Question

已知y=f（x+1）是定义在R上的偶函数，当x属于【1,2】时，f（x）=2^X,设a=f（1/2），b=f（4/3）

c=f(1),求啊a，b，c的大小关系

Answer 1

f（x+1）是定义在R上的偶函数，即f(x+1)=f(-x+1),即f(x)关于x=1对称。f(x)=f(2-x)
a=f(1/2)=f(2-1/2)=f(1.5)
b=f(4/3)
c=f(1)
1<=x<=2时，f(x)=2^x是递增函数
所以，f(1)<f(4/3)<f(1.5)
即c<b<a

Answer 2

let y = g(x)= f(x+1)
g(x) 是定义在R上的偶函数

f(1/2) = a, f(4/3) =b, f(1) = c

f(x) = 2^x x属于[1,2]
f(1) = 2 = c

g(x) = f（x+1)
g(-x) = g(x) = f(x+1)

put x= -1/2
f(-1/2+1) = g(-1/2)
=>f(1/2) = g(-1/2)
= g(1/2)
= f(3/2)
= 2^(3/2) = a

put x= 1/3
f(1/3+1) = g(1/3)
=> f(4/3) = g(1/3)
= g(-1/3)
= f( 2/3)
= 2^(2/3) = b

=> a > c> b