能力升级练(十六)导数及其综合应用(2)1.(2019湖北荆州质检)已知函数f(x)=ax-lnx.(1)讨论f(x)的单调性;(2)若a∈-∞,-1e2,求证:f(x)≥2ax-xeax-1.(1)解由题意得f'(x)=a-1x=ax-1x(x>0),①当a≤0时,则f'(x)<0在(0,+∞)上恒成立,∴f(x)在(0,+∞)上单调递减.②当a>0时,则当x∈1a,+∞时,f'(x)>0,f(x)单调递增,当x∈0,1a时,f'(x)<0,f(x)单调递减.综上,当a≤0时,f(x)在(0,+∞)上单调递减;当a>0时,f(x)在0,1a上单调递减,在1a,+∞上单调递增.(2)证明令g(x)=f(x)-2ax+xeax-1=xeax-1-ax-lnx,则g'(x)=eax-1+axeax-1-a-1x=(ax+1)eax-1-1x=(ax+1)(xeax-1-1)x(x>0),设r(x)=xeax-1-1(x>0),则r'(x)=(1+ax)eax-1(x>0), eax-1>0,∴当x∈0,-1a时,r'(x)>0,r(x)单调递增;当x∈-1a,+∞时,r'(x)<0,r(x)单调递减.∴r(x)max=r-1a=-1ae2+1≤0a≤-1e2,∴当0
-1a时,g'(x)>0,∴g(x)在0,-1a上单调递减,在-1a,+∞上单调递增,∴g(x)min=g-1a.设t=-1a∈(0,e2],则g-1a=h(t)=te2-lnt+1(00,T为关于x的增函数;当169π+123;当a>-3时,M(a)≥F(-2)=|g(-2)-a|=6+a>3;当a=-3时,M(a)=3.综上,当M(a)最小时,a=-3.4.已知函数f(x)=12ax2-lnx,a∈R.(1)求函数f(x)的单调区间;(2)若函数f(x)在区间[1,e]上的最小值为1,求a的值.解函数f(x)的定义域是(0,+∞),f'(x)=ax-1x=ax2-1x.(1)①当a=0时,f'(x)=-1x<0,故函数f(x)在(0,+∞)上单调递减.②当a<0时,f'(x)<0恒成立,所以函数f(x)在(0,+∞)上单调递减.③当a>0时,令f'(x)=0,又因为x>0,解得x=❑√1a.(ⅰ)当x∈(0,❑√1a)时,f'(x)<0,所以函数f(x)在(0,❑√1a)上单调递减.(ⅱ)当x∈(❑√1a,+∞)时,f'(x)>0,所以函数f(x)在(❑√1a,+∞)上单调递增.综上所述,当a≤0时,函数f(x)的单调减区间是(0,+∞),当a>0时,函数f(x)的单调减区间是(0,❑√1a),单调增区间为(❑√1a,+∞).(2)①当a≤0时,由(1)可知,f(x)在[1,e]上单调递减,所以f(x)的最小值为f(e)=12ae2-1=1,解得a=4e2>0,舍去.②当a>0时,由(1)可知,(ⅰ)当❑√1a≤1,即a≥1时,函数f(x)在[1,e]上单调递增,所以函数f(x)的最小值为f(1)=12a=1,解得a=2.(ⅱ)当1<❑√1a
