| ²é¿´: 476 | »Ø¸´: 2 | ||
[½»Á÷]
¡¾ÌÖÂÛ¡¿¿¼¿¼Êýѧ´óÏÀ»ù´¡ÖªÊ¶£¬ÓÐÐËȤµÄ½ø
|
2Â¥2011-01-27 17:57:26
jiangwu8888(½ð±Ò+5): 2011-01-28 12:57:53
|
g(x)=2^x-a,f(x)=x^2+ax; g(0)=1-a,f(0)=0. ËùÒÔg(0)>f(0) g(2)=4-a,f(2)=4+2a,ËùÒÔg(2) g(-inf)=-a f(-inf)=inf ËùÒÔg(-inf) ËùÒÔ lg f(x) = lg g(x) µÄʵ¸ùΪ2»òÕß3¸ö¡£ ²¹³ä£º ¿¼ÂÇ ¼«¶Ëµã µ±a=1/2 g(x)=2^x-1/2,f(x)=x^2+1/2x g(-1)=0 f(-1)=1/2>0. f(-1)>g(-1) ËùÒÔx3ÔÚ£1ºÍ0¼äÇÒf(x3)=g(x3)>0ËùÒÔÔ·½³Ì¹²ÓÐÈýʵ¸ù¡£ µ± a=1/sqrt(2) g(x)=2^x-1/sqrt(2),f(x)=x^2+1/sqrt(2)x g(-1/2)=0 f(-1/2)=1/4-1/(2*sqrt(2))<0 ËùÒÔ g(-1/2)>f(-1/2) ÇÒ g(-inf) [ Last edited by cronozhang on 2011-1-28 at 08:35 ] |
3Â¥2011-01-28 08:00:08









»Ø¸´´ËÂ¥
20