¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýy=f£¨x£©£¬Èç¹û´æÔÚÇø¼ä[m£¬n]⊆D£¬Í¬Ê±Âú×ãÏÂÁÐÌõ¼þ£º¢Ùf£¨x£©ÔÚ[m£¬n]ÄÚÊǵ¥µ÷µÄ£»¢Úµ±¶¨ÒåÓòÊÇ[m£¬n]ʱ£¬f£¨x£©µÄÖµÓòÒ²ÊÇ[m£¬n]ʱ£¬Ôò³Æ[m£¬n]ÊǸú¯ÊýµÄ¡°ºÍÐ³Çø¼ä¡±£®
£¨1£©ÅжϺ¯ÊýÊÇ·ñ´æÔÚ¡°ºÍÐ³Çø¼ä¡±£¬²¢ËµÃ÷ÀíÓÉ£»
£¨2£©Èç¹û[m£¬n]ÊǺ¯ÊýµÄÒ»¸ö¡°ºÍÐ³Çø¼ä¡±£¬Çón-mµÄ×î´óÖµ£»
£¨3£©ÓÐЩº¯ÊýÓÐÎÞÊý¸ö¡°ºÍÐ³Çø¼ä¡±£¬Èçy=x£¬ÇëÄãÔÙ¾ÙÒ»ÀࣨÎÞÐèÖ¤Ã÷£©
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©¸ÃÎÊÌâÊÇÒ»¸öÅжÏÐÔÎÊÌ⣬´ÓÕýÃæÖ¤Ã÷ÓÐÒ»¶¨µÄÄѶȣ¬¹Ê¿É²ÉÓ÷´Ö¤·¨À´½øÐÐÖ¤Ã÷£¬¼´ÏȼÙÉèÇø¼ä[m£¬n]Ϊº¯ÊýµÄ¡°ºÍÐ³Çø¼ä¡±£¬È»ºó¸ù¾Ýº¯ÊýµÄÐÔÖʵõ½Ã¬¶Ü£¬½ø¶øµÃµ½¼ÙÉè²»³ÉÁ¢£¬Ô­ÃüÌâ³ÉÁ¢£®
£¨2£©Éè[m£¬n]ÊÇÒÑÖªº¯Êý¶¨ÒåÓòµÄ×Ó¼¯£¬ÎÒÃÇ¿ÉÒÔÓÃa±íʾ³ön-mµÄȡֵ£¬×ª»¯Îª¶þ´Îº¯ÊýµÄ×îÖµÎÊÌâºó£¬¸ù¾Ý¶þ´Îº¯ÊýµÄÐÔÖÊ£¬¿ÉÒԵõ½´ð°¸£®
£¨3£©¸ù¾Ý¡°ºÍÐ³Çø¼ä¡±µÄ¶¨Ò壬ÎÒÃÇ»¹¿ÉÒÔд³öÒÔϺ¯Êý£ºy=a-x£¨aΪ³£Êý£©£¬£¨k£¾oΪ³£Êý£©Âú×ãÓÐÎÞÊý¸ö¡°ºÍÐ³Çø¼ä¡±£®
½â´ð£º½â£º£¨1£©Éè[m£¬n]ÊǺ¯ÊýµÄ¡°ºÍÐ³Çø¼ä¡±£¬ÔòÔÚ[m£¬n]Éϵ¥µ÷£®
ËùÒÔ[m£¬n]⊆£¨-¡Þ£¬0£©»ò[m£¬n]⊆£¨0£¬+¡Þ£©
Òò´Ë£¬ÔÚ[m£¬n]ÉÏΪÔöº¯Êý£®
Ôòf£¨m£©=m£¬f£¨n£©=n£®¼´·½³ÌÓÐÁ½¸ö½âm£¬n
ÓÖ¿É»¯Îªx2-3x+4=0£¬¶øx2-3x+4=0ÎÞʵÊý½â£®
ËùÒÔ£¬º¯Êý²»´æÔÚ¡°ºÍÐ³Çø¼ä¡±
£¨2£©ÒòΪÔÚ[m£¬n]ÉÏÊǵ¥µ÷µÄ£¬
ËùÒÔ[m£¬n]⊆£¨-¡Þ£¬0£©»ò[m£¬n]⊆£¨0£¬+¡Þ£©
Ôòf£¨m£©=m£¬f£¨n£©=n
ËùÒÔm£¬nÊǵÄÁ½¸öͬºÅµÄʵÊý¸ù
¼´·½³Ìa2x-£¨a2+a£©x+1=0ÓÐÁ½¸öͬºÅµÄʵÊý¸ù£¬×¢Òâµ½
Ö»Òª¡÷=£¨a2+a£©2-4a2£¾0£¬½âµÃa£¾1»òa£¼-3
ËùÒÔ
ÆäÖÐa£¾1»òa£¼-3£¬ËùÒÔ£¬µ±a=3ʱ£¬n-mÈ¡×î´óÖµ
£¨3£©´ð°¸²»Î¨Ò»£¬Èç¿Éд³öÒÔϺ¯Êý£ºy=a-x£¨aΪ³£Êý£©£¬£¨k£¾0Ϊ³£Êý£©
µãÆÀ£º±¾ÌâÖ÷ÒªÒÔж¨ÒåÎªÔØÌ壬×ۺϿ¼²éÁ˺¯ÊýµÄµ¥µ÷ÐÔ¡¢º¯ÊýµÄ×îÖµ·½³ÌµÄ¸ùµÄÇé¿ö¡¢¶þ´Îº¯ÊýµÄ×îÖµµÄÇó½â£¬¿¼²éÁËÀûÓÃÒÑѧ֪ʶ½â¾öÐÂÎÊÌâµÄÄÜÁ¦£¬¿¼²éÁËÍÆÀíÔËËãµÄÄÜÁ¦£¬±¾Ìâ×ÛºÏÐÔ½ÏÇ¿£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýy=f£¨x£©£¬Èç¹û´æÔÚÇø¼ä[m£¬n]⊆D£¬Í¬Ê±Âú×㣺
¢Ùf£¨x£©ÔÚ[m£¬n]ÄÚÊǵ¥µ÷º¯Êý£»
¢Úµ±¶¨ÒåÓòÊÇ[m£¬n]ʱ£¬f£¨x£©µÄÖµÓòÒ²ÊÇ[m£¬n]£®Ôò³Æ[m£¬n]ÊǸú¯ÊýµÄ¡°ºÍÐ³Çø¼ä¡±£®
£¨1£©ÇóÖ¤£ºº¯Êýy=g(x)=3-
5
x
²»´æÔÚ¡°ºÍÐ³Çø¼ä¡±£®
£¨2£©ÒÑÖª£ºº¯Êýy=
(a2+a)x-1
a2x
£¨a¡ÊR£¬a¡Ù0£©ÓС°ºÍÐ³Çø¼ä¡±[m£¬n]£¬µ±a±ä»¯Ê±£¬Çó³ön-mµÄ×î´óÖµ£®
£¨3£©Ò×Öª£¬º¯Êýy=xÊÇÒÔÈÎÒ»Çø¼ä[m£¬n]ΪËüµÄ¡°ºÍÐ³Çø¼ä¡±£®ÊÔÔÙ¾ÙÒ»ÀýÓС°ºÍÐ³Çø¼ä¡±µÄº¯Êý£¬²¢Ð´³öËüµÄÒ»¸ö¡°ºÍÐ³Çø¼ä¡±£®£¨²»ÐèÖ¤Ã÷£¬µ«²»ÄÜÓñ¾ÌâÒÑÌÖÂÛ¹ýµÄy=x¼°ÐÎÈçy=
bx+c
ax
µÄº¯ÊýΪÀý£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýf£¨x£©£¬Èô´æÔÚÇø¼äM=[a£¬b]⊆D£¨a£¼b£©£¬Ê¹µÃ{y|y=f£¨x£©£¬x¡ÊM}=M£¬Ôò³ÆÇø¼äMΪº¯Êýf£¨x£©µÄ¡°µÈÖµÇø¼ä¡±£®¸ø³öÏÂÁÐÈý¸öº¯Êý£º
¢Ùf(x)=(
12
)x
£»   ¢Úf£¨x£©=x3£»    ¢Ûf£¨x£©=log2x+1
Ôò´æÔÚ¡°µÈÖµÇø¼ä¡±µÄº¯ÊýµÄ¸öÊýÊÇ
2
2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýy=f£¨x£©£¬ÈôͬʱÂú×ãÏÂÁÐÌõ¼þ£º¢Ùf£¨x£©ÔÚDÄÚµ¥µ÷µÝÔö»òµ¥µ÷µÝ¼õ£»¢Ú´æÔÚÇø¼ä[a£¬b]⊆D£¬Ê¹f£¨x£©ÔÚ[a£¬b]ÉϵÄÖµÓòΪ[a£¬b]£»ÄÇô°Ñy=f£¨x£©£¨x¡ÊD£©½Ð±Õº¯Êý£®
£¨1£©Çó±Õº¯Êýy=-x3·ûºÏÌõ¼þ¢ÚµÄÇø¼ä[a£¬b]£»
£¨2£©ÅжϺ¯Êýf£¨x£©=
3
4
x+
1
x
£¨x£¾0£©ÊÇ·ñΪ±Õº¯Êý£¿²¢ËµÃ÷ÀíÓÉ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

£¨2012•³çÃ÷ÏØÒ»Ä££©¶¨Ò壺¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýf£¨x£©£¬Èç¹û´æÔÚt¡ÊD£¬Ê¹µÃf£¨t+1£©=f£¨t£©+f£¨1£©³ÉÁ¢£¬³Æº¯Êýf£¨x£©ÔÚDÉÏÊÇ¡°T¡±º¯Êý£®ÒÑÖªÏÂÁк¯Êý£º
¢Ùf£¨x£©=
1x
£»¡¡
¢Úf£¨x£©=log2£¨x2+2£©£»
¢Ûf£¨x£©=2x£¨x¡Ê£¨0£¬+¡Þ£©£©£»¡¡
¢Üf£¨x£©=cos¦Ðx£¨x¡Ê[0£¬1]£©£¬ÆäÖÐÊôÓÚ¡°T¡±º¯ÊýµÄÐòºÅÊÇ
¢Û
¢Û
£®£¨Ð´³öËùÓÐÂú×ãÒªÇóµÄº¯ÊýµÄÐòºÅ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

¶ÔÓÚ¶¨ÒåÓòΪDµÄº¯Êýf£¨x£©£¬ÈôͬʱÂú×ãÏÂÁÐÌõ¼þ£º¢Ùf£¨x£©ÔÚDÄÚÓе¥µ÷ÐÔ£»¢Ú´æÔÚÇø¼ä[a£¬b]⊆D£¬Ê¹f£¨x£©ÔÚÇø¼ä[a£¬b]ÉϵÄÖµÓòҲΪ[a£¬b]£¬Ôò³Æf£¨x£©ÎªDÉϵġ°ºÍг¡±º¯Êý£¬[a£¬b]Ϊº¯Êýf£¨x£©µÄ¡°ºÍг¡±Çø¼ä£®
£¨¢ñ£©Çó¡°ºÍг¡±º¯Êýy=x3·ûºÏÌõ¼þµÄ¡°ºÍг¡±Çø¼ä£»
£¨¢ò£©ÅжϺ¯Êýf(x)=x+
4
x
(x£¾0)
ÊÇ·ñΪ¡°ºÍг¡±º¯Êý£¿²¢ËµÃ÷ÀíÓÉ£®
£¨¢ó£©Èôº¯Êýg(x)=
x+4
+m
ÊÇ¡°ºÍг¡±º¯Êý£¬ÇóʵÊýmµÄȡֵ·¶Î§£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸