Question

一道函数题

设Y=F(X)是定义在实数集R上的函数，满足条件Y=F(X+1)为偶函数，且当X大于等于一时，F(X)=2^X -1，则F(2/3),F(3/2),F(1/3)的大小关系，请详解

Answer 1

f(x+1)=f(-x+1)

f(2/3)=f(-1/3+1)=f(1/3+1)=f(4/3)=2^(4/3)-1
f(3/2)=2^(3/2)-1
f(1/3)=f(-2/3+1)=f(2/3+1)=f(5/3)=2^(5/3)-1

所以f(2/3)<f(3/2)<f(1/3)

Answer 2

F(x+1) = F(-x+1)

x>=1, F(x) = 2^x - 1

x>=0, F(x+1) = 2^(x+1) - 1

x<=1, 1-x >= 0, F(x) = F((x-1)+1) = F(-(x-1)+1) = F(1-x + 1)

= 2^(2-x) - 1 单减

F(3/2) = F(1/2)

F(1/3) > F(1/2) = F(3/2) > F(2/3)