第16講　導(dǎo)數(shù)與函數(shù)的單調(diào)性

第16講導(dǎo)數(shù)與函數(shù)的單調(diào)性【備選理由】例1考查導(dǎo)函數(shù)的圖象與原函數(shù)圖象的關(guān)系,從圖象角度認(rèn)識(shí)導(dǎo)數(shù)的符號(hào)與函數(shù)單調(diào)性的關(guān)系;例2考查含參函數(shù)的單調(diào)性問題,需要對(duì)參數(shù)進(jìn)行分類討論;例3考查根據(jù)函數(shù)的單調(diào)性求參數(shù)的取值范圍,一般是通過參變分離轉(zhuǎn)化為恒成立問題,但當(dāng)函數(shù)比較復(fù)雜時(shí),需要構(gòu)造新函數(shù),正向分類討論,具有較強(qiáng)的綜合性;例4考查根據(jù)函數(shù)的單調(diào)性求參數(shù)的取值范圍;例5考查構(gòu)造法解決f(x)與f'(x)共存不等式問題,解題的關(guān)鍵是根據(jù)題意構(gòu)造函數(shù),求導(dǎo)后判斷函數(shù)的單調(diào)性,從而利用函數(shù)的單調(diào)性解不等式,考查數(shù)學(xué)轉(zhuǎn)化思想,屬于較難題.例1[配例1使用](多選題)設(shè)函數(shù)f(x)在定義域內(nèi)可導(dǎo),y=f(x)的圖象如圖所示,則下列不可能是導(dǎo)函數(shù)y=f'(x)的圖象的是 (ABC) A B C D[解析]由題圖可知,當(dāng)x<0時(shí),函數(shù)f(x)單調(diào)遞增,所以導(dǎo)函數(shù)f'(x)的值不為負(fù)數(shù),故A,C不可能是導(dǎo)函數(shù)y=f'(x)的圖象;當(dāng)x>0時(shí),函數(shù)f(x)先增、后減、再增,所以導(dǎo)函數(shù)f'(x)的值先正、后負(fù)、再正,故B不可能是導(dǎo)函數(shù)y=f'(x)的圖象;D可能是導(dǎo)函數(shù)y=f'(x)的圖象.故選ABC.例2[配例2使用]已知函數(shù)f(x)=(x-2)(aex-x).(1)當(dāng)a=4時(shí),求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)討論f(x)的單調(diào)性.解:(1)由f(x)=(x-2)(aex-x),得f'(x)=aex-x+aex(x-2)-(x-2)=(x-1)(aex-2).當(dāng)a=4時(shí),f(0)=-8,f'(0)=-2,則曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y+8=-2x,即2x+y+8=0.(2)由(1)知,f'(x)=(x-1)(aex-2).①當(dāng)a≤0時(shí),aex-2<0,所以當(dāng)x∈(-∞,1)時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x∈(1,+∞)時(shí),f'(x)<0,f(x)單調(diào)遞減.②當(dāng)a>0時(shí),由f'(x)=(x-1)(aex-2)=0,得x1=1,x2=ln2a(i)若0<a<2e,則x1<x2所以當(dāng)x∈(-∞,1)∪ln2a,+∞時(shí),f'(x)>0,f(x)在(當(dāng)x∈1,ln2a時(shí),f'(x)<0,f(x)(ii)若a=2e,則x1=x2=1,f'(x)≥0,f(x)在R上單調(diào)遞增(iii)若a>2e,則x1>x2所以當(dāng)x∈-∞,ln2a∪(1,+∞)時(shí),f'(x)>0,f(x)在-∞,ln2當(dāng)x∈ln2a,1時(shí),f'(x)<0,f(x)綜上可得,當(dāng)a≤0時(shí),f(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減;當(dāng)0<a<2e時(shí),f(x)在(-∞,1),ln2a,+∞當(dāng)a=2e時(shí),f(x)在R上單調(diào)遞增當(dāng)a>2e時(shí),f(x)在-∞,ln2a,(1,+∞)上單調(diào)遞增,例3[配例3使用][2023·全國(guó)乙卷]已知函數(shù)f(x)=1x+aln(1(1)當(dāng)a=-1時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若函數(shù)f(x)在(0,+∞)上單調(diào)遞增,求a的取值范圍.解:(1)當(dāng)a=-1時(shí),f(x)=1x-1ln(x+1),x>-1,則f'(x)=-1x2×ln(x+1)+1x據(jù)此可得f(1)=0,f'(1)=-ln2,所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-0=-ln2(x-1),即xln2+y-ln2=0.(2)由函數(shù)f(x)的解析式可得f'(x)=-1x2ln(x+1)+1x+a×1因?yàn)閒(x)在(0,+∞)上單調(diào)遞增,所以f'(x)≥0在(0,+∞)上恒成立.令-1x2ln(x+1)+1x+a1x+1≥0,則-(令g(x)=ax2+x-(x+1)ln(x+1),x>0,則原問題等價(jià)于g(x)≥0在(0,+∞)上恒成立.易知g'(x)=2ax-ln(x+1),當(dāng)a≤0時(shí),2ax≤0,因?yàn)閘n(x+1)>0,所以g'(x)<0,g(x)在(0,+∞)上單調(diào)遞減,此時(shí)g(x)<g(0)=0,不符合題意,故a>0.令h(x)=g'(x)=2ax-ln(x+1),x>0,則h'(x)=2a-1x當(dāng)2a≥1,即a≥12時(shí),因?yàn)閤∈(0,+∞),所以1x+1<1,故h'(x)>0,h(x)在(0,+∞即g'(x)在(0,+∞)上單調(diào)遞增,所以g'(x)>g'(0)=0,則g(x)在(0,+∞)上單調(diào)遞增,所以g(x)>g(0)=0,符合題意.當(dāng)0<a<12時(shí),由h'(x)=0,得2a-1x+1=0,可得x=當(dāng)x∈0,12a-1時(shí),h'(x)<0,則h(x)在0,12a-因?yàn)間'(0)=0,所以當(dāng)x∈0,12a-1時(shí),g'(x)<g'(0)=0,此時(shí)因?yàn)間(0)=0,所以當(dāng)x∈0,12a-1時(shí),g(x)綜上可知,實(shí)數(shù)a的取值范圍是12例4[配例3使用][2024·滄州模擬]已知函數(shù)f(x)=lnx-ax2-x的定義域?yàn)?e,e,任取x1,x2∈1e,e(其中x1>x2)都有x1f(x2)-x2f(x1)<x1x2(x1-x2)成立,則實(shí)數(shù)a的取值范圍為 (A)A.(-∞,1] B.[1,+∞)C.(-∞,e] D.[e,+∞)[解析]因?yàn)閤1,x2∈1e,e,所以由x1f(x2)-x2f(x1)<x1x2(x1-x2),可得f(x2)x2+x2<f(x1)x1+x1,又x1>x2,所以函數(shù)h(x)=f(x)x+x在1e,e上單調(diào)遞增.函數(shù)h(x)=f(x)x+x=lnxx-ax-1+x,則h'(x)=1-lnxx2-a+1≥0在1e,e上恒成立,所以1-lnxx2+1≥a對(duì)任意x∈1e,e恒成立.設(shè)函數(shù)g(x)=1-ln例5[配例5使用][2024·重慶一中模擬]已知函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),且滿足當(dāng)x>0時(shí),lnx·f'(x)+1xf(x)<0,則不等式(x-985)f(x)>0的解集為A.(985,+∞) B.(-985,985)C.(-985,0) D.(0,985)[解析]令函數(shù)g(x)=lnx·f(x)(x>0),則g'(x)=1x·f(x)+lnx·f'(x)<0,所以函數(shù)g(x)在(0,+∞)上單調(diào)遞減,因?yàn)間(1)=0,所以當(dāng)0<x<1時(shí),g(x)>0,當(dāng)x>1時(shí),g(x)<0.因?yàn)楫?dāng)0<x<1時(shí),lnx<0,當(dāng)x>1時(shí),ln