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 March 16th, 2009, 03:20 AM #1 Newbie   Joined: Mar 2009 Posts: 20 Thanks: 0 Domains and composites. Hi, For the function$q(x)= \frac{5x-6}{(x+1)(x-2)(x+11)$ would the domain be any real number except -1, 2 and -11? I'm not sure if I expand the brackets out first or not. For the functions $f: R -> R^+ : x -> x^2$ and $h: R^+ -> R : x -> sqrt{x}$ are the composites; f.h = $x^2(sqrt{x})$ h.f = $sqrt{x}(x^2)$ which would then result in both being $x$ ? And would the codomain be $R \geq 0$ ?
 March 16th, 2009, 08:59 AM #2 Newbie   Joined: Mar 2009 Posts: 21 Thanks: 0 Re: Domains and composites. For the first function, the domain would be $\mathbb{R} - \{-1,2,11\}$ For the second problem: $h \circ f:\mathbb{R}\rightarrow\mathbb{R} \mapsto |x|" /> $f \circ h:\mathbb{R}^+\rightarrow\mathbb{R}^+ \mapsto x" />
 March 16th, 2009, 02:22 PM #3 Newbie   Joined: Mar 2009 Posts: 20 Thanks: 0 Re: Domains and composites. Not -11? Also, how did you get the first of the composites?
 March 16th, 2009, 05:25 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Domains and composites. Yes, -11. For h of f: first you square, then take the (positive) square root. Zero goes to zero, positive x goes to x, negative x goes to -x (which is positive). For f of h: first you take the square root, then square. Zero goes to zero, positive x goes to x, negative x has no square root and so can't be in the domain. The codomain for both is all non-negative real numbers.
 March 16th, 2009, 06:41 PM #5 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Domains and composites. Aswoods, you mean the range! You don't get to specify the codomain.
 March 17th, 2009, 04:23 AM #6 Newbie   Joined: Mar 2009 Posts: 21 Thanks: 0 Re: Domains and composites. Good ol' codomain/range war =]

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