函数的概念
x , (d ) f ( x + 2) = 2 x + 3,2
解： 令 t = x + 2 , 则 x = t 2 .当 t≤2 时 x≤0
当 x≤0 , 求 f (2 x 4) . 当 x>0
f (t ) = f ( x + 2 ) = x 2 = (t 2 ) 2 ,x >0
x≤0
当 t >2 时 x>0
f (t ) = f ( x + 2) = 2 x + 3 = 2(t 2) + 3 = 2t 1 ;2 (t 2) , 当 t ≤ 2 ; f (t ) = 2t 1, 当 t > 2 . 2 2 ((2 x 4) 2) = (2 x 6) , 当 2 x 4 ≤ 2 f ( 2 x 4) = 当 2x 4 > 2 2(2 x 4) 1 = 4 x 9,
2 (2 x 6) , 当 x ≤ 3 f (2 x 4) = 当 x>3 4 x 9,