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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^2x6)/(x3) and g(x)=x5. 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)=(x3)(x+2)/(x3)=(x+2) composite function f(g(x))=x3 with the domain x cannot be 3 method 2 first do the composite function f(g(x))=((x5)^2(x5)6)/(x53) =(x^211x+24)/(x ....... step A =(x3) 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: 20,627 Thanks: 2077 
For clarity, I will use the variable u in defining f. At no stage will I exclude x = 3. Consider f(u)=(u²  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. 

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