# Domain of a function

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• Sep 23rd 2012, 06:48 AM
M670
Domain of a function
http://webwork.mathstat.concordia.ca...3ee9b0cf31.png
I am trying to figure out how to write the domain in interval notation
i reduced the denominator to (x-4)^2 but am stuck now
any help would be great

Thanks
• Sep 23rd 2012, 06:57 AM
Siron
Re: Domain of a function
What you have said is not entirely correct, you can reduce the denominator to $\displaystyle (x+4)^2$ thus
$\displaystyle f(x) = \frac{x-2}{x^2+8x+16} = \frac{x-2}{(x+4)^2}$

For which value is the fraction not defined? i.e when is the denominator zero?
Can you make a conclusion now?
• Sep 23rd 2012, 07:02 AM
M670
Re: Domain of a function
I believe the denominator is = to zero when x = to -4 or 4
so in interval notation I got (infinity,-4) U (4, inifinity)
• Sep 23rd 2012, 07:07 AM
Siron
Re: Domain of a function
Quote:

Originally Posted by M670
I believe the denominator is = to zero when x = to -4 or 4
so in interval notation I got (infinity,-4) U (4, inifinity)

$\displaystyle f(4) = \frac{4-2}{(4+4)^2} = \frac{2}{64} = \frac{1}{32}$
thus $\displaystyle x=4$ is not a problem. Only $\displaystyle x=-4$ is a problem because then the denominator will be equal to 0.

Thus the domain is ... ?
• Sep 23rd 2012, 07:09 AM
skeeter
Re: Domain of a function
correction ...

$\displaystyle (-\infty,-4) \cup (-4, \infty)$

only x = -4 is not in the domain.
• Sep 23rd 2012, 07:09 AM
M670
Re: Domain of a function
The domain is all real number excluding -4
• Sep 23rd 2012, 07:13 AM
Siron
Re: Domain of a function
Quote:

Originally Posted by M670
The domain is all real number excluding -4

Exactly! $\displaystyle \mathbb{R} \setminus \{-4\}$
• Sep 23rd 2012, 07:16 AM
M670
Re: Domain of a function
Thank You to both of you
I have been stuck on this for over a day now