已知函數(shù)f(x)=alnx+(a+1)/2x^2+1討論函數(shù)f(x)的單調(diào)性

數(shù)學人氣：563 ℃時間：2019-10-18 03:21:38

優(yōu)質(zhì)解答

f(x)=alnx+(a+1)/2·x²+1,x＞0
f'(x)=a/x+(a+1)x=[(a+1)x²+a]/x
(1) a=-1,f'(x)=-1/x＜0,f(x)在(0,+∞)單調(diào)減；
(2) a＜-1,(a+1)x²+a＜0恒成立,f'(x)＜0,f(x)在(0,+∞)單調(diào)減；
(3) -1＜a＜0,令f'(x)=0,即考慮到x＞0,解得x=√[-a/(a+1)],
在(0,√[-a/(a+1)]),f'(x)＜0,f(x)單調(diào)減;
在(√[-a/(a+1)],+∞),f'(x)＞0,f(x)單調(diào)增;
(4)a≧0,(a+1)x²+a≧0恒成立,f'(x)≧0,f(x)在(0,+∞)單調(diào)增.
綜上,.