tincho
花朵
共回答了12个问题采纳率:91.7% 举报
(1)函数f(x)的值域为(-1,+∞),
由y=2 x -1得x=log 2 (y+1),
所以f -1 (x)=log 2 (x+1)(x>-1)(4分)
(2)证明:任取-1<x 1 <x 2 ,
f -1 (x 1 )-f -1 (x 2 )=log 2 (x 1 +1)-log 2 (x 2 +1)=log 2
x 1 +1
x 2 +1
由-1<x 1 <x 2 得0<x 1 +1<x 2 +1,因此
0<
x 1 +1
x 2 +1 <1得log 2
x 1 +1
x 2 +1 <0
所以f -1 (x 1 )<f -1 (x 2 )
故f -1 (x)在(-1,+∞)上为单凋增函数.(9分)
(3)f -1 (x)≤g(x)即
log 2 (x+1)≤log 4 (3x+1) ⇔
x+1>0
3x+1>0
(x+1 ) 2 ≤3x+1 ⇔
x+1>0
(x+1)2≤3x+1 (11分)
解之得0≤x≤1,所以x的取值范围是[0,2](13分)
1年前
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