2000Çг⵵ ¼ö´É¹®Á¦ ¿¹Ã¼´É°è 10¹ø

<º¸±â>ÀÇ ÇÔ¼ö f(x) Áß (fofof)(x) = f(x) °¡ ¼º¸³ÇÏ´Â °ÍÀ» ¸ðµÎ °í¸¥ °ÍÀº? [3Á¡]


<º¸±â>   ¤¡. f(x) = x + 1
  ¤¤. f(x) = - x
  ¤§. f(x) = - x + 1

¨ç ¤¡ ¨è ¤¤ ¨é ¤§
¨ê ¤¡, ¤§ ¨ë ¤¤, ¤§


°üÂûÇϱâ

  • ¤¡ÀÇ °æ¿ì (fof)(x) = (x + 1)+1 = x + 2

  • ¤¤ÀÇ °æ¿ì (fof)(x) = -(-x) = x   ¡Å fof ´Â Ç×µîÇÔ¼ö

  • ¤§ÀÇ °æ¿ì (fof)(x) =  -(-x + 1)+1 = x ¡Å fof ´Â Ç×µîÇÔ¼ö

Ç®À̺¸±â

¤¡ÀÇ °æ¿ì (fofof)(x) = x + 3
¤¤ÀÇ °æ¿ì (fofof)(x) = (fo(fof))(x) =  f(x)
¤§ÀÇ °æ¿ì (fofof)(x) = (fo(fof))(x) = f(x)

Á¤´ä ¨ë

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Update : 2000³â 01¿ù 02ÀÏ  ¼öÇм±»ý´Ô®  ¼öÇб³À°¿¬±¸©   mathel@unitel.co.kr