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 ▼ john littlefield Junior Member Posts: 8 Threads: 4 Joined: Jan 1970 02-08-2005, 01:09 PM I need help finding the domain of the function f(x)=1/1-ln(9-sqrrt(x^2-9)) ▼ Junior Member Posts: 10 Threads: 3 Joined: Jan 1970 02-08-2005, 07:23 PM x < 3sqrt(10) && x != sqrt(90-18e-e^2) ?? Hugh Evans Senior Member Posts: 302 Threads: 34 Joined: Aug 2007 02-08-2005, 07:33 PM I assume you're just solving this by hand and not asking how to do this with a 49G+/other graphing model. Remember that the domain of a function is simply the set of x values for which the function is defined. To find where the function is not defined in this case you must find solutions that result in 1/0. If you set the denominator equal to 0 and solve for x you will find the domain. To get you started: 1-ln(9-sqrt(x^2-9))=0 Some relatively basic algebra will give you the answer. Keep the properties of the squared term in mind. Good luck! Vieira, Luiz C. (Brazil) Posting Freak Posts: 4,027 Threads: 172 Joined: Aug 2005 02-08-2005, 08:34 PM Hi, John; Hugh is correct, one of the first restrictions is the one he pointed out. Now, let me add some guidance that I hope will help you a bit. When defining/checking domain, discontinuities, poles and, if applicable, undefined points must be found so you can set all conditions that define your domain. For each particular function you'll have a series of restrictions. I guess that you may find your way to get the answer yourself, just take some guidelines for this particular case: 1 - denominator different of zero, meaning:```1-ln(9-sqrrt(x^2-9)) =! 0 ,hence ln(9-sqrt(x^2-9)) =! 1 9-sqrt(x^2-9) =! e (2.7183...) sqrt(x^2-9) =! 9-e (now you isolate x... your turn!)``` 2 - because it is a ln operand:```9-sqrt(x^2-9) > 0 , hence: sqrt(x^2-9) > 9``` 3 - sqrt parameter should be greater than zero as well:```x^2-9 > 0 , hence x^2 > 9 and |x| > 3``` Can you put all of these conditions together in one only? If so, you'll have your domain. (at least I hope this is correct; have someone found anythng wrong here, please add corrections) Cheers. Luiz (Brazil) ▼ Hugh Evans Senior Member Posts: 302 Threads: 34 Joined: Aug 2007 02-08-2005, 11:21 PM Well put, Luiz. I was somewhat hesitant to elucidate since this looks to be a homework problem. You did an excellent job of laying the logical groundwork and demonstrating optimal technique to determine the solution without giving much away. Regards, HDE ▼ Vieira, Luiz C. (Brazil) Posting Freak Posts: 4,027 Threads: 172 Joined: Aug 2005 02-08-2005, 11:35 PM Hi, Hugh; thanks for your comments. You see, that's the technique I always try to use when teaching: "teasing" to an answer. Not always possible, depends on the problem itself and the subject, but when applicable, it is rewarding for both teacher and students. I, as a teacher, must "feel" where to stop (sometimes going further, sometimes stopping earlier), and the students, in their turn, must feel confident going ahead by their own. I don't remember attending classes like the ones I propose to my students today, but I guess we must all evolve, enhance. Your comments show me I'm on the right path. I neither blame my teachers for their classes, I am confident they were doing the best they could. You see, I tell my students that when the job is done by both teacher and students, learning is a consequence of teaching and examination is just a matter of bureaucratic evidence. If students search for knowledge and do not feel satisfied until their doubts are gone, (good) teaching is always effective. Thanks! Luiz (Brazil) Edited: 8 Feb 2005, 11:41 p.m.

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