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HMMT 十一月 2014 · 团队赛 · 第 2 题

HMMT November 2014 — Team Round — Problem 2

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 5 ] Let f ( x ) = x + 6 x + 7. Determine the smallest possible value of f ( f ( f ( f ( x )))) over all realnumbers x .
解析

英文解析

  1. [ 5 ] Let f ( x ) = x + 6 x + 7. Determine the smallest possible value of f ( f ( f ( f ( x )))) over all realnumbers x .
    2 2
    Answer: 23 Consider that f ( x ) = x + 6 x + 7 = ( x + 3) − 2. So f ( x ) ≥ − 2 for real numbers x .
    Also, f is increasing on the interval [ − 3 , ∞ ) . Thereforef ( f ( x )) ≥ f ( − 2) = − 1 ,
    f ( f ( f ( x ))) ≥ f ( − 1) = 2 ,
    f ( f ( f ( f ( x )))) ≥ f (2) = 23 .and
    Thus, the minimum value of f ( f ( f ( f ( x )))) is 23 and equality is obtained when x = − 3 .