# Absolute Values Help

• March 23rd 2013, 08:33 PM
JohnyyMal
Absolute Values Help
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<img src="attachment.php?attachmentid=27676&amp;stc=1" attachmentid="27676" alt="" id="vbattach_27676" class="previewthumb">

Hello Please see the above picture for the question this is a question im stuck with any help will be highlyappreciated
• March 23rd 2013, 09:15 PM
princeps
Re: Absolute Values Help
Hint :

Domain :

$x^2-16 \neq 0$

and :

$|x+4| =\begin{cases} -x-4, & \text{if }x<-4 \\ x+4, & \text{if }x>-4\end{cases}$
• March 23rd 2013, 09:26 PM
Cesc1
Re: Absolute Values Help
$\frac{\mid x+4\mid}{x^2-16}=\frac{\mid x+4\mid}{(x+4)(x-4)}=\frac{\mid x+4\mid}{x+4}\frac{1}{x-4}$

Hint:
Note that if x>-4 then x+4 is always positive, so

$\frac{\mid x+4\mid}{x+4}=1$

Now, what happens when x<-4?

About the other question: when is the denominator zero?