函数的概念
3x 2, 当 0 ≤ x ≤ 1 1 (e) f ( x 1) = , 求 f ( f ( x)) , x ∈ [ 2, ]. 3 当 1 ≤ x < 0 x 1,
解： 令 t = x 1, 则 x = t + 1 .当 1 ≤ t ≤ 0 时 0 ≤ x ≤ 1 f (t ) = f ( x 1) = 3 x 2 = 3(t + 1) 2 = 3t + 1,0 ≤ x ≤1
当 2 ≤ t ≤ 1 时 1 ≤ x ≤ 0 ,
1 ≤ x ≤ 0 f (t ) = f ( x 1) = x 1 = t , 3t + 1, 当 1 ≤ t ≤ 0 ; f (t ) = t , 当 2 ≤ t < 1 . 3x + 1, 当 1 ≤ x ≤ 0 ; f ( x) = x, 当 2 ≤ x < 1 . f (3x + 1), 当 1 ≤ x ≤ 0 f ( f ( x)) = . f ( x), 当 2 ≤ x < 1
1≤ x ≤ 0