Set Builder Notation: Domain of f, g & h

roam
Messages
1,265
Reaction score
12
Hi!

We have 3 functions; f=http://img59.imageshack.us/img59/8682/fovgdt3.png ,[/URL] g=\frac{1}{\sqrt{(2x2 - 1)(x2-1)}} and h= \frac{\sqrt{2+x}+x(x^2-1)}{(x^2-1).\sqrt{2+x}}

And we want to write the domain of these functions in the set builder notation, which I'm not very familiar with.

Well, I know that for the Domain of g, x ≠ 0 and ±1. How do we write it in that notation? is it [0,∞]\{±1}?

For the domain of f, x must be greater than or equal to -2 and it cannot equal 0 & ±1

In h, domain: x≥-2, & it can't be ±1 as well.


I need to write the domains of these functions in the set builder notation. Not knowing something doesn't mean you're stupid, just asking about it ... :biggrin:

 
Last edited by a moderator:
Physics news on Phys.org
You could write the domain of x as [-2,∞)\{0,±1}
or as
Dom(f) = {x |x \in [-2,∞)\{0,±1}}

Make sense?

Notice the open parentheses on the infinity, since x cannot actually infinity (as it is not a number)
 
OK, Nick. How should I express the domains of g and h?

Where g = \frac{1}{\sqrt{(2x^2 - 1)(x^2-1)}}

Would the domain of g be written as: dom(g)= R\{0, ±1} ?


And... how would the domain of h be in this notation?
 
Last edited:
Yes that would be a way to write the domain of g.

For h you could write
Dom(h) = [-2,∞)\{±1}
 
Right :wink:
Thanks a lot! Makes sense now...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top