
x a22.解:(1)g'(x)=1 0= “(x>0)当a≥0时,,g'(x)>0,g(x)在(0, ∞)上单调递增……···当a<0时,令g'(x)>0,解得x>-a,令g(x)<0,解得0
作者:网络 来源: 2022-08-05 18:44:37 阅读:次
x a22.解:(1)g'(x)=1 0= “(x0)当a≥0时,,g'(x)0,g(x)在(0, ∞)上单调递增……···当a0时,令g'(x)0,解得x-a,令g(x)0,解得0
x a22.解:(1)g'(x)=1 0= “(x>0)当a≥0时,,g'(x)>0,g(x)在(0, ∞)上单调递增……···当a<0时,令g'(x)>0,解得x>-a,令g(x)<0,解得0