培優(yōu)課-函數(shù)的單調(diào)性與導(dǎo)數(shù)關(guān)系的應(yīng)用課件-高三數(shù)學(xué)一輪復(fù)習(xí)

培優(yōu)課——函數(shù)的單調(diào)性與導(dǎo)數(shù)關(guān)系的應(yīng)用第五章GAOZHONGTONGBUXUEANYOUHUASHEJI2022探究點(diǎn)二根據(jù)函數(shù)的單調(diào)性求參數(shù)值(范圍)【例2】

(1)若函數(shù)f(x)=(x2-cx+5)ex在區(qū)間[,4]上單調(diào)遞增,則實(shí)數(shù)c的取值范圍是(

)A.(-∞,2] B.(-∞,4] C.(-∞,8] D.[-2,4]【例2】

(1)若函數(shù)f(x)=(x2-cx+5)ex在區(qū)間[,4]上單調(diào)遞增,則實(shí)數(shù)c的取值范圍是(

)A.(-∞,2] B.(-∞,4] C.(-∞,8] D.[-2,4]探究點(diǎn)三函數(shù)單調(diào)性的應(yīng)用【例3】

(1)已知f(x)為R上的可導(dǎo)函數(shù),其導(dǎo)函數(shù)為f'(x),且對于任意的x∈R,均有f(x)+f'(x)>0,則(

)-2019f(-2019)<f(0),e2019f(2019)>f(0)-2019f(-2019)<f(0),e2019f(2019)<f(0)-2019f(-2019)>f(0),e2019f(2019)>f(0)-2019f(-2019)>f(0),e2019f(2019)<f(0)(2)已知f(x)的定義域?yàn)?0,+∞),f'(x)為f(x)的導(dǎo)函數(shù),且滿足f(x)<-xf'(x),則不等式f(x+1)>(x-1)f(x2-1)的解集是(

)A.(0,1) B.(2,+∞) C.(1,2) D.(1,+∞)解析

(1)構(gòu)造函數(shù)h(x)=exf(x),因?yàn)閒(x)+f'(x)>0在R上恒成立,所以h'(x)=exf(x)+exf'(x)=ex[f(x)+f'(x)]>0,所以函數(shù)h(x)在R上單調(diào)遞增,故h(-2

019)<h(0),即e-2

019f(-2

019)<e0f(0),即e-2

019f(-2

019)<f(0).同理,h(2

019)>h(0),即e2

019f(2

019)>f(0),故選A.【例3】

(1)已知f(x)為R上的可導(dǎo)函數(shù),其導(dǎo)函數(shù)為f'(x),且對于任意的x∈R,均有f(x)+f'(x)>0,則(

)-2019f(-2019)<f(0),e2019f(2019)>f(0)-2019f(-2019)<f(0),e2019f(2019)<f(0)-2019f(-2019)>f(0),e2019f(2019)>f(0)-2019f(-2019)>f(0),e2019f(2019)<f(0)(2)構(gòu)造函數(shù)y=xf(x),x∈(0,+∞),則y'=f(x)+xf'(x),又f(x)+xf'(x)<0,所以函數(shù)y=xf(x)在(0,+∞)上單調(diào)遞減.又因?yàn)閒(x+1)>(x-1)f(x2-1),所以(x+1)f(x+1)>(x2-1)f(x2-1),所以不等式f(x+1)>(x-1)f(x2-1)的解集是(2,+∞).故選B.規(guī)律方法

用函數(shù)單調(diào)性比較大小或解不等式時(shí)常構(gòu)造函數(shù),常見的有:(1)對于f'(x)>g'(x),構(gòu)造h(x)=f(x)-g(x).(2)對于f'(x)+g'(x)>0,構(gòu)造h(x)=f(x)+g(x).(3)對于f'(x)+f(x)>0,構(gòu)造h(x)=exf(x).(4)對于f'(x)>f(x),構(gòu)造h(x)=.(5