēng)為極值.①函數(shù)fx在x0處有極值的必要不充分條件是f′x0＝0，極值點(diǎn)是f′x＝0的根，但f′x＝0的根不都是極值點(diǎn)例如fx＝x3，f′0＝0，但x＝0不是極值點(diǎn).②極值反映了函數(shù)在某一點(diǎn)附近的大小情況，刻畫(huà)的是函數(shù)的局部性質(zhì).極值點(diǎn)是函數(shù)在區(qū)間內(nèi)部的點(diǎn)，不會(huì)是端點(diǎn).2．函數(shù)的最值(1)在閉區(qū)間[a，b]上連續(xù)的函數(shù)f(x)在[a，b]上必有最大值與最小值．(2)若函數(shù)f(x)在[a，b]上單調(diào)遞增，則f(a)為函數(shù)的最小值，f(b)為函數(shù)的最大值；若函數(shù)f(x)在[a，b]上單調(diào)遞減，則f(a)為函數(shù)的最大值，f(b)為函數(shù)的最小值．3常用結(jié)論1.對(duì)于可導(dǎo)函數(shù)f(x)，“f′(x0)＝0”是“函數(shù)f(x)在x＝x0處有極值”的必要不充分條件.2.求最值時(shí)，應(yīng)注意極值點(diǎn)和所給區(qū)間的關(guān)系，關(guān)系不確定時(shí)，需要分類(lèi)討論，不可想當(dāng)然認(rèn)為極值就是最值.3.函數(shù)最值是“整體”概念，而函數(shù)極值是“局部”概念，極大值與極小值之間沒(méi)有必然的大小關(guān)系.考點(diǎn)四利用導(dǎo)數(shù)研究生活中的優(yōu)化問(wèn)題1．生活中經(jīng)常遇到求利潤(rùn)最大、用料最省、效率最高等問(wèn)題，這些問(wèn)題通常稱(chēng)為優(yōu)化問(wèn)題．2．利用導(dǎo)數(shù)解決優(yōu)化問(wèn)題的實(shí)質(zhì)是求函數(shù)最值．3．解決優(yōu)化問(wèn)題的基本思路是什么？答案上述解決優(yōu)化問(wèn)題的過(guò)程是一個(gè)典型的數(shù)學(xué)建模過(guò)程．4.對(duì)于優(yōu)化問(wèn)題，建立模型之后需要對(duì)模型進(jìn)行最大值最小值的求解，從而轉(zhuǎn)化為導(dǎo)數(shù)求極值最值問(wèn)題.高頻考點(diǎn)一導(dǎo)數(shù)的概念及其意義例1、函數(shù)SKIPIF1<0的圖象如圖所示，SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，則下列數(shù)值排序正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由圖知：SKIPIF1<0，即SKIPIF1<0.故選：A【變式訓(xùn)練】1、若函數(shù)SKIPIF1<0在點(diǎn)（1，f（1））處的切線的斜率為1，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由已知SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立．故選：A．高頻考點(diǎn)二導(dǎo)數(shù)的運(yùn)算例1、已知SKIPIF1<0，求SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0得SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0，因此SKIPIF1<0，故選：D【變式訓(xùn)練】1、函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】對(duì)函數(shù)SKIPIF1<0求導(dǎo)，得SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線斜率為2，所以函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0.故選：A高頻考點(diǎn)三導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用例1、已知函數(shù)SKIPIF1<0，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】當(dāng)過(guò)原點(diǎn)的直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象相切時(shí)，設(shè)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，可得過(guò)點(diǎn)SKIPIF1<0的切線方程為SKIPIF1<0，代入點(diǎn)SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0，此時(shí)切線的斜率為SKIPIF1<0，由函數(shù)SKIPIF1<0的圖象可知，若直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，直線的斜率SKIPIF1<0的取值范圍為SKIPIF1<0．故答案選：D【變式訓(xùn)練】1、已知函數(shù)SKIPIF1<0，若SKIPIF1<0是函數(shù)SKIPIF1<0的唯一極值點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值集合是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】函數(shù)SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0，由題意可得，SKIPIF1<0是方程SKIPIF1<0唯一變號(hào)的根，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上沒(méi)有變號(hào)零點(diǎn)，令SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)