Question

find domain f(x)= (2x+1) / (2x-1) and domain f(x)=(2x^2+5x+3) / (2x^2-5x-3)

Answer 1

f(x) = (2x+1) / (2x-1) What values cannot be x.

Answer 2

1 nd -1 ?

Answer 3

Literally what is the domain? The domain is just all the possible x-values you're allowed to pick, that's it! So if you look at the domain of:

1/x

You'll see that you can pick any number EXCEPT for 0. Why? Well, you can't divide by zero, can you?

So any time you have something at the bottom of a fraction with x's in it, set the bottom equal to zero and solve for what makes the bottom =0. Then you know you can't ever pick those numbers.

Answer 4

but the bottom doesn't have any x's

Answer 5

Yes they both do.

Answer 6

oh yea I see the bottom one the way it was typed I forgot it is on the bottom

Answer 7

\[ f(x)=\frac{ (2x+1) }{ (2x-1) }\] So when 2x-1=0 we know that will be dividing by zero. So we have to solve for x here to see when that is so we know that it's NOT in the domain.

Answer 8

so we solve 2x-1=0?

Answer 9

(-infty,1/2)u(1/2,infty)

Answer 10

I got that for the bottom is that correct and do I need domain for both top and bottom or only bottom? Some1 plz help and I still need the second problem too