PUMaC 2021 · 代数（B 组） · 第 4 题

PUMaC 2021 — Algebra (Division B) — Problem 4

For a bijective function g : R → R , we say that a function f : R → R is its superinverse if − 1 − 1 it satisfies the following identity ( f ◦ g )( x ) = g ( x ), where g is the inverse of g . Given 3 2 g ( x ) = x + 9 x + 27 x + 81 and f is its superinverse, find | f ( − 289) | . 2 3 4 2 πi/ 5 2 π 2 π

解析

For a bijective function g : R → R , we say that a function f : R → R is its superinverse if − 1 − 1 it satisfies the following identity ( f ◦ g )( x ) = g ( x ), where g is the inverse of g . Given 3 2 g ( x ) = x + 9 x + 27 x + 81 and f is its superinverse, find | f ( − 289) | . Proposed by: Henry Erdman Answer: 7 − 1 − 1 − 1 − 1 − 1 Applying each side of the identity to g gives f ◦ g ◦ g = g ◦ g . Noting that g ◦ g is − 1 − 1 3 just the identity function, we have f = g ◦ g . Computing from g ( x ) = ( x + 3) + 54, we 1 − 1 − 1 − 1 3 have g ( x ) = ( x − 54) − 3. Since g ( − 289) = − 10, and g ( − 10) = − 7, | f ( − 289) | = 7. 1 2 π 2 π 2 3 4 2 πi/ 5