HMMT 二月 2005 · CALC 赛 · 第 4 题

HMMT February 2005 — CALC Round — Problem 4

Let f : R → R be a smooth function such that f ( x ) = f ( x ) f ( x ) for all x . Suppose (4) ′ f (0) = 1 and f (0) = 9. Find all possible values of f (0).

解析

Let f : R → R be a smooth function such that f ( x ) = f ( x ) f ( x ) for all x . Suppose (4) ′ f (0) = 1 and f (0) = 9. Find all possible values of f (0). √ Solution: ± 3 ′ ′′ 2 Let f (0) = a . Then the equation gives f (0) = a . Differentiating the given equation gives ′ ′′ ′′′ ′ ′′ 2 f ( x ) f ( x ) = f ( x ) f ( x ) + f ( x ) f ( x ) , ′ ′′ ′′′ or f ( x ) f ( x ) = f ( x ) f ( x ). Differentiating once more gives ′ ′′′ ′′ 2 (4) ′ ′′′ f ( x ) f ( x ) + f ( x ) = f ( x ) f ( x ) + f ( x ) f ( x ) 1 √ ′′ 2 (4) (4) 4 or f ( x ) = f ( x ) f ( x ), giving 9 = f (0) = a . Thus a = ± 3. These are indeed √ ± x 3 both attainable by f ( x ) = e . ′′ ′ f ( x ) f ( x ) Alternative Solution: Rewrite the given equation as = . Integrating both ′ f ( x ) f ( x ) ′ ′ sides gives ln f ( x ) = ln f ( x ) + C , and exponentiating gives f ( x ) = Cf ( x ). This has 1 Cx solution f ( x ) = Ae for constants A and C . Since f (0) = 1, A = 1, and differentiating √ 4 (4) ′ we find that C = f (0) = 9, yielding f (0) = C = ± 3.