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问题: 反函数问题

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解答:

f(x)的反函数为y=log(2)[x+1].
(1)log(2)[x+1]≤log(4)[3x+1]
log(2)[x+1]≤log(4)[3x+1]
2lg(x+1)/lg(2)≤lg(3x+1)/lg(2)
2lg(x+1)≤lg(3x+1)
(x+1)²≤3x+1
x²-x≤0...0≤x≤1...D∈[0,1]
(2)H(x)=g(x)-f^-1(x)/2=log(4)[3x+1]-log(2)[x+1]/2
H(x)=lg(3x+1)/lg(4)-lg(x+1)/2lg(2)=[lg(3x+1)-lg(x+1)]/2lg2
H(x)=[lg[(3x+1)/(x+1)]/2lg2=[lg(3-2/(x+1)]/2lg2
H(x)单调递增所以当x=1时取最大值1/2
当x=0时取最小值0
H(x)的值域为[0,1/2]