函数的概念
3)求函数值问题
1 (a) f ( x) = 2 x + x + 1 , 求 f ( ) , f ( f ( x)) . x 1 1 2 1 2 1 2 + x + x2 解： f ( ) = 2( ) + + 1 = 2 + + 1 = . 2 x x x x x x2
f ( f ( x)) = f (2 x + x + 1) = 2( x + x + 1) + ( x + x + 1) + 1 .2 2 2 2
1 , 当 x ≤1 (b) 设 f (x) = , 求 f [ f [ f (x)]] (01 年考研试题) 0 , 当 x >1 解： 当 x ≤ 1 时 ， f [ f [ f ( x )]] = f [ f (1)] = f (1) = 1
而当
x > 1 时 ， f [ f [ f ( x )]] = f [ f ( 0 )] = f (1) = 1x ∈ ( ∞ , +∞ ) ,
f [ f [ f ( x )]] = 1 .
∴ 对于一切