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Jordan's solution
- 1. 1+ x ( ) f ( f ( x)) = 1 ∗ 1 = 1+ x = x +1 1 1+ x 1+1+ x x + 2 (1 + ) ( ) 1+ x 1 So x ≠ 2 − 1 1+x (( ) +1) ( ) 1+x 1 1 +1 + x 2 +x f ( f ( f ( x ))) = ∗ = = 1 1+x 2 +1 +2 x 3 +2 x (( ) +2) ( ) 1+x 1 So −3 x≠ 2 And from original problem, x ≠ −1