Domain of a function

**M670** · September 23rd 2012, 06:48 AM

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

**Siron** · September 23rd 2012, 06:57 AM

What you have said is not entirely correct, you can reduce the denominator to $(x+4)^2$ thus
$f(x) = \frac{x-2}{x^2+8x+16} = \frac{x-2}{(x+4)^2}$

For which value is the fraction not defined? i.e when is the denominator zero?
Can you make a conclusion now?

**M670** · September 23rd 2012, 07:02 AM

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

**Siron** · September 23rd 2012, 07:07 AM

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)

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

Thus the domain is ... ?

**skeeter** · September 23rd 2012, 07:09 AM

correction ...

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

only x = -4 is not in the domain.

**M670** · September 23rd 2012, 07:09 AM

The domain is all real number excluding -4

**Siron** · September 23rd 2012, 07:13 AM

Originally Posted by M670

The domain is all real number excluding -4

Exactly! $\mathbb{R} \setminus \{-4\}$

**M670** · September 23rd 2012, 07:16 AM

Thank You to both of you
I have been stuck on this for over a day now

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