log2(x-1)+log2(x+2)反函数为g(x) g(2)=?

题目

log2(x-1)+log2(x+2)反函数为g(x) g(2)=?
log2(x-1)+log2(x+2)反函数为g(x) g(2)=?

答案

log2(x-1)+log2(x+2)=log2[(x-1)(x+2)]=log2(x²-x-2)
反函数为g(x),g(2)即为log2(x²-x-2)＝2时x的值
log2(x²-x-2)=2
x²-x-2=2²
x²-x-6=0
(x-3)(x+2)=0
x=3或x=-2
即g(2)=3或g(2)=-2
又g(x)的值域为原函数的定义域
log2(x-1)+log2(x+2)定义域为
x-1>0且x+2>0
解得x>1
∴舍去g(2)=-2
∴g(2)=3