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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 18:01:01 IST
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f(x)=log3+x(x2-1) <marquee>find the domain of this function </marquee>
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There are two ways to live: you can live as if nothing is a miracle; you can live as if everything is a miracle.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 18:01:41 IST
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There are two ways to live: you can live as if nothing is a miracle; you can live as if everything is a miracle.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 18:06:10 IST
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===================================================== ======== =============== =========================
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There are two ways to live: you can live as if nothing is a miracle; you can live as if everything is a miracle.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 18:15:17 IST
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x^2 - 1>0 x^2 > 1 -1>x>1 also x + 3 not equal 0,1 x not equal to -3 ,-2 domain (-infin.,-3) u (-3,-2) u (-2,-1) u (1,infin.) i think im rite .correct me if im wrong cheers rate me if im rite
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kaushik krishna .R
bits pilani
mech engg |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 18:28:36 IST
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I want to ask: y r u using HTML tags here and there??!!??!! 
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 19:03:50 IST
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I would like to add a little to pottermania's solution. It is slightly incorrect. One more condition must be observed x+3>0 (As the base of the logarithm must be positive and unequal to unity) x> -3 The solution then becomes x (-3,-2) (-2,-1) (1, )
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 20:48:36 IST
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i forgot elessar thanks man
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kaushik krishna .R
bits pilani
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2007 20:55:43 IST
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answer for this should be ==== x =1/2+root 5/2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jul 2007 08:59:06 IST
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pottermania's is worang
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logarithm function will be defined when x 2-1>0 &3+x>0 &3+x  1 i.e (x+1)(x-1)>0 & x>-3 & x  -2 i.e x  (-  ,-1)  (1,  ) &x>-3&x  -2 so the intersection of these domains is the ANSWER THAT IS (-3,-2)  (-2,-1)  (1,  )
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