高中人教版数学复习讲义（必修一）：映射与函数（PDF版）.pdf

设( ) 2 ( ) fxax gxkx=? = ( ( )) ( ) 2 2 3 2 3 f g x ag x akx x ak ∴ = ?= ?=?∴= ①( ( )) ( ) ( 2) 2 3 2 2 2 gfx kfx kax akx k x k =?= ?= ?=?∴= ②? - 第 7 页 - ①②得 1 3 k a = ??= ? 32 yx yx = ??∴?= ?得 1 xy == 交点( 1, 1 ) 例 7 .略例 8 . B 例 9 . ( 1 )令 1 x t x + = 1 1 x t ∴= ? 2 2 2 1 1 1 ( ) 1 ( ) 1 1 1 1 t fttftttt t ??+ ?????=+?∴=?+≠??????? 2 ( ) 1( 1)fx x x x ∴=?+≠(2)解: 2 2 1 2() ( 0)fx f xx x ??+=> ???? ① 以 1 x 代替 x , 2 2 11 2()ffx x x ??+= ???? ② ①×2 ?② 2 1 3() 2fx x x = ? 2 21 () 33 fx x x ∴=?则 21 () ( 0) 3 3 fx x x x =?> ( 3 )解: 1 2 () 1 x fx x = ? 2 2 222 2 () 1 () (()) 1 1 x fx x x fx ffx f xxx x ?==== ???? 2 2 32 222 2 2 () 12 () ( ()) 1() 13 1 12 x fx x x fx ffx f xx x x ?=== = ??????????…… 2 () 1 n x fx nx ∴= ?