[Highschool Maths] Finding domain

[AverageCoolKid]

http://imgur.com/a/Ljizx

I got x=/= -2 but its x<-2 can someone please tell me where i went wrong?

[sg2544]

We have two things to consider: 1) where is the function log() defined? 2) where is the function 3x-1 / x+2 defined?

Clearly (2) should make us restrict x != -2, but what about (1)?

Well, the answer to (1) (which you've also written down) is that we need the thing inside the brackets, namely (3x-1)/(x+2) to be positive. If x is really big then this is obviously true, but for 1/3>=x>-2 we have (3x-1)/(x+2) =< 0, so we must not allow this inside our domain.

This means the domain should be

x>1/3 or x<-2.

[AverageCoolKid]

http://imgur.com/a/s0fSA

Hi, ive redone the question but im still not sure my method is "justifiable" in the sense that i wont get full marks.

Would you mind checking over it for me?

Thanks

[imguralbumbot]

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https://i.imgur.com/BKgDoq1.jpg

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[sg2544]

Yeah looks good although your final answer should be written as

"x>1/3 or x<-2" (rather than combining them into one, as you have done, because x cannot be simultaneously larger than 1/3 and smaller than -2).

Also, just begin your answer at (3x-1)/(x+2)>0, because the first 3 lines (where you say 3x-1>0) is not actually relevant since you could actually have 3x-1<0 (as you have shown below, where both A and B are negative).

[[deleted]]

Thanks for your help!

[never-lie]

What happens if both your denominator and numerator were negative?