Question

log2(x-1)+log2(x+2)反函数为g(x) g(2)=?

Answer 1

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