Question

(理)若函数f(x)的导函数为f′(x)=-x(x+1),则函数g(x)=f(log_ax)(0＜a＜1)的单调递减区间是

A.［-1,0］        B.［,+∞),(0,1］        C.［1,］        D.(-∞,］,［,+∞)

Answer 1

答案：(理)C  由f′(x)=-x(x+1)≤0,得x≤-1或x≥0,即f(x)的递减区间为(-∞,-1］或［0,+∞),f(x)的递增区间为［-1,0］.又∵0＜a＜1,∴y=log_ax在(0,+∞)上为减函数.由复合函数的单调性知,-1≤log_ax≤0,即1≤x≤时g(x)为减函数,∴单调递减区间为［1,］.