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 November 6th, 2012, 09:52 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Do you find the restrictions before or after simplifying? Consider f(x)=(x^2-x-6)/(x-3) and g(x)=x-5. find f(g(x)) and the domain. method 1 x cannot be 3 for f(x) now factoring the numerator of f(x) gives f(x)=(x-3)(x+2)/(x-3)=(x+2) composite function f(g(x))=x-3 with the domain x cannot be 3 method 2 first do the composite function f(g(x))=((x-5)^2-(x-5)-6)/(x-5-3) =(x^2-11x+24)/(x- ....... step A =(x-3) so, the domain x cannot be 3 and 8. We have to delete x=8 from the domain because of step A. The book says x cannot be 3 only. Does it mean that you should simplify f(x) first before doing the composite? Please explain. November 6th, 2012, 10:42 PM #2 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Do you find the restrictions before or after simplifying Good question, I think 8 should be deleted from the domain because f(g( ) = f(3) and f(3) is undefined. I'm not 100% sure though...  November 7th, 2012, 02:07 AM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Do you find the restrictions before or after simplifying I think the same. Also, f(g(3)) = f(-2) = 0 so x can be 3. November 7th, 2012, 04:56 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 For clarity, I will use the variable u in defining f. At no stage will I exclude x = 3. Consider f(u)=(u� - u - 6)/(u - 3), where u is not 3, and g(x) = x - 5, where x can have any value. As has been discovered already, f(u) simplifies to u + 2, but its domain still excludes u = 3. Now g(x) has an unrestricted domain, but when we substitute u = g(x) = x - 5 to find f(u) in terms of x, we still need to exclude u = 3, which means we need to exclude x - 5 = 3, which is equivalent to excluding x = 8. Thus f(g(x)) = x - 3, where x = 8 is excluded. Simplifying after doing the substitution, one first gets f(g(x)) = (x - 8)(x - 3)/(x - 8), where x is not 8, then that simplifies to the result I already obtained above. Tags find, restrictions, simplifying Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post HuskO Algebra 1 July 27th, 2013 09:17 AM restin84 Applied Math 2 February 4th, 2012 05:20 PM playthious Algebra 1 September 3rd, 2011 06:52 PM spitzie Applied Math 1 March 16th, 2011 03:57 PM HuskO Calculus 1 December 31st, 1969 04:00 PM

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