Domains and composites.

funsize999 · March 16th, 2009, 04:20 AM

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 $R \geq 0$ ?

501622731 · March 16th, 2009, 09:59 AM

For the first function, the domain would be $\mathbb{R} - \{-1,2,11\}$
For the second problem:
$h \circ f:\mathbb{R}\rightarrow\mathbb{R}<img src=$ \mapsto |x|" />
$f \circ h:\mathbb{R}^+\rightarrow\mathbb{R}^+<img src=$ \mapsto x" />

funsize999 · March 16th, 2009, 03:22 PM

Not -11?

Also, how did you get the first of the composites?

aswoods · March 16th, 2009, 06:25 PM

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.

dman315 · March 16th, 2009, 07:41 PM

Aswoods, you mean the range! You don't get to specify the codomain.

501622731 · March 17th, 2009, 05:23 AM

Good ol' codomain/range war =]

March 16th, 2009, 04:20 AM	#1
funsize999 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 $R \geq 0$ ?
$funsize999 is offline$

March 16th, 2009, 09:59 AM	#2
501622731 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: $h \circ f:\mathbb{R}\rightarrow\mathbb{R}<img src=$ \mapsto \|x\|" /> $f \circ h:\mathbb{R}^+\rightarrow\mathbb{R}^+<img src=$ \mapsto x" />
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March 16th, 2009, 03:22 PM	#3
funsize999 Newbie Joined: Mar 2009 Posts: 20 Thanks: 0	Re: Domains and composites. Not -11? Also, how did you get the first of the composites?
$funsize999 is offline$

March 16th, 2009, 06:25 PM	#4
aswoods 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.
$aswoods is offline$

March 16th, 2009, 07:41 PM	#5
dman315 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.
$dman315 is offline$

March 17th, 2009, 05:23 AM	#6
501622731 Newbie Joined: Mar 2009 Posts: 21 Thanks: 0	Re: Domains and composites. Good ol' codomain/range war =]
$501622731 is offline$

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