来自党红梅的问题
已知函数f(x)=(x+1)lnx-x+1.(Ⅰ)若xf′(x)≤x2+ax+1,求a的取值范围;(Ⅱ)证明:(x-1)f(x)≥0.
已知函数f(x)=(x+1)lnx-x+1.
(Ⅰ)若xf′(x)≤x2+ax+1,求a的取值范围;
(Ⅱ)证明:(x-1)f(x)≥0.
1回答
2020-04-14 15:53
已知函数f(x)=(x+1)lnx-x+1.(Ⅰ)若xf′(x)≤x2+ax+1,求a的取值范围;(Ⅱ)证明:(x-1)f(x)≥0.
已知函数f(x)=(x+1)lnx-x+1.
(Ⅰ)若xf′(x)≤x2+ax+1,求a的取值范围;
(Ⅱ)证明:(x-1)f(x)≥0.
(Ⅰ)函数的定义域为(0,+∞)
求导函数,可得f′(x)=x+1x+lnx−1=lnx+1x