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）

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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所以,f(1)即c======以下答案可供参考======
供参考答案1:
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