(1)y=3^x-1
y+1=3^x,即x=log(3)(y+1)
所以反函数为y=log(3)(x+1)=log(9)(x+1)^2 (x>-1)
所以(x+1)^2≤3x+1
解得D=[0,1]
(2)h(x)=log(9)(3x+1)-log(9)(x+1)=log(9)[3-2/(x+1)]
当0≤x≤1时,-2/(x+1)∈[-2,-1]
3-2/(x+1)∈[1,2]
所以h(x)的值域为[0,log(9)2]
(1)y=3^x-1
y+1=3^x,即x=log(3)(y+1)
所以反函数为y=log(3)(x+1)=log(9)(x+1)^2 (x>-1)
所以(x+1)^2≤3x+1
解得D=[0,1]
(2)h(x)=log(9)(3x+1)-log(9)(x+1)=log(9)[3-2/(x+1)]
当0≤x≤1时,-2/(x+1)∈[-2,-1]
3-2/(x+1)∈[1,2]
所以h(x)的值域为[0,log(9)2]