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Mathematics 11 Online
OpenStudy (anonymous):

How do you find the Domain, I just want the steps

OpenStudy (anonymous):

The domain is x

OpenStudy (anonymous):

\[\frac{ \sqrt{x+3} }{ (x+8)(x+2) }\]

OpenStudy (anonymous):

Of the x coordinate like (x, y)

OpenStudy (anonymous):

Oh that domain.

OpenStudy (anonymous):

NVM then.

OpenStudy (anonymous):

lol okay, do you know how to find this domain?

OpenStudy (anonymous):

There are some restrictions in the domain: Hints: Look in the numerator, it is \(\sqrt{x+3}\), this will be imaginary if the values of x is less than -3. This rational function will be undefined if the value of the denominator is equal to zero. Thus, the zeros on the denominator were the restrictions in the domain. Hope this helps!

OpenStudy (anonymous):

@Data_LG2 I'm a bit confused now

OpenStudy (anonymous):

which part are you confused with?

OpenStudy (anonymous):

I'm just completely lost

OpenStudy (anonymous):

okay let's do it step by step. In the numerator, which values of x will make the value an imaginary number?

OpenStudy (anonymous):

-3

OpenStudy (anonymous):

not actually -3 because if you substitute x=-3 in the numerator it will be \(\sqrt{x+3}=\sqrt{-3+3}=\sqrt{0}=0\). If you will try x=-4 , you will get imaginary number. so you can say that \(x \ge-3, x \epsilon\ real\ numbers\)

OpenStudy (anonymous):

did you get the first part?

OpenStudy (anonymous):

Yeah a it, why would you use -4?

OpenStudy (anonymous):

just an example to prove that any value of x below -3 will give an imaginary number.

OpenStudy (anonymous):

Oh okay, so I'm trying to turn the numerator into a imaginary number?

OpenStudy (anonymous):

yeah, to find the restrictions in the domain. But we are not done yet

OpenStudy (anonymous):

Okay, I have to do something with denominator right?

OpenStudy (anonymous):

yes, you have to find the zeros of the denominator.

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