第11節(jié)導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用第一課時導(dǎo)數(shù)與函數(shù)的單調(diào)性【選題明細(xì)表】知識點、方法題號判定函數(shù)的單調(diào)性、求單調(diào)區(qū)間2,5,6,8由單調(diào)性理解導(dǎo)函數(shù)圖象1比較大小或解不等式3,10,11由單調(diào)性求參數(shù)的取值范圍4,7,12由導(dǎo)數(shù)研究函數(shù)單調(diào)性的綜合問題9,13,14基礎(chǔ)鞏固(時間:30分鐘)1.已知函數(shù)y=f(x)的圖象是下列四個圖象之一,且其導(dǎo)函數(shù)y=f(x)的圖象如圖所示,則該函數(shù)的圖象是(B)解析:由導(dǎo)函數(shù)的圖象知,在-1,1上f(x)0,故函數(shù)f(x)在-1,1上是單調(diào)遞增的.又因為在-1,0上f(x)的值逐漸增大,在0,1上f(x)的值逐漸減小,所以在-1,0上,f(x)的增長率逐漸增大,在0,1上 f(x) 的增長率逐漸變小.故選B.2.函數(shù)f(x)=x-ln x的單調(diào)遞減區(qū)間為(A)(A)(0,1) (B)(0,+)(C)(1,+)(D)(-,0)(1,+)解析:函數(shù)的定義域是(0,+),且f(x)=1-=,令f(x)0,解得0xf(3)f()(B)f(3)f(2)f()(C)f(2)f()f(3)(D)f()f(3)f(2)解析:因為f(x)=1+x-sin x,所以f(x)=1-cos x,當(dāng)x(0,時,f(x)0,所以f(x)在(0,上是增函數(shù),所以f()f(3)f(2).4.(2018山東淄博桓臺二中月考)若函數(shù)f(x)=kx-ln x在區(qū)間(2,+)上單調(diào)遞增,則k的取值范圍是(B)(A)(-,-2(B),+)(C)2,+)(D)(-,)解析:f(x)=k-,因為函數(shù)f(x)=kx-ln x在區(qū)間(2,+)上單調(diào)遞增,所以f(x)0在區(qū)間(2,+)上恒成立.所以k,而y=在區(qū)間(2,+)上單調(diào)遞減,所以k,所以k的取值范圍是,+).5.(2018湖南長沙長郡中學(xué)月考)求形如y=f(x)g(x)的函數(shù)的導(dǎo)數(shù),我們常采用以下做法:先兩邊同取自然對數(shù)得ln y=g(x)ln f(x),再兩邊同時求導(dǎo)得y=g(x)ln f(x)+g(x)f(x),于是得到y(tǒng)=f(x)g(x)g(x)ln f(x)+g(x)f(x),運用此方法求得函數(shù)y=的單調(diào)遞增區(qū)間是(C)(A)(e,4)(B)(3,6)(C)(0,e)(D)(2,3)解析:由題設(shè),y=(-ln x+)=(x0).令y0,得1-ln x0,所以0x0,則(-x2+2)ex0,因為ex0,所以-x2+20,解得-x0,解得a-3,所以實數(shù)a的取值范圍是(-3,0)(0,+).答案:(-3,0)(0,+)8.已知函數(shù)f(x)=x3+ax2-x+c,且a=f().(1)求a的值;(2)求函數(shù)f(x)的單調(diào)區(qū)間.解:(1)由f(x)=x3+ax2-x+c,得f(x)=3x2+2ax-1.所以a=f()=3()2+2a-1,解得a=-1.(2)由(1)可知f(x)=x3-x2-x+c,則f(x)=3x2-2x-1=3(x+)(x-1),令f(x)0,解得x1或x-;令f(x)0,解得-x0在f(x)的定義域上恒成立,即f(x)+f(x)0在f(x)的定義域上恒成立.對于選項A,f(x)+f(x)=2-x-2-xln 2=2-x(1-ln 2)0,符合題意.經(jīng)驗證,選項B,C,D均不符合題意.故選A.10.(2018惠州調(diào)研)已知函數(shù)f(x)=xsin x+cos x+x2,則不等式f(ln x)+f(ln )2f(1)的解集為(D)(A)(e,+) (B)(0,e)(C)(0,)(1,e)(D)(,e)解析:f(x)=xsin x+cos x+x2是偶函數(shù),所以f(ln )=f(-ln x)=f(ln x),所以f(ln x)+f(ln )2f(1)可變形為f(ln x)0,所以f(x)在(0,+)上單調(diào)遞增,在(-,0)上單調(diào)遞減,所以f(ln x)f(1)等價于-1ln x1,所以xe.11.(2018重慶市一模)已知函數(shù)f(x)的導(dǎo)函數(shù)為f(x),且f(x)f(x)對任意的xR恒成立,則下列不等式均成立的是(A)(A)f(ln 2)2f(0),f(2)2f(0),f(2)e2f(0)(C)f(ln 2)e2f(0)(D)f(ln 2)2f(0),f(2)e2f(0)解析:令g(x)=,則g(x)=0,20,故g(ln 2)g(0),g(2)g(0),即,即f(ln 2)2f(0),f(2)e2f(0).12.(2018安徽江南十校聯(lián)考)設(shè)函數(shù)f(x)=x2-9ln x在區(qū)間a-1,a+1上單調(diào)遞減,則實數(shù)a的取值范圍是.解析:f(x)的定義域為(0,+),且f(x)=x-.由f(x)=x-0,解得0x3.因為f(x)=x2-9ln x在a-1,a+1上單調(diào)遞減,所以解得10,即x(x+2)ex0,得f(x)在區(qū)間(-,-2),(0,+)上單調(diào)遞增,在區(qū)間(-2,0)上單調(diào)遞減.(3)由(2)知,f(x)在區(qū)間(-2,0)上單調(diào)遞減,在區(qū)間(0,2)上單調(diào)遞增,fmin(x)=f(0)=0.當(dāng)x-2,2時,不等式f(x)2a+1能成立,須2a+1fmin(x),即2a+10,故a-.故a的取值范圍為-,+).14.已知函數(shù)f(x)=exln x-aex(aR).(1)若f(x)在點(1,f(1)處的切線與直線y=x+1垂直,求a的值;(2)若f(x)在(0,+)上是單調(diào)函數(shù),求實數(shù)a的取值范圍.解:(1)f(x)=exln x+ex-aex=(-a+ln x)ex,f(1)=(1-a)e,由(1-a)e=-1,得a=2.(2)由(1)知f(x)=(-a+ln x)ex,若f(x)為單調(diào)遞減函數(shù),則f(x)0在x0時恒成立,即-a+ln x0在x0時恒成立.所以a+ln