函數(shù)單調(diào)性與導(dǎo)數(shù)

函數(shù)單調(diào)性與導(dǎo)數(shù)(2)yx0判斷函數(shù)單調(diào)性的常用方法：（1）定義法（2）導(dǎo)數(shù)法復(fù)習(xí)回顧增函數(shù)減函數(shù)設(shè)函數(shù)y=f(x)在某個(gè)區(qū)間內(nèi)可導(dǎo)，用導(dǎo)數(shù)法確定函數(shù)的單調(diào)性時(shí)的步驟是：（1）求出函數(shù)的導(dǎo)函數(shù)（2）求解不等式f/(x)>0，求得其解集，再根據(jù)解集與定義域?qū)懗鰡握{(diào)遞增區(qū)間（3）求解不等式f/(x)<0，求得其解集，再根據(jù)解集與定義域?qū)懗鰡握{(diào)遞減區(qū)間A課前練習(xí)一、綜合應(yīng)用:解:(1)函數(shù)的定義域是R,令,解得令,解得因此f(x)的遞增區(qū)間是:遞減區(qū)間是:例1:確定下列函數(shù)的單調(diào)區(qū)間:(1)f(x)=+sinx;解:函數(shù)的定義域是[0,a],且當(dāng)x≠0,a時(shí),有:由及解得0<x<3a/4,故f(x)的遞增區(qū)間是(0,3a/4).由及解得3a/4<x<a,故f(x)的遞減區(qū)間是(3a/4,a).說明:事實(shí)上在判斷單調(diào)區(qū)間時(shí),如果出現(xiàn)個(gè)別點(diǎn)使得導(dǎo)數(shù)為零,不影響包含該點(diǎn)的某個(gè)區(qū)間上的單調(diào)性,只有在某個(gè)區(qū)間內(nèi)恒有導(dǎo)數(shù)為零,才能判定f(x)在這一區(qū)間內(nèi)是常數(shù)函數(shù).例2:設(shè)f(x)=ax3+x恰有三個(gè)單調(diào)區(qū)間,試確定a的取值范圍,并求其單調(diào)區(qū)間.解:若a>0,對(duì)一切實(shí)數(shù)恒成立,此時(shí)f(x)只有一個(gè)單調(diào)區(qū)間,矛盾.若a=0,此時(shí)f(x)也只有一個(gè)單調(diào)區(qū)間,矛盾.若a<0,則,易知此時(shí)f(x)恰有三個(gè)單調(diào)區(qū)間.故a<0,其單調(diào)區(qū)間是:單調(diào)遞增區(qū)間:單調(diào)遞減區(qū)間:和例3.已知函數(shù)f(x)=lnx-ax+-1(a∈R).當(dāng)a≤時(shí),討論f(x)的單調(diào)性；解:(1)f(x)=lnx-ax+-1(x>0)，當(dāng)x∈(1,+∞),h(x)<0,f′(x)>0,函數(shù)f(x)單調(diào)遞增.f′(x)=令h(x)=ax2-x+1-a(x＞0),(1)當(dāng)a=0時(shí),h(x)=-x+1(x＞0),當(dāng)x∈(0,1),h(x)＞0,f′(x)＜0,函數(shù)f(x)單調(diào)遞減；(2)當(dāng)a≠0時(shí),由h(x)=0,即ax2-x+1-a=0,解得:x1=1,x2=(i)當(dāng)a=時(shí),x1=x2,h(x)≥0恒成立,此時(shí)f′(x)≤0,

函數(shù)f(x)單調(diào)遞減；x∈(,+∞)時(shí),h(x)＞0,f′(x)＜0,函數(shù)f(x)單調(diào)遞減.(ii)當(dāng)0＜a＜時(shí),＞1＞0，x∈(0,1)時(shí),h(x)＞0,f′(x)＜0,函數(shù)f(x)單調(diào)遞減；x∈(1,)時(shí),h(x)＜0,f′(x)＞0,函數(shù)f(x)單調(diào)遞增；(ⅲ)當(dāng)a＜0時(shí),＜0，當(dāng)x∈(0,1)時(shí),h(x)＞0,f′(x)＜0,函數(shù)f(x)單調(diào)遞減；當(dāng)x∈(1,+∞)時(shí),h(x)＜0,f′(x)＞0,函數(shù)f(x)單調(diào)遞增.綜上所述:當(dāng)a≤0時(shí),函數(shù)f(x)在(0,1)單調(diào)遞減,(1,+∞)單調(diào)遞增；當(dāng)a=時(shí),函數(shù)f(x)在(0,+∞)單調(diào)遞減；當(dāng)0＜a＜時(shí),函數(shù)f(x)在(0,1)單調(diào)遞減,(1,)單調(diào)遞增,(,+∞)單調(diào)遞減例4:當(dāng)x>1時(shí),證明不等式:證:設(shè)顯然,當(dāng)x>1時(shí),,故f(x)是[1,+∞)上的增函數(shù).所以當(dāng)x>1時(shí),f(x)>f(1)=0,即當(dāng)x>1時(shí),說明:利用函數(shù)單調(diào)性證明不等式是不等式證明的一種重要方法.2:證明方程只有一個(gè)根x=0.證:設(shè)則>0恒成立.故f(x)是R上的增函數(shù).而f(0)=0,故原方程有唯一根x=0.思考?結(jié)論:(1)若f(x)在區(qū)間(a,b)內(nèi)遞增→f’(x)≥0在(a,b)內(nèi)恒成立.(2)若f(x)在區(qū)間(a,b)內(nèi)遞減→f’(x)≤0在(a,b)內(nèi)恒成立.例5.(宅1)已知凝函數(shù)f(x)=a盯x3+3x2+x攪+1在R上是辨增函盤數(shù),求a的取謊值范準(zhǔn)圍.(2厲)函數(shù)f(x)=x2+2(a-1祖)x+2在區(qū)掉間(-尚∞,洗4]是減函數(shù),求a的取泡值范財(cái)圍。[－2,＋∞)練習(xí)：(1)若函數(shù)y＝x3+x2+mx+1是R上的單調(diào)函數(shù),則實(shí)數(shù)m的取值范圍是_________.(2)已知函數(shù)f(x)＝alnx+x在區(qū)間[2,3]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是_________.小值結(jié)