ÏÂÁÐ˵·¨ÖУº
¢Ùº¯Êýf(x)=
x-1
x+1
Óëg£¨x£©=xµÄͼÏóûÓй«¹²µã£»
¢ÚÈô¶¨ÒåÔÚRÉϵĺ¯Êýf£¨x£©Âú×ãf£¨x+3£©=-f£¨x£©£¬Ôò6Ϊº¯Êýf£¨x£©µÄÖÜÆÚ£»
¢ÛÈô¶ÔÓÚÈÎÒâx¡Ê£¨1£¬3£©£¬²»µÈʽx2-ax+2£¼0ºã³ÉÁ¢£¬Ôòa£¾
11
3
£»
¢Ü¶¨Ò壺¡°Èôº¯Êýf£¨x£©¶ÔÓÚÈÎÒâx¡ÊR£¬¶¼´æÔÚÕý³£ÊýM£¬Ê¹|f£¨x£©|¡ÜM|x|ºã³ÉÁ¢£¬Ôò³Æº¯Êýf£¨x£©ÎªÓн緺º¯£®¡±Óɸö¨Òå¿ÉÖª£¬º¯Êýf£¨x£©=x2+1ΪÓн緺º¯£®
ÔòÆäÖÐÕýÈ·µÄÊÇ
¢Ù¢Ú¢Û
¢Ù¢Ú¢Û
£®
·ÖÎö£ºÁªÁ¢Á½¸öº¯ÊýµÄ½âÎöʽ£¬¹¹Ôì·½³Ì×飬·½³Ì×é½âµÄ¸öÊý¼´Îªº¯ÊýͼÏó½»µã¸öÊý£¬ÓÉ´Ë¿ÉÅжϢٵÄÕæ¼Ù£¬¸ù¾ÝÖÜÆÚº¯ÊýµÄ¶¨Òå¿ÉÅжϢڵÄÕæ¼Ù£¬¸ù¾Ý¶þ´Îº¯ÊýµÄÐÔÖÊ£¬ÎÒÃǿɹ¹Ôì¹ØÓÚaµÄ²»µÈʽ×飬½â²»µÈʽ×飬¼´¿ÉÇó³öaµÄÈ¡Öµ·¶Î§£¬½ø¶øÅжϢ۵ÄÕæ¼Ù£¬¸ù¾Ý¶þ´Îº¯ÊýµÄÐÔÖʽáºÏÓн緺º¯µÄ¶¨Ò壬¿ÉÒÔÅжϢܵÄÕæ¼Ù£¬½ø¶øµÃµ½´ð°¸£®
½â´ð£º½â£ºÁªÁ¢Á½¸öº¯ÊýµÄ½âÎöʽ£¬
y=
x-1
x+1
y=x
£¬Ò׵ø÷½³Ì×éÎ޽⣬
Ôòº¯Êýf£¨x£©=
x-1
x+1
Óëg£¨x£©=xµÄͼÏóûÓй«¹²µã£¬¹Ê¢ÙÕýÈ·£»
¶¨ÒåÔÚRÉϵĺ¯Êýf£¨x£©Âú×ãf£¨x+2£©=-f£¨x-1£©£¬
Ôòf£¨x+3£©=-f£¨x£©£¬f£¨x+6£©=-f£¨x+3£©=f£¨x£©£¬
ËùÒÔf£¨x£©µÄÖÜÆÚΪ6£¬¹Ê¢ÚÕýÈ·£»
Èô¶ÔÓÚÈÎÒâx¡Ê£¨1£¬3£©£¬²»µÈʽx2-ax+2£¼0ºã³ÉÁ¢£¬
Áîf£¨x£©=x2-ax+2£¬Ôòf£¨1£©£¼0ÇÒf£¨3£©£¼0£¬
½âµÃa£¾
11
3
£¬¹Ê¢ÛÕýÈ·£»
µ±x£¾0ʱ£¬²»´æÔÚÕý³£ÊýMʹ|x2+1|=x2+1¡ÜM|x|=Mxºã³ÉÁ¢£¬¹Ê¢Ü´íÎó£»
¹Ê´ð°¸Îª£º¢Ù¢Ú¢Û£®
µãÆÀ£º±¾Ì⿼²éµÄ֪ʶµãÊÇÃüÌâµÄÕæ¼ÙÅжÏÓëÓ¦Ó㬺¯ÊýµÄÖÜÆÚÐÔ£¬º¯Êýºã³ÉÁ¢ÎÊÌ⣬ÊìÁ·ÕÆÎÕº¯ÊýµÄÐÔÖʼ°²»µÈʽÓë¶ÔÓ¦º¯ÊýÖ®¼äµÄ±çÖ¤¹ØϵÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÖУº
¢Ùº¯Êýf(x)=
x-1
x+1
Óëg£¨x£©=xµÄͼÏóûÓй«¹²µã£»
¢ÚÈô¶¨ÒåÔÚRÉϵĺ¯Êýf£¨x£©Âú×ãf£¨x+2£©=-f£¨x-1£©£¬Ôòº¯Êýf£¨x£©ÖÜÆÚΪ6£»
¢ÛÈô¶ÔÓÚÈÎÒâx¡Ê£¨1£¬3£©£¬²»µÈʽx2-ax+2£¼0ºã³ÉÁ¢£¬Ôòa£¾
11
3
£»
¢Üº¯Êýy=log2£¨x2-ax-a£©µÄÖµÓòΪR£¬Ôòa¡Ê£¨-4£¬0£©£»
ÆäÖÐÕýÈ·ÃüÌâµÄÐòºÅΪ
 
£¨°ÑËùÓÐÕýÈ·ÃüÌâµÄÐòºÅ¶¼ÌîÉÏ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÖУº
¢Ùº¯Êýf(x)=
1
lgx
ÔÚ(0£¬+¡Þ)
ÊǼõº¯Êý£»
¢ÚÔÚƽÃæÉÏ£¬µ½¶¨µã£¨2£¬-1£©µÄ¾àÀëÓëµ½¶¨Ö±Ïß3x-4y-10=0¾àÀëÏàµÈµÄµãµÄ¹ì¼£ÊÇÅ×ÎïÏߣ»
¢ÛÉ躯Êýf(x)=cos(
3
x+
¦Ð
6
)
£¬Ôòf£¨x£©+f'£¨x£©ÊÇÆ溯Êý£»
¢ÜË«ÇúÏß
x2
25
-
y2
16
=1
µÄÒ»¸ö½¹µãµ½½¥½üÏߵľàÀëÊÇ5£»
ÆäÖÐÕýÈ·ÃüÌâµÄÐòºÅÊÇ
¢Û
¢Û
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÖУº
¢Ùº¯Êýf(x)=
x-1
x+1
Óëg£¨x£©=xµÄͼÏóûÓй«¹²µã£»
¢ÚÈô¶¨ÒåÔÚRÉϵĺ¯Êýf£¨x£©Âú×ãf£¨x+2£©=-f£¨x-1£©£¬Ôò6Ϊº¯Êýf£¨x£©µÄÖÜÆÚ£»
¢ÛÈô¶ÔÓÚÈÎÒâx¡Ê£¨1£¬3£©£¬²»µÈʽx2-ax+2£¼0ºã³ÉÁ¢£¬Ôòa£¾
11
3
£»
¢Ü¶¨Ò壺¡°Èôº¯Êýf£¨x£©¶ÔÓÚÈÎÒâx¡ÊR£¬¶¼´æÔÚÕý³£ÊýM£¬Ê¹|f£¨x£©|¡ÜM|x|ºã³ÉÁ¢£¬Ôò³Æº¯Êýf£¨x£©ÎªÓн緺º¯£®¡±Óɸö¨Òå¿ÉÖª£¬º¯Êýf£¨x£©=x2+1ΪÓн緺º¯£®
ÔòÆäÖÐÕýÈ·µÄ¸öÊýΪ
3
3
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£ººÓÄÏÊ¡¹Ìʼ¸ßÖÐ2011½ì¸ßÈýµÚÒ»´ÎÔ¿¼ÎÄ¿ÆÊýѧÊÔÌâ ÌâÐÍ£º022

ÏÂÁÐ˵·¨ÖУº

¢Ùº¯Êýf(x)£½Óëg(x)£½xµÄͼÏóûÓй«¹²µã£»

¢ÚÈô¶¨ÒåÔÚRÉϵĺ¯Êýf(x)Âú×ãf(x£«2)£½£­f(x£­1)£¬Ôòº¯Êýf(x)ÖÜÆÚΪ6£»

¢ÛÈô¶ÔÓÚÈÎÒâx¡Ê(1£¬3)£¬²»µÈʽx2£­ax£«2£¼0ºã³ÉÁ¢£¬Ôòa£¾£»

¢Üº¯Êýy£½log2(x2£­ax£­a)µÄÖµÓòΪR£¬Ôòa¡Ê(£­4£¬0)£»

ÆäÖÐÕýÈ·ÃüÌâµÄÐòºÅΪ________(°ÑËùÓÐÕýÈ·ÃüÌâµÄÐòºÅ¶¼ÌîÉÏ)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸