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2Â¥2011-01-27 17:57:26
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cronozhang

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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)>f(inf)¡£Ôò ÓÐһʵ¸ùx2ÔÚ2ºÍinf¼ä¡£Ò×Öªg(x)>0 ºÍ f(x)>0 µ±x>0¡£ÖÁÉÙÓÐÁ½Êµ¸ù¡£

g(-inf)=-a f(-inf)=inf ËùÒÔg(-inf) µ«g(x3)=f(x3)>0 »òÕß g(x3)=f(x3)<0 ¸ù¾ÝaµÄÖµÀ´¾ö¶¨

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
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