# Range and Domain

• Sep 28th 2008, 03:56 PM
Vivianp
Range and Domain
I need a little help here-not sure if I got it right:
f(x)= 2/x+1, g(x)= x/x+1
domain of f is {x is part of a set where x >0} and domain of g is {x is part of a set where x is greater than or equal to 0} [0, infinity)U(0,infinity).
• Sep 28th 2008, 04:01 PM
Prove It
OK

Both of the functions are rational functions... in other words, of the form

$f(x) = \frac{g(x)}{h(x)}$, where $g(x)$ and $h(x)$ are continuous functions of $x$.

$f(x)$ is also continuous at all points except where the denominator $(h(x))=0$.

So where does the denominator equal 0? In both cases, where $x = -1$.

So the domain of each function is $\mathbf{R}\backslash \{-1\}$.
• Sep 28th 2008, 04:03 PM
Vivianp
Range and domain
The first one...
• Sep 28th 2008, 04:33 PM
Prove It
Sorry for the double post, didn't realise I edited the first one instead of adding a new one.

OK

Both of the functions are rational functions... in other words, of the form

http://www.mathhelpforum.com/math-he...518d3190-1.gif, where http://www.mathhelpforum.com/math-he...482c35c3-1.gif and http://www.mathhelpforum.com/math-he...e8ba0833-1.gif are continuous functions of http://www.mathhelpforum.com/math-he...155c67a6-1.gif.

http://www.mathhelpforum.com/math-he...a378be62-1.gif is also continuous at all points except where the denominator http://www.mathhelpforum.com/math-he...1881b207-1.gif.

So where does the denominator equal 0? In both cases, where http://www.mathhelpforum.com/math-he...16e275fc-1.gif.

So the domain of each function is $\mathbf{R} \backslash \{-1\}$.
• Sep 28th 2008, 04:37 PM
Vivianp
range and domain
wow..so i wasn't even close, huh?
• Sep 28th 2008, 04:42 PM
Prove It
Quote:

Originally Posted by Vivianp
wow..so i wasn't even close, huh?

No, but that's ok, that's what we're here for.

Just always have in the back of your mind "You can't divide by 0" and you'll be fine (Happy)