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November 6th, 2012, 09:52 PM   #1
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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.
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November 6th, 2012, 10:42 PM   #2
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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...

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November 7th, 2012, 02:07 AM   #3
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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.
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November 7th, 2012, 04:56 AM   #4
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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.
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