Question

(2005•海淀区二模)已知函数f(x)=1+2x-1，则其值域为_;f-1(5...

(2005•海淀区二模)已知函数f(x)=1+2x-1，则其值域为{x|x＞1}{x|x＞1};f-1(5)=33.

Answer 1

解答:解:∵2x-1＞0
∴f(x)=1+2x-1＞1即f(x)的值域为{x|x＞1}
令1+2x-1=5解得x=3
∴f(3)=5根据反函数的性质可知f-1(5)=3
故答案为:{x|x＞1}，3